Board Exam Problems

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Problem 10: (etermie the curret i the 16 ohm resistor usig ortos ortos theorem A. 1.5$ A &. 1.25 A %. 1.0 A (. 1.-- A Problem 110: (etermie the curret i the 106 ohm resistor usig heveis theorem. A. 0.8-- A &. 0. A %. 0.0 A (. 0.508 A Problem 111: (etermie the curret i the 6 ohm resistor by Duperpositio theorem. A. -.$ A &. $.2 A %. $.0 A (. -.8 A Problem 112: (etermie the loo3ig bac3 resistace bet"ee termials a M b. A. 11  &.  %. 10  (. 12  Problem 11-: (etermie l. A. $.50 A &. $.28 A %. 5.-$ A (. -.5 A Problem 11$: (etermie the voltage ,. A. 1$ , &. 12 , %. 10 , (. 1 ,

TEST " # Electrostatics Electrostatics Problem 115: 77 &oard Carch 18 hree eual positive charges of 10 statcoulomb each are located at the vertices of a euilateral triagle of 2 cm leg. +hat is the magitude of the force actig o each charge# A. 12.5 dyes &. 25 dyes %. $-.- dyes (. 50 dyes Problem 11: 77 &oard ctober 1

"o "o poit charges 10 cm apart exert a force of 1 x 10 raised to the >6-? grams o each other. If the charges are the same value! "hat is the value of each charge i statcoulomb# A. 8.0 stat coulomb &. .0 stat coulomb &. .0 stat coulomb (. .0 stat coulomb Problem 11: 77 &oard April 1 "o "o electros i a vacuum experiece a force of 2 x 10 615 . /o" far apart are the electros# A. 0.$8 x 106 m &. 2.8$ x 10612 m %. -.- x 106 m (. 5.05 x 10612 m Problem 118: "o "o poit charges of 2 % ad 5 % are 10 meters apart. A third poit poit charge of  % is placed bet"ee them. At "hat distace from the 5 % charge must the  % charge is placed! i such a "ay that it "ill stay i euilibrium# A. .125 m &. 5.-5 m %. .025 m (. 5.25 m Problem 11: 77 &oard ctober 1 "o "o spheres separated from each other by 10 m have charges of 0.001 coulomb ad 0.00- coulomb respectively. respectively. I bet"ee the t"o spheres is a poit of Bero electric field. +hat is the distace from the 0.001 coulomb sphere# A. .-$ m &. .8 m %. 2.2$ m (. -. C Problem 120: %alculate the magitude of the electric field at a poit that is -0 cm from a poit charge Q< 6-.2 x 10 6 %. A. -.2 x 10  % &. -.0 x 105 % %. 2.8 x 10  % (. 2.$ x 105 % Problem 121: 77 &oard ctober 18 "o "o metallic plates separated by 1 cm are coected across a 12 , battery. A certai plastic material is iserted completely fillig the space bet"ee the plates ad the charge o each plate observed to double. +hat is the dielectric costat of the plastic material# A. 0.5 &. 2 %. $ (.  Problem 122: 77 &oard ctober 11 %alculate the capacitace bet"ee t"o parallel plates of "hich is 100 cm 2 ad 2 mm apart i air. A. 0.$$- RRE &. 0.515 RRE %. 0.$52 RRE (. 0.502 RRE Problem 12-: A capacitor has t"o parallel metal plates measurig 0.$5 m suare each. he distace bet"ee the plates is 1 cm i free air. If a dielectric material "hose permittivity is $ ad 5 mm thic3 is placed i the upper plate! leavig a air gap bet"ee the bottom plate ad the dielectric! "hat is the capacitace of the capacitor# A. 20 RRE &. 25$ RRE %. 1- RRE (. 28 RRE Problem 12$: A capacitor is composed of t"o plates separated by a -6mm isulatio "hose dielectric costat is $. A additioal piece of 6mm isulatio is iserted bet"ee the plates. +hat is the dielectric costat of the additioal isulatio if the capacitace of the capacitor decreases to half as much of its origial value# A. 5 &. 8 %.  (.  Problem 125:

A parallel6 plate capacitor has a euivalet capacitace of 850 pE. he he area of each plate is  cm 2 ad the thic3ess of the dielectric is 0.02 cm. If the material used as the dielectric has a costat of 8! ho" may sectios are there i parallel# A. $ &. 5 %. (.  Problem 12: 77 &oard ctober 1$ A parallel plate capacitor is made of -50 plates! separated by paraffied paper 0.0010 cm thic3 >3< 2.5?. he effective siBe of a plate is 15 by -0 cm. +hat is the capacitace of the capacitor# A. -5 RE &. 15 RE %. 1$0 RE (. 0 RE Problem 12: 77 &oard ctober 1 A capacitor is charged "ith 0.2- "att6 secod of eergy at a voltage of $8 volts. +hat is its capacitace# A. 180 RE &. 2$0 RE %. 200 RE (. 220 RE Problem 128: 77 &oard ctober 1 A 206 RE capacitor is charged by a 126 , battery. +hat is the stored eergy i the capacitor# A. 1.0 x 106- S &. 2.88 x 1062 S %. 2.88 x 106- S (. 1.$$ x 106- S Problem 12: 77 &oard ctober 1$ A parallel plate capacitor "ith air bet"ee its plate is charged util the potetial differece , appears across it. Aother capacitor havig a hard rubber >dielectric costat < -? bet"ee its plates but other"ise idetical! is also charged to the same potetial differece. If the eergy of the first capacitor is +! that of the secod is A. + &. +%. + (. -+ Problem 1-0: A capacitor "hose plates is 20 cm x -.0 c m ad is separated by a 1.06 mm air gap is coected across a 126 , battery. battery. (etermie the charge accumulated o each plate after a log time. A. 5.- x 10612 % &. .2 x 106 % %. 8.1 x 10611 % (. .$ x 10610 % Problem 1-1: 77 &oard ctober 10 he result of capacitace % 1 <  microfarads ad % 2 coected i series is - microfarads. %apacitor % 2 i microfarads is A. &. $ %. 8 (.  Problem 1-2: he euivalet capacitace of t"o capacitors i series is 0.0- RE ad "he coected i parallel! 0.1 RE. If the capacitor "ith smaller capacitace is replaced "ith a capacitor "hose capacitace is double as much "hat "ill be the e" euivalet series capacitace of the combiatio. A. 0.0$8 RE &. 0.005 RE %. 0.0$2 RE (. 0.050 RE Problem 1--: hree capacitors of 1! 15 ad 12 RE respectively are coected i series. +hat is the maximum voltage that ca be supplied to the combiatio if the voltage drop across ay of the three capacitors must ot to exceed 100 ,. A! 20 , &. 2$5 , %. 255 , (. 250 , Problem 1-$: hree capacitors of 20! 10 ad 15 RE respectively are coected i series across a 100 , supply. supply. Eid the voltage across the 10 RE capacitor. A. 5$.52 , &. $.15 ,

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Problem 1-5: hree codesers of ! 10 ad 15 RE are coected i series across a 200 , supply. supply. If the codesers are discoected! ho" much eergy ca the charged codeser provide if coected across each other "ith li3e polarities together# together# A. 0.052 S &. 0.00 S %. 0.0 S (. 0.08 S Problem 1-: "o "o capacitors "ith capacitace of  RE ad $ RE are coected i series across a 100 , dc source. If the supply voltage is cut6 off ad the t"o capacitors are coected i parallel! "hat "ill be the fial charge accumulated i the $ RE capacitor# A.  R% &. 12 R% %. 12 R% (. 88 R% Problem 1-: A $0 RE capacitor is charged to store 0.2 S of eergy. eergy. A ucharged 0 RE capacitor is coected i parallel "ith the first oe through a perfectly coductig leads. +hat is the eergy lost due to the "or3 doe by the charges i spreadig out over the t"o capacitors# A. 0.08 S &. 0.12 S %. 0.80 S (. 0.012 S Problem 1-8: Eour capacitors A! &! % ad (are coected across a 1506, (% source. he charges accumulated i A! & ad % are eual to 2000! 50 ad 1500 R%! respectively. respectively. If the total capacitace of the combiatio is 3o" to be -5 RE! ho" much charged is the accumulated i capacitor (# A. 1200 R% &. 1000 R% %. 1150 R% (. 1$00 R% Problem 1-: 77 &oard ctober 1A codeser codeser of 1 microfarad capacitace is charge to 100 volts ad the discoected from the po"er supply. supply. A secod but ucharged capacitor of - microfarad is the coected across the first capacitor. he voltage across the parallel combiatio is A.  2- volts &. $5 volts %. 25 ,olts (. 110 volts Problem 1$0: 77 &oard ctober 1$ hree capacitors A! & ad % are charged as follo"s: A< 10 RE! 100 ,T &< 15 RE! 150 , ad %< 25 RE! 200 ,. they are coected i parallel "ith termials of li3e polarity together. +hat is the voltage across the combiatio# A. 150 A &. 120 , %. 185 , (. 15 , Problem 1$1: A 100 RE capacitor is charged from a 2006 , supply. supply. he charged capacitor is the coected across three ucharged capacitors coected coected i parallel. he charges o these capacitors after a log time become $000! 5000 ad 000 R%. Dolve the voltage across the combiatio. A. $ , &. 50 , %.  , (. 80 , Problem 1$2: "o "o capacitors of 200 ad 800 E! respectively respectively are coected i parallel. he combiatio is the coected i series "ith aother t"o capacitors of % ad 00 E coected i parallel. If the total capacitace of the set6 up is 500 E! determie the capacitace %. A. 50 E &. 550 E %. $00 E (. 500 E

Problem 1$-:77 &oard April 1 A coaxial cable has a 106mm diameter iside coductor ad a metallic sheath "ith a iside diameter of 20 mm. If the isulatig medium has a dielectric costat! 3< 2! ad a permittivity of 1.111.11- × 10610! "hat is the capacitace bet"ee the coductor ad the shield per 3m# A. 0.105 µE3m &. 0.80- µE3m %. 0.080- µE3m (. 0.0$0- µE3m Problem 1$$:77 &oard ctober 1$ A lead6sheath cable for udergroud service has a copper coductor >diameter<0.-50 ich? surrouded by 0.206ich "all of rubber isulatio. Assumig Assumig the dielectric costat of $.- for rubber! calculate the capacitace of the cable per mile legth. A. 1.01 µEmile &. 0.50$ µEmile %. 0. µEmile (. 0.252 µmile

TEST $ – %anetic Circuits Problem 1$5:77 &oard ctober 1$ he force actig o a pole of - +b is 12 . he magetic itesity of the magetic field is A. $  per +b &. 12 12  per +b %. - -  per +b (. o oe 0f these Problem 1$: "o "o straight straight parallel parallel "ires "ires 2 m log ad - mm apart carries a curret of 8.0 A i opposite opposite directio. %alculate the force bet"ee these coductors. A. 8.-- × 106-  &. 8.25 × 106-  %. 8.5- × 106-  7. 8.5 × 106-  Problem 1$: A "ire 12 cm log ad carryig a curret of -0 A is placed i bet"ee the pole face of a maget maget "hose magetic flux desity is 0. tesla. If the "ire is iclied at a agle 0 degrees from the plae of the magetic field! "hat is the force exerted o the "ire# A. 2.8  &. 2.5  %. -.1  (. 2.2  Problem 1$8:77 &oard ctober 12 he reluctace of a o6magetic circuit is 12 uits. /o" much flux "ill be set up if surrouded by a coil 00 turs carryig a curret of - A. A. 150 &. 1. %. 50 (. 200 Problem 1$: he coil i a magetic cotactor reuires 0.5 A to provide a magetiBig magetiB ig force of 500 A. /o" may turs are ecessary# A. 250 &. 500 %. 1000 (. 2000 Problem 150: A soleoid -06cm log is "oud "ith -00 turs! "hat is the value of its field stregth iside the soleoid! "he the coil is carryig a curret of 2 amperes#

A. 1500 Am &. 2000 Am %. 1800 Am (. oe of these these Problem 151: %alculate the magetic field itesity to produce a flux desity of 10 ×106- tesla at the ceter of a log straight soleoid. A. 58 Am &. $2- Am %. 8-2- Am (. 8 Am Problem 152:77 &oard ctober 1 If a curret curret of 5 A flo"s through a log "ire radius 0.00$ meter! "hat is the flux desity >/? produced 0.02 meter a"ay from the surface of the "ire. A. 1.0 Am&. m &. -.80 Am %. .-$ Am (. --.1 Am Problem 15-:77 &oard ctober 1! 77 &oard ctober 18 he relative permeability of a certai silico steel is $500. A certai magetic loop cosists of a silico steel of 10 cm suare! 20 cm log ad a air gap of U cm. +hat is the reluctace of the magetic circuit# A. 2.5$ × 1062 &. .$$ × 106- %. .$$ × 106$ (. 2.5$ × 106Problem 15$:77 &oard Carch 18 A certai lamiated steel core has a permeability of -000. he legth is 5 cm ad the cross sectio is 2 s. cm. +hat is its reluctace# A. 500 &. 8.-- × 106$ %. 1200 (. 2.5 Problem 155: A coil "ith 00 turs is "oud over a magetic core "ith a reluctace reluctac e of 10!000 A+b. If a curret of 0.5 A is pass through the coil! ho" much is the magetic flux that the coil geerates# A. 0.02 +b &. 0.058 +b %. 0.0.0-2 +b (. 0.0$ .0$5 +b Problem 15: A air6core circular coil has 5000 turs. he ier ad outer diameters are 20 ad 2$ cm respectively. respectively. If a curret of 2 A is passed through the coil! determie the flux desity iside the coil. A. 0.018  &. 0.021  %. 0.01-  (. 0.015  Problem 15: A give magetic circuit has a magetic magetic field itesity of $00 Am. m. If the legth of the magetic path is doubled maitaiig the same magetomotive force! ho" much is the e" magetic field itesity# A. 800 Am &. $00 $00 Am %. 200 Am (. 00 00 Am Problem 158: 77 &oard ctober 18 A magetic circuit cosists cosists of silico steel of -000 permeability! of 10 cm legth ad a cross sectio of 1.5 s. cm ad a air gap of the same cross sectio ad of 2 cm legth. A 4 6 ampere curret flo"s through the 50006 tur coil. +hat is the field itesity at the air gap# A. 250 oersted &. 5 oersted %. 222 oersted (. 15 oersted Problem 15: 77 &oard Carch 18 A magetomotive magetomotive force is supplied by a curret of oe ampere through 100 turs. he magetic circuit cosists of a steel core of 1000 permeability! permeability! 10 cm log ad $ s. cm area ad a air gap oe cm log. +hat is the field itesity at the air gap# A. 12.25 oersted &. $. oersted %. 100.5- oersted (. 12$.$ oersted Problem 10: A circular rig of iro havig a cross sectioal area of 5 cm 2! a average diameter of -0 cm is "oud "ith a coil of 1000 turs. If a curret of - A is pass through the coil! determie the flux geerated# Assume the relative permittivity of iro iro is 250.

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5 x 106$ +b +b .5 x 106$ +b

&.  x 106$ +b (. $.5 x 106$ +b

Problem 11: A o6 magetic rig havig a cross sectioal area of 10 cm 2 is uiformly "oud "ith -00 turs of a give "ire. If a curret of 1 A is passed through the coil! 2.$ µ+b of flux is geerated iside the rig. (etermie the average diameter of the rig. A. 50 cm &. $2 cm %. $8 cm (. 5$ cm Problem 12: A coil "ith 150 turs is "oud over a 200 cm a cylidrical iro core "hose relative permittivity is 250. if a curret of 2 A is pass through the coil! determie the flux desity i the core. A. 0.0$  &. 0.0  %. 0.12  (. 0.10  Problem 1-: 77 &oard Carch 18 A magetic circuit cosists cosists of silico steel of -000 permeability ad a air gap. he legth of the steel core is 10 cm ad the air gap is 2 cm. &oth have the same cross sectio of 1.5 s. cm. c m. A curret of 4 ampere flo"s through the "idigs to produce 2-51 max"ells flux. /o" may turs are there i the coil# A. $120 turs &. 5000 turs %. 2500 turs (. 1250 turs Problem 1$: A toroidal toroidal core "ith a mea circumferece of 1000 cm ad a cross sectioal area of 10 cm 2 is "oud "ith 500 turs of "ire. +hat curret "ould be reuired to geerate a flux of 1 m+b i the core# Assume the core has a relative permeability of 800. A. 1.0 A &. 2.0 A %. 1.5 A (. 2.5 A Problem 15: A magetic rig >relative permittivity permittivity < 800? has a mea radius of 10 cm ad a cross sectioal area of 5 cm2. A air gap measurig 1.5 mm is cut i the rig. (etermie the reuired mmE i order to produce a flux of 0.25 m+b i the air gap. A. 80 &. 20 %. 0 (. 2 Problem 1: A magetic rig "ith a mea diameter of 25 cm ad a cross sectioal area of 5 cm 2 is "oud "ith a coil of 00 turs. A air gap of $ mm is made by cuttig a sectio of the rig. A curret curret of 10 A is passed through the coil. (etermie the eergy stored i the air gap. Assume relative permittivity of the rig to be 1000. A. 2.08 S &. 1. S %. 1.5 S (. 1. S Problem 1: 77 &oard April 15 A permaet permaet maget loudspea3er is euipped "ith a 10 ohm 10 tur voice coil "oud o a tube 1 ich i diameter. he he flux desity i the air gap is 10 000 lies per suare ich. If the impedace of the voice coil is a pure resistace! "hat "ill be the maximum thrust delivered to the core if the impressed sigal is 20 d& above 0.005 "att# A. 0.0--  &. 0.011  %. 0.02  (. 0.05-2  Problem 18: 77 &oard ctober 1 A small sigle6 phase trasformer has 10.2 "atts at o load. he core has a volume of 50 cubic cm. he maximum flux desity is 10 000 gauss ad the hysteresis costat of the core is 5 x 10 6$. 9sig the DteimitB la" to fid the hysteresis! determie the eddy curret loss. A. $.55 + &. 5.55 + %. -.55 + (. 2.55 +

Test ' – Electro!anetic (nduction Problem 1: 77 &oard ctober 18 A magetic coil produces 100 000 max"ells "ith 2000 turs ad "ith a curret of 2 A. the curret is cut6 off ad the flux collapses i 0.01 sec. +hat is the average voltage that "ill appear across the coil# A. 20 3, &. 200 , %. 2000 3, (. 2 3, Problem 10: 77 &oard ctober 18 A maget! "hich geerates 800 max"ells of flux! is iserted through a coil of 1000 turs. It is the "ithdra" "ithi 0.15 sec. +hat is the voltage that appears across the coil# A. 0.22 , &. 5-.- , %. 5-- , (. 0.05-- , Problem 11: he armature coil of a geerator "ith 50 coductors o it rotates i a magetic field "ith a magetic flux desity 1.5 . the the effective area of the coil is 1000 suare cm. If this coil is move across the pole i 0.5 sec! determie the average emf iduced i the coil. A. 20 , &. 15 , %. 10 , (. 12 , Problem 12: 77 &oard April 1 A coductor! coductor! 50 mm log! moves at a velocity of 2.5 meters per secod across a magetic field of 0.0 2 +bm . +hat is the voltage geerated# A. 0.125 , &. 0.0$8 , %. 0.1125 , (. 0.02 , Problem 1-: A airplae travels at a speed of 1000 3ph i a regio "here the earths magetic magetic field is 5.0 x 10 65  ad is early vertical. +hat is the potetial differece iduced bet"ee the "ig tips of the airplae! "hich is 20 m apart# A. 0.28 , &. 0.2- , %. 0.- , (. 0.15 , Problem 1$: 77 &oard ctober 18 A piece of coductor 10 cm log moves across a magetic field of 10! 000 gauss at a velocity of 120 cmsec. +hat voltage appears across the coductor# A. 0.2$ , &. 0.12 , %. 0 , (. 120 , Problem 15: A coductor coductor 50 cm i legth moves at a agle of -0 o to the directio of a magetic field "ith a velocity of 100 cms. If the field has the flux desity of 2 +bm 2! calculate the emf iduced i it. A. 0.5 , &. 2.0 , %. 1.0 , (. 1.5 , Problem 1: A metal dis3 of -0 cm radius rotates at a agular velocity of 1200 rpm across a uiform vertical magetic field of flux desity 0.5 +bm 2. %alculate the voltage iduced bet"ee the rim ad the ceter of the dis3. A. $. , &. 5.8 , %. 5.1 , (. $.$ , Problem 1: A coductig coductig disc 206cm i diameter pivoted by a vertical spidle "hich is lyig i uiform vertical magetic field of flux desity 0.- . A fixed brush is coected at the edge of the disc. +he a $ Ω resistace is coected across the brush ad the spidle! a curret of 15 mA "as observed to flo" through the resistor. (etermie the velocity i "hich the disc is rotatig. A. -$ rpm &. -82 rpm %. -02 rpm (. 228 rpm Problem 18:

/o" much is the iductace of a coil that iduces 500 , "he the curret chages at the rate of 5 mA i 2 µs# A.

12.5 µ/

&.

0. 2 /

%.

0.1 /

(.

2 m/

Problem 1: he field "idigs of a (% geerator is "oud "ith 00 turs ad has a resistace of 50 ohms. If the is excited from a 2$0 , dc source! the magetic flux li3ig the coil is $ m+b. (etermie the self6 iductace of the coil. A. 0.5 / &. 0.8- / %. 0.2 / (. 0.5 / Problem 180: A coil of 1000 turs is uiformly "oud o a rig of o6 magetic material! the mea diameter beig 20 cm. he cross sectioal area of the coil is $ suare cm. (etermie the self6 iductace of the coil. A. 0. m/ &. 0. m/ %. 0. m/ (. 0.8 m/ Problem 181: A iro core 0.$ m log ad 5 suare cm i cross sectio! is "oud "ith -00 turs. +he a curret of 0.5 A flo"s i the coil! ho" much is the iductace of the coil. Assume the core has a permeability of 2500. A. 0.-5 / &. 0.$2 / %. 0.2 / (. 0.-1 / Problem 182: 77 &oard ctober 10 +hat is the coefficiet of couplig of t"o coils "hose mutual iductace is 1 / ad "hose self6 iductace are 1.2 ad 2 /# A. 1.5$ &. 2.$00 %. 0.$5 (. 0.12 Problem 18-: A coil of iductace 8 m/! produces 80 µ+b of magetic flux. f this total flux! 0 µ+b are li3ed to a secod coil of iductace 2 m/! ho" much is the mutual iductace bet"ee coils# A. - m/ &. $ m/ %. 5 m/ (. 2 m/ Problem 18$: 77 &oard April 12 he total iductace of t"o coils! A ad & "he coected i series aidig is 0.5 / ad 0.2 / "he coected i series opposig. %oil A "he isolated from coil & has a self6 iductace of 0.2 /. calculate the self6 iductace if coil &. A. 0.1 / &. 0.12 / %. 0.20 / (. 0.15 / Problem 185: "o idetical coils "ith self6 iductace 250 µ/ each are coected i series. +he coected series aidig! the euivalet iductace is 550 µ/ ad $50 µ/ "he coected series opposig. /o" much is the couplig coefficiet of t"o coils# A. 0.2 &. 0.%. 0.1 (. 0.$ Problem 18: A cylider cylider "hose radius is 2 cm ad altitude of 100 cm is uiformly "oud "ith 2500 turs of "ires i s sigle layer. A secod layer of 100 turs ad of much fier "ires is "oud over the first ear its ceter. %alculate the mutual iductace of the t"o coils. A. 0.$0 m/ &. 0.-8 m/ %. 0.52 m/ (. 0.- m/ Problem 18:

"o "o coils havig 200 ad -00 turs respectively are "oud side by side o a commo closed core "hose cross sectio is 100 cm 2 ad mea legth of 200 cm. +hat "ill be the voltage iduced i the secod coil if the curret i the first coil chages from 0 to 5 A i i 0.05 secod. Assume relative permittivity of the core to be 1500. A. 52.- , &. .8 , %. 2. , (. 58. , Problem 188: 77 &oard ctober 1 he eergy >+o? >+o? stored i a coil is depedet i the iductace >;? of the coil ad the curret flo"ig. If the iductace "ere doubled "ith the same curret flo"ig! "hat "ould be the resultig stored eergy# eergy# A. 2.0 +o &. 0.5 +o %. 1.0 +o (. $.0 +o Problem 18: 77 &oard ctober 10 A .0 / coil "hose resistace is 12 ohms is coected "ith series "ith a 2$ ohms resistor ad to a 1$$ , battery ad a s"itch. s"itch. he s"itch is closed closed at t < 0. (etermie the eergy eergy stored i the magetic magetic field at steady state. state. A.

50 S

&.

$8 S

%.

52 S

(.

$ S

Problem 10: "o "o coils havig self6 iductace ad mutual iductace of -/! 2/ ad 1/ respectively are coected i series. If a curret of $ A is is passed through the coil! evaluate the ratio of the eergy stored i the magetic field "he the fluxes >self ad mutual? are i the same directio to the eergy stored "he the said fluxes are agaist each other. A. -.2$ &. 2.-%. 1.(. 2. Problem 11: 77 &oard April 15 "o "o coils of iductaces $ ad  hery are coected i parallel. If their mutual iductace is - hery! calculate the euivalet iductace of the combiatio if mutual iductace assist ad if mutual iductace opposes the self6 iductace respectively. respectively. A. 2.$ / M 1.$ / &. -.5 / M 0.$ / %. .25 / M 2 5 / (. 1 / M $ / Problem 12: A 506 m/ iductor is charged to store 1 S of eergy. eergy. If a ucharged -06 m/ iductor is the coected across its termials! ho" much eergy is trasferred to the secod coductor# A. 0.15$ S &. 0.22 S %. 0.2-$ S (. 0.1$1 S Problem 1-: A soleoid soleoid 500 cm log ad 10 cm i diameter is "oud "ith 1500 turs. Eid the eergy stored i the magetic field "he a curret of 5 A flo"s. A. 0.$5 S &. 0.55 S %. 0.$ S (. 0.2 S Problem 1$: A coil of 250 turs is "oud o a closed rig iro "hose cross sectioal area is 20 cm 2 ad mea legth of 100 cm. (etermie the eergy stored i the magetic field if a curret of 2 A is passed through it. Assume Assume the permeability of iro is is 00. A. 2.8- S &. -.12 S %. 2.0- S (. 1.88 S

Test Test ) # Sinle#Pase *C Syste!+ Part ( Problem 15: 77 &oard ctober 1 +hat is the "avelegth of a carrier "ave "ith a freuecy of 100 megahertB# A. -.0 m &. .5 m %. 1.5 m (.

.0 m

Problem 1: 77 &oard April 1 A chart speed of a recordig istrumet is 25 mmsec. e c ycle of the sigal beig recorded exteds over 5 mm. +hat is the fr euecy of the sigal# A. 20 cps &. 2 cps %. 50 cps (. 5 cps Problem 1: 77 &oard April 10 A 2$06 ,! 25 /B siusoidal geerator is coected to a 20 ohms resistor. (etermie the istataeous curret "he elapsed time is 0.01 secod. A. 15.$- A &. 1.-0 A %. 1. A (. 12.00 A Problem 18: 77 &oard April 1 A "ire carries a curret! I < - cos -1$t amperes. +hat is the average curret over  secods# A. 0A &. 1.5 A %. -.0 A (. 0.52- A Problem 1: "o "o curret sources deliver a commo load. he first source delivers a curret "hose euatio is 25 si 100π 100πt amperes "hile the secod delivers a curret "hose euatio is 15 cos 100 πt amperes. +hat is the rms value of the curret i the load# A. 2.15 A &. $0 A %. 20. A (. 10 A Problem 200: "o "o alterators A ad & delivers 100 ad 150 A! respectively to a load. If these currets are out of phase by -0 electrical degrees! determie the total curret dra" by the load. A. 201.5 A &. 250.0 A %. 215.$ A (. 2$1.8 A Problem 201: 77 &oard Sue 10 hree loads! uits A! & ad % are coected i parallel ad ta3e currets that are respectively 12! 10 ad 15 A respectively. Assumig l a to be the referece phasor. l b leads la by -0o ad lc lags behid l a by 5 o! calculate the total >resultat? curret. A. 28.-- A &. -0.21 A %. 2.$ A (. -2.10 A Problem 202: 77 &oard ctober 18$ A idustrial coil has a resistace of -2 ohms ad a reactace of 2$ ohms ad rated $$0 volts at 0 /B. A factory "ill coect the coil to a $$0 ,! 50 /B supply. Dolve for the value of a series resistor eeded to avoid overcurret coditio. A. 2.0 Ω &. 2.$ Ω %. 2.$$ Ω (. 2.25 Ω Problem 20-: 77 &oard April 1 Across a 2-0 ,! 0 /B po"ers supply is a 156 ohm o6 iductive resistor. +hat is the euatio of the resultig curret# A. 21.8 si -t &. 2.55 si -t %. 15.-- si -t (. 28.1 si -t Problem 20$: 77 &oard ctober 18 "o "o relays each "ith 20 ohms resistace ad 0.1 hery iductace are coected i series. +hat is the euivalet impedace# A. 20 = * 102.2 ohms &. 20 =* 5.-2 ohms %. $0 = * 120.- ohms (. $0 = 25.-2 ohms Problem 205:77 &oard ctober 10

A iductive coil ta3es a curret of 2 A ad cosumes 10 + "he coected coecte d to a 2$0 , A% supply. A secod coil "he coected across the same supply ta3es - A ad 500 +. Eid the total po"er "he the t"o coils are coected i series to this supply. A. 1$$.5 + &. 1-$.-1 + %. 150.22 .22 + (. 128.-5 .-5 + Problem 20: A 4 hp! 1106,! 0 /B! sigle phase iductio motor motor has a efficiecy efficiec y of 88J ad a po"er factor of 0.0 laggig at rated load. his motor motor is to be coected temporarily o a 220 ,! ,! 0 /B lie. (etermie the resistace reuired to be placed i series "ith the motor i order to prevet the machie from experiecig overcurret# A. 25.2 25.2 ohms ohms &. 2-.5 2-.5 ohm ohms %. 1.5 1.5 ohm ohms (. 2. 2. ohms ohms Problem 20: "o "o coils A ad & 3o" to have the same resistace resistac e are coected coecte d i series across a 1106,! 1106,! 06cycle lie. he curret ad po"er delivered by the source are respectively $.1 A ad -00 +. If the voltage across coil A is t"ice that across coil &! calculate the iductace of coil &. A. 8.- m/ &. .- m/ %. .02 m/ (. $.$ m/ Problem 208:77 &oard ctober 185 A coil dra"s 185 "atts "he coected to a 150 , (% source. It cosumes -02 "atts "he used o 2$0 ,! 0 /B A% source. Eid the iductace of the coil. A. 0.0255 / &. 0.0-$1 / %. 0.015- / (. 0.02$0 / Problem 20:77 &oard ctober 1$ A curret of 10 A ad a po"er factor of 0.8 laggig is ta3e from a sigle phase 250 volt supply. he reactive po"er of the system system is A. 1500 &. 2000 %. 2500 (. oe of these Problem 210:77 &oard ctober 1 A resistor of  ohms ad a u3o" impedace impedac e coil i series dra"s 12 A from a 1206,! 06/B lie. If the real po"er ta3e from the lie is 1152 "atts! "hat is the coil iductace# A. 15. m/ &. 10.0 m/ %. 20.0 m/ (. 1.5 m/ Problem 211:77 &oard April 1 (etermie the po"er agle i the series circuit "hich cosists of @<25 ohms! ;<0.2 / across a po"er supply of 200 volts! -0 /B# A. -.$° -.$° &. 52.$° 52.$° %. $.$° $.$° (. 5.$° 5.$° Problem 212: +he a -0 , (% source is applied to a give coil! the coil dra"s 150 +. +he 2-0 , (% is applied to the same coil! it dra"s -1$ +. (etermie (etermie the operatig po"er factor of the coil. A. 0. &. 0. %. 0. (. 0.8 Problem 21-:77 &oard April 1he impedace coil absorbs 250 "atts "he coected across 220 ,! 0 /B mais. It is the coected across 110 ,! 25 /B mais ad also absorbs 250 "atts. +hat is the coductace of the coil# A. 0.125 / &. 0.1$ / %. 0.15$ / (. 0.1- / Problem 21$:77 &oard April 15 A impedace coil ta3es10 A ad absorbs 250 + "he coected across a 220 ,! 0 /B source. +hat po"er "ill it absorb absorb "he coected across 110 110 ,! ,! 25 /B /B mais# A. 5- + &. 2- + %. $- + (. -- + Problem 215: 77 &oard ctober 18$

A idustrial coil has a resistace of -2 ohms ad a reactace of 2$ ohms ad rated $$0 volts at 0 /B. A factory "ill coect the coil to a $$0 ,! 50 /B supply. /o" much percetage overcurret "ill the coil suffer# A. 5J &. 10J %. J (. 8J Problem 21: he ope circuit voltage across the output termials of a A% po"er supply is 15.2 ,. +he these termials are short6 circuited! a curret of -.2 A flo"s. +he a 56 ohm resistace is placed across these termials! the curret through it is 1.8 A. %alculate the series iteral resistace of the po"er supply. A. 2.-$ ohms &. 2.2$0 ohms %. 2.125 ohms (. 2.0$5 ohms Problem 21: A sigle6 phase load ta3es 55 3+ at 0J pf laggig from a 2$0 ,! 50 /B supply. If the supply is made 0 /B! "ith the voltage remaiig the same! "hat "ill be the 3+ load at 0 /B# A. $2 3+ &. $8 3+ %. $0 3+ (. $5 3+ Problem 218: 77 &oard Carch 18 A 256 ohm resistor coected i series "ith a coil of 506 ohm resistace ad 150 m/ iductace. +hat is the po"er factor of the circuit# A. 85 J &. 80 J %. 0 J (. 5 J Problem 21: 77 &oard ctober 18 he ohmic resistace of a large magetic cotractor is measured to be 20 ohms. A 2-0 , is impressed o the cotractor ad the curret is ta3e as -.2 A. eglectig core loss! determie the iductace if the cotractor i m/# A. 21 &. -15 %. 18(. 251 Problem 220: A 86 ohm resistor is coected i series "ith a iductive coil ad the combiatio is coected across a 120 ,! ,! 0 /B source. A voltmeter coected across the resistor reads -2 , ad 10$ , "he coected across the coil. (etermie the resistace of the coil. A. 8Ω &. Ω %. 12 Ω (. 10 Ω Problem 221: +he coected to a 2206 ,! 50 /B supply! a impedace coil circuit ta3es 5 A! but this curret falls to $.$ A "he a 106 ohm resistor is added i series. Eid the iductace of the coil. A. 0.20 / &. 0.15 / %. 0.2- / (. 0.12 / Problem 222: A impedace coil! i series "ith a resistor ta3es a curret of 5 A "he coected to a 1006 , A% source. 9sig a voltmeter! the voltage across the coil "as foud out eual to 80 , ad that across the resistor eual to -0 ,. Eid the operatig po"er factor of the coil. A. 0.5 &. 0.8%. 0.(. 0.2 Problem 22-: 77 &oard Carch 18 A load of 20 = *-5 ohms is coected across a220 volts source. (etermie the po"er factor ad the ,A@D. ,A@D. A. $. J! 10$2 ,A@D &. 85.- J! 5 ,A@D c. 52.2 J! 102- ,A@D (. $2.- J! 108 ,A@D Problem 22$: A load is rated at 115 ,! 0 /B! 10 A ad 1150 +. A coil is to be "oud "ith a ratio of K ; to @ of 5! such that "he coected i series "ith the load o 2-06 ,! 0 /B lie! the load may operate ormally. Dolve for @. A. 2.$50 ohms &. -.$8 ohms %. 2.215 ohms (. -.82 ohms

Problem 225: 77 &oard ctober 10 Eid the total impedace i rectagular for the follo"ig three impedaces: 12 ∠10o! 25∠ 25∠15o ad -$∠ -$∠2o A. .52 = * 2-.$ &. 8.-$ = * 20.5$ %. $.-1 = * 21.5 (. .0 = * 22.0$ Problem 22: 77 &oard ctober 1 A impedace dra"s a curret I < 10 cos >"t ' -0 o? from a voltage! v < 220 si >"t = -0 o?. +hat is the impedace# A. 15. ' * 15. &. 15. = * 15. %. 1.1 ' * 11.1 (. 11.0 = * 1.1 Problem 22: 77 &oard April 10 A series resistace6 capacitace capacitace >@6 %? circuit is coected to a 2-0 volt 0 cycle source. If the po"er ta3e by the circuit is $! 800 800 "atts ad the voltage voltage drop across the resistor resistor is 115 volts! volts! calculate the capacitace capacitace of the capacitor. A. 5$0 µE &. 5-0 µE %. 55 µE (. 50- µE Problem 228: 77 &oard Carch 18 A 500 ad 1006 microfarad capacitors are coected i series ad across a 100 si >"t = -0 o? voltage. +rite the euatio of the curret. A. 1.2 si >"t = 120 o? &. 1.2 si >"t = 0o? %. 5.5 si >"t =120 o? (. 5.5 si >"t = 0 o? Problem 22: 77 &oard ctober 1 A series circuit composed of 1006 ohm resistor ad a 206 microfarad capacitor coected across a 2$06 ,! 0 /B lie. +hich of the follo"ig as"er is +@O# A. he he im impeda edace ce of th the ci circu rcuit is 1 1 oh ohms &. Agl Aglee bet bet"e "ee e the the cur curre ret t ad ad the the volt voltag agee vec vecto tors rs is 5-.1 5-.1 degr degree eess %. he re resultig cu curre rret is is 0. 0.2 2- am ampere (. he he vol volta tage ge acro across ss the the res resis ista tac cee is is 1$$ 1$$. . volt voltss Problem 2-0: 77 &oard April 1$ A capacitace is coected to a 1156 ,! 256 /B mais ad ta3es 5 A. +hat curret "ill it ta3e "he the capacitace ad the freuecy are both doubled# A. 2.5 A &. 5A %. 20 A (. 10 A Problem 2-1: A ideal iductor of 50 m/ is coected i series "ith a 156 µE capacitor. he combiatio is the coected across a 1206 , variable freuecy A% source. At "hat freuecy "ill this circuit ta3e a leadig curret of 5 A# A. 25 /B &. -0 /B %. 28 /B (. 22 /B Problem 2-2: A series cosistig cosistig of a variable resistor i series "ith a capacitace of 80 µE is coected across a 1206 ,! 50 /B supply. o "hat value should @ is ad*usted so that the po"er absorbed by the series circuit "ill be 100 +# A. 20 ohms &. 18 ohms %. 15 ohms (. 12 ohms Problem 2--: 77 &oard ctober 1 A capacitor is rated 100 3,A@! -80 volts! 50 /B. +hat "ill its ratig be at 0 /B! 220 volts# A. 50 3,A@ &. $0 3,A@

%.

0. 3,A@

(.

5. 3,A@

Problem 2-$: 77 &oard ctober12 A resistor ad a capacitor are coected i series across a supply of 250 ,. +he +he the supply freuecy is 50 /B the curret i the circuit is 5 A. +he the supply freuecy is 0 /B! the curret is 5.8 A. Eid the value of the capacitace. A. 58.- µE &. 0.2 µE %. .1 µE (. 0.2 µE Problem 2-5: 77 &oard ctober 1A series circuit composed of a 0.2 hery iductor ad $6 microfarad capacitor is coected to a 0 , variable freuecy source. At "hat freuecy is the curret be $ amperes "ith a laggig po"er factor# A. 50 /B &. 51 /B %. $8 /B (. $ /B Problem 2-: A sigle phase laggig po"er factor load ta3es -00 + ad 5A at 120 ,. fid the reactace of a pure capacitor that maybe placed i series "ith this load so that it "ill operate ormally from a 2$06 , source. A. 2.-2 ohms &. .25 ohms %. 8. 15 ohms (. $.$8 ohms Problem 2-: A coil >@;? is coected i series "ith a capacitor across a 220 ,! 0 /B A% supply. he circuit is desiged such that the voltage across the coil ad the capacitor are umerically eual. If this circuit operates at 0.8 leadig pf! determie the magitude magitude of this voltage. A. 212 , &. 228 , %. 205 , (. 22- , Problem 2-8: A series circuit coected across a 1006 , A% source cosist of a 106 Ω resistor! a 56 Ω capacitor! a resistace @ that ta3es 50 + ad a reactace K ta3ig 100 iductive ,A ,A@D. (etermie the miimum value of curret reuired satisfyig the above coditios. A. 1.20 A &. 0.2 A %. 1.0$ A (. 1.1- A Problem 2-: 77 &oard ctober 18 he maximum istataeous voltage ad curret outputs of a alterator are -00 , ad 20 A! respectively. respectively. o +hat is the po"er output i "atts if the voltage leads the curret by -0 # A. 258 &. 000 %. 51 (. -000 Problem 2$0: 77 &oard ctober 18 A 506 microfarad is coected "ith a coil havig 50 ohms resistace ad 1506 m/ iductace. he source voltage is 100 si >"t6 -0 o? ,. +hat is the maximum po"er# A. 1 "atts &. 1$ "atts %. 212 "atts (. 15 "atts Problem 2$1: 77 &oard ctober 1 A impedace dra"s a curret I < 10 cos >"t6 -0 o? from a voltage v < 220 si "t. +hat is the maximum po"er# A. 2200 "atts &. 1100 "atts %. 10.5 "atts (. 1-20 "atts Problem 2$2: 77 &oard April 15 A icadescet lamp load geerally cosidered to be made up of resistors ta3e $.8 3+ from a 1206 , A% source. he istataeous maximum of po"er is A. $800 + &. 2$00 + %. $80 + (. 00 +

Test Test , # Sinle#Pase Sinle#Pas e *C Syste!+ Part (( Problem 2$-: 77 &oard ctober 181 A circuit cosists of K a < *5 ohms! K c < 6*5 ohms ad @ < 5 ohms all are coected i parallel. Eid the euivalet impedace. A. 5.5 Ω &. 5.0 Ω %. $.8 Ω d. 5.2 Ω Problem 2$$: 77 &oard ctober 185 Oive: )1 < *2.5 ohmsT ) 2 < *$ ohmsT ) - < 5 ohmsT ) $ < 1 = *5 ohms. If the four impedaces i mpedaces are coected i parallel! fid the euivalet impedace i ohms. A. $.1 = *0.2 &. $.- = *0.$5 %. $.2 = *0.-5 (. $.0 = *0. Problem 2$5: 77 &oard April 18$! 77 &oard April 18 hree impedaces ) a < - = *$ ohms! ) c < $ ' *$ ohms ad ) r < 0 = *- ohms are coected i parallel. Dolve for the pf of the combiatio. A. 0.5- leadig &. 0.55$ laggig %. 0.50- leadig (. 0.20 laggig Problem 2$: 77 &oard ctober 18A pure capacitace of 5-0.515 x 10 6 farad ad a iductace of 5-0.515 × 1061$ hery are coected i parallel across a A% po"er source. Dolve Dolve for their resultat impedace impedace assumig that the freuecy freuecy is -0 /B. A. 10Ω 10Ω &. Ifiite %. )ero (. 9defied Problem 2$:77 &oard Carch 18 A coil of a 506ohm resistace ad of 150 m/ iductace is coected i parallel "ith a 50 µE capacitor. +hat is the po"er factor of the circuit# A. 80J &. 50J %. 0J (. 0J Problem 2$8:77 &oard April 182 hree impedaces ) a! )c! )r are are coected i parallel. If at 0 /B! ) a<0=*8! )c<06*2 ad ) r <5=*0 <5=*0 ohms. Dolve for the resultat po"er factor. A. 0. 0.$1 laggig &. 0. 0.$1 leadig %. 0. 0.5- laggig (. 0. 0.5- leadig Problem 2$:77 &oard ctober 1 A resistor of 50 ohms ad a impedace of 100 = * 50 ohms are coected i parallel across a 220 volts supply. supply. +hat is the po"er factor of the load# A. J &. J %. 8J (. 5J Problem 250:77 &oard April 10 A capacitor! capacitor! a electric resistace heater! ad a impedace are coected i parallel to a 120 ,! 0 /B system. he capacitor dra"s 50 ,A@! the heater dra"s 100 + ad the impedace coil dra"s 2 ,A at a pf of 0.$ laggig. (etermie the system po"er factor. A. 0.-- laggig &. 0.28 laggig %. 0.1 laggig (. 0.11 laggig Problem 251: A give load ta3es $0 3,A at 50J laggig po"er factor! "hile aother load coected i parallel across the same source ta3es 80 3,A at 8.J laggig po"er factor. Eid the apparet po"er. A. 11$ 11$.5 .5 3,A 3,A &. 11.2 1.2 3,A 3,A %. 11.$ 1.$ 3,A 3,A (. 115. 115.- 3,A 3,A Problem 252:

A $$0 ,! ,! 0 /B geerator! geerat or! supplies t"o parallel6coecte parallel6 coected d loads A ad &. ;oad A dra"s a apparet appar et po"er of 100 3,A at 0.80 pf laggig ad load & dra"s a apparet po"er of 0 3,A at uity pf. (etermie the 3,A@ supplied by the geerator A. 10.5 3,A@ 3,A@ &. 120.2 3,A@ 3,A@ %. 10$.- 3,A@ 3,A@ (. 115. 3,A@ 3,A@ Problem 25-:77 &oard April 12 "o sigle6 phase motors are coected i parallel across a 1206volt! 06cycle source of supply. Cotor A is a split6phase iductio type ad motor & is a capacitor type: (etermie total po"er factor. A. 0.88 lag &. 0.8$ lag %. 0.81 lag (. 0.825 lag Problem 25$: A 25063,A! 0.5 laggig po"er load is coected i parallel to a 1806+! 0.8 leadig po"er factor load ad to a -006,A! 100 ,A@ iductive load. (etermie the total apparet po"er i 3,A. A. 28 &. 5$8 %. 15 (. 0$ Problem 255:77 &oard April 12 A 250 ,! -0 /B geerator supplies po"er to a parallel circuit cosistig of a 20 hp motor "hose efficiecy is 0J at 0.80 pf laggig ad a secod load that dra"s a apparet po"er of  3,A at uity pf. (etermie the system po"er factor. A. 0. 0.828 laggig &. 0. 0.8-1 laggig %. 0. 0.802 laggig (. o. o.88$ laggig Problem 25:77 &oard April 185 A resistace of 5 ohms is coected i series "ith a capacitor of $$2.1 µE. he combiatio is the coected i parallel "ith a iductace of 21.22 m/. Dolve for the resultat curret if the circuit is coected across a 120 ,! 0 /B A% source. A. .$$ A &. 10.8 A %. 11.2 A (. 10.-$ A Problem 25: "o impedaces ) 1 < - = * $ ad ) 2 < 5 ' * 8. ohms respectively are coected i parallel. If the combiatio is coected across a 2$0 , A% source! ho" much is the total curret# A. $$.$ A &. $2.1 A %. $0. A (. -. A Problem 258: A coil havig a resistace of -0 ohms ad a iductace ; is coected i parallel "ith a resistor havig a resistace of 1006ohm. he combiatio is coected across a 1006,! 0 /B source. If the po"er delivered by the source is $00 +! fid the value of ;. A. 0.20$ / &. 0.011 / %. 0.02 / (. 0.10 / Problem 25: 77 &oard April 1A iductor ; 1is coected i series "ith a parallel combiatio of iductor ; 2 ad capacitor %. he impedace of the circuit "< $00 rasec is *100 ohms. he circuit is to yield ifiite impedace at "< 1! 000 radsec ad Bero impedace at "< 2000 radsec. (etermie the value of %. A. 1.2 µE &. 1. µE %. 2.0 µE (. 1.-2 µE Problem 20: A resistor ad a iductor are coected i parallel across a 1206 ,! 0 0 /B source. if the total curret ad po"er are respectively 15 A ad 100 "atts! solve solve for the iductace of the the iductor. A. 50.2 m/ &. $8.- m/ %. $-.2 m/ (. $.- m/ Problem 21:

A iductive coil "ith a resistace @ ad iductace ; is coected i parallel "ith a -06 ohm resistor. he combiatio is the coected across a 0 /B A% source. If the curret i the coil! resistor ad the total are ! $ ad 8 A! respectively! determie the iductace ; of the coil. A. 51.- m/ &. $8.-2 m/ %. 0.-5 m/ (. 5-.2 m/ Problem 22: 77 &oard April 12 A siusoidal siusoidal curret source! 10 cos 1000t! is i parallel both "ith a 206 ohm resistor ad the series combiatio of a 106 ohm resistor ad a 106 m/ iductor. Eid the euatio of the voltage across the 106 ohm resistor. A. -.25 cos >1000t ' 18.$- o? &. 1.-2 cos >1000t ' 20.-$ o? %. 5.- cos >1000t ' 1.$5 o? (. 0.12 cos >1000t ' 1.-8 o?

Problem 2-: 77 &oard April 1A 16hp! 220 ,! 0 /B capacitor6 start motor has mai ad auxiliary "idig impedace at startig of -.5 = *2.5 ohms ad 8. = *2.5 ohms! ohms! respectively. respectively. (etermie the value of of the startig capacitace capacitace that "ill place the mai mai o ad auxiliary "idig currets 0 apart from startig. A. 18.5 µE &. 1$.-5 µE %. 182.$- µE (. 10. µE Problem 2$: 77 &oard ctober 10 "o impedaces A ad & are coected i parallel across a 1206 , A% supply. he total curret ad the curret i each impedace are ad*usted to 20 A. he po"er dra" by A is doubled that of & ad the po"er factor is laggig. (etermie po"er factor of A. A. 0.50 laggig &. 0.0$ laggig %. 0.8$1 laggig (. 0. laggig Problem 25: 77 &oard Carch 18 A coil of 506 ohm resistace ad of 1506 m/ iductace is coected i parallel "ith a 506 µE capacitor. he source voltage is 100 si >"t = -0 o?. +hat is the euatio of the lie curret# A. 1.1 si >"t = 52.5 o? &. 1.25 si >"t = 5.5o? %. 1.82 si >"t ' 2 o? (. 1.-2 si >"t ' 5.5 o? Problem 2: 77 &oard ctober 18$ A resistor @ is coected i parallel "ith a 106 ohm iductive reactace. he combiatio is the coected i series "ith a $6 ohm capacitive reactace. he "hole combiatio is coected across a 1006 volt! 0 /B supply. supply. /o" much is @ if the agle bet"ee the supply voltage ad the total curret is $5 degrees. A. 12 Ω &. 25 Ω %. 1 Ω (. 20 Ω Problem 2: A resistace of 20 ohms ad a u3o" capacitace are coected across a 110 ,! variable freuecy A% source. +he the freuecy is 0 /B! the curret dra" by the circuit is  A. At "hat freuecy "ill the curret dra" falls to 5.8 A# A. $2.-- /B &. 50.12 /B %. $.02 /B (. $$.18 /B Problem 28: A sigle6 phase capacitor6 start suirrel6 cage iductio motor ta3e 2.5 A from a 2206 , lie. he curret i the startig "idig is 1.- A ad the curret i the mai "idig is 1.$5 A. he total po"er iput is 550 +. +hat is the resistace of the mai +idig# +idig# Assume the auxiliary "idig ta3es a leadig curret "hile the mai "idig ta3es a laggig curret.

A. %.

1-.$- Ω 1$2.$5 Ω

&. (.

125.8 Ω 151. Ω

Problem 2: A iductive reactace of 10 ohms is coected i parallel "ith a capacitive reactace of -0 ohms. If the combiatio is coected i series "ith a 10 ohm resistace! solve for the euivalet po"er factor of the "hole combiatio. A. 0.555 laggig &. 0.58 laggig %. 0.0- laggig (. 0.2- laggig Problem 20: A resistor @ ad a variable capacitor % are coected i parallel. he combiatio is the coected i series "ith a iductive reactace of 25 Ω. he circuit is eergiBed by 1106,! 0 /B source. If the capacitace is varied util the voltage across the resistor ad that across the iductor are eual to the supply voltage! determie the real po"er supplied by the source. A. 5- + &. $1 + %. 20 + (. $02 + Problem 21:77 &oard April 180 hree impedaces ) 1<16*$! )2< 6* ad ) -< $= *- ohms respectively are coected i series parallel. ) 1 is coected i series "ith the parallel combiatio of ) 2 ad )-. (etermie the euivalet impedace of the combiatio. A. $. $.-2 ' *1.21 &. 2. 2.2- ' *-.-2 %. . . ' *5.8 (. 5. 5.$2 ' *.21 Problem 22: A iductive reactace of - ohms is coected i parallel "ith a capacitive reactace of $ ohms. If the combiatio is coected i series "ith a $6ohm resistace! solve for the po"er factor of the "hole combiatio. A. 0.--&. 0.$0 %. 0.5 (. 0.-1 Problem 2-T A iductive reactace of 10 ohms is coected i parallel "ith a capacitive reactace of -0 ohms. If the combiatio is coected i series "ith a 106ohm resistace! solve for total real po"er if the "hole circuit is coected across a 2$0 ,! 0 /B supply. A. 2.05 3+ &. 2.8 3+ %. 1. 3+ (. 1.25 3+ Problem 2$:77 &oard ctober 18$ A 56ohm resistor is coected i parallel "ith a 106ohm iductive reactace. he combiatio is the coected i series "ith a $6ohm capacitive reactace. he "hole combiatio is coected across a 1006volt! 0 /B supply. /o" much is the total curret dra" by the circuit# A. 22.- A &. 20.$5 A %. 2-.1 A (. 1.8 A Problem 25:77 &oard April 18A o6iductive o6iductive resistor is coected i parallel "ith a iductive reactace of 10 ohms. he combiatio is the coected i series "ith a capacitive reactace of 5 ohms. he "hole combiatio is coected across a 1006,! 1006,! 0 /B A% source. If @ is eual to 5 ohms! solve for the voltage across the parallel combiatio. A. 8.5- , &. 88.-$ , %. 8.$$ , (. 1.8 , Problem 2:77 &oard April 180 hree impedaces ) 1 < 1 ' *$! )2 < 6* ad ) - < $=*- ohms respectively are coected i series parallel. ) 1 is coected i series "ith the parallel combiatio of ) 2 ad )-. If this circuit is coected across a 2-0 ,! /B source! determie the voltage across the parallel combiatio of ) 2 ad )-. A. 15.- , &. 1$.8 , %. 1-5. , (. 1-.2 , Problem 2:

A -06ohm resistor is coected i parallel "ith a 0.106/ iductor. iductor. he parallel combiatio is the coected i series "ith a 86E capacitor. If the "hole circuit is coected across a 125 ,! ,! 0 /B! fid the curret through the iductor. A. $.51 A &. -.2 A %. 2.2 A (. 2.-$ A Problem 28:77 &oard ctober 180 Oive three impedaces: ) 1<10=*0! ) 2<-=*$ ad ) -<86*. Impedace ) 2 ad )- are coected i parallel ad the combiatio is coected i series "ith impedace ) 1 across 120 , sigle6phase 0 /B source. Eid the total po"er dra" by the impedaces. A. 1008 + &. 120$ + %. 10-8 + (. 110- +

Test Test - # Sinle#Pase *C Syste!+ Part ((( Problem 2:77 &oard ctober 1If admittace L < 0.0 ' * 0.08 mho! the coductace O euals! A. '0.0 &. 0.0 %. 0.08

(. '0.08

Problem 280:77 &oard ctober 18! 77 &oard April 1A parallel circuit cosists cosists of a resistor havig a coductace of $ mhos! a iductive reactor havig a susceptace of 8 mhos ad a capacitive reactor havig a susceptace of 5 mhos. +hat is the impedace of the circuit# A. 0.11 0.11 = *0.1*0.1- &. 0.10.1- = *0.11 *0.11 %. 0.12 0.12 =*0.1 =*0.1 (. 0.1 0.1 = *0.12 *0.12 Problem 281: "o "o parallel braches have admittaces 0.- = *0.$ ad 0.2 ' *0.25! respectively! respectively! If the curret i the first brach is 10 A! determie the total total curret supplied to to the parallel combiatio. combiatio. A. 10.$$ A &. 12.10 A %. 15.-2 A (. 11.2$ A Problem 282:77 &oard ctober 1$ A capacitor brach havig a ratio of K c to @ of 5 is paralleled "ith a impedace cosistig of a $ Ω resistace ad a - Ω iductive reactace. he po"er factor of the resultig circuit is 0.8 leadig. Eid the siBe of the capacitor i microfarads if the freuecy is 0 /B. A. 8. &. 108. %. 8. (. 8. Problem 28-: "o impedaces ) 1 < 2 = * $ ad ) 2 < @ = *0 are coected i parallel. (etermie @ so that the po"er factor of the circuit is 0. laggig. A. -.2 ohms &. 2.1 ohms %. 2.8 ohms (. 2.5 ohms Problem 28$: +hat value of resistace should be placed i parallel "ith a 50 µE capacitor i order to have a total po"er factor of 0.8 o a 06cycle A% A% system# A. 50 Ω &. $0 Ω %. $2 Ω (. $ Ω Problem 285:77 &oard April 1%apacitor of -06microfarad capacitace is i series "ith a coil across a 8!0006cycle supply. supply. +hat iductace is reuired for resoace# A. 1$.-$ µ/ &. 10.$5 µ/ %. 1-.1 µ/ (. 12.55 µ/ Problem 28: A series @;% circuit is coected across a 1106,! 1106,! variable freuecy source. +he the freuecy is varied! the maximum effective curret is 5.5 A ad occurs at a freuecy of $5 cps.. At this freuecy freuecy!! the voltage across the capacitor is 200.8 ,. Eid the pf of the circuit at 0 /B.

A. 0. 0.8 laggig

&. 0. 0.0 laggig

%. 0. 0.- laggig

(. 0. 0.02 laggig

Problem 28:77 &oard ctober 18 e leg of a radio6tued circuit has a capacitace of oe times te to the mius ie farad. It is tured at 200 3/B! "hat is the iductace of the other leg i hery# A. .-- × 106$ &. 8.25 × 1065 %. 20 × 106(. 120 × 106Problem 288:77 &oard April 18 A loud spea3er "hose iductace is 1.15 heries is coupled to a po"er tube through a codeser of 2 µE capacity. o "hat freuecy "ill the combiatio be resoat# A. 110 /B &. 108 /B %. 105 /B (. 100 /B Problem 28:77 &oard April 15 +hat capacitace must be placed i series "ith a iductace of 0.05 / so that at 100 /B! the i mpedace becomes eual to the ohmic ohmic resistace# A. 50. µE &. -5.5 µE %. 0. µE (. 8.0 µE Problem 20: 77 &oard April 18 A coil has a resistace of 50 ohms ad a reactace of 100 ohms! is shuted by a capacitor! "hich has practically o losses losses i order that the voltage voltage across the coil be i the phase "ith the total total curret supplied to the parallel combiatio. combiatio. +hat is the impedace impedace of the parallel combiatio combiatio uder the give give coditio# A. 250 ohms &. 200 ohms %. 2-0 ohms (. 220 ohms Problem 21: 77 &oard April 18A o6 iductive resistor resistor @ is coected i parallel "ith a iductive reactace of 10 ohms. he combiatio is the coected i series "ith a capacitive reactace of 5 ohms. Dolve for @ at "hich the po"er factor of the give circuit "ould be uity. A. 10 Ω &. 12 Ω %. 1- Ω (. 11 Ω Problem 22: A 106 ohm resistor is coected i parallel "ith a iductive reactace of 10 ohms. he combiatio is the coected i series "ith a capacitive reactace of  ohms. Dolve for the freuecy at resoace. he idicated values have bee derived at 50 /B. A. 8.08 /B &. 1.22 /B %. .51 /B (. $.2- /B Problem 2-: A -006 ohm iductive reactace is coected across a series combiatio of a resistor @ ad a capacitive reactace Kc. Dolve for @ if the euivalet impedace of the combiatio is a pure resistace of 100 ohms. A. 2 ohms &. 88 ohms %. 8 ohms (. 0 ohms Problem 2$: 77 &oard ctober 182 "o impedaces ) 1 < 15 = *20 ad ) 2 < 5 ' * K c are coected i parallel. Dolve for the value of K c so that the total curret dra" by the combiatio "ill be i phase "ith ay supply voltage. A. 28.5$ Ω &. -0.$- Ω %. --.12 Ω (. 2.55 Ω Problem 25: A coil "hose resistace is 10 Ω ad a iductace of 50 m/ is coected i parallel "ith a series combiatio of 5 Ω resistace ad a variable capacitor %. he circuit is supplied by a 2206 ,! 0 /B freuecy. %alculate the po"er >miimum value for K c? delivered by a source at resoace.

A. %.

10.-1 3+ 1$.5 3+

&. (.

15.-5 3+ 11.0 3+

Problem 2: 77 &oard April 185 A resistace of 5 ohms is coected i series "ith a capacitor of $$2.1 µE. he combiatio is the coected i parallel "ith a iductace of 21.22 m/. Dolve for the freuecy of the impressed voltage "ith "hich the iductive reactace is eual to the capacitive reactace i magitude. A. 50 /B &. 51 /B %. 52 /B (. oe of these Problem 2: A impedace of $ ' *- ohms is coected i parallel "ith a ideal iductor ; across a 110 ,! 0 /B mais. (etermie ; if the total voltage supplied to the circuit "ill be i phase "ith the total curret dra". A. 1. m/ &. 1.2 m/ %. 22.1 m/ (. 2.8 m/ Problem 28: 77 &oard April 18 A coil has a resistace of 50 ohms ad a reactace of 100 ohms! is shuted by capacitor! "hich has practically o losses. losses. +hat must be the reactace of the capacitor i order order that the voltage across across the coil be i phase "ith the total total curret supplied to to the parallel combiatio# combiatio# A. 120 ohms &. 12 ohms %. 125 ohms (. 1-2 ohms Problem 2: 77 &oard April 182 hree impedaces ) a! )c! )r are are coected i parallel. If at 0 /B! ) a < 0 = *8! ) c < 0 ' *2 ad ) r < < 5 = *0 ohms. Dolve for the freuecy at resoace. A. -0 /B &. -$ /B %. - /B (. 28 /B Problem -00: A resistace @ is coected i parallel "ith a variable iductive reactace K ;. he combiatio is coected i series "ith impedace of 5 ' *2 ohms. Dolve for @! such so there is oly oe coditio of resoace as K; as varied. A. 2 ohms &. 5 ohms %. - ohms (. $ ohms Problem -01: 77 &oard April 181 A resistor @ is coected i parallel "ith a 206 ohm iductive reactace. he combiatio is the coected i series "ith 56 ohm capacitive reactace. Dolve for the value of @ at "hich the po"er factor of the resultat impedace is uity. A. 10.05 Ω &. .15 Ω %. 11.55 Ω (. 10.- Ω Problem -02: A coil cosumes 2! 000 "atts! "hile dra"ig a curret of 25 A from a 1106 ,! 1006 /B A% source. (etermie the capacitace of the capacitor to be coected i parallel "ith the coil i order to ma3e the total curret be i phase "ith the supply voltage. A. 2- µE &. 2$0 µE %. 2$8 µf (. 205 µE Problem -0-: A coil havig a impedace of 10 = *5 ohms is coected i parallel "ith a series combiatio of a variable resistace @ ad a capacitor havig a reactace of 8 ohms. he combiatio is coected across a 2$6 , A% source. (etermie the total curret dra" if @ is ad*usted i order to brig the total pf of the circuit to uity. A. -.-2 A &. $.5$ A %. 2.28 A (. $.0 A

Problem -0$: 77 &oard ctober 18 A coil has resistace of 50 ohms ad the reactace of 0 ohms. A capacitor is coected i parallel to produce resoace. he he source voltage is 120 120 ,. ,. +hat +hat is the po"er dra" by the the circuit# A. 12 + &.  + %. 1-2 + (. 52 + Problem -05: A coil "hose resistace is 100 Ω ad iductace of ; 1 is coected i parallel "ith a series combiatio of iductace ; 2 ad capacitace %. +he " < 1 x 10 5! the impedace of the "hole combiatio is -00 ohms! purely resistive. +he "< 2 x 10 5 o curret flo"s to"ards the coil. Eid the value of %. A. -8.2- E &. -$.02 E %. -5.- E (. -2. E Problem -0: A series @;% circuit has a Q of 5.0 at its resoace freuecy of 100 3/B. Assumig Assumig the po"er dissipatio of the circuit is 100 + "he dra"ig a curret of 0.8 A! determie the capacitace % of the circuit. A. 2.0$ E &. 2.-2 E %. -.02 E (. 2.5$ E Problem -0: A impedace coil dra"s a apparet po"er of 50 volt6 amperes ad a active po"er of $0 "atts. Dolve for the Q6 factor of the coil. A. 0. &. 1.25 %. 0.8 (. 0.5 Problem -08: 77 &oard April 15 A coil is supplied "ith 200 volts ad ta3es a curret >rms? of 2 amperes at 0.0 laggig. he uality factor >Q? of the coil is A. 25 &. 1 %. 10 (. 100 Problem -0: A o6 iductive resistor of 10 ohms reuires a curret of 8 A ad is to be feed from a 200 ,! 50 /B supply. If a cho3ig coil of effective resistace 1.2 ohms is used to cut do" the voltage! fid the reuired Q6 factor of the coil. A. 18. &. 1$.2 %. 20.(. 1. Problem -10: he total curret i the give circuit has the euatio: I < 10 =1 si"t ' 8 si"t = 5 cos -"t. %alculate the effective value of the curret. A. 18.- A &. 12.2 A %. 1.5 A (. oe of these Problem -11: 77 &oard April 12 (etermie >rms? value of the curret dra" by a 2 µE codeser! "hich is coected across a source of potetial. he he potetial has a third ad fifth harmoic compoets! compoets! "hich are 20 J ad -0 J respectively of the fudametal. he fudametal siusoidal compoet has a pea3 value of 1000 volts ad 0 /B freuecy. freuecy. A. 0.8 A &. 0.5 A %. 0.1 A (. 0.8$ A Problem -12: 77 &oard ctober 12 A capacitor of -.8 microfarads is coected i parallel "ith a resistace of 2000 ohms. he combiatio is further coected i series "ith a iductace of 5 m/ ad resistace of 100 ohms across a supply give by e < $0 si"t = 80 si >-"t = 0 o?. Assume " < -1$ radiassec. (etermie the rms value of the total curret. A. 0.$0 A &. 0.-- A %. 0.5 A (. 0.$5 A Problem -1-: 77 &oard April 11

(etermie the effective value of the circuit curret if a emf of 151 si -t is coected i series "ith a (% emf of 110 volts. &oth supply a load of 10 = *8 ohms. A. 10.- A &. 12.5 A %. 1-.8 A (. 11.$ A Problem -1$: 77 &oard April 1$ A alteratig curret ad a direct curret flo"s simultaeously i the same coductor. If the effective value of the alteratig curret is 5 A ad the direct curret is 10 A! "hat "ill a A% ammeter read "he coected i circuit# A. .5 A &. 15 A %. 11.18 A (. oe of these Problem -15: 77 &oard April 1A 100∠ 100∠0o , 120 /B geerator ad a 80∠ 80 ∠0o , 0 /B geerator are coected i series "ith a 0 , battery ad coil. he resistace ad iductace of the coil are - Ω ad 2.5 m/! respectively. respectively. (etermie the rms curret of the coil. A. $2.5$ A &. $$ 2$ A %. $-.55 A (. $0 $$ A Problem -1:77 &oard April 1 If e < 100 si >"t = -0 °? ' 50 cos -"t = 25 si >5"t = 150 °? ad I < 20 si >"t = $0 °? = 10 si >-"t = -0° -0 °? ' 5 si >5"t 6 50° 50 °?. %alculate the po"er i "atts. A. 11 &. 18.5 %. 10$d. 22.$ Problem -1: Eid the average po"er i a resistace @ < 10 ohms! if the curret i the Eourier6 series form is I < 12 si"t = 8 si"t =- si 5"t amperes. A. 1085 + &. 120- + %. 1150 + (. 102 + Problem 218:77 &oard ctober 12 A capacitor of -.18 microfarads is coected i parallel "ith a resistace of 2!000 ohms. he combiatio is further coected i series "ith a iductace of 5 m/ ad resistace of 100 ohms across a supply give by e < $00 si"t = 80 si >-"t = 0 °?. Assume " < -1$ radiassec. (etermie the po"er dissipated. A. $. + &. 8.05 + %. 80.28 + (. 5. + Problem -1: A series @; circuit i "hich @< 5 Ω ad ;< 20 m/ has a applied voltage e<100 = 50 si "t = 25 si -"t! "ith "< 500 radias per sec. (etermie the po"er dissipated i the resistor of the circuit. A. 2510 + &. 22-$ + %. 2052 + (. -202 + Problem -20: hree siusoidal geerators ad a battery are coected i series "ith a coil "hose resistace ad iductace are 8 Ω ad 2.5- m/! respectively. respectively. he freuecy ad rms voltages of the respective geerators are 15 ,! 20 /BT -0 ,! 0 /B ad $0 ,! 100 /B. he ope circuit of the battery is  ,. eglect iteral resistace of the battery. battery. Eid the apparet po"er po"er delivered by the circuit. circuit. A. 1$.$ ,A &. 18.5 ,A %. 18.- ,A (. 182. ,A Problem -21: A series circuit cotaiig a 25 µE capacitor ad a coil "hose resistace ad iductace are - Ω ad $.$2 m/! respectively are supplied by the follo"ig series coected geerators: -5 , at 0 /B! 10 , at 180 /B ad 8 , at 2$0 /B. (etermie the po"er factor of the circuit. A. 0.$8 &. 0.$18 %. 0.$5 (. 0.$- Problem -22:77 &oard 12

A capacitor of -.18 microfarads is coected i parallel "ith a resistace of 2!000 ohms. he combiatio is further coected i series "ith a iductace of 5 m/ ad resistace of 100 ohms across a supply give by e< $00 si"t = 80 si >-"t = 0 °0. Assume "<-1$ radias per sec. (etermie the circuit po"er factor. f actor. A. 0.02 &. 0.50 %. 0.-(. 0.12

TEST 1. # Tree#Pase *C Syste!+ Part ( Problem -2-:77 &oard April 1 A 10 3,! -6phase electric ele ctric source delivers deli vers 200 C,A to a balaced load! "hich "hic h has a po"er of 0J laggig. +hat is the lie curret# A. 25 A &. 502 A %. $02 A (.  A Problem -2$:77 &oard ctober 1 A three6phase three6phase motor is rated 50 hp! $$0 volts ad 85J po"er factor. +hat is its rated curret# A. 1.5 A &. 5.5 A %. 55 A (. 5 A Problem -25:77 &oard April 185 A balaced -6phase load dra"s 120 amperes lie curret at 2-0 volts lie to lie! 0.8$8 pf laggig curret. Dolve for the real po"er. A. $0.5$ 3+ &. $2.-5 3+ %. $1.$5 3+ (. $$-.15 3+ Problem -2:77 &oard Carch 18 A geerator supplies three6phase po"er to balaced load. he voltage is 2-0 volts! the curret is 18 A ad the po"er factor is 85J. +hat is the po"er# A. -. 3+ &. 1. 3+ %. .1 3+ (. 1.$ 3+ Problem -2:77 &oard April 18$ A balaced -6phase load dra"s 5 amperes lie curret at 2-0 , lie to lie ad 0.8$8 laggig po"er factor. Dolve for the reactive po"er beig dra". A. 15.8- 3,A@ &. &. 15.2 3,A@ %. 15.-5 3,A@ (. 15.$ 3,A@ Problem -28:77 &oard April 10 he iput po"er factor to a three6phase! 6poles! $0 volts! 0 /B! 506hp iductio motor is 0.2 as 20 A is dra" by the motor. Eid the po"er iput to the motor. A. !880 + &. !5 + %. !85 + (. !$8 + Problem -2:77 &oard April 10 A $0 volt! three6phase motor dra"s 208 A "ith a po"er factor of 0.1 laggig. %alculate the 3+ iput to the motor. A. 150.8 &. 15.%. 152.$ (. 10.Problem --0: (etermie the ratio of the lie curret dra" by three idetical impedaces coected i "ye to the lie curret dra" by the same impedaces coected i delta. &oth coectios beig impressed "ith the same balace -6φ -6φ voltages. A. 1:2 &. -:$ %. 2:(. 1:Problem --1T77 &oard April 1A "ye6coected "ye6coected load has a 5 ∠20° 20° ohm impedace per phase ad is coected across a 1206, three6phase source. %alculate the lie curret. A. 2$ A &. 1-.85 A %. $1.5 a (. 15.$5 A

Problem --2: A balaced delta coected load havig impedace per phase of 20 ∠-.8° -.8° ohms is supplied from a balaced -6phase! 2$0 , source. (etermie (etermie the total real po"er. po"er. A. 82$ + &. 12 + %. 015 + (. $0 + Problem ---:77 &oard April 1hree codesers! each havig capacity of 5 microfarads are coected i star to a $$0 volts! -6phase! 50 cycles supply. supply. %alculate the capacitace of each of the three codesers so that "he they are coected i delta to the same supply the lie curret remais the same. A. 20 µE &. 28 µE %. 25 µE (. -0 µE Problem --$:77 &oard Carch 18 hree impedaces! 6*10! *10 ad 10 ohms are "ye coected. (etermie the impedace of a euivalet delta. A. 12.5! *12.5! 612.5 ohms &. 10! *10! 6*10 ohms %. *8.5! 6*12.5! 8 ohms (. 5! *5! 6*5 ohms Problem--5: A system system cosistig of three eual resistors coected coected i star is fed from a three6phase supply. supply. &y ho" much is the po"er reduced if oe of the three resistors is discoected# A. 1&. -$ %. 12 (. 2Problem --:77 &oard April 1A balaced three6phase load is "ye6coected ad has a impedace ) p < $ ' * - ohms per phase. Eid the lie curret if this load is coected across a 220 , three6phase source. A. 25.$ A &. 22.- A %. 20.5 A (. 2. A Problem --:77 &oard ctober 18 hree 106ohm resistaces are coected delta o a balaced three6phase source. If the euatio of the phase ,a <120 si "t. +hat is the euatio of the lie curret i lie a# A. 20.8 si >"t = -0° -0°? &. 1-.15 si > "t 6 -0 °0 %. 12 si >"t ' 5.5° 5.5°? (. - - si "t Problem --8:77 &oard April 182 Oive a balaced -6"ire! three6phase system servig the follo"ig loads: ,ab <200∠ <200∠0° , )ab <10∠ <10∠5-.1-° 5-.1-° Ω , bc <200∠ <200∠2$0° 2$0° , ) bc <10∠ <10∠0° Ω ,ca <200∠ <200∠120° 120° , )ca <10∠ <10∠-0° -0° Ω (etermie lie curret o lie b. A. 20.-$ A &. 22.0$ A %. 2$.- A (. 21.5 A Problem --:77 &oard April 182 Oive the follo"ig lie volages ad t"o load impedacesT ,ab <11∠ <11∠0° , )ac <5=*12 ohms , bc <11∠ <11∠2$0° 2$0° , ) bc < 1-<*0 ohms ,ca <11∠ <11∠120° 120° , Dolve for the lie curret i lie c. A. 1.$1 A &. 1.5 A %. 1.2 A

(. 18.$ A

Problem -$0:77 &oard April 18 hree ueual sigle6phase loads so coected across the lies of a balaced! -6phase! 2-0 volts circuit. he first ta3es 10 A at 0.8 pf laggig ad is coected across lies 1 M 2. he secod ta3es 1$2 A! at 0.82 pf

laggig ad is coected across a lies 2 M -. Ad the third ta3es 28.$ 3+ at 0. pf laggig ad is coected across lies - M 1. fid the three lies currets. A. 25$.$0 A! 211.-8 A! 252 A &. 2-1.2 A! 215.20 A! 28 A %. 25$.$0 A! 215.20 A! 252 A (. 2-1.2 A! 211.-8 A! 28 A Problem -$1:77 &oard ctober 12 A 1206, per phase! three6phase L coected source deliveres po"er to the follo"ig delta6coected load: Phase 1 < $0∠ $0 ∠0° Phase 2 < 20∠ 20 ∠6 0° 0° Phase - < 15∠ 15 ∠$5° $5° (etermie the three lie currets. A. 12.$5 A!  A! 22.$5 A &. 1-.$ A!  A! 22.$5 A %. 1-.$ A! 10 A! 20.22 A (. 12.$5 A! 10 A! 20.22 A Problem -$2:77 &oard ctober 185 Oive: ,ab < 2$0∠ 2$0∠0° , )ab <  = *8 ohms , bc < 2$0∠ 2$0∠2$0° 2$0° , ) bc < 5 = *8. ohms ,ca <2$0∠ <2$0∠120° 120° , )ca < 10 = *0 ohms Dolve for the three lie currets l a! l b ad lc. A. la < $5 a! l b < $- A! lc <20 A &. la < $8 A! l b < $2 A! lc <2$ A %. la < $5 A! l b < $2 A! lc <20 A (. la < $8 A! l b < $- A! lc < 2$ A Problem -$-:77 &oard April 185 A three6phase 2-06, circuit serves t"o sigle phase loads! A ad &. ;oad A is is a iductio motor rated 8 hp! 2-0 v! 0.0 pf! 0.0 efficiecy! efficiecy! "hich is coected across lies a ad b. ;oad & dra"s 5 3+ at 1.0 pf ad is coected across lies b ad c. Assume a seuece of a6b6c! solve for the lie curret o lie b. A. $2.1 A &. 2.$ A %. $1.08 A (. -$.$ A Problem -$$: A four "ire 2086, three6phase system is used to supply po"er to a three6phase 56hp iductio motor ad a sigle phase  3+ heater coected bet"ee lie c ad the eutral lie. he operatig efficiecy ad po"er factor of the motor are 81 ad 1 percet! respectively. %alculate the curret i lie c. Assume a phase seuece of a6b6c. A. 1 A &. 5 A %. $ A (.  A Problem -$5:77 &oard April 180 A factory is supplied by a three6phase! -6"ire system "ith the follo"ig characteristics: ,ab < 2-0∠ 2-0∠0° , la < 110∠ 110∠ 6 -.8° -.8° A , bc < 2-0∠ 2-0∠2$0° 2$0° , lc < 125∠ 125∠5-.1-° 5-.1-° A Eid lie curret l b. A. 1$5.- A &. 1-.- A %. 18$. A (. 1.5 A Problem -$: Oive a balaced -6"ire! three6phase system servig a ubalaced "ye6 coected load. (etermie the lie curret i lie a. ,ab < 210∠ 210∠0° , )a < 10 , bc < 210∠ 210∠120° 120° , ) b <10 < *10 ,ca < 210∠ 210∠2$0° 2$0° , )c < 10 A. 15 A &. 12 A %. 1$ A (. 1 A Problem -$:77 &oard ctober 1$ hree resistors 10! 15 ad -0 ohmic values are coected i "ye6cofiguratio to a balaced 2086volt three6phase supply. supply. %alculate the total po"er of the system. A. 2$$ + &. 2880 + %. -080 + (. -280 +

Problem -$8:77 &oard march 18 hree impedaces each 10 = *5 ohms are coected i delta o a balaced three6phase source. If the euatio of the phase voltage is , a < 120 si "t. +hat is the euatio of the curret through the impedace coected across phase A ad &# A. 20.0 si >"t 622° 622°? &. 18.5 si >"t = -.$$ °? %. 1.21 si >"t = 5.5° 5.5 °? (. 21.-2 si >"t ' 8.15 °? Problem -$:77 &oard Sue 10 "o parallel coected loads A ad & are supplied by a $$0 ,! -6phase! 0 /B geerator. ;oad A dra"s a apparet po"er of 100 3,A at 0.80 pf laggig ad load & dra"s a apparet po"er of 0 3,A at uity pf. (etermie the feeder curret. A. 208 A &. 212 A %. 21$ A (. 202 A Problem -50: "o differet loads A ad & are coected across a balaced 2-0 ,! -6 φ lies. ;oad A is a three6phase load that dra"s 10 3,A at 0J laggig pf. ;oad & is a sigle6phase load coected across lies a ad b dra"ig 5 3,A at uity pf. (etermie lie curret i lie a. Assume a phase seuece of a6b6c. A. - A &. $2 A %. -1 A (. $ A Problem -51:77 &oard April 10 A three6phase motor ta3es 10 3,A at 0. pf laggig from a source of 2-0 volts. It is i parallel "ith a balaced delta load havig havig 1 ohms resistace resistace ad 12 ohms capacitive capacitive reactace i series i each phase. phase. (etermie the total po"er factor. A. 0. laggig &. 0.8 laggig %. 0.1 laggig (. 0. laggig Problem -52: A balaced delta load "hose impedace is $5∠ $5 ∠0° 0° ohms per brach! a three6phase motor that dra"s a total of 10 3,A at 0.5 pf laggig ad a "ye coected load "hose impedace is 10 ohms >resistace? per brach are supplied from a three6phase! three6"ire! 208 ,! 0 /B feeder. (etermie (etermie the overall po"er factor of the feeder loads. A. 0.5&. 0.0 %. 0.82 (. 0.2 Problem -5-: 77 &oard April 185 A three6 phase 2-06 , circuit serves t"o sigle phase loads! A ad &. ;oad A is a iductio motor r ated 8 hp! 2-0 ,! 0.0 pf! 0.0 efficiecy! "hich is coected across lies a ad b. ;oad & dra"s 5 3+ at 1.0 pf ad is coected across lies b ad c. Assume a seuece of a6b6c! solve for the total po"er factor of the load. A. 0.0 &. 0.0$ %. 0.8$ (. 0.88 Problem -5$: A $$06 , three6 phase! -6 "ire "ire! 0 /B feeder supplies po"er to t"o -06 hp iductio motors. e motor is operatig at full load! 85J efficiecy ad has a po"er factor of 0J. he other motor is operatig at oe ' half of its rated horsepo"er! 8J efficiecy ad has a po"er factor of 8J. +hat is the overall po"er factor of the system of motors# A. 0.82 laggig &. 0.82 laggig %. 0.8-8 laggig (. 0.801 laggig Problem -55: 77 &oard April 188 hree ueual sigle6 phase loads so coected across the lies of a balaced! -6 phase! 2-0 volts circuit. he first ta3es 10 A at 0.8 pf laggig ad is coected across lies 1 M 2. he secod ta3es 1$2 A! at 0.82 pf laggig ad is coected across lies 2 M -. Ad the third ta3es 28.$ 3+ at 0. pf laggig. (etermie the total apparet po"er. A. $ 3,A &. 8- 3,A %. 8 3,A (. 101 3,A Problem -5:

"o6balaced L6 coected loads A ad & are coected across a three6 phase -6 "ire balaced system. ;oad A dra"s  3+ at 80J po"er factor laggig "hile load & dra"s 12 3+ at 8-.5J po"er factor leadig. If the curret i the three lies are each eual to 8 A! determie the curret dra" by load A. A. $.$5 A &. -.58 A %. $.15 A (. -.28 A Problem -5: 77 &oard Carch 18 he phase & lie voltage ad the phase A lie curret of a balaced three6 phase system are ,< 220 si >"t o = 210 ? ad l < 10 si >"t ' -0 o?! respectively. +hat is the po"er of the system# A. 105 + &. --00 + %. 51 + (. -810 + Problem -58: 77 &oard ctober 18 he follo"ig iformatio is give for a delta6 coected load of three umerically eual impedaces that differ i po"er factor: ;ie voltage < 120 voltsT ) ab < 15∠ 15∠-0oT ) bc < 15∠ 15∠0oT )ca < 15∠ 15∠6-0 phase seuece of voltages is a6b6c. 9sig the phase seuece as a guide! calculate the total po"er dra" by the load. A. 2!2$ + &. 2!$2 + %. 2!5$ + (. 2!-1 + Problem -5: 77 &oard April 10 A three6 phase motor ta3es 10 3,A at 0. pf laggig from a source of 2-0 volts. It is i parallel "ith a balaced delta load havig havig 1 ohms resistace resistace ad 12 ohms capacitive capacitive reactace i series i each phase. phase. (etermie the total lie curret. A. -0.05 A &. -$.55 A %. -2.0$ A (. -1.2$ A P@&;7C -0: 77 &A@( AP@I; 1I A% circuits! fid the total po"er i 3+ i a ubalaced three6 phase circuit loaded as follo"s: Phase l < 120 ,! ,! 100 A uity pf. Phase ll < 100 ,! 2-0 A! 80 J pf ad phase lll < 110 ,! 85 A!  J pf. A. -. 3+ &. -5.- 3+ %. -2.8 3+ (. -1.2$ 3+ Problem -1: 77 &oard April 18"o "o sigle6 phase trasformers are coected i , >ope delta? ad servig a delta coected impedace load! each impedace is eual to 1 ∠-.8o ohms. If the trasformer voltages impressed o the impedaces are: ,ab < 2$0∠ 2$0∠0 o! ,ab < 2$0∠ 2$0∠2$0o! ,ab < 2$0∠ 2$0∠120o volts Dolve for the total 3,A dra" by the load. A. .2&. 8.-1 %. 10.8 (. 11.Problem -1: 77 &oard ctober 180! 77 &oard ctober 182 hree impedaces! ) a < 20 = *0! ) b < 1 = *12! ) c < 5 ' *15 ohms are coected i "ye across a 2-0 , >lie to lie?! -6 phase! $6 "ire source. he phase seuece is a6b6c! coutercloc3"ise. (etermie the curret passig thru the eutral. eutral. A. .5$ A &. .12 A %. 8.81 A (. 8.02 A Problem --: 77 &oard April 181 A "ye6 coected trasformer "ith eutral coectio has balaced voltages of 25 , bet"ee lies ad eutral. he trasformer is servig t"o sigle6 phase motors. Cotor A >rated $ hp! 0.0 efficiecy! efficiecy! 0.80 po"er factor laggig? is coected across lie a ad eutral. Cotor & >rated - hp! 0.85 efficiecy! efficiecy! 0.85 po"er factor! laggig? is coected across lie b ad eutral. Dolve for the eutral curret! usig , a as referece vector. A. 20.$2 A &. 25.- A %. 22.$5 A (. 2-.1$ A Problem -$:

"o "o sigle6 phase motors are coected bet"ee the lies ad eutral termials of a -6 phase! $6 "ire! 2-0 , secodary side of a three6 phase trasformer. ;oad A is is coected across lie a ad eutral "hile & is across lie b ad eutral. he currets currets l a ad l b are $0 A! 0.0 pf laggig! ad 25 A! uity pf! respectively. respectively. (etermie the eutral curret ! l . A. 5$.25 A &. 50.-$ A %. 51.8 A (. 52.- A Problem -5: "o "o sigle6 phase motors are coected to a -6 phase! $6 "ire "ye feeder havig 220 , bet"ee phase ad the eutral. e motor rated 5 /P! 0.8 pf laggig ad 0J eff is coected across phase A ad eutral. he secod motor is coected across phase % ad eutral! ad rated - /P! 0.0 pf laggig ad 0J eff. Dolve for curret through the eutral. A. 20.5 A &. 2.5$ A %. 21. A (. 2-.$- A Problem -: A 2-06 ,! ,! three6 phase! $6 "ire balaced system supplies po"er to group of lamp loads. If the lie currets are respectively 0 A! 8 A ad $0 A respectively! solve for the curret i the eutral "ire. Assume the po"er factor of lamps to be uity. A. $0 A &. 1$ A %.  A (. - A Problem -: 77 &oard ctober 18 he loads of a "ye6 coected trasformer are: Ia < 10 cis >6-0 o? I b < 12 cis 215 o Ic < 15 cis 82 o +hat is the eutral curret# A. 1.0$ cis 2.8o &. 2.21 cis >6-0o? %. 0.2 cis 2.5o (. -.11 cis 2.8o

TEST 11 # Tree#Pase *C Syste!+ Part (( Problem -8: 77 &oard April 180 A factory is supplied by a three6 phase! -6 "ire system "ith the follo"ig characteristics: ,ab < 2-0∠ 2-0∠0o , Ia < 110∠ 110∠ 6 -.8° -.8° A , bc < 2-0∠ 2-0∠2$0° 2$0° , Ic <125∠ <125 ∠5-.1-° 5-.1-° A (etermie the total po"er cosumed by the load. A. $2. 3+ &. $8.8 3+ %. $0.2- 3+ (. $5.12 3+ Problem -: A -6phase! - "ire feeder has the follo"ig lie currets ad lie voltages: ,ab <220∠ <220∠0° , Iaa < -06*-0 A , bc <220∠ <220∠2$0° 2$0° v Icc <2=*-.8 A (etermie the reactive po"er supplied by the feeder. A. .5 3,A 3,A@ &. .-$ 3,A 3,A@ %. 8.-2 3,A 3,A@ (. 8.$5 3,A 3,A@ Problem -0: Oive a balaced three6phase system "ith the follo"ig loads: ,ab <2-0∠ <2-0∠0° , )ab < 0 , bc <2-0∠ <2-0∠120° 120° , ) bc < -0=*50 ,ca <2-0∠ <2-0∠2$0° 2$0° , )ca < 6*80 (etermie the readig of the t"o "attmeters properly coected usig lie c as the commo poit. A. $$0. +! 0.5 + &. -$.1 +! 51.2 +

%. $$0. +! 51.2 +

(. -$.1 +! 0.5 +

Problem -1: Oive a balaced -6"ire! three6phase system servig the follo"ig loads: ,ab <220∠ <220∠0° , )ab <11∠ <11∠0° Ω , bc <220∠ <220∠2$0° 2$0° , ) bc <11∠ <11∠$5° $5° Ω ,ca <220∠ <220∠120° 120° , )ca <11∠ <11∠-0° -0° Ω (etermie the readig of the t"o "attmeters "ith lie b as commo. A. 21.5 +! $820 + &. 21.5 +! $$00 + %. -$2.5 +! $50 + (. -$2.5 +! $$00 + Problem -2: he follo"ig voltages ad lie currets "ere measured to a -6phase! -6"ire feeder servig a commercial buildig. ,ab <2!$00∠ <2!$00∠0° , )ab <85∠ <85∠--0° --0° A , bc <2!$00∠ <2!$00∠2$0° 2$0° , lc <100∠ <100∠80° 80° A Dolve for the real po"er i 3+ dra" by the commercial buildig. A. $02.2 &. $0$.5 %. $1.(. $21.5 Problem --:77 &oard April 181 A -6phase! -6"ire load dra"s the follo"ig lie currets: Ia < 0∠ 0∠ --0o A I b < 8.$∠ 8.$∠ 21$o A Ic < 5∠ 5∠ 80o A If the voltages impressed o the load are balaced -6 phase. /avig a magitude of $1$0 volts lie to lie! solve for the total po"er i 3+. A. 55.15 &. 50.85 %. 5-.5$ (. 520.18 Problem -$: Oive the follo"ig voltages ad load impedac es: ,ab < 220∠ 220∠ 0o , )ab < 10∠ 10∠ -.8o Ω , bc < 220∠ 220∠ 2$0o , ) bc < 10∠ 10∠ 0o Ω ,ca < 220∠ 220∠ 120o, )ca < 10∠ 10∠ $5o Ω (etermie the readig of a 3+ meter coected "ith its curret coil coected i lie b "hile the potetial coil coected across lies b ad c. A. 5.12 3+ &. $.2 3+ %. $.08 3+ (. 5.$5 3+ Problem -5: 77 &oard ctober 18Oive the follo"ig load impedaces i delta ad is impressed voltages as follo"s: ,ab < 220∠ 220∠ 0o , )ab < 8 = * o , bc < 220∠ 220∠ 2$0 , ) bc < 8. ' *5 ,ca < 220∠ 220∠ 120o , )ca < 10 = *0 +hat "ill be the readig of the t"o "attmeters coected to measure total po"er. 9se lie a as the commo potetial poit. A. -.8 3+! .0-1 3+ &. 2.5$ 3+! 8.-5 3+ %. -.125 3+! .8 3+ (. $.055 3+! .8$8 3+ Problem -: 77 &oard ctober 18$ he -6 phase po"er supply to a factory has the follo"ig measuremets:

,ab < 2$0∠ 2$0∠ 0o , , bc < 2$0∠ 2$0∠ 2$0o , ,ca < 2$0∠ 2$0∠ 120o , Dolve for the total po"er dra". A. 0.2 3+ &. %. 58.8 3+ (.

la < 120∠ 120∠ --0o A l b < 15∠ 15∠ 21$o A lc < 150∠ 150∠ 80o A 5.5 3+ 2.$ 3+

Problem -: Oive ,ab < 220∠ 220∠0o , , bc < 220∠ 220∠ 6120o , (etermie the readig of the "attmeter as sho". A. &. %. (.

,ca < 220∠ 220∠ 62$0 ,

$05.$5 + $8.-2 + $$2.88 + $.5$ +

Problem -8: 77 &oard April 185 A balaced -6 phase load dra"s 120 amperes lie curret at 2-0 volts lie to lie 0.8$8 0.8$ 8 pf laggig curret. Dolve for the readigs of the t"o "attmeters used to measure the -6 phase po"er. A. 25.5$- 3+! 15.08 3+ &. 28.155 3+! 12.-85 3+ %. 2$.-5 3+! 1.15 3+ (. 2.58- 3+! 12.5 3+ Problem -: A umber of icadescet lamps are coected i delta across the lies of a balaced 2-0 ,! -6 phase! -6 "ire system. A sigle6 phase "attmeter is coected i the circuit. he curret coil of the the meter is coected i lie 1 "hile the potetial coil is coected across lies 1 ad -. his "attmeter registers 8.$ 3+. If the potetial termial of the meter! "hich "as coected i lie -! is trasferred to lie 2! the "attmeter reads .5 3+. If each lamp rated 100 +! ho" may lamps are coected across lies 1 ad 2. Assume lamp pf to be uity. A. $2 &. $5 %. $(. $$ Problem -80: 77 &oard April 188 C7@A;% used t"o "attmeters to measure the balaced -6 phase dyatro elevator motor drive. he curret coil of the "attmeters are coected to the curret trasformers! "hich are i lies 1 ad 2 respectively. respectively. he lie potetial are 2-0 , ad the lie currets are each 150 A. he "attmeters each idicate 1. 3+. Assume Assume load is "ye6 coected. +hat is the total po"er supplied# A. $.15 3+ &. $8.2- 3+ %. $5.$1 3+ (. $.-50 3+ Problem -81: 77 &oard April 12 A $0 volt! three6 phase motor dra"s 208 A "ith a po"er factor of 0.1 laggig. %alculate the idicatios of +1 ad +2 for the give coditio. A. 5.$0 3+! 5.$0 3+ &. 1.2- 3+! 5.58 3+ %. 8.5 3+! 1.25 3+ (. 5.2$ 3+! 55.5 3+ Problem -82: "o "o "attmeters are used to measure the po"er delivered to a balaced delta coected load. +attmeter +a ttmeter 1 reads 1000 + ad "attmeter 2 reads 500 + after the curret coil termials are reversed. If the lie to lie voltage is 208 ,! solve for the lie curret dra" by the load.

A. %.

8.2- A 8.08 A

&. (.

.5 A .-$ A

Problem -8-: he po"er dra" by a three6 phase balaced load is measured usig t"o "attmeters properly coected. At "hat po"er factor "ill the readig of oe of the "attmeters be Bero# A. 0.8 &. 1 .0 %. 0 .5 (. 0.0 Problem -8$: "o "o "attmeters coected to a three6 phase $$0 , suirrel cage iductio motor idicate a total of 12 3+. If the operatig po"er factor of the motor is 0.$! determie the idividual readig of the "attmeters. A. 10.$ 3+! 1.0 3+ &. .2$ 3+! 2. 3+ %. 1-.$ 3+! 61.$ 3+ (. oe of these Problem -85: A -6 phase iductio motor is coected across a 2206 , three6 phase source. A "attmeter reads 00 + "he its curret coil is coected i lie 1 ad its potetial coil across lie 1 ad 2. 2 . if the potetial coil of the "attmeter is trasferred to lies 2 ad -! the readig of the "attmeter is the same. %alculate the po"er factor of the motor. A. 0. &. 0.82 %. 0.8 (. 0.0 Problem -8: "o "attmeters used to measure the total po"er dra" by a three6 phase circuit ad are coected properly. properly. he first "attmeter "attmeter reads $800 "atts "hile the other reads $00 "atts after reversig its curret curret coil. Dolve for the load po"er factor. A. 0.5$ &. 0.5-2 %. 0.$ (. 0.$- Problem -8: 77 &oard Sue 10 "o6 "attmeter method is used to test a 25 /P! 2-0 volt! 1800 rpm! 0 cycle! -6 phase iductio motor. +he the lie voltages are 2-0 volts! oe "attmeter reads =1-!$00 "atts ad the other =! $00 "atts. (etermie motor po"er factor. A. 0.1 &. 0.88 %. 0.8$ (. 0.80 Problem -88: 77 &oard ctober 18 A "oud rotor motor! .5 /P! 2-0 volts! -6 phase ta3es a lie curret of 18.$ amperes! "he operatig at rated output at a efficiecy of 88J. %alculate the idicatio of the "attmeter "he this is iserted to measure po"er by the 6 method. A. -.1 3+ &. $.$01 3+ %. -.-1 3+ (. $.0$2 3+ Problem -8:77 &oard ctober 1$ A "attmeter "attmeter "ith its curret coil i lie 2 ad a d potetial coil across lies 2 ad - is coected to a balaced -6 phase system. he oly load supplied is a sigle phase oe coected to lies 1 ad 2. his load is 3o" to be iductive. If the "attmeter reads Bero "atts! determie the po"er factor of the sigle6 phase load. A. 0.0 &. 0.8 %. 0.800 (. 0.00 Problem -0:

hree ueual sigle6 phase motor loads are coected across the lies of a balaced 2-06 ,! 0 /B! three6 phase system. Cotor V1 is coected across lies 1 ad 2 ad dra"s 10 A at 80J pf. Cotor V2 is coected across lies 2 ad - ad ta3es 1$2 A at 8.J pf. Ad Ad motor V- is coected across a cross lies ad 1 ta3es 1$ A at 5J pf. 9sig t"o "attmeters properly coected "ith the commo p otetial poit i lie 2! "hat are the readig of these istrumets# A. 2.$2 3+! 1-. 3+ &. 50.1 3+! 25.- 3+ %. $8.21 3+! 2.8 3+ (. $.25 3+! 2.8- 3+ Problem -1: 77 board April 18$ A balaced -6 phase load dra"s 5 amperes lie curret at 2-0 volts lie to lie! ad 0.8$8 laggig po"er factor. If the t"o6 "attmeter method is used! solve for the readigs of the the "attmeters. A. 15.-2 3+! 10.02 3+ &. 1.8 3+! .$8 3+ %. 1.$2 3+! 8.2 3+ (. 1.2$ 3+! 8.10 3+ Problem -2: 77 &oard ctober 180 hree eual impedaces! each havig a resistace of 8 ohms ad a iductive reactace of  ohms are coected i delta to lies a! b ad c of a 2$0 ,! -6 phase! -6 "ire lie! phase seuece: seuece : a6b6c. +hat is the readig of a sigle6 phase "attmeter coected "ith its curret coil i lie a ad potetial coil across lies b ad c# A. !180 + &. ! -2$ + %. ! 51 + (. ! 00$ + Problem --: A balaced three6 phase load dra"s 20 3+ at 0.$$6 pf laggig from a 2-06 ,! 0 /B three6 phase trasmissio lie. Eid the readigs of the t"o "attmeters properly coected to measure po"er# A. 18.$5 3+! 1.55 3+ &. 1$.25 3+! 5.5 3+ %. 21.55 3+! 61.55 3+ (. 25.5 3+! 65.5 3+ Problem -$: (etermie the operatig po"er factor of the three6 phase balaced load if the readig of the t"o "attmeters properly coected to measure po"er is -:1. A. 0. &. 0.55 %. 0.5 (. 0.81 Problem -5: 77 &oar ctober 12 A three6 phase feeder carries t"o laggig balaced loads. he po"er observed by each is measured by t"o "attmeters method! givig the follo"ig readigs: Eirst load: +1 < 10 3+ +2 <  3+ Decod load: +1 < 0 3+ +2 < $8 3+ +hat is the combied 3,A load o the feeder# A. $-$.8 &. $2.- %. 50$.2$ (. $20.12 Problem -: A sigle6 phase "attmeter is used to determie the po"er ta3e by a three6 phase suirrel cage iductio motor. he he curret coil is placed i lie 1 "hile the potetial coil is shifted from lies 1 a d 2! the from lies 1 ad - respectively. If the "attmeter registers 50 ad --80 "atts respectively! "hat is the lie curret dra" by the motor# ;ie to lie voltage is 2-0 ,. A. 25.1 A &. 2-.$ A %. -$.1 A (. -.8- A

Problem -: 77 &oard ctober 12 atioal Po"er %orporatio used t"o "attmeters to measure -6 phase po"er of a balaced L6 L6 coected laggig po"er factor motor loads. 7ach "attmeter idicates 15.$ 3+. he he voltage coils are coected across lie 2 ad - ad across lies 1 ad -! respectively. respectively. he lie6to6lie voltages are 2-0 volts "ith ,12 leadig ,2- ad the lie currets are each ea ch 120 A. %alculate the total po"er supplied. A. -.$$ 3+ &. -0.2 3+ %. -. 3+ (. -$.88 3+ Problem -8: A three6 phase! 00 , balaced system supplies eergy to a ubalaced delta coected load. he three lie currets la! l b! lc are 1$1.$ A! 100 A ad 100 A respectively. respectively. A "attmeter "attmeter "ith its curret coil i lie a ad potetial coil across a ad b idicates 8$.8$ 3+ 3 +. Dolve for the total po"er dra" by the load. Assume a seuece a6b6c. A. 102.$5 3+ &. 100.- 3+ %. 10.25 3+ (. 110.12 3+ Problem -: hree eual o6 iductive resistaces are co ected i star across a balaced $00 ,! ,! 0 /B system. A sigle6 phase "attmeter is coected o the system "ith its curret coil i lie 1 ad potetial coil i lies 1 ad -. his "attmeter registers 00 +. if lie lie 2 is accidetally ope circuited! "hat "ill be the "attmeter readig uder this coditio# A. 00 + &. 1200 + %. 00 + (. -00 + Problem $00: A balaced three6 phase load havig a resistace of 10 ohms ad a reactace of 10 ohms are coected i star across the lies of a balaced 2-0 , three6 phase supply. Eid the readig o each t"o "attmeters properly coected to measure the po"er d ra". A. 2055 +! 58 + &. 2085 +! 55 + %. 215 +! $ + (. 2105 +! 5- + Problem $01: A ubalaced delta6 coected load is coected across the lies of a $6 "ire balaced three6 phase supply "ith eutral coectio. "o "o "attmeters +1! +2 are coected i lies 1 ad - "ith a commo potetial coectio i the eutral e utral "ire. hese "attmeters read 1.5 ad -000 "atts! respectively. respectively. he curret lies 1 ad - are 20 ad 25 A! respectively. If the commo potetial p otetial coectio is trasferred to lie 2! "hat "ill be the readig of the t"o "attmeters# he system voltage per phase is 208 ,. Assume a phase seuece of 1626- "ith ,1 as referece. A. +1 < $01- "atts! +2 < $50 "atts &. +1 < $0$- "atts! +2 < $5-0 "atts %. +1 < $0- "atts! +2 < $510 "atts (. +1 < $0- "atts! +2 < $500 "atts Problem $02: "o "o "attmeters are used to measure the total po"er dra" by a balaced delta coected load. If the "attmeter reads 1200 + ad 800 + respectively! ad the lie to lie voltage is 208 ,! solve for the impedace per phase of the load. A. 5.-1 = *2-. ohms &. 5.$ = *20.0 ohms %. 50.-1 = *-1.2 ohms (. 52.$ = *2-.82 ohms Problem $0-:

"o "o "attmeters used to measure the total po"er dra" by the three6 phase circuit ad are coected properly. e of the "attmeters read 10 3+ "he the po"er factor o f the load is uity. +hat does each "attmeter read "he the po"er factor fall to 0.8 laggig# Assume the total three6 phase po"er remais the same. A. 12.$5 3+! .55 3+ &. 1-.-- 3+! . 3+ %. 15.25 3+! $.5 3+ (. 10. 3+! .-- 3+ Problem $0$: 77 &oard April 180 A factory is supplied by a three6 phase! -6 "ire system "ith the follo"ig characteristics. (etermie the average po"er factor of the load. ,ab < 2-0∠ 2-0∠0o , la < 110∠ 110∠-.8o A , bc < 2-0∠ 2-0∠ 2$0o , lc < 125∠ 125∠ 5-.1-o A Problem $05: 77 &oard ctober 181 A three6 phase! -6 "ire load dra"s the follo"ig lie currets: ;a < 0∠ 0∠ --0o A l b < 8.$∠ 8.$∠ 21$o A lc < 5∠ 5∠ 80o A If the voltages impressed o the load are balaced three6 phase! havig a magitude of $1$0 volts lie to lie! solve for the po"er factor of o f the load. A. 0. &. 0. %. 0.82 (. 0.0

Test 12 – Electrical Transients Problem $0: 77 &oard April 1! 77 &oard ctober 182 I a @; circuit! circuit! Wirchhoff Wirchhoff s ;a" gives gives the follo" follo"ig ig relatio relatio:: 7 < ; didt = @i "here: "here: supply voltage >200 volts? @ < resistace >20 ohms? ; < iductace >1 hery? t < time i secods i < curret i amperes If i < 0 "he t < 0.02 secod. A. -. - A &. -.1 A %. -.2 A (. -.0 A Problem $0: 77 &oard ctober 180 I a @; @; circ circuit uit!! Wirchho Wirchhoff ffss ;a" ;a" gives gives the the follo follo"i "ig g relati relatio: o: 7< 7< ; did didtt = @i "her "here: e: supply voltage >200 volts? @ < resistace >20 ohms? ; < iductace >1 hery? t < time i secods i < curret i amperes If i < 0 "he t < 0! fid i after a log time. A. 10 A &. 11.2 A %. 0 (. ifiite

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Problem $08: 77 &oard ctober 10 A .0 / coil "hose resistace is 12 ohms is coected i series "ith a 2$ ohms resistor ad to a 1$$ , battery ad a s"itch. he s"itched is closed at t< 0. (etermie the time costat of the circuit. A. 0.- sec &. 0.$5 sec %. 0. sec (. 0.50 sec Problem $0:

he curret i the coil decays as follo"s: "he t < 0.002 sec! the curret is 10 mAT "he t< 0.00 0 .00 s! the curret drops to $ mA. he time costat of the coil is  is  . A. $.50 ms &. $.28 ms %. $.- ms (. 2.15 ms Problem $10: 77 &oard April 10 A time time of 10 millisecods is reuired for the curret o a series @; dc circuit to reach 0J of its fial steady state value. Assume at t< 0! i>0? < 0. +hat is the time costat i secods for the circuit# A. $.25 ms &. -.8 ms %. -.- ms (. $.0$ ms Problem $11: A coil has a resistace of 10 ohms ad a time costat of 1 secod. If the coil is coected co ected across a steady potetial of 2$ ,! ho" much is the curret after 2 ms# A. $.0 mA &. $. mA %. $.1 mA (. $.0$ mA Problem $12: 77 &oard April 15 he shut "idig of machie has a resistace of 80 ohms ad a iductace of $ / is suddely s"itched o to a 2206 , supply. Eid the time ta3e for the curret to rise to half its steady state value. A. 0.05 sec &. 0.0-$ sec %. 0.0251 sec (. 0.012 sec Problem $1-: A dc voltage 7 is applied app lied across a @; circuit at t< 0. At "hat time after! is the voltage vo ltage across the resistor eual to that across the iductor# A. 0.- % &. 0.- % %. 0.0 % (. 0. % Problem $1$: A 5006 ohm relay coil has a iductace of 10 m/. his relay reuires a curret of $0 mA i order to trigger. /o" log after the s"itched is closed "ill the relay trigger if a 2$ , (% source is applied across its termials. Assume Assume at t< 0! i< 0. A. 2-.5$ µsec &. 1.50 µsec %. -5.8- µsec (. -.22 µsec Problem $1: A coil havig a resistace of 10Ω 10Ω ad iductace of 10 / is coected i series "ith a 15Ω 15Ω resistor. resistor. A 2006 , (% source is coected to the circuit at t< 0. (etermie the voltage across the coil at t< 0.05 sec. A. 10.8 , &. 188.51 , %. 18.2 , (. 185.0 , Problem $1: A series @; circuit is coected across a dc source through a s"itch. hree >-? millisecods after the s"itch is closed! the voltage across the iductace is 20 , ad drops to 5 , after 20 ms. If @ < 100 ohms! fid ;. Assume at t< 0!l < 0. A. 1.-0 / &. 1.2 2 / %. 1 .5  / (. 1.25 /

Problem $18: A 1Ω 1Ω resistor is coected i series "ith a 26 m/ iductor. If a 126 , dc source is suddely coected across its termials! ho" log "ill it ta3e! such that the voltage across the resistor "ill be t"ice as that of the iductor. A. 2.5 ms &. 1. ms %. 2.2 ms (. 1. ms Problem $1: A resistace resistace of 100 1 00 3 Ω is coected i series "ith a 100 µE capacitor. If the combiatio is suddely coected across a 125 , (% source! determie the curret oe secod after the s"itch is closed. A. 1.1- mA &. 1.25 mA %. 1.05 mA (. 1.12 mA Problem $20: 77 &oard ctober 12 A ucharged codeser i series "ith a 1206 1 206 volt voltmeter of 10! 000 ohms o hms resistace is suddely coected to a 100 , battery. battery. e secod later! the voltmeter reads 0 volt. (etermie the capacitace of the codeser. A. 18.5$ µE &. 10.2 µE %. 15. µE (. 12.2- µE Problem $21: A 120 µE capacitor has a iitial charge of 00 µ%. It is discharged through a $0 Ω resistor. /o" log "ill it ta3e i order for a capacitor to discharge at 150 µ%# A. .5 ms &. .82 ms %. $. ms (. 5.85 ms Problem $22: A 50 µE capacitor is coected across a $00 , dc source. It is the discharged through a voltmeter "hose resistace is 1006 3 Ω. /o" log "ill it ta3e for a capacitor to discharge at 1.8 S of stored ee rgy# A. 2. - s &. 1. s %. 2.0 s (. 1. s Problem $2-: 77 &oard April 1 A 206 ohm resistace @ ad a 0.001 farad cap acitace % are i series. A direct curret voltage 7 of 100 volts is applied across the series circuit at t< 0 ad the iitial curret i>0? < 5 A. he applicable differetial euatio is: @ (didt? = ic < 0 Dolve the differetial euatio ad determie the resultig curret i>t? at t< 0.0 1 secod. A. -.-$ A &. -.  A %. 2 . 8 A (. -.0- A Problem $2$: 77 &oard ctober 11 A series @% circuit cosist of @ < 2 CΩ CΩ ad a ucharged capacitor % < 5 µE. he circuit is coected cross a 100 ,(% source at t< 0. (etermie the the voltage across the resistor 5 secods later. A. -.-1 , &. 0.5 , %. .2$ , (. .22 , Problem $25: 77 &oard ctober 12

A ucharged codeser i series "ith a 1206 1 206 voltmeter of 10!000 ohms resistace is suddely coected to a 100 , battery. battery. e secod later! the voltmeter reads 0 volt. (etermie the rate at "hich the voltage across the codeser is chargig. A. 51 e60.55 t &. 51 e60.51 t %. 55 e60.55 t (. 55 e0.51 t Problem $2: 77 &oard ctober 181 I a circuit cosistig of a resistace! @ ohms! i series "ith a capacitace! % farad! ad coected to a dc source "ith emf 7 volts! the electrical charge! Q coulombs ad the curret i! amperes! maybe determied by solvig the simultaeous euatios: Q dQ @i = 7 ad l <   < % dt 6 If @ < 20 ohms! % < 250 x 10 farad ad 7 < 100 volts! fid i after a log time. A. 1A &. 0A %. ifiity (. 5A Problem $2: A 80 µE capacitor i series "ith a voltmeter of 10! 000 ohm resistace is coected suddely across a 1006 , supply. /o" log "ill it ta3e for the voltmeter to register 15 ,# A. 1.8$ sec &. 1.52 sec %. 1.- sec (. 1.25 sec Problem $28: 77 &oard April 1A 100 µE capacitor iitially charged to 2$ , is discharged across a series combiatio of a 1 3 Ω resistor ad a 200 µE capacitor. Eid the curret after 1 sec. A. .-$ A &. .2$ A %. 8.$- A (. 8.8$ A Problem $2: A 80 µE capacitor i series "ith a 10006 ohm resistor is coected suddely across a 110 , (% supply. Eid Eid the value of the curret after oe time costat. A. 0.0$05 A &. 0.052 A %. 0.05$- A (. 0.0-22 A Problem $-0: 77 &oard ctober 11 A series @% circuit cosist of @ < 2 CΩ CΩ ad a ucharged capacitor % < 5 µE. the circuit is coected across a 100 ,(% source at t < 0. +hat is the iitial rate of chage of voltage across the resistor# A. 610.0 ,s &. 10.0 ,s %. 612.$ ,s (. oe of these Problem $-1: A series series @;% circuit "ith @ < 1 3 Ω! ; < 1 /! % < .25 µE is suddely coected across a 2$ , dc source. At t< 0! i < 0 ad < 0. (etermie the curret after 0.01 sec. A. -.$5 mA &. $.1 mA %. 5.$0 mA (. 5.05 mA Problem $-2:

A series @;% circuit has @< 200 Ω! ;< 0.1 / ad a capacitor %< 10 µE. If a 100 , dc source is coected across the ed termials of the series circuit at t< 0! determies the curret after 1 millisecod. Assume Bero iitial coditios. A. 0.-5- A &. 0.22 A %. 0.25- A (. 0.-8 A Problem $--T A series series @;% has @< $0 ohms! o hms! ;<100 m/ ad %< 50 µE. he he circuit is coected across a 100 , dc source at t< 0 through a s"itch. (etermie the curret 0.02 secod after the s"itch is closed. Assume all iitial coditio to be Bero. A. .$0 mA &. 5.-1 mA %. .22 mA (. $.85 mA Problem $-$:77 &oard Carch 18 A 106ohm resistace @ ad a 1.06 hery iductace ; are coected i series. A A% A% voltage e>t? < 100 si -t is applied across the series circuit. he applicable differetial euatio is: @i = ; >didt? < e>t? Dolve for the particular solutio >"ithout the complimetary solutio? ad determie the amplitude of the resultig siusoidal curret I i>t?. A. 0.-21 A &. 0.22 A %. 0.2$1 A (. 0.25 A Problem $-5:77 &oard April 11 A certai certai electric "elder has a basic circuit euivalet to a series @; "ith @< 0.1 Ω ad ; < 1 m/. It is coected to a A% source GeH through a s"itch GsH operated by a automatic au tomatic timer! "hich closes the circuit at ay desired poit o the 0 cycle! siusoidal "ave GeH. %alculate the magitude of the trasiet curret resultig "he GsH closes as GeH is passig through its pea3 value of 100 volts. A. 25.$1 A &. 5.$ A %. 80.5$ A (. .-2 A Problem $-:77 &oard ctober 1 Oive the differetial euatio: ; didt=@i< 100 si >-t? +here ;< 0.1 hery! hery! @< 10 ohms ad i>0?< i>0?< 0. (etermie the curret curret at t< 0.01 secod. A. 2.8$ &. 2.-01 %. 2.5-1 (. -.02 Problem $-: A series @% circuit "ith @< -0 ohms ad %< 250 µE is coected across a A% voltage source of emf 100 si 100t through a s"itch. he s"itch is closed at t< 0. (etermie the curret "he t< 0.01 secod. Assume Assume at t< 0! voltage v oltage across the capacitor is Bero. A. 1.$5 A &. 1.2$ A %. 0.- A (. 2.05 A Problem $-8:77 &oard April 1 A series series @% circuit has @< 10 ohms! ;< 0.1 hery ad %< 0.0001 farad. A A% A% voltage e< 100 si -t is applied across the series circuit ad the applicable differetial euatio is: ; >d2idt2?=@>didt?=>1%?I< ?=@>didt?=>1%?I< dedt Dolve for the particular solutio >"ithout the complimetary solutio? ad determie the amplitude of the resultig siusoidal curret i>t?. A. 5.51 A &. .0 A %. . A (. .-$ A Problem $-:77 &oard ctober 1 A 106ohm resistace @ ad a 0.001 farad capacitace % are i series. A A% A% voltage e < 100 si -t is applied across the series circuit. he applicable differetial euatio is:

@>didt?=>1%?I< dedt. Dolve for the particular solutio >"ithout the complimetary solutio? ad determie the amplitude of the resultig siusoidal curret i>t?. A. . A &. 10.- A %. 8.   A (. 11.0 A Problem $$0: A series series @; circuit has @ < 10 ohms ad ; < 1 hery. he he circuit is coected across a voltage 610t source e>t? < 10 e through a s"itch that is iitially ope. he s"itch is closed at t < 0. (etermie the curret after 0.01 secod. Assume iitial curret to be Bero. A. 8.8 mA &. 0.5 mA %. 8.- mA (. 2.2 mA Problem $$1: 77 &oard ctober 1 he differetial euatio of a series @% circuit "ith (% applied vo ltage is: @i = 1% / idt < 7. If @ < 50 ohms! % < 0.0001 farad! 7 < 100 volts ad i>0? < 2.0 amperes! determie the laplace trasform expressio for l>s?. A. A. l>s? < 2 s = 200 &. l>s? < 2 s >s = 2? %. l>s? < 2 s = 50 (. l>s? < 2 s = 2 Problem $$2: A series series @% circuit has @ < 13 ad % < 100 RE "ith a iitial charge of 2000 R%. he combiatio is coected across a 100 , dc source ad the s"itch is closed at t < 0. If the capacitor is coected i the circuit such that it "ill be dischargig! determie the laplace trasform expressio of the curret dra". A. i>s? < 0.12 s = 10 &. i>s? < 0.12 s = 12 %. i>s? < 0.10 s = 10 (. i>s? < 0.10 s = 12 Problem $$-: 77 &oard Carch 18 (etermie the iverse laplace trasform of i>s? < 200 s2 = 50s = 1025. A. i>t? < 2 e625tsi 100t &. i>t? < 2t e625tsi 100t %. i>t? < 2 e625tcos 100t (. i>t? < 2t e625tcos 100t Problem $$$: 77 &oard Carch 18 A (% ammeter has a iteral resistace of 0.1 ohm. A shut shut of 1.010 x 10 1 0- ohm is coected to the ammeter. +hat is the multiplier of the set up# A. 80 &. 100 %. 50 (. 10 Problem $$5: A 1J dArsoval meter movemet has coil ad s"ampig resistace addig to 200 ohms. he full scale voltage is 50 m,. m,. (etermie the shut resistace reuired producig 1 A full full scale curret meter# A. 52.02 m &. $.22 m %. 50.01 m (. $8.$ m Problem $$: A galvaometer "ith a 206 ohm coil resistace has full scale deflectio curret of 10 mA. If a 0.02 ohm is placed across the meter to icrease its capacity! "hat is the approximate e" full scale curret of the meter# A. 10.10 A &. 11.01 A %. 10.11 A (. 10.01 A Problem $$:

A movig iro meter gives full scale deflectio "ith 100 , applied across the coil. he coil has 20! 000 turs ad a resistace of 2.5 3. If this istrumet is to used as a ammeter "ith a full scale deflectio of 10 A! ho" may turs are eeded# A. 80 &. 100 %. 50 (. 0 Problem $$8: A movig iro ammeter is "oud "ith 50 turs ad gives full scale deflectio "ith $ A. If it is desired to exted the full scale readig to 10 A! ho" may turs are eeded o the same bobbi# A. $0 &. 20 %. 10 (. -0 Problem $$: A galvaometer has a coil resistace of 50 ohms ad a curret sesitivity of 0.001 RA per mm. /o" much is the voltage across its termials "he the deflectio is 15 cm at full scale# A. .50 R, &. 10.0 R, %. 8.25 R, (. .05 R, Problem $50: 77 &oard ctober 1$ A 06 10 mA galvaometer "ith a coil resistace of 20 ohms is coverted to a 06 50 , voltmeter by usig A. 50 ohms shut resistor &. $80 ohms series resistor %. 5000 ohms series resistor (. 5020 ohms series resistor Problem $51: 77 &oard ctober 12 A certai 50 RA meter movemet has a resistace of 0 ohms. +hat value of series multiplier is eeded to ma3e the istrumet to read 5 , >maximum?. A. 102 3 &. .1 3 %. 110 3 (. 8 3 Problem $52: A voltmeter "ith a iteral resistace of 5000 ohms is calibrated to read 10 volts at full scale. /o" much series resistace is eeded i order to measure 150 volts at full scale# A. 0 3 &. 5 3 %. 2 3 (. 8 3 Problem $5-: A 1000 ohms per volt! 2J meter movemet has a c oil resistace of 200 ohms. (etermie the exteral resistace ecessary to provide full scale readig of 5 ,. A. $. 5 3 &. 5.2 3 %. 5.0 3 (. $.8 3 Problem $5$: A ammeter rated 5 A! havig a resistace of 0.5 ! is to be coverted ito a 200 , voltmeter by coectig a resistor i series "ith the ammeter. %alculate the value of this resistace. A. -.$  &. -$.  %. -.5  (. -5.5  Problem $55: A direct curret "attmeter "ith a potetial coil coected across the load side of the istrumet reads 250 +. If the voltage across the load is rated 2-0,! "hat is the actual po"er dra" by the load. he resistace of the potetial coil is 3o" to be 1 3. A. 1$.2 + &. 1.1 + %. 1-.5 + (. 18. +

Problem $5: 77 &oard April 12

he stator of a $$0 ,! 506 hp shut motor is tested at 5o% for its isulatio resistace bet"ee termials coected together ad the groud frame. +he a 50!0006 ohm voltmeter is coected i series "ith a 006 , (% source ad the isulatio resistace! the meter idicates 125 ,. %alculate the isulatio resistace. A. 10 3 &. 10 3 %. 180 3 (. 200 3 Problem $5: It is desired to measure the resistace of the isulatio bet"ee the motor "idig ad the motor frame. A -006 ,! ,! 50!000 ohm voltmeter is used i ma3ig the measuremets. he voltmeter "he coected across the source reads 2-0 , ad "he coected i series "ith the isulatio! it reads 5 ,. Eid the isulatio resistace. A. 2.$2 C &. 2.25 C %. 2. C (. 2.-5 C Problem $58: +he a 150 3 voltmeter is coected across a costat source! it registers 2-0 ,. +he oe termial is coected to the core of a cable "hile the rubber isulatio of the cable is coected to the other termial of the lie! the meter registers 15 ,. ,. (etermie (etermie the isulatio resistace of the cable. A. 2150 3 &. 2000 3 %. 2202 3 (. 2$80 3 Problem $5: A u3o" voltage is to be measured by meas of a 250 , scale voltmeter havig a iteral resistace of 10!000 ohms. Eor safety purposes a 1000 ohms resistor is coected i series "ith oe o f the voltmeter termials. If the meter reads 200 volts! determie the u3o" voltage. A. 250 , &. 220 , %. 2-0 , (. -00 , Problem $0: "o "o idetical 1000 ohm per volt (% voltmeter are coected i series to measure a u3o" supply voltage. he first voltmeter is coected to its 200 , scale "hile the secod voltmeter is coected to its 250 , scale. +hat is the voltage of the supply if the first voltmeter reads 100 ,# A. 225 , &. 220 , %. 2-0 , (. 2$2 , Problem $1: A 20 3 per volt voltmeter idicated a readig of 250 , o its -00 , scale "he used to measure a u3o" voltage. If a 50 3 resistor "as coected i series "ith the istrumet durig the test! determie the value of the u3o" voltage. A. 2-8 , &. 2-5 , %. 2$8 , (. 225 , Problem $2: A -006 , voltmeter dra"s 26 mA curret for full scale deflectio. his voltmeter is used to measure the voltage across a 50 3 resistor i series "ith a 25 3 resistor. he combiatio of the resistors is beig coected across a 1506 , source. +hat is the voltmeter readig# A. 8 , &. 102 , %. 0 , (. 88 , Problem $-: 77 &oard ctober 18 he C7@A;% test of a 106 A "attmeter havig a costat of 0.$! the dis3 ma3es $0 revolutios i 5-. secods. he average volts ad amperes durig this period of test are 11 volts ad .$ A. +hat +hat is the percet accuracy of the meter at this load# A. .$5J &. 8.58J %. 8.0J (. .$$J Problem $$: 77 &oard April 12

I a test of a 106 ampere "attmeter havig a costat of 0.$0! the dis3 ma3es $0 revolutios i 55 secods. he average volts ad amperes durig d urig this period are 11 volts ad 10 amps. (etermie the percet accuracy of this meter at this load. A. 2.-$J &. 1.0-J %. 8-.$J (. 0.28J Problem $5: A give ampere6 hour meter is uder test by coectig it across a 2-06, (% source. Eor a duratio of 180 miutes! a costat curret of 20 A flo"s. his meter registers 50.5 ad 582.- 3+6hr before ad after the test respectively. respectively. %alculate the percetage accuracy of the meter. A. 2.-1J &. 0.02J %. 1.$$J (. 8.20J Problem $: A 15 A! A! 1206 , "atthour meter has a dis3 costat of 2. +he tested o a uity po"er factor load! 2$ dis3 revolutios are couted i a period of 2 miutes. /o" may dis3 revolutios "ould be couted per miute if the po"er factor "ere chaged to 0.5 laggig. Assume Assume the same lie curret ad voltage i both coditios. A.  &. 10 %. 8 (. 12 Problem $: A sigle6 phase three "ire! 0 /B! 220 , "atthour6 meter has a disc costat of 1.2. (etermie the eergy recorded by the meter after 18!000 revolutios of the disc. A. 21. 3+6hr &. 20.5 3+6hr %. 25.$ 3+6hr (. 2. 3+6hr Problem $8: A 1156 1156 ,! 10 A! three6 phase "att6hour meter! havig a basic meter costat of 3 < 2- is coected to a three6 phase! three6 "ire circuit through a 100:5 10 0:5 ampere % ad 2-00:115 volt P. A time chec3 sho"s that the meter dis3 is ma3ig 15 revolutios i 50 secods. +hat is actual 3+ of the load# A. 20 3+ &. 288 3+ %. 52$ 3+ (. -80 3+ Problem $: 77 &oard Carch 18 A po"er plat customer dra"s po"er at 220 volts from trasformers o a pole. %urret trasformers "ith ratio of 2005 are used to meter the electrical usage. +hat is the multiplier of the 3+6hr 3+6hr ad demad meters# A. $0 &. 200 %. 100 (. 80 Problem $0: A curret trasformer trasformer "ith a turs ratio of 100:5 ad a potetial trasformer "ith a turs ratio of 10:1 are coected to the curret ad voltage coils of a sigle phase "attmeter measurig po"er delivered to a load. If the "attmeter readig is 2$0 +! "hat is the actual po"er measured delivered d elivered to the load# A. 50!000 "atts &. $2!000 "atts %. $8!000 "atts (. $5!000 "atts Problem $1: 77 &oard ctober 1 At a 115 3, substatio! the P ratio is 1000 ad the % ratio is 12005. he potetial goig ito the "attmeter is 115 volts. +hat is the C+ idicated "he the "attmeter reads 800 "atts# A. 12 C+ &. 15.0 C+ %. 1.2 C+ (. 150 C+ Problem $2: A 15 A! A! 1206 , "att6 hour meter has a dis3 costat of 2. +he tested o a 0.8 po"er factor load! 2$ dis3 revolutios are couted i a period of - miutes. %alculate the curret dra" from the load. A. 10 A &. 8A %. A (. 12 A

Problem $-: I a give test! it "as foud out that "ith a curret of $0 A! the disc of a give ampere6 hour meter made 10 revolutios i $ miutes. If the meter costat of the istrumet is 50 coulombs per revolutio! determie the multiplyig factor to be used i order to give the correct eergy cosumptio of the load. A. 0.8-&. 0.0 %. 0.22 (. 0.8 Problem $$: A u3o" resistor @ x as sho" is to be measured by meas of a +heatstoe bridge! "he the ratio arms are arraged @ 1 < 10 ohms ad @ 2 < 1200 ohms! balace is obtaied "he rheostat arm is eual to 125 ohms! "hat is the value of the u3o" resistor# A. 10.$ &. 11.2$ %. 1-.22 (. 1$.05 Problem $5: A u3o" resistace x as sho" is measured by a slide "ire bridge >a simple form of +heatstoe bridge? A% is a uiform "ire 100 cm log ad ( is a slidig cotact. &alaced is obtaied "he A( euals -2.5 cm. Eid the value of the u3o" resistace x. A. $5.-2 &. $8.1$ %. 50.0$ (. $-.$$ Problem $: A t"o6 coductor udergroud cable 2800 ft log is used to itercoect a home base statio to aother statio. e coductor is grouded at some poit. A Curray loop test "as coducted at the home base "ith a 100 cm slide "ire bridge. If a balace o the bridge "as obtaied at  cm! at "hat distace is the fault from the home base# A. 128 ft &. 1$15 ft %. 1-$$ ft (. 1-02 ft A.

0.0$ ohm

&.

0.-2 ohm

%.

0.12 ohm

(.

0.- ohm

Problem 5-: Problem $: A four6pole! four6pole! (% geerator "ith lap "idig has h as $8 slots ad $ elemets per slot. /o" may coils does it have# A. $8 &. -8$ %. 12 (.  Problem $8: If the armature of a eight6pole machie "ere "oud "ith a simplex "ave "idig! ho" may parallel paths "ould there be# A. 1 paths &. 8 paths %. 2 paths (. $ paths Problem $: he armature of a four6pole shut geerator is lap "oud ad geerates 21 volts "he ruig at 00 rpm. he armature has 1$$ slots! "ith six coductors per slot. If this armature is re"oud! "ave coected! fid the emf geerated at the same speed ad flux pole. A. $12 , &. $2 , %. $5 , (. $-2 ,

Problem $80: A t"o6pole (c geerator has a armature cotaiig a total of $0 coductors coected i t"o parallel paths. he flux per pole is .$8 x 108 lies ad the speed o f the prime mover is -0 rpm. he resistace of each coductor is 0.01 ohm ad the curret carryig capacity cap acity of each coductor is 10 A. calculate the termial voltage of the geerator. A. 12. , &. 125.2 , %. 1-0.8 , (. 128.$ , Problem $81: A four6pole four6pole (c geerator has a armature "idig cotaiig a total of $8 coductors coected i t"o parallel paths. If the flux per pole is 0.-21 x 10 max"ells ad the speed of rotatio of the armature is 1!800 rpm. %alculate the ampacity reuired i each armature coductor if the po"er developed by the armature is 5 3+. A. $0 A &. -0 A %. 20 A (. 25 A Problem $82: A four6pole four6pole (% geerator "ith duplex lap "idig has $8 slots ad four elemets per slot. he flux per pole is 2.5 x 10 max"ells ad it rus at 1500 rpm. +hat is the output voltage# A. 0 , &. -0 , %. 225 , (. 120 , Problem $8-: he armature of a $6pole! 128 slots shut geerator is lap "oud "ith four coductors per slot. he flux per pole is 50 m+b. he armature ad shut field resistaces are 0.0$ ohm ad 50 ohms respectively. respectively. (etermie the speed of the machie "he supplyig $00 A at a termial voltage of 250 ,. A. 15 rpm &. 22 rpm %. -$ rpm (. 2$ rpm Problem $8$: A four6pole four6pole shut coected geerator ge erator has a lap coected armature "ith 28 coductors. he flux per pole is 25 m+b. If the geerator supplies t"o hudred fifty 110 110 ,! ,! 5 + icadescet lamp. he field ad armature resistaces are 110 ohms ad 0.025 ohm respectively. respectively. (etermie the speed i rpm of the geerator. A. -$ &. -8 %. - (. -Problem $85: A speed6idicatig device cosists of a 26pole permaet maget (c geerator "ith a 20006ohms (% voltmeter coected across its termials. he geerator has a armature "ith 1200 coductors havig t"o parallel paths ad a total resistace of 1000 ohms bet"ee brushes. he air gap flux is $0!0 00 lies per pole. If the voltmeter reads 10 ,! ,! at "hat speed "ill this readig idicates# A. 15-$ rpm &. 152 rpm %. 1512 rpm (. 1582 rpm Problem $8: A shut geerator supplies a load of 5!500 + at 110 , through a pair of feeder coductors havig a resistace of 0.02 ohm each. he armature! shut field resistaces are 0.15 ohm ad 50 ohms respectively. respectively. Eid the emf geerated. A. 11 , &. 118 , %. 121 , (. 120 , Problem $8: he o6load voltage of a 5063+! 220 volts! shut geerator i the E79 laboratory is 2-2 volts ad rated load voltage is 220 volts. he armature resistace icludig brushes! is 0.02 ohm ad the resistace of the shut field resistace is 52 ohms. +hat is the iduced electromagetic force at rated load#

A.

221.- ,

&.

22.02 ,

%.

22$.5 ,

(.

22.05 ,

Problem $88: he field curret of a 180 3+! 250 volts (% shut geerator is  A "he ruig full load at rated termial voltage. If the combied armature ad brush resistace is 0.02 ohm! solve for the electrical efficiecy of the geerator. A. 2.1-J &. $.05J %. -.-J (. 1.8J Problem $8: 77 &oard Sue10 he field circuit of a 200!000 "atts! 2-06 , shut geerator is 8 A "he ruig full load at rated termial voltage. If the combied brush ad armature resistace is 0.0- ohm! solve for the electrical efficiecy of the geerator. A. 85.-$J &. 8.1$J %. 88.1J (. 8.82J Problem $0: 77 &oard ctober 180 he field curret of a 100 3+! 250 volt shut geerator is 5 amperes "he deliverig po"er at rated termial voltage ad at rated load curret. he combied armature ad brush resistace is 0.01 ohm. (etermie the efficiecy of the geerator. A. .2J &. $.-J %. 8.2J (. $.J Problem $1: 77 &oard April 1 he armature of the shut geerator has 0.05 ohm effective resistace. 7ach brush has a effective resistace of 0.01 ohm. he termial voltage is 100 volts "hile the curret is $0 amperes. +hat is its efficiecy# A. 5.15J &. .5J %. .-0J (. .52J Problem $2: A shut geerator is coected to a load that dra"s 28.8 3+. @a < 0.0$  ad @sh < $8 . If the po"er developed i the armature is 10.-J of the po"er delivered to the load. (etermie the termial voltage of the geerator. A. 2-8 , &. 2$0 , %. 2-$ , (. 2- , Problem $-: A 15 3+! 2-0 ,! shut geerator "as ru light as a motor to determie its stray po"er losses at rated load. he applied voltage across the armature! computed for test "as 25$ , ad the armature curret dra" is 2 A. he armature ad shut field resistaces are 0.25 ad 100  respectively. %alculate %alculate the geerator efficiecy at half load. Assume stray po"er losses to be costat. A. 81.-$J &. 8$.8-J %. 8-.J (. 8.0J Problem $$: A shut geerator supplies 100 A at 200 ,. ,. he armature ad shut resistaces are 0.02 5  ad 100 ! respectively. If the copper losses are eual to the iro losses at this load! fid the bra3e horsepo"er of the egie drivig the geerator. A. 2.$- hp &. 28.58 hp %. -0.12 hp (. -2.2$ hp Problem $5: A 20 3+! 3+! 200 , shut geerator is operated at rated load. If the drivig egie is developig -0 bhp! determie the iro ad frictio losses. Armature Armature resistace is 0.05 ohm ad shut field resistace is 50 ohms. A. 1-$0 + &. 10- + %. 5$ + (. 10-$ +

Problem $: A 220 , shut geerator has a armature resistace of 0.02  ad a shut resistace of $$ . At rated load! the drivig egie develops d evelops 2.25 hp. If the iro ad frictio losses amout to 100 +! ho" much is the curret delivered to the load at rated coditio# A. 18 A &. 200 A %. 15 A (. 12 A Problem $: 77 &oard ctober 1 1  ad 77 &oard April 1 A (% geerator has o6 load output voltage of 120 volts. Its armature circuit resistace is 0.5 ohm ad its field coils are separately eergiBed. e ergiBed. If the load is rated 2 3+ at 115 ,. ,. eglectig the effect of armature reactio! "hat po"er could be delivered to the load# A. 1!518 + &. 2!180 + %. 1!0 + (. 2!000 + Problem $8: he armature of a 6 pole separately excited (% geerator is lap "oud "ith 5-$ coductors. his machie delivers po"er to the load at 250 , "hile beig drive at 1000 rpm. At this load! the armature circuit dissipates $0 +. If If the flux per pole of this geerator is -06m+b! determie the 3+ ratig of the load served. Assume a total brush cotact co tact drop of 2 ,. ,. A. 10. 3+ &. 11.2$ 3+ %. 12.5 3+ (. 15.$$ 3+ Problem $: A separately excited geerator "he ruig at 1200 rpm supplies 20 0 A at 125 , to a circuit of costat resistace. Armature Armature circuit resistace is 0.10 ohm. (etermie the termial voltage "he the speed is dropped to 1000 rpm# Assume that the field curret is ualtered. A. 10$ , &. 125 , %. 118 , (. 11$ , Problem 500: A dc series geerator is supplyig a curret of 5 A to a series lightig system through a feeder of total resistace! 1.0 ohm. he termial voltage is 2500 2 500 ,. he armature ad series field resistaces are respectively 18 ad 15 ohms respectively. A -06ohm diverter resistace is shuted across the series field. (etermie the po"er developed i the armature of the geerator# A. 1-2.25 3+ &. 1-.50 3+ %. 12-.2 3+ (. 1-.80 3+ Problem 501: A -0 3+ series geerator has a efficiecy of 88 percet "he operatig at rated load. If the stray po"er loss is 15J of the full load losses! losses! calculate the maximum efficiecy of the geerator. Assume that the stray po"er loss is costat ad the other losses vary as the suare of the load. A. 0.$-J &. 2.0$J %. 1.12J (. 8.2J Problem 502: he hysteresis ad eddy curret losses of a (% machie ruig at 1000 rpm are 250 "atts ad 100 "atts! respectively. If the flux remais costat! at "hat speed "ill the total iro loss be halved# A. 500 rpm &. 550 rpm %. 50 rpm (. -0 rpm Problem 50-: he eddy curret loss i a (% machie is 00 "atts "he the total flux is 2!000!000 max"ells per pole ad the machie is ruig at 1000 rpm. (etermie the loss "he the excitatio of the machie is ad*usted to produce 2!500!000 max"ells per pole ad the speed is icreased to 1200 rpm. A. 00 "atts &. 1025 "atts %. 1280 "atts (. 1-50 "atts

Problem 50$: I a (% geerator! the iro losses at 1!000 rpm are 10 3+ at a give field curret. At a speed of 50 rpm ad at the same field curret! the total iro losses become  3+ 3 +. Assume Assume the hysteresis loss is directly proportioal to the speed ad the eddy curret loss is proportioal to the suare of the speed. (etermie the iro losses at 500 rpm. A. $ 3+ &. 5 3+ %. - 3+ (. 2 3+ Problem 505: A compoud dyamo supplies a load of 20 3+ at $00 , through a feeder havig a resistace eual to 0.1 ohm. he shut ad series field resistaces are 85 ohms ad 0.0 oh m respectively. respectively. (etermie the total electromotive force "he the geerator is coected log shut. Assume a brush cotact drop of 1 , per brush ad a armature resistace of 0.1 ohm. A. $18.-$ , &. $12.50 , %. $15. , (. $21.$$ , Problem 50: A log shut compoud geerator has the follo"ig parameters: @a < 0.1 ! @se < 0.05  ad a d @sh < 120 . his machie supplies po"er aggregate dc load of 10 3+. At At this load the armature geerates 2$ volts. (etermie the termial voltage. A. 251.55 , &. 25.8 , %. 25-.82 , (. 2$.-- , Problem 50: A compoud geerator is to be operated! as a flat6compouded log shut compoud geerator. o obtai the same rated voltage at full load ad at o load "he operated as a shut geerator! it is ecessary to icrease the field curret by 0.20 A. he shut field "idig has 1000 turs per pole! ad the series field "idig has $ turs per pole. If the full load armature curret of the compoud geerator is 5 A ad the series field resistace is 0.025 ohm! calculate the diverter resistace reuired. A. 0.08  &. 0.0$2  %. 0.028  (. 0.0-$  Problem 508: he armature! series field ad shut field resistaces of a 125 3+! 2506, log shut cumulative compoud (% geerator are 0.025 0.0 25 ! 0.01  ad -0 ! respectively. If the stray po"er loss at rated loss is 5 3+! determie the or bra3e horsepo"er reuired to drive this machie. Assume a brush drop of 2 , ad a stray load loss of 1J of the output. A. 182.-2 hp &. 201.-- hp %. 10.85 hp (. 210.5 hp Problem 50: A log shut compoud geerator delivers 100 A! 250 , at rated load. he resistaces of the various "idigs are: @a < 0.1! @ip < 0.02! @sh < 100ad @se < 0.025. If the stray po"er losses amout to 1.8 3+! solve for the overall o verall efficiecy of the geerator. A. 0.52J &. 8.$-J %. 88.-2J 8.-J Problem 510: A certai certai 5.5 3+! 1256, (% log shut compoud geerator "as operated as a shut motor at o load for determiatio of stray po"er losses. +ith 1-5 , eual to the geerated emf at full load applied to the armature ad at ormal speed of o f 100 rpm! the curret iput to the armature "as foud to be 2.5 A. he armature! shut field ad series field resistaces are 0.2 ! 25  ad 0.25 ! respectively. (etermie (etermie the efficiecy of the geerator at rated load. A. 8$.52J &. 8-.5J %. 80.21J (. 82.2-J Problem 511:

A compoud dyamo supplies a load of 20 3+ at $00 , through a feeder havig a resistace eual to 0.1 ohm. he shut ad series field resistaces are 85 ohms ad 0.0 oh m respectively. respectively. (etermie the total electromotive force "he the geerator is coected short shut. Assume a brush cotact drop of 1 , per brush ad a armature resistace of 0.01 ohm. A. $15.$8 , &. $11.$2 , %. $1.2- , (. $12.$2 , Problem 512: A short shut compoud geerator has the follo"ig has a armature! series field ad shut field resistaces are 0.0$ ohm! 0.0- ohm oh m ad 80 ohms respectively. he geerator serves a 20 3+! 200 , load load through a feeder "ith a euivalet resistace of 0.01 ohm. Dolve for the po"er geerated by the armature. A. 2-.8$ 3+ &. 21.-0 3+ %. 22.2 3+ (. 21.2 3+ Problem 51-: A short shut compoud geerator has a armature! shut field ad series field resistaces of 0.02 ! 80  ad 0.00.0 -  respectively. he geerators iduced iduced emf is 510 , "he the termial voltage is 500 ,. ho" much load curret is delivered uder this coditio. co ditio. A. 1.$ A &. 10.$ A %. 1$.5- A (. 12.2 A Problem 51$: A short shut field compoud geerator delivers 150 at a termial voltage is 2-0 ,. he shut field curret is 2.5 A. he resistace of the armature! series field ad diverter are 0.0-2 ! 0. 015  ad 0.0- ! respectively. respectively. %alculate the electrical e lectrical efficiecy of the geerator at this load. A. .$J &. -.8J %. $.2J (. 5.J

Problem 515: "o "o shut (% geerators ge erators O1 ad O2 are rated at 100 ad 150 3+ at 110 , respectively. heir exteral load characteristics may be cosidered as straight lie. he drop i the termial voltage from o6 load to full6 load is 10 volts for O 1 ad 20 volts for O2. %alculate the o6 load voltages of these geerators "he they are operated i parallel to supply a load of 2000 2 000 A! "hich is divided bet"ee them i proportio of their ratigs. A. O1 < 12-. ,! O2 < 1--. , &. O1 < 12-. ,! O2 < 1-1. , %. O1 < 121. ,! O2 < 1--. , (. O1 < 121. ,! O2 < 1-1. , Problem 51: "o "o shut geerators operatig i parallel are givig a total output of -20 3+. Oeerator A &

Armature @e @esistace 0.02 ohm 0.025 ohm Eid the bus bar voltage. eglect shut field curret. A. $50.- , &. $$.5 , %. $2-.$ ,

Iduced em emf $50 , $0 , (.

$$.2 ,

Problem 51: "o idetical shut geerators ruig i parallel supply a total load curret of 2500 A. 7ach machie has a field resistace of 50 ohms ad a armature resistace of 0.01 ohm. he iduced emfs geerated are 2-0 ad 2-5 volts! respectively. %alculate the commo operatig voltage of each machie. A. 220 , &. 222 , %. 22- , (. 218 ,

Problem 518: hree shut geerators operatig i parallel supply a total load curret of 2000 amperes. 7ach machie has a field resistace of $0 ohms ad a armature resistace of 0.02 ohm. he emfs geerated are 2$0! 2$2 ad 2$5 volts respectively. (etermie (etermie curret delivered by each geerator. A. 550 A! 00A! 850 A &. 550 A! 50 A! 800 A %. 500 A! 50 A! 850 A (. 00 A! 00 A! 800 A Problem 51: "o "o shut geerators ruig i parallel deliver a total curret of -!00 A. 7ach machie has a field resistace of 0 ohms ad a armature resistace of 0.01 ohm. he emfs geerated b y each machie are $55 ad $0 ,! respectively. respectively. %alculate the 3+ 3 + share of each machie. A. 52 3+! 108 3+ &. 0 3+! 10 3+ %. $1 3+! 108 3+ (. 82 3+! 105 3+ Problem 520: A shut shut geerator ad a battery are coected i parallel p arallel servig a 500 A bus load. he geerator has a exteral characteristics such that the termial voltage drops uiformly from $0 volts o o6 load to $$0 volts o rated load "hile deliverig 125 A of curret. he battery has a ope circuit emf is $20 , ad iteral resistace of 0.01 ohm. +hat percetage of the load curret does the battery deliver# A. 5-J &. $J %. $J (. -J Problem 521: A 50 3+! 3+! 220 , (% geerator has a voltage regulatio of 10J. It is co ected i parallel "ith a battery of 115 cells! each "ith a emf of 2, ad iteral resistace of 0.01 ohm. +hat curret "ill the battery carry if by accidet the load is trip6 off from from the bus# A. 0A &. 8 .0 2 A %. .1 5 A (. .20 A Problem 522: "o "o idetical 00 3+! 2-0 , dc geerators are operatig i parallel ad ta3e eual shares of a 800 3+! 250 , bus load. he voltage regulatio of each machie is 5J. If oe of the geerators is accidetally tripped off from the lie! "hat is the voltage of the remaiig machie# A. 250 , &. 2$- , %. 2$5 , (. 2$ , Problem 52-: "o "o 250 , dc geerators are operatig i parallel p arallel ad ta3ig eual shares of a -000 A! 250 , busload. he ratig of the geerators are as follo"s: Cachie Parameters @egulatio >J? @ated Po"er >3+?

Oeerator 1 2 500

Oeerator 2 $ $00

If the load drops to 2100 A! determie the e" operatig bus voltage. A. 252 , &. 25 , %. 25$ , (.

25- ,

Problem 52$: "o "o shut geerators are operatig i parallel ad ta3ig eual shares of a 2$00 2$ 00 A! 2-0 , busload. he ratigs of the geerators are as follo"s:

Cachie Parameters @egulatio >J? @ated Po"er >3+? @ated ,oltage

Oeerator 1 500 250

Oeerator 2 $ 00 250

If a additioal 006 A load load is coected to the bus! determie the curret delivered by each geerator at this coditio. A. 151 A! 1$8$ A &. 1525 A! 1$5 A %. 112 A! 1-88 A (. 155 A! 1-$5 A Problem 525: A simplex lap "oud armature has 580 coductors ad carries a curret of 125 amperes per armature curret path. If the flux per pole is 20 m+b. %alculate the electromagetic torue d eveloped by the armature# A. 210.$6m &. 252.2 6m %. 2-0. 6m (. 20.- 6m Problem 52: A 6pole lap "oud shut motor ta3es -00 A "he "he the speed is 500 rpm. he flux per pole is 0.0 +b ad the armature has 80 turs. eglectig the shut field curret! calculate the bra3e horsepo"er of the motor. Assume Assume a costat loss of $J. A. 15 &. 18 %. 12 (. 15 Problem 52: A 220 , shut motor drivig its ormal load dra"s a armature curret of 50 A from a 220 , dc source. he armature resistace of this motor icludig brushes is 0.25 ohm. /o " much armature curret >miimum? "ill this motor dra" from a 200 , dc source "he the same load "ith the field ad*usted to maitai the same speed. A. 55. A &. 58.2$ A %. 0.12 A (. -.$8 A Problem 528: A (% shut motor has a full load ratig of 15 hp! 2-0 volts! 5.1 amperes! 1$00 rpm. he armature circuit resistace is 0.1- ohm ad the field circuit resistace is 115 ohms. eglectig the effect of armature reactio! determie the o6load lie cu rret. A. .$ A &. 5. 2 A %.  .$ 5 A (. 8.$ A Problem 52: At rated load! a shut motor dra"s d ra"s a armature curret of 50 A from from a 2-0 , dc mais "hile ruig at 1000 rpm. At o load the armature curret drops to 5 A. If the effect of armature reactio has "ea3e the flux by 2J from o6load to full load! determie speed of the motor at o6load. Assume the armature resistace to be 0.15 ohm. A. 1202 rpm &. 1221 rpm %. 1122 rpm (. 1010 rpm Problem 5-0: he ameplate ratig of a shut shu t motor is 150 hp! 00 volts! 205 A! 100 rpm. he resistace of the shut field circuit is 2$0 ohms ad the total armature circuit resistace is 0.15 ohm. %alculate the speed regulatio of the motor. A. $.0J &. $ .8  J %. 5 .2 5 J (. 5.18J

Problem 5-1: A (c shut motor rus at 00 rpm o a 2$06volt 2$0 6volt supply "hile dra"ig a lie curret of -0 A. Its armature ad field resistaces are 0.5  ad 120  respectively. +hat resistace resistace should be placed i series "ith the armature circuit i order to reduce the speed to $00 rpm. Assume o chages i the armature or field curret. A. 2.  &. 2 .$ 1  %. 2 .0   (. 2.8-  Problem 5-2: A 1206, shut motor has a armature euivalet of 0.5  ad a field resistace of 0 . At full load! the motor ta3es 10 A ad the speed is 1000 rpm. At "hat speed must this motor maybe drive to deliver 10 A to a exteral load at 120 ,# A. 1158 rpm &. 108 rpm %. 122 rpm (. 10$5 rpm Problem 5--: A belt6drive belt6drive 150 3+ shut "oud (% geerator is ruig at $50 rpm ad a d is supplyig full load to a bus bar at 2$0 ,. At "hat speed "ill it ru if the belt brea3s ad the machie cotiues to ru ta3ig 8.5 3+ from the bus bar# he armature ad field resistaces are 0.05 ohm ad 85 ohms respectively. he he brush cotact drop is 1.5 volt per brush. eglect armature reactio. A. -5 rpm &. -88 rpm %. $02 rpm (. -58 rpm Problem 5-$: A shut motor dra"s a curret of $0 A from a 120 , source ad rus at 1200 rpm at rated load. he armature ad field circuit resistaces are 0.1  ad 0 ! respectively. (etermie the speed of the motor at half load. Assume a brush drop of 2 , at rated load ad 1 , at half load. A. 1180 rpm &. 12-8 rpm %. 12-0 rpm (. 102 rpm Problem 5-5: A (% shut motor has a full load ratig of 15 hp! 2-0 volts! 5.1 amperes! 1$00 rpm. he armature series circuit resistace is 0.1- ohm ad the field circuit resistace is 115 ohms. eglectig the effect of armature reactio! determie the o6load speed. A. 1-2 rpm &. 12$ rpm %. 1$$1 rpm (. 1-80 rpm Problem 5-: A 2206, shut motor delivers $0 hp o full load at 50 rpm ad has a efficiecy of 88J. he armature ad field resistaces are 0.2 ohm ad 110 ohms! respectively. (etermie (etermie the startig resistace such that the startig armature curret does ot exceed 1. times the full load curret. A 1206, (% motor rated at 56hp has a full load efficiecy of 8J. he field ad armature resistaces are 0 ohms ad 0.5 ohm! respectively. +hat +hat starter resistace "ill be reuired to limit the armature curret at startig to 200J of its rated armature curret# A. 1.02 ohms &. 1.12 ohms %. 1.00 ohms (. 1.22 ohms Problem 5-8: A 106hp 2-0 , (% motor of 85J full load efficiecy is located $50 ft from the supply mais. If the motors startig curret is 1.5 times the full load curret! "hat is the smallest cross6sectioal area of copper "ire reuired "he the allo"able voltage drop i the feeder at startig is limited to 2$ volts# A. -0 C%C &. 28 C%C %. 2 C%C (. 2 C%C Problem 5-: he ameplate ratig of a shut shu t motor is 150 hp! 00 volts! 205 A! 100 rpm. he resistace of the shut field circuit is 2$0 ohms ad the total armature circuit resistace is 0.15 ohm. If the motor is to be deliver full6load torue at 1!200 rpm! "hat value of resistace must be added to the armature circuit#

A.

0.8-- ohm

&.

0.802 ohm

%.

0.82 ohm

(.

0.82 ohm

Problem 5$0: A 120 , shut motor dra"s a curret of  A at rated load. he armature ad shut field resistaces are 0.2  ad 0 ohms respectively. A multi6tap multi6tap starter rheostat is iserted i the armature circuit to limit the startig curret. If the rheostat has a resistace of 1.-2 ohms! determie the percetage tap used such that the startig curret "ill be limited to 150J of its full load value# A. 0J &. 50J %. 55J (. $5J Problem 5$1: A 106hp po"er 220 , (c shut motor has a armature ad field resistace of 0 .25 ohm ad 100 ohms respectively. respectively. he full load efficiecy is 8-J. (etermie the value of startig resistace i order that the startig "ill ot exceed 200 percet of full6load value. A. 2.88  &. 2 .5   %. 2 .1 5  (. 2.$5  Problem 5$2: A 506/P! 506/P! 550 , shut "oud motor dra"s a lie curret of $.5 A at o load. he shut field resistace is 25 ohms ad the armature resistace exclusive o f brushes! is 0.- ohm! the brush drop at full load is 2 ,. at full load! the motor dra"s a lie curret of 8$ A. %alculate the efficiecy at full load. A. 2.0J &. 88.5J %. 1.2J (. 8.J Problem 5$-: A 1006volt shut motor is developig  hp "hile operatig at a overall efficiecy of 8J. he armature ad shut field resistaces are 0.0 ad 50 ohms respectively. (etermie stray po"er losses. A. -8 + &. - + %. -2 + (. -$ + Problem 5$$: he shaft po"er of a shut motor is .8 hp. It dra"s 50 A from 120 ,. ,. he field "idig dra"s 1.2 A. +hat is the efficiecy of the motor# A. $.J &. -J %. .J (. 8.2J Problem 5$5: A 500 , shut motor ruig at 20 rpm ta3es a armature curret of 50 A. he euivalet armature resistace is 0.$ ohm. +hat resistace must be placed i series "ith the armature to reduce the speed to 5$0 rpm "hile maitaiig the same torue# A. 2.2 ohms &. 2. ohms %. 2.$ ohms (. 2.8 ohms Problem 5$: A certai shut motor has a armature resistace of 0.05 ohm. It dra"s 50 A at a termial voltage of 120 ,. Assume other miscellaeous losses at 1J. (etermie the output horsepo"er of the motor. A. 8.5 hp &. .1 hp %. .8 hp (. .8 hp Problem 5$: he armature ad field resistaces of a shut machie are 0.2 ohm ad 110 ohms respectively. respectively. +hile ruig as a geerator! the geerated emf is 250 , at 1200 rpm. If the machie is ru as a shut motor! it ta3es $ A at 220 ,. At a certai load the motor ta3es -0 A at 220 ,. /o"ever o load! the armature reactio "ea3es the field by 2J. Eid the motor speed at this load. A. 1120 rpm &. 1050 rpm %. 10$2 rpm (. 1025 rpm Problem 5$8:

A shut motor! motor! "hich has a field resistace of 220 ohms ad a armature resistace of 0.8 ohm ta3es 2 A from 20 , supply "he ruig at 500 rpm o full load. I order to cotrol the speed of the motor a 1.26ohm resistor is coected i series "ith the armature. %alculate the speed at "hich the motor "ill ru "he supplyig full load torue. A. $-8 rpm &. $$5 rpm %. $2$ rpm (. $2 rpm Problem 5$: he iput ad output po"ers of a 220 ,! shut motor are -0 - 0 ad 28 3+ respectively. respectively. he field ad armature circuit resistaces are $0 ohms ad 0.0125 ohm respectively. (etermie (etermie the efficiecy of the motor if the output po"er is reduced to 10 3+. A. 8-.5J &. 88.J %. 8$.-J (. 8.$85 Problem 550: A 220 , series motor ta3es 15 A ad rus at 00 rpm. he armature resistace icludig the resistace of the series "idig is 0.8 ohm. At "hat speed "ill it ru if a 26ohm resistace is coected i series "ith supply voltage# Assume the motor is ta3ig the same curret. c urret. A. 5 rpm &. 01 rpm %. 52 rpm (. 58- rpm Problem 551: he euivalet armature resistace of a series motor is 0.1 ohm. +he coected across a 1106 , mais! the armature ta3es 20 A ad its speed is 1000 rpm. (etermie its speed "he the armature ta3es 50 A from the same mais! "ith the field icreased by 10J. A. 80 rpm &. 5$ rpm %. 85 rpm (. 88$ rpm Problem 552: A $00 , series motor motor has a field resistace of 0.2 ohm ad a armature resistace of 0.1 ohm. he motor ta3es -0 A of curret at 1000 rpm "hile developig full load torue. +hat is the motor speed "he this motor develops 0J of full load torue# A. 1-02.$ rpm &. 125.2 rpm %. 12. rpm (. 11-5.5 rpm Problem 55-: A 106 hp! 2-06 ,! 1200 rpm series motor havig rated load efficiecy of 85.5J. he armature resistace icludig brushes is 0.- ohm. he field resistace is 0.25 oh m. Assumig Assumig the flux varies directly as the armature curret! "hat value of resistace should be placed i series "ith this motor "he startig! i order that the startig curret maybe limited to the value that "ill exert a startig torue eual to150J of its rated torue# A. $. ohms &. $.$ ohms %. 5.1 ohms (. -.8 ohms Problem 55$: he field ad armature resistace of a 220 2 20 , series motor are 0.2  ad 0.1 ! respectively. he motor ta3es -0 A of curret "hile ruig at 00 rpm. If the total iro ad frictio losses are -50 +! determie the motor efficiecy. A. 0.J &. 1.5J %. 8.$J (. 2.2J Problem 555: A 1006 , series series motor is used to drive a load through a pulley. his machie has a armature resistace of 0.2 ohm ad a series field resistace of 0.25 ohm. +he a torue of 25 6m is applied to the pulley! the speed is 00 rpm. If stray po"er losses at this load load is -00 +! +! calculate the armature curret. A. 20.2 A &. 21.8 A %. 22.2 A (. 2-.$1 A

Problem 55: A $00 , series motor! havig a armature euivalet resistace of 0.5 ohm! o hm! ta3es $$ A of curret "hile ruig at 50 rpm. +hat is the motor speed for a lie curret of - A# A. 80- rpm &. 822 rpm %. 812 rpm (. 80 rpm Problem 55: he resistace of each of the t"o coils of a series motor is 0.0$ ohm! armature resistace! 0.0$ ohm. he motor ta3es 50 A at 100 , "hile ruig at 800 rpm ad the coils are i series. +hat "ill its speed if the coils are re6coected i parallel "hile the load torue is doubled# A. 800 rpm &. $00 rpm %. 200 rpm (. 50 rpm Problem 558:  full load! a 500 , series motor ta3es 100 A ad rus at 820 rpm. he armature ad series field resistaces are 0.1 ad 0.0$ ohm! o hm! respectively. +hat "ill be its speed "he developig half6 h alf6 full load torue ad "ith a 0.086ohm diverter coected across the series field "idig. Assume flux is proportioal to the field curret. A. 15- @PC &. 1-$ rpm %. 1-5 rpm (. 1$2 rpm Problem 5: A $00 , series motor "or3ig "ith usaturated field is ta3ig 0 A ad ruig at 8$0 rpm. he total resistace of the motor is 0.1 ohm. At "hat speed "ill the motor ru "he developig half6full load torue. A. 11- rpm &. 1202 rpm %. 11 rpm (. 1-52 rpm Problem 50: A log shut compoud motor ta3es a curret of $2 A from from 2-0 , source. he geerator parameters are: @a < 0.1 ! @se < 0.2  ad rsh < 50 . If the frictio ad "idage losses amout to $00 +! determie the overall efficiecy e fficiecy of the machie. A. 82.2$J &. 80.5J %. 81.22J (. .5-J Problem 51: A 2-0 ,! ,! log log shut machie has the follo"ig parameters: @a < 2.0 ! @sh < $0  ad @se < 0.25 . +he the machie is ru at o6load at its ormal speed ad rated voltage! the armature dra"s 0. A. (etermie the armature curret dra" if the machie delivers a output of 5 bhp. A. 2$.-5 A &. 21.22 A %. 20.- A (. 22.8- A Problem 52: A 25 hp! 250 , log shut compoud motor ta3es 5 A "he ruig light ad rus at 1150 rpm. Armature resistace! 0.15 ! Dhut field resistace! 200  a d series field resistace! 0.05 . At rated load the motor ta3es 8.5 A ad rus at 20 rpm. (etermie the ratio of torue developed from o6load to full load. A. 1:2 &. 1:-$ %. 1:22 (. 1:-8 Problem 5-: A 200 volt shut motor has a armature ad shut field resistaces of 0.15  ad 125 ! respectively. respectively. +he ruig light! it ta3es  A ad rus at 1200 rpm. A series "idig of 0.05 ohm is added to ma3e it log shut cumulatively compouded. his "idig icreases the flux per pole by 25J "he the motor is ta3ig its full load curret of 100 A. eglectig armature reactio! solve for the operatig speed as a compoud motor. A. 88 rpm &. 80 rpm %. 88 rpm (. 85 rpm

Problem 5$: A 220 , (% motor has a armature resistace of 0.$ ohm! a shut field resistace of 1 5 ohms ad a series field resistace of $ ohms. his machie is coected as a short shut compoud motor to 220 22 0 , mais. %alculate the po"er developed by the armature if the armature curret is -0 A. eglect the brush drop. A. 225$. + &. 252.8 + %. 252. + (. 2150.5 + Problem 55: he iput curret to a 220 ,! shut compoud motor at o load is  A. he shut field circuit resistace is 220 ohmsT the armature resistace is 0.10 ohm a d the series field resistace is 0.08 ohm. +hat is the stray po"er loss# A. 115-.-1 + &. 10.22 + %. 12-2.5$ + (. 10-.25 + Problem 5: he poy bra3e test of a elevator door drive shut motor! the ammeter ad voltmeter measurig the iput read -$ A ad 220 ,. ,. he he speed of the motor is foud to be 10 rpm ad the balace o a 2 ft bra3e arm read 2.2 pouds. he tare "eight of the arm is foud to be positive 2.- pouds. (etermie the efficiecy of the motor of this load. A. 8.0-J &. 8.05J %. 8$.$5J (. 88.-2J Problem 5: I a bra3e test of a elevator door motor! the ammeter ad voltmeter measurig the iput read -$ A at 220 ,. ,. he he speed of the motor is foud fou d to be 10 rpm ad the balaced o a 0.1 m bra3e arm reads 11. 3g. he dead "eight of the arm is foud to be 1.0 3g. +hat is the output hp of the door motor# A. .$0 hp &. 10.-8 hp %. 8.2 hp (. .58 hp Problem 58: A shut motor "ith a armature ad field resistace of 0.055 ad -2 ohms! respectively! is to be tested for its mechaical efficiecy by meas of a rope bra3e. +he the motor is ruig at 1$00 rpm! the logitudial pull o the 6ich diameter pulley is 5 pouds. he readigs of the lie ammeter ad voltmeter are -5 ad a d 105! respectively. %alculate the efficiecy. A. .10J &. 5.-2J %. 8.28J (. .12J Problem 5: A shut motor "as tested by meas of a poy bra3e havig a legth arm of -.5 -. 5 feet ad tare "eight of 5. lbs. he curret dra" by the machie from a 2$0 , lie "as 50. A "he the scale readig "as 2$ lbs ad the speed of the motor "as 1215 rpm. %alculate the rotatioal losses of the motor. he armature ad shut field resistaces of the machie are 0.25  ad 120 ! respectively. A. $ + &. 2 + %. 20 + (. 05 + Problem 50: %alculate the force that "ill be exerted o the scale i a poy bra3e test "he a 206hp! 1!$00 rpm motor is operatig at full load. he legth of the bra3e arm is - ft. ad the tare "eight of the bra3e is -.5 - .5 lb. A. -0.$5 lbs &. 28. lbs %. 21.5$ lbs (. -.-- lbs

TEST 1' – *LTE0N*T0S

Problem 51: +hat is the speed of a 100 3+! 2-06 2-06 volt! three6phase! four 'pole! 0 /B alterator# A. $50 rpm &. 00 rpm %. 1200 rpm (. 1800 rpm Problem 52: A six6 pole! -6 phase A% geerator has 2 slots! the coil spa is 12. +hat is the pitch factor# A. 1.000 &. 0.- %. 0.85 (. 0. Problem 5-: A three6 phase alterator has  slots per pole per phase. (etermie the distributio factor. A. 0. &. 0.5 %. 0.(. 0.$8 Problem 5$: A -0 C,A! 15 3, three6 phase alterator "ill have a per phase omial impedace of XXXXXX A. .5 ohms &.  ohms %. 15 ohms (. oe of these Problem 55: A t"o6 pole three phase phas e 50 /B L6 coected A% geerator has  slots per pole ad  coductors per slo t. +hat is the flux per pole if the voltage o ope circuit is 1.1 3,. Assume a coil spa of uity. A. 0.15 +b &. 0.1$2 +b %. 0.150 +b (. 0.128 +b Problem 5: A 6 pole! -6 phase! 0 cycle alterator has 12 slots per pole ad $ coductors per slot. he "idig is 5 pitch. here are 2!500!000 2!500!000 max"ells >0.025 "eber? "eber? eterig the armature from each orth orth pole! ad this flux flux is siusoidally distributed distributed alog the air gap. he armature coils are all coected i series. he "idig is "ye6 "ye6 coected. (etermie the ope circuit emf of the alterator. alterator. A. 5-2.1 , &. 50$.2 , %. 512.$ , (. 52.$ , Problem 5: A four6 pole! three6 phase! 50 /B! star coected A% geerator has 2$ stator slots. here are  coductors per slot ad a siusoidally siusoidally distributed flux per pole of - m+b. m+b. Eid the termial voltage o ope ope circuit! if the coil spa factor is uity. A. 5$.05 , &. 5.$$ , %. 58.$8 , (. 558.2 , Problem 58: A three6 phase! 6 pole! 2 slot armature of a sychroous geerator is "oud usig a t"o6 layer lap coil havig 20 coductors per coil "ith a pitch of 8-.--J. he flux per pole is $!800!000 max"ells. If the rotor of the geerator is drive at a speed of 1200 rpm! calculate the iduced emf per phase. A. 580 , &. 5-0 , %. 55$ , (. 585$ , Problem 5: A 1 C,A! 11 3,! three6 phase "ye6 coected sychroous geerator has a sychroous reactace of 5 ohms ad a egligible resistace per phase. At a certai field curret the geerator delivers rated load at 0. laggig po"er factor at 11 3,. 3,. Eor the the same excitatio! "hat is trhe termial voltage at 0. 0. leadig po"er factor full full load# A. 11. 3, &. 11.$ 3, %. 11.- 3, (. 11.5 3, Problem 580: A three6 phase star coected A% geerator is rated at 2.5 C,A! 11 3,! 0 /B. he geerator has a egligible resistace ad a sychroous reactace 20 ohms per phase. o "hat value "ill the termial voltage rise "he full load at 80J pf laggig is s"itch6off# A. 1$.2 3, &. 1-.5 3, %. 15.- 3, (. 12. 3, Problem 581.

A 1200 3,A! 006volt! - phase L coected alterator has a effective resistace of 0.$0 ohm ad a reactace of  ohms per phase. It delivers full load curret at 0. 80 laggig pf at rated voltage. +hat "ill be the termial voltage for the same excitatio ad load curret! if the po"er factor is 0.80 leadig# A. $50 , &. 88 , %. 88, (. $250 , Problem 582 A 2500 3,A! 00 , sychroous turbo alterator is operatig at full load ad 0.8 pf laggig. his machie has a sychroous reactace per phase is 10.$ ohms ad a egligible resistace. %alculate the percetage voltage regulatio. A. $8.5J &. $2.-5J %. $5.8$J (. $-.J Problem 58A 50 3,A! 5506volt sigle phase alterator dra"s a field curret 10 A at rated load. +ith the same field curret! the ope circuit emf ad short6circuit curret are -00 , ad 150 A. he ohmic resistace of the armature bet"ee termials is 0.18 ohm. he ratio of of effective to ohmic resistace resistace maybe ta3e as 1.5. (etermie (etermie the regulatio at 0.8 pf laggig. A. 2.8J &. 2.$J %. 25.5J (. 2$.J Problem 58$ A geerator is rated 100 C+! 1-.8 3,! ad 0J pf. he effective resistace is 1.5 times the ohmic resistace. he ohmic resistace is obtaied by coectig t"o termials to (% source. he curret ad voltage are 8. A ad  , respectively. +hat is the effective resistace per phase# A. 0.051-  &. 0. 0.051-  %. 0.051-  (. 0.051-  Problem 585 o get the armature resistace of a 100 C+! 1-.8 3, ad 0J po"er factor geerator! t"o termials are coected to a (% source. he measured curret ad voltage are 8. amperes ad  volts respectively. respectively. +hat is the (% resistace per phase# A. 0.022- ohm &. 0.0-$2 ohm %. 0.085 ohm (. 0.01 ohm Problem 58: A 2 C,A! 2.-6 3, three6 phase L6 coected alterator operates at rated 3,A at a po"er factor of 80J. he dc armature "idig resistace bet"ee termials is 0.08 . he field ta3es 0 A at 125 , from a exciter euipmet. Erictio ad "idage loss is 20 3+! iro losses - 3+ ad stray load losses are 2.0 3+. %alculate the efficiecy of the alterator. Assume Assume the effective armature6 "idig resistace is 120J of the (% resistace. A. $.0J &. 2.1J %. -.2J (. 8.5J Problem 58: (etermie the efficiecy of a 1500 3,A! 2-00 ,! -6 phase! L6 coected alterator! "hich operates at rated output "ith a po"er factor of 80J. he (% armature resistace at 0 o % bet"ee termials is 0.08 0. 08 ohm. he field ta3es 0 A at 120 , from the exciter euipmet. Erictio ad "idage losses are 15 3+! iro loss is -5 3+ ad stray load losses is 1.5 3+. Assume the effective armature "idig resistace is 1.5 times the (% value. A. $.5J &. 0.-1J %. -.-5J (. 2.2J Problem 588: A 2500 3,A 3,A! three6 phase! 0 /B! . 3, "ye6coected alterator has a field resistace of 0.$5  ad a armature resistace of 0.05  per phase. he field curret at full load 0.85 pf! is 200 A. he stray po"er losses amout to 82 3+. %alculate %alculate the efficiecy of the alterator at full load! 0.85 pf laggig. A. -.J &. $.8J %. 5.2J (. .-J Problem 58 A . 3,! L6coected! -6phase alterator has a sychroous reactace of   per phase ad a egligible resistace. At At a certai field curret the armature is 3o" to iduce 5.8 3, per phase. %alculate the maximum po"er that ca be developed developed by the alterator alterator at this excitatio.

A.

12 C+

&.

11 C+

%.

.8 C+

(.

10 C+

Problem 50 A 11!000 volts three6phase "ye6coected "ye6coected turbo alterator has a sychroous reactace of  ohms ad a egligible resistace per phase. +he the field curret is 8 A! the ope circuit voltage is 12!000 volts. (etermie the armature curret "he alterator developed maximum po"er. A. 1$5 A &. 112 A %. 1$- A (. 15 A Problem 51 A three6phase! 11 3, "ye coected sychroous alterator has a sychroous reactace of 8 ohms per phase but egligible egligible resistace. If the excitatio excitatio is such that the the ope circuit voltage voltage is 1$ 3,! 3,! determie determie the po"er factor at the maximum output. A. 0.8 &. 0.2 %. 0.(. 0.08 Problem 52 A -6phase -6phase A% geerator is supplyig supplyig po"er to a load of -200 3+ at 2-00 volts ad a pf 0f 0J. Assume Assume that the loss of the lie! the geerator armature ad the load is eual to 10J of the load! "hat "ould be the savigs i "atts if the pf "ere raised to 80J# A. 100!000 &. 1$0!000 %. 80!000 (. 2-0!000 Problem 5A -6phase 50 C,A! 1-.8 3,! -00 rpm! 0 /B grid coected geerator "ith a sychroous reactace of -.8 ohms per phase operatig "ith a laggig po"er factor of 0.88 at rated voltage ad curret. (etermie e" po"er factor if the motor motor curret is decreased by 5J. A. 0.20 &. 0.-5 %. 0.0 (. 0. Problem 5$ A -6phase 50 C,A! 1-.8 3,! -00 rpm! 0 /B grid coected geerator "ith a sychroous reactace of -.8 ohms per phase operatig "ith a laggig po"er factor of 0.88 at rated voltage ad curret. If the load decreases such that its prime mover po"er is decreased by 25J ad the rotor curret decreases b y the same percetage! "hat "ill be the e" pf# A. 0.28 &. 0.8 %. 0.8 (. 0.0 Problem 55 A -6phase 50 C,A! C,A! 1-.8 3,! -00 rpm! 0 /B grid coected coecte d geerator "ith a sychroous sychroou s reactace reacta ce of -.8 ohms per phase operatig "ith a laggig po"er factor of 0.88 at rated voltage ad curret. If the machie is to provide a laggig reactive reactive po"er of $0 C,A C,A@ ad still operatig at at its rated curret! "hat percet percet chage i the rotor curret is reuired# A. 10.$8J &. 10.2$J %. 11.-0J (. 12.28J Problem 5 A three6phase three6phase turbo alterator has a reactace of 15 ohms ad a egligible resistace! the machie dra"s a armature curret of 250 A at 0.8 pf laggig "he ruig o 12!000 volts ifiite busbars. If the steam admissio is costat but the emf is raised by 20J! calculate the e" operatig pf. A. 0.5$ &. 0.58 %. 0.50 (. 0.55 Problem 5 A three6phase! 20 3,! 3,! sychroous sychroous alterator has a reactace of 10 ohms ad a egligible resistace. he machie dra"s a armature curret of 150 A! 0.80 pf laggig. If the steam admissio is costat ad field curret is ad*usted raisig the iduced emf by 25J! determie the value of the e" armature curret. A. $$2 A &. $52 A %. $21 A (. $-- A Problem 58 A stadby (iesel geerator set "ill have the follo"ig loads: iductive load dra"ig 50 3,A at 0.8 pf T lightig load dra"ig 20 3,A at 1.0 pf. At "hat pf "ill the geerator operate#

A.

0.855 laggig

&.

0.82 laggig

%.

0.821 laggig

(.

0.8$ laggig

Problem 5 A 250 ,! -0 /B geerator supplies po"er to a parallel circuit cosistig of a 20 hp motor "hose efficiecy is 0J at 0.80 pf laggig ad a secod load that dra"s a apparet po"er of  3,A at uity pf. (etermie the system reactive po"er. A. 2-.582 3,A@ &. 12.$-5 3,A@ %. 10.-$1 3,A@ (. 20.-8$ 3,A@ Problem 00 "o -6phase! $10 ,! 0 /B alterators are operated i parallel. he total load of the system is 1050 3+ at 5J laggig pf. If the alterator A is carryig 00 3+ at 80J pf laggig! determie 3,A@ 3,A@ of alterator &. A. $01 3,A@ &. $05 3,A@ %. $11 3,A@ (. $08 3,A@ Problem 01 "o alterators A ad & are operatig i parallel supplyig a load dra"ig 1000 3,A at 0.80 po"er factor laggig. If alterator A cotributes cotributes 500 3,A at 0.0 po"er factor laggig! determie the po"er factor of alterator &. A. 0.28 laggig &. 0.852 laggig %. 0.01 laggig (. 0. laggig Problem 02 "o idetical three6phase L6coected alterators A ad & share eually a load of 10 C+ at -- 3, ad 80J laggig pf. he sychroous reactace of each machie is $ ohms per phase ad the armature r esistace is egligible. Alterator Alterator A has its field excitatio ad*usted to carry 125 A laggig curret! "hat is the curret supplied by alterator &# A.  A &.  A %. 8 A (. 5 A Problem 0Alterator A > 100 3,A! -6Y! 2$0 ,! 0 /B! 1800 rpm? is operatig i parallel "ith alterator & > 125 3,A! -6Y! 2$0 ,! ,! 0 /B! 1800 rpm?. he load of alterator A is 0 3+ at 0J pf laggig ad the load of alterator & is 80 3+ at 0J pf laggig. (etermie pf of load. A. 0.5 laggig &. 0.8$ laggig %. 0.8- laggig (. 0.8 laggig Problem 0$ "o -6phase! L6coected turbo alterators supply a load of -000 3+ at . 3, ad at 0.8 pf laggig. he excitatio of machie A is ad*usted so that it delivers 150 A at a laggig po"er factor! ad the goverors are so set that the load is shared eually bet"ee the machies. (etermie the operatig po"er factor of machie & A. 0.2 &. 0.2 %. 0.0$ (. 0.8 Problem 05 "o "o alterators are operatig i parallel supplyig a commo load of $250 3+ at 0.85 po"er factor laggig. Alterator A cotributes 00 3,A@ at 0.2 po"er factor laggig. Dolve for the po"er factor of alterator &. A. 0.81- laggig &. 0.8-$ laggig %. 0.80$ laggig (. 0.820 laggig Problem 0 "o L6coected alterators A ad & ruig i parallel supply the follo"ig loads at -.- 3,: ;oad 1: 800 3+ at uity pf ;oad 2: 00 3+ at 0.8 pf laggig ;oad -: $00 3+ at 0.0 pf laggig If alterator A is ad*usted to carry a armature curret of 150 A at 0.85 pf laggig "hat is the armature curret of alterator &# A. 208 A &. 180 A %. 200 A (. 10 A Problem 0

"o A% geerators ruig i parallel supply the follo"ig loads: $00 3+ at uity pfT 500 3+ at 0. pf laggigT 00 3+ at 0.8 pf laggig. e machie is loaded to 1000 3+ at a pf of 0.2 laggig! "hat is the operatig po"er factor of the secod machie. A. 0.$ laggig &. 0.21 laggig %. 0.-2 laggig (. 0.51 laggig Problem 08 "o alterators are coected i parallel. he total load is $250 3+ at 0.85 pf laggig. Alterator A operates "ith a load of 2125 3+ at 0.0 pf laggig. (etermie the 3,A load of alterator &. A. 22$5 &. 208 %. 2185 (. 2$10 Problem 0 Dipalay Cies! has t"o -6phase! 0 /B A% geerators operatig i parallel. he first uit has a capacity of 1000 3,A ad the secod uit has a capacity of 1500 3,A. he first is drive by a prime mover so ad*usted that the freuecy fall from 1 /B at o6load to 5. /B at full load. he secod has a differet speed6load characteristics! the freuecy fall from 1.$ /B at full load. +he these alterators are *oitly deliverig 2000 3+! "hat "hat is the load of each geerator# A. 88$ 3+! 111 3+ &. 8 3+! 11-- 3+ %. 15 3+! 1085 3+ (. 80$ 3+! 11 3+ Problem 10 "o "o alterators are drive by shut motors. he shut motors have speed6load droop characteristics of -J ad $J respectively. he alterators are i parallel ad each carryig 50 3+. here is o automatic speed6load cotrol. A additioal 50 3+ is s"itched o. +hat ar e the resultig loads of the alterators assumig that the speed6load cotrol of each ad*usted# A. 8.5 3+  1.$- 3+ &. 82.51 3+  .$ 3+ %. 5 3+  5 3+ (. 80.-1 3+  . 3+ Problem 11 here are t"o alterators! 100 3+! -6phase -6phase i parallel are drive by shut motors "hose speed6load characteristics are as follo"s: Alterator Alterator o.1! o6load speed 00 rpm ad the full6load speed 5-0 rpm! "hile alterator o. 2! o6load speed 50 rpm ad full6load speed 550 rpm. Eor "hat load "ill the alterator divide the load eually# A. 2.$5 3+ &. 8.8 3+ %. .5$ 3+ (. $. 3+ Problem 12 here are t"o alterators! 100 3+! -6phase i parallel are drive by shut motor "hose speed are drive by shut motor "hose speed6load characteristics are as follo"s: Alterator Alterator o.1! o6load speed 00 rpm ad the full6 load speed 5-0 rpm! "hile alterator o.2! o6load speed 50 r pm ad full6load speed 550 rpm. +hat is the greatest load that ca be delivered "ithout overloadig either alterator# A. 11.$- 3+ &. 18.8 3+ %. 1.5$ 3+ (. 1$. W"

TEST 1) – T0*NS0%E0 '13 # ''. Problem 1he maximum flux desity i the core of a -0002$06,! -0002$06,! 50 /B sigle6phase distributio trasformer is 1.25 esla. esla. If the voltage iduced per tur is 8 volts! determie the cross sectioal of the core i cm 2. A. 22 &. -5 %. 288 (. -2 Problem 1$ he et cross sectio of a trasformer is 20 suare iches. he maximum flux desity is 11 3ilogausses. he freuecy is 0 /B. If there are 1!000 turs i the primary! primary! determie the primary iduced emf. A. -80 , &. -$5 , %. -50 , (. -850 ,

Problem 15 %alculate the total magetic flux i a 06cycle trasformer i "hich the iduced emf per tur of the "idig is eual to 2 ,. A. .50 m+b &. .02 m+b %. .50 m+b (. .$2 m+b Problem 1 he lo" side of a certai "eldig trasformer has 2 turs ad delivers -00 A. if the high voltage side has -0 turs! ho" much curret flo"s through it# A. 10 A &. 12 A %. 20 A (. 15 A Problem 1 A sigle6phase sigle6phase trasformer is rated $!10 , primary ad 250 , secodary. It has 1!500 turs i the primary ad a voltage regulatio of 8J. +hat should be the umber of turs i the secodary "idig# A. 88 turs &. 2 turs %. 0 turs (. 8 turs Problem 18 he curret desities i the primary ad secodary "idigs of a step do" trasformer are 1250 ad 1500 A per per suare ich! respectively. respectively. he ratio of trasformatio is 10:1 ad the mea legth per tur of the primary is 15J greater tha that of the secodary. secodary. If the primary "idig resistace is 20 ohms! calculate the resistace of the secodary "idigs. A. 0.202 ohm &. 0.21 ohm %. 0.20 ohm (. 0.220 ohm Problem 1 A certai trasformer trasformer has a ratio of 8 is to 1. he primary "idig resistace is 0.0028- ohm per tur ad has 1155 more turs tha the secodary "idig. If the copper losses i the primary is 20J less tha i the secodary! secodary! fid resistace of the secodary "idig. A. 0.0- ohm &. 0.00 ohm %. 0.08 ohm (. 0.08$ ohm Problem 20 A 10 3,A! 3,A! 2$002$06, sigle6phase trasformer has the follo"ig resistaces ad reactaces. Eid the primary voltage reuired reuired to produce 2$0 , at the secodary termials termials at full load! "he the the po"er factor is 0.8 laggig. @ 1 < - ! K1 < 15  @2 < 0.0 ! K2 < 0.150  A. 2$50 , &. 252 , %. 2$ , (. 2-2 , Problem 21 A 1063,A 1063,A! 2!000$00 , sigle6phase distributio trasformer trasformer has a primary ad ay secodary "idig impedace of 5.0 = * 10 ad 0.2 = * 0.$5 ohms! respectively. respectively. %alculate the voltage across the secodary termials at full load! 80J po"er factor laggig! "he the primary is supplied "ith 2!000 volts# A. $00 , &. -$ , %. -85 , (. - , Problem 22 A sigle6phase trasformer rated 5 3,A! 8000 volts primary 2$0 volts secodary! is give the short circuit test. +ith the secodary termials short6circuited! $$0 volts >0 /B? is impressed o the primary! primary! "hich the dra"s rated curret ad 1125 "atts. Dolve for the percetage impedace of the trasformer. A. 5.2J &. $.-J %. 5.50J (. 5.1-J Problem 2A short circuit test "as performed upo a 1063,A! 2-002-0 volt trasformer "ith the follo"ig results: 7D% < 1- voltsT P D% < 12 +T ID% < $.-$ A. %alculate i primary terms the reactace of the trasformer. A. -1.5 ohms &. 2.88 ohms %. 10.1 ohms (. 12.$5 ohms Problem 2$

A sigle6phase! 25 3,A! 3,A! $8002$0 volt trasformer "as give a short circuit tests: +ith the secodary termials short6circuited! short6circuited! -.5 volts "as applied o the primary so that the ammeter reads rated curret ad the "attmeter reads 12.$ "atts! all o the primary side. Dolve for the percetage resistace of the trasformer. A. 1.5J &. -.-J %. -.02J (. 2.2-J Problem 25 he follo"ig data "ere obtaied "he a short circuit test "as performed upo a 100 3,A 3,A! 2$002$0 volts distributio trasformer: 7 D% < 2 voltsT I D% < $1. AT P D% < 1!180 +. All istrumets are o the high side durig the short circuit test. %alculate the euivalet resistace ad reactace of the trasformer. A. 0.82 ! 1.5  &. 0.82 ! 1.-  %. 0.52 ! 1.-  (. 0. 0.52 ! 1.5  Problem 2 A 1-.8 3,$80 ,! 10 C,A three6phase trasformer has 5J impedace. +hat is the impedace i ohms referred to the primary# A. 0.52 ohm &. 0.0- ohm %. 5.125 ohms (. .01 ohms Problem 2 A short circuit test "as performed o a 20 3,A! 2000200 , sigle6phase trasformer. +ith the secodary "idig short6circuited! full load curret flo"s i the primary "idig "ith 100 , applied o it. he "attmeter coected i the primary read -00 +. (etermie the termial voltage o full load uity pf. A. 18. , &. 1-.$ , %. 1. , (. 15.2 , Problem 28 he follo"ig data "ere obtaied "he a short circuit test "as performed upo a 100 3,A 3,A! 2$002$0 volts distributio trasformer: 7 D% < 2 voltsT I D% < $1. AT P D% < 1!180 +. All istrumets are o the high side durig the short circuit test. %alculate the percet regulatio at a po"er factor of 0.5 laggig. A. 2.5$J &. 2.8J %. 2.2J (. 2.1J Problem 2 A 20063,A trasformer "ith impedace of 5J ad the po"er factor of the load is 85J. he primary voltage is !000 volts "hile the copper loss is 5 3+. Eid the percetage regulatio. A. 5.0-J &. 5.21J %. $.$-J (. $.0-J Problem -0 A 150 3,A! three6phase trasformer supplies 100 3,A at 50J laggig po"er factor. he trasformers "idig impedace is 2J resistace ad 5J reactace.! based o 150 3,A. 3,A. (etermie the voltage regulatio of the trasformer at this load. Assume 100J voltage. A. -.55J &. 5.--J %. $.2$J (. $.05J Problem -1 A short circuit test "as performed upo a 1063,A! 2-002-0 volt trasformer "ith the follo"ig results: 7D% < 1- voltsT P D% < 12 +T I D% < $.-$ A. %alculate the percet regulatio at a laggig po"er factor of 0.0. A. 5.28J &. 5.55J %. 5.-J (. 5.2J Problem -2 A 100 3,A! 0000 ,! 0 /B sigle6phase trasformer operatig at rated coditio has the follo"ig losses: hysteresis loss < 520 +T eddy curret loss < -0 + ad copper loss < 2000 +. if this trasformer is to be used o a 50 /B system! determie its e" 3,A ratig. Assume the same total losses ad maximum core flux. A. 8-.8 &. $.%. 8.$ (. 1.8 Problem -+he 120 ,! 0 /B is applied to the primary "idig of a trasformer! the core loss is 2$0 +. +he 50 ,! 25 /B is applied the core loss is 50 +. (etermie the core loss "he 2-0 ,! 50 /B is applied to the primary "idig.

A.

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Problem -$ he eddy curret ad hysteresis losses of a $!$006,! 06/B 06/B sigle6phase distributio trasformer are 280 ad 50 "atts! respectively. respectively. (etermie the core loss "he the trasformer is coected across a $!006volt! 506/B source. A.  + &. 8- + %. $ + (.  + Problem -5 I a $00 ,! 50 cs trasformer! the total iro loss is 2500 +. +he the supplied p.d. is 220 , at 25 cs! the correspodig loss is 850 +. %alculate the eddy curret loss at ormal freuecy ad p.d. A. 2-- + &. 215 + %. 2010 + (. 221 + Problem - +he a 220 ,! 0 /B is impressed o a certai trasformer at o load! the total core loss is 200 +. +he the freuecy of the impressed voltage is chaged to 25 /B ad the magitude of the voltage is made such as to maitai the same maximum flux desity as before! the core loss falls to 5 +. %alculate the hysteresis losses at 0 /B. A. 1- + &. 1 + %. 11 + (. 1 + Problem - A 10063,A! $102-0 ,! 0 /B sigle6phase trasformer has percetage impedace of 1.5 = * $.0. its efficiecy "he operatig at rated coditios at 5J pf laggig is J. (etermie the efficiecy! efficiecy! if the trasformer is loaded at rated load ad voltage but at 50 /B. Assume the ratio of eddy curret losses to hysteresis losses at 0 /B to be 2.5. A. .1-J &. .J %. .1J (. .$2J Problem -8 he full load copper loss of a 5 3,A sigle6phase trasformer is thrice the iro losses. At half load uity pf! the efficiecy is J. J. (etermie the full load efficiecy at 0.8 pf laggig. laggig. A. 5.J &. 8.$5J %. $.8J (. .8J Problem - A 50 3,A! 3,A! 2-002-0 volt! 06cycle trasformer is tested i the laboratory so that its characteristics maybe determied. he stadard test reuires a ope circuit test ad short circuit test. pe circuit test l < .5 A 7 < 2-0 , P < 18 + Dhort circuit test l < 21. A 7 < 115 , P < 50 + (etermie the efficiecy of the trasformer at half load! uity pf. A. .55J &. 8.0J %. 8.$5J (. .80J Problem $0 he iro loss of a 2563,A! 2563,A! 0 /B sigle6phase distributio trasformer is 1J of the rated output at uity po"er factor ad the full full load copper loss is -0 -0 +. +. (etermie (etermie the efficiecy "he operatig operatig oe half of the rated output at 0J po"er factor laggig. A. 5.--J &. .82J %. $.2$J (. .2J Problem $1 Oive that full load copper losses are exactly t"ice the iro losses i a 5063,A trasformer! ad that the uarter load efficiecy is .5J. %alculate the full load efficiecy at uity po"er factor. A. .$8J &. 8.12J %. .$J (. .25J Problem $2 A 15 3,A! 2-002-06volt trasformer is uder test. +ith the lo" side supplied "ith rated voltage ad the high side ope circuited! iput po"er through a "attmeter reads 2$5 +. +ith the high side supplied "ith 5 ,

"hile the side short circuited! iput po"er at rated curret reads -50 +. (etermie the maximum efficiecy that ca be expected from this trasformer at uity pf. A. .2J &. .25J %. .8-J (. 8.21J Problem $A 1063,A 1063,A! 2-002-0 trasformer has a euivalet resistace referred to the secodary "idig of 0.05 ohm. he core loss of this trasformer is 5 +. At "hat 3,A load "ill this trasformer operate at maximum efficiecy# A. 8.1 3,A &. .08 3,A %. .-2 3,A (. 8.0 3,A Problem $$ he core loss of a 5 3,A sigle6phase trasformer "ith ormal voltage applied to the primary is 5 "atts. he maximum efficiecy of the trasformer of 0.80 po"er factor# A. 5.1J &. -.-8J %. 8.52J (. 1.-J Problem $5 A 10063,A trasformer has $J efficiecy at full load ad 50J full load "ith uity pf i both cases. (etermie the maximum efficiecy of the trasformer at uity pf load. A. .2J &. 5.8J %. -.$$J (. $.-2J Problem $ A 10063,A 10063,A trasformer has a maximum efficiecy of 8J at 10J uderload ad at uity pf. 7valuate the efficiecy at uarter load! 0.80 pf laggig. A. .-J &. 5.-J %. $.J (. .J Problem $ A 1 C,A! 11 3,! 0 /B trasformer has a euivalet series impedace of 1.0 = * 5 ohms referred to the lo" voltage side ad a o6load loss of 5500 "atts at rated termial termial voltage. +hat is the maximum efficiecy at 0.8 pf laggig# A. 8.-$J &. .$2J %. 8.02J (. .-8J Problem $8 he all day efficiecy of a 10 3,A sigle6phase trasformer is $.J "he loaded as follo"s: full load uity pf for $ hrs ad a o load the rest of the day. If the full load copper loss at uity pf is $00 "atts! calculate the core loss. A. 5 3+ &. 0 3+ %. 2 3+ (. 8 3+ Problem $ A 25 3,A distributio trasformer trasformer operates at 20J overload! 0.- pf laggig for 1 hourT at full load! 0. laggig pf for - hours! 50J load at uity pf for 8 hoursT ad 10J load at 0.8 leadig pf for the rest of the day. If core loss is 150 "atts ad full load copper loss is 500 +! "hat is the all day efficiecy of the trasformer# A. $.2-J &. 2.8J %. .1$J (. .8J Problem 50 Eid the all6day efficiecy of a trasformer havig a maximum efficiecy of 8 percet at 15 3,A at uity pf ad loaded as follo"s: follo"s: 12 hours 66662 3+ at 0.5 pf laggigT  hours hours 6666 12 3+ at 0.8 pf laggigT laggigT  hours 6666 at o6load. A. 5.-J &. 1.J %. $.J (. -.-J Problem 51 A -0 3,A! 2$002$0 ,! 0 /B trasformer has a full load po"er of uity over the period of 2$ hours. he maximum efficiecy is 5J ad it occurs at f ull load. %alculate the all6day efficiecy if loaded as follo"s:  hours at full load ad  hours at uarter load. A. 1.5-J &. 2.$5J %. 0.-J (. -.8J

Problem 52 A 1000 3,A 3,A! 00$00 volts! 0 cycle sigle phase core type trasformer has the follo"ig average daily load: full load at 0.8 pf for 8 hrsT half load at 0.0 pf for 10 hrsT o6load for  hrs. %alculate the ratio of full load copper loss to the iro loss for the trasformer to be most ecoomical for the above loadig. A. 2.05 &. -.0 %. 2.1 (. 2.28 Problem 5he full load copper loss ad iro loss of a 5 3,A trasformer are eual to 1 3+. (urig a give day! the trasformer is loaded as follo"s: full load at uity pf for 8 hours! oe6half load at uity for 8 hours ad o6load the rest of the day. %alculate the all day efficiecy. A. .-J &. .2-J %. .$J (. 8.-2J Problem 5$ A . 3,00 ,! ,! multi6tap! sigle6phase distributio distributio trasformer is coected to the far ed of a distributio lie for "hich the rear ed voltage is maitaied at . 3,. he euivalet impedace of the trasformer referred to lo" voltage side is 0.05 = *0.25  per phase. he impedace of each lie is 1 = *2 . It is reuired to maitai a termial voltage of 00 , "he the lie curret of -00 A at 80J pf laggig is delivered to the load. +hat approximate percetage tappig is reuired i the primary side# eglect chages i impedace due to chages i turs ratio. A. $J &. 2J %. 0J (. 88J Problem 55 A 0 /B alterator geerates a sigle6phase voltage of 120 , ad has a reactace of 0.25 ohm. he alterator feeds a lie! "hich is essetially resistive "ith a impedace of 0.1 ohm. he lie is termiated "ith a 2:1 step up trasformer. +hat should should be the load impedace to be coected to the secodary side to have maximum po"er factor trasfer to it# it# A. 0.5 = * 1.2 &. 0.$ ' * 1.0 %. 0.5 ' * 1.2 (. 0.$ = * 1.0 Problem 5 A trasformer trasformer cosist of a primary "idig "ith 500 turs ad t"o secodary "idigs of 125 turs ad - turs. he 1256tur secodary "idig has 0 ohms coected to its termials ad the -6tur secodary "idig has - ohms coected to its termials. If the primary is coected to a 1206,! 1206,! 0 /B source! determie the curret i the primary "idigs. A. 0.--2 A &. 0.$12 A %. 0.2-5 A (. 0.25 A Problem 5 A -6phase! 0 /B trasformer has primary! primary! secodary ad tertiary "idigs. he primary "idig is delta coected ad rated !00 ,. he secodary "idig is delta coected ad rated 1000 ,. he tertiary "idig is "ye coected ad rated $$0 ,. If the secodary "idig has a -6phase balaced load of 100 3,A at 80J pf laggig ad the tertiary "idig has a balaced -6phase load of 50 3+ at uity pf! "hat is the curret i the primary "idig ad its operatig pf# A. 12.52 A! 0.08 &. .2- A! 0.02 %. 12.52 A! 0.02 (. .2- A! 0.08 Problem 58 A 2063,A 2063,A! 5006, load is to be supplied by a ideal step6up autotrasformer from a $006, source. Eid the curret i the commo "idig. A. -0 A &. $0 A %. 50 A (. oe of these Problem 5: he t"o "idigs of a 1063,A! 1063,A! 2-002-0 volt! distributio trasformer is coected i series to form a autotrasformer givig a small reductio to potetial differece from a 2-006volt lie. (etermie its 3,A ratig as a ideal autotrasformer! operatig i the above coditios. A. 10 3,A &. 100 3,A %. 110 3,A (. oe of these

Problem 0: A 10 3,A! $$0110 volt! sigle6phase trasformer has a efficiecy of J at a rated load of uity po"er factor. he he full load copper loss is 250 "atts. If this trasformer is to be coverted as a 550$$06volt autotrasformer! autotrasformer! determie its efficiecy "he supplyig full load at 85 percet po"er factor. A. .25J &. .0-J %. .$5J (. 8.1J Problem 1: +hat should be the 3,A 3,A ratig of each t"o trasformers i a ope6 delta ba3 "he the three6 phase balaced load is 20.$ 20.$ 3,A 3,A# A. 155.0 3,A &. 1-.$0 3,A %. 150.25 3,A (. 1. 3,A Problem 2: A , or ope6 delta coected trasformer ba3! composed of t"o idetical uits serves a balaced three6 phase load of 1 3,A 3,A at 2-0 volts. Dolve for for the miimum siBe i 3,A 3,A of the trasformers eeded. eeded. A. 8 3,A &. .2$ 3,A %. 1 3,A (. 8.25 3,A Problem -: A , >ope6 delta? coected trasformer ba3 serves a balaced -6 phase load of $0 3,A at 2-0 , ad 0.8 pf laggig. Dolve for the miimum 3,A of each trasformer eeded to accommodate the load "ithout overloadig. A. 25 &. 2$ %. 2(. 20 Problem $: "o "o sigle6 phase distributio trasformers are ba3 i , >ope delta?. If each trasformer is rated -.5 3,A 3,A! solve for the maximum three6 phase load that the ba3 ca carry "ithout overloadig! cosiderig that the load has a po"er factor of 0.85 laggig. A. 52.5 3+ &. $.5 3+ %. 5-. 3+ (. 55.2 3+ Problem 5: A ope6 delta ba3! cosistig of t"o sigle6 phase trasformers is operatig "ith a balaced three6 phase load of 50 3,A! 3,A! $$0 , at 0.8 pf laggig ad a sigle6 phase load of 10 3+ resistive coected across c6a leg. (etermie the miimum ratigs of the t"o trasformers. Assume a phase seuece of a6b6c. A. -8.80 3,A! -$.0 3,A &. -$.-2 3,A! --.$0 3,A %. -5.$8 3,A! -.-$ 3,A (. -0.8- 3,A! -.52 3,A Problem : "o "o >2? sigle6 phase distributio trasformers! coected i ope6 delta "ill supply po"er to a 200 hp! -6 phase iductio motor! operatig operatig at 0.0 po"er factor ad 0.0 0.0 efficiecy. efficiecy. Dolve for the miimum siBe i 3,A 3,A of each trasformer eeded to supply the po"er "ithout beig overloaded. A. 1-.- 3,A &. 12$.$5 3,A %. 1-0.-8 3,A (. 12.$8 3,A Problem : he 9tility %ompay supplies t"o sigle6 phase trasformers ba3ed i ope delta. If each trasformer is rated 563,A capacity! "hat is the maximum -6 phase load i 3ilo"atts that the ba3 ca carry "ithout sufferig overload cosiderig that the load has a laggig po"er factor of 0.80. A. 10$ 3+ &. 112 3+ %. 1-0 3+ (. 11$ 3+ Problem 8: "o "o idetical 16 phase trasformers coected i ope delta "ill serve a three6 phase motor load of 250 hp! 0.0 pf laggig! 0.0 efficiecy. efficiecy. Dolve for the miimum siBe i 3,A 3,A of each trasformer reuired to serve the load "ithout overloadig.

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(.

185.1-

Problem : A ope6 delta coected trasformer ba3 serves a balaced -6 phase load of $0 3,A at 2-0 , ad 0.8 pf laggig. Dolve for the real po"er ad pf at "hich each trasformer trasformer is operatig. operatig. A. 11.5$ .5$ 3+ 3+ aatt 0. 0.5 pf pf! 22-.0 3+ 3+ aatt u uity pf &. 12.22 .22 3+ 3+ at at 0. 0.5 pf pf! 22 22.$2 3+ 3+ at at u uity pf %. 11.5$ .5$ 3+ at 0.8 0.8  pf! pf! 2-.0 -.0 3+ at 0.0 .0 pf (. 12.2 12.22 2 3+ 3+ at 0.8 0.8  pf! pf! 22.$2 2.$2 3+ at 0. 0.0 0 pf Problem 0: A ope6 delta! three6 phase trasformer cosistig of t"o sigle6 phase trasformers is operatig "ith a balaced three6 phase load load of 50 3,A 3,A! $$06, at a laggig pf of of 0.8. (etermie the operatig pf pf of each trasformer. trasformer. A. 0.-- laggig! 0.- laggig &. 0.-8- laggig! 0.88 laggig %. 0.-8- laggig! 0.- laggig (. 0.-- laggig! 0.88 laggig Problem 1: "o "o sigle6 phase trasformers are coected i , >ope delta? ad servig a delta coected impedace load! each impedace is eual to 1Z-.8 0 ohms. If the trasformer voltages impressed o the impedaces are: ,ab < 2$0Z00! ,ab < 2$0Z2$00! ,ab < 2$0Z1200 volts Dolve for the 3,A delivered by each trasformer. A. .28! 5.20 &. 5.8! .1 %. .8$! 5.$ (. .2$! .2$ Problem 2: "o "o sigle6 phase trasformers i a ope delta ba3 served a three6 phase balaced load of 20.$ 3,A. 3,A. If a third trasformer is added for delta operatio! "hat percet icrease i load does this represet# A. -.2J &. 8$.J %. 80.5J (. .$J Problem -: hree sigle6 phase trasformers each rated 5 3,A are ba3ed i delta ad supplyig a three6 phase load dra"ig 10 3,A at 0.8 laggig po"er factor. If oe trasformer is removed for repairs! solve for the amout of overloadig of the remaiig uits. A. -0 3,A &. 10 3,A %. 20 3,A (. 25 3,A Problem $: A 006 volt! 06 /B! 26 phase et"or3 is coected to a three6 phase! $6 "ire system by a Dcott6 coected trasformer combiatio. If there are 1500 turs o the t"o6 phase side! calculate the umber of turs o the three6 phase side reuired for for the mai ad teaser trasformers. trasformers. ,o ,oltage per phase o the -6phase -6phase side is 2$0 ,. ,. A. $ turs! 8- turs &. - turs! 8$ turs %. 5 turs! 82 turs (. 2 turs! 85 turs Problem 5: A balaced three6 phase trasformer ba3 has delta6 coected primary ad "ye6 coected secodary "idigs. he primary to secodary "idig voltage ratio is 2$002. Dolve for the primary lie curret if the lo"6side >secodary? load dra"s a lie curret of 1000 A. A. 200 A &. 115 A %. 150 A (. 15 A Problem : A three6 phase 150 3,A! $10 ,! 0.8 laggig pf load is to be supplied by three sigle6 phase trasformers havig a turs ratio of 2. %alculate the curret that flo"s i the primary "idigs if the trasformers are coected delta6delta. A. 10.$1 A &. .01 A %. 20.82 A (. 12.02 A Problem :

+hat should be the turs ratio of a three phase trasformer to trasform 10!000 3,A from 2-0 3, to $10 , if the trasformer is to be coected [ 6 L # A. $ &. 8 %.  (.  Problem 8: A ba3 of three 50:1 sigle6 phase trasformers coected L6[ L6[ is used to step do" the /, trasmissio lie rated 1-8 3,. (etermie the magitude of the step do" voltage. A. 12 3, &. .0 3, %. 1-8.55 3, (. 15.-5 3, Problem : he potetial trasformer of a lie has a ratio 1-2!000. hese are coected "ye6 delta. he voltmeter reads $ volts. +hat is the voltage of the lie# A. 221. 3, &. 112.5 3, %. 1-5.2 3, (. 1-2 3, Problem 80: A po"er trasformer rated 50!000 3,A! 3,A! -$.5 3,1-.8 3, is coected "ye6 grouded primary ad delta o the secodary. (etermie (etermie the full load phase curret o the secodary side. A. 202 A &. 125 A %. 1$$ A (. 1208 A Problem 81: A 1-.8 3,! 10 C,A three6 phase trasformer is coected "ye6 grouded delta. +hat is the 3, ratig of the secodary "idig# A. . &. 15 %. 2-. (. 1-.8 Problem 82: he Ps of a 220 3, trasmissio lie have ratio of 1-2. 3,.-5 3, ad are coected "ye6"ye. A voltmeter coected lie to groud reads  volts. +hat is the trasmissio lie voltage# A. 228.- 3, &. 220.10 3, %. 22-.15 3, (. 225.25 3, Problem 8-: A po"er trasformer rated 50!000 3,A! 3,A! -$.5 3,1-.8 3, is coected "ye6"ye. +hat are the lie currets at full load# A. 525.2 A! 125.$ A &. 8-. A! 202 A %. $8-.1 A! 120. A (. 1$$.- A! -2-.2 A Problem 8$: "o trasformers are coected i parallel to supply a commo load of 15 3,A. rasformer A is rated 100 3,A "ith euivalet impedace of . ohms "hile trasformer & is rated 5 3,A "ith euivalet impedace of 12.5 ohms. Eid the 3,A load of each trasformer. A. 8! &. ! %. 80!5 (. 5!100 Problem 85: "o distributio trasformers are coected i parallel supplyig a load of 125 3,A. If trasformer A ia rated 50 3,A! "ith 1$ ohms euivalet reactace ad trasformer & is rated 5 3,A "ith 11 ohms euivalet reactace. Dolve for the loadig i 3,A of each trasformer. A. 55 3,A! 0 3,A &. 50 3,A! 5 3,A %. 58 3,A!  3,A (. 52 3,A! - 3,A Problem 8: A 1256 3,A distributio trasformer "ith $J impedace is coected i parallel "ith aother trasformer rated 563,A 563,A ad "ith -J impedace. &oth have the same voltage ratio. eglect resistace of each trasformer. trasformer. he total load is 1$0 3,A at 85 percet po"er factor. /o" much load does each carry# A. 80.5 3,A 5.5 3,A &. .8 3,A 2.22 3,A %. 80 3,A 0 3,A (. 8.50 3,A 52.5 3,A

Problem 8: "o parallel trasformers have 8J reactace each. e is rated 25 C,A ad the other at 15 C,A. Eor a give -0 C,A load "ith a laggig pf of 0.8! determie the load carried by each trasformer# A. 18.50 C,A! 11.50 C,A &. 18.5 C,A! 11.25 C,A %. 15.25 C,A! 1$.5 C,A (. 1.$5 C,A! 1-.55 C,A Problem 88 A 50063,A sigle6phase trasformer GAH "ith percetage impedace of 0.010 =*0.05 is coected i parallel "ith a 25063, 25063,A trasformer G&H "ith percetage percetage impedace of 0.015 = *0.0$. *0.0$. If the trasformers are servig a sigle6phase load rated 80063,A 80063,A! 80J laggig pf! determie the operatig pf of each trasformer. Assume Assume the trasformers have the same tur ratio. A. A < 0.2! & < 0.80&. A < 0.2! & < 0.85 %. A < 0.2! & < 0.85 (. A < 0.2! & < 0.80Problem 8 he C7@A;% has t"o sigle6phase trasformers "ith eual turs ratio ad ratigs ad are operated i parallel to supply a load of 280 3+ at 0.80 laggig pf. rasformer rasformer A has a resistace of 2J ad a reactace of 8J. rasformer rasformer & has a resistace of 1J ad a reactace of J. (etermie the po"er delivered by the trasformer A ad & to the load. A. A < 12- 3+! & < 15 3+ &. A < 125 3+! & < 155 3+ %. A < 122 3+! & < 158 3+ (. A < 12 3+! & < 15- 3+ Problem 0 "o "o sigle6phase trasformer "ith eual ratigs ad turs ratio are operated i parallel to supply a load of 180 3+ at a laggig pf of 0.0. rasformer rasformer A has a resistace of 1 J ad a reactace of J. rasformer & has a resistace of 2J ad a reactace of 5J. %alculate the po"er delivered by each trasformer. A. 8- 3+!  3+ &.  3+! 101 3+ %. 88 3+! 2 3+ (. 80 3+! 100 3+ Problem 1 "o "o sigle6phase trasformers A ad & are coected i parallel to serve a 006A sigle6phase sigle6phase load at 0. 80 pf laggig. he euivalet euivalet impedace impedace of the trasformers are 2 = *- ad 2.2.- = * 5 ohms! respectively. respectively. (etermie the ratio of the 3+ output of trasformer A to the 3+ output of trasformer &. Assume the trasformer have eual ratigs ad turs ratio. A. 1.5 &. 1.88 %. 1.8 (. 1.$

TEST 1- – 3# P*SE (ND4CT(N %T0S Problem 2: A sigle phase iductio iductio motor is rated 5 hp! 5J po"er factor ad 220 volts. +hat is its full load curret# A. 22.5 A &. 20.5 A %. 1. A (. 18.5 A Problem -: A 5 hp! -6 phase motor dra"s curret "he coected to 2-0 ,. ,. (etermie (etermie the curret dra" by the motor if the po"er factor is 0.0 ad efficiecy of 8-J. A. 12.5- A &. 15.1- A %. 1-.25 A (. 1$.05 A Problem $: A -6 phase motor is rated 50 hp at $0 volts lie to lie! 0.85 pf laggig ad 0.5 efficiecy. efficiecy. Dolve for the rated curret. A. 8.$ A &. 02.2- A %. $05.$- A (. 1.-2 A Problem 5:

A $0 hp! -6 phase! $6 pole! 2-06 volt! 0 /B iductio motor operatig at 0.0 efficiecy! 0.85 po"er factor ad -J slip! drives a "ater pump. (etermie the curret at "hich the motor is operatig. A. $.5-1 A &. .2-$A %. .15 A (. -.-51 A Problem : %alculate the speed of a 06 cycle! 1$6 pole motor if the slip is 5J. A. 51$.2 rpm &. $88.5 rpm %. $5.-2 rpm

(.

$2.- rpm

Problem : he rotor speed of a 6 pole! 506 cycle iductio motor is 0 rpm. %alculate the percet slip. A. 5J &. 2J %. $J (. -J Problem 8: +hat is the speed of a iductio motor of six poles if the percet slip is 2.5J# A. 1$2 rpm &. 110 rpm %. 8 rpm (.

155 rpm

Problem : A suirrel cage iductio motor motor "ith ameplate data of: 150 hp! -6 phase! $06 ,! 0 /B! 6 pole! 0.85 pf! "as sub*ected to certai performace tests. he test results readigs "ere as follo"s: Eull load curret < 202 A Eull load torue < .8 lb6ft. Dolve the percetage slip. A. -.28J &. $.15J %. -.05J (. 2.25J Problem 00: A six6 pole! three6 phase suirrel6 cage iductio motor is coected to a 06 cps supply. supply. At full load! the rotors iduced emf ma3es 2 complete cycles i 1 miute. Eid the rotor speed. A 11 rpm &. 1200 rpm %. 105$ rpm (. 112$ rpm Problem 01: A 506 hp! three6 phase motor! rated 220 ,! 5J po"er factor has  poles. he slip at full load is -J. +hat is the freuecy of the rotor curret at full load# A. 1.8 /B &. 0 /B %. 5.$ /B (. 58.2 /B Problem 02: A 1506 hp! -6 phase! 6 pole! $0 ,! 0 /B iductio motor dra"s 15 A lie lie curret at full load speed of 110 rpm ad 0.85 pf. Dolve for the percet slip at full load. A. 2.5J &. -.0J %. 2.8J (. 2.0J Problem 0-: he deep "ell pump motor is 50 hp! -6 phase! $6 poles! 2-0 volts! 0 /B iductio motor! operatig at 0.0 efficiecy! efficiecy! 0.85 po"er factor ad -J slip. (etermie the curret at "hich the motor is operatig. A. 112. A &. 122.$ A %. 120.5 A (. 12$. A Problem 0$: he rotor of a 220 ,! ,! 0 /B! $6 pole iductio motor has a po"er iput of 80 3+ is observed that the rotor emf ma3e 0 cycles c ycles per miute. %alculate the rotor copper loss. A. 2 3+ &. 5 3+ %. - 3+ (. $ 3+ Problem 05: A iductio motor carries a rotor curret per phase of 10 A "he loaded util the slip is 5J. If the rotor effective resistace is 0.1 ohm per phase! determie the total po"er trasferred across the air gap per phase. A. 210 + &. 2$0 + %. 10 + (. 200 + Problem 0:

A 10 hp >ameplate?! 6 pole! 0 /B! three6 phase iductio motor delivers . hp "ith a iput 200 "atts. he core loss is $50 +. he stator copper loss is 50 "atts ad the rotatioal losses is 150 "atts. +hat is the motor speed# A. 111 rpm &. 1200 rpm %. 1210 rpm (. 1125 rpm Problem 0: A three6 phase suirrel cage iductio motor is dra"ig $5 3+ from a $$0 ,! 06 /B! -6 phase lie. If stator losses amout to 00 +! determie total mechaical po"er developed. Assume a slip of -J. A. $0.-2 3+ &. $$.$ 3+ %. $8.22 3+ (. $2.8 3+ Problem 08: A 106 hp! $6 pole! 25 /B! -6 phase iductio motor is ta3ig 100 "atts from the lie. %ore loss 20 "attsT stator copper loss is 58 "attsT rotor copper loss is $$5 "atts ad the frictio ad "idage loss 121 "atts. (etermie the output torue i e"to6meters. A. 108.$ &. 112.5 %. 101.2 (. 10-.Problem 0: A $06 hp! 2-06 ,! ,! 86 pole! 25 /B! three6 phase iductio motor is ruig at -55 rpm at rated load. he torue lost by frictio is 2$ 6m. If total stator losses amout to 1000 +! evaluate evaluate the efficiecy of the motor at this load. A 1.$2J &. 88.51J %. 0.-1J (. 8.1J Problem 10: he rotor of 2206 ,! 0 0 /B! $6 pole iductio motor has a po"er iput of 80 3+ is observed that the rotor emf ma3es 0 cycles per miute. %alculate the gross mechaical po"er developed. A. 2 3+ &. 8 3+ %. 5 3+ (.  3+ Problem 11: A 6 pole! 50 /B! -6 phase iductio motor rus at 0 rpm "hile deliverig a shaft torue of 120 6m. if frictio ad "idage loss amout to 180 +! determie determie rotor copper loss. A. 521 + &. 518 + %. 510 + (. 52 + Problem 12: (etermie the speed i rpm of the motor at full load "ith the follo"ig circuits elemets per phase. @otor lea3age reactace < 0.5 @otor resistace < 0.10 Dtator lea3age reactace < 1.12 Dtator resistace < 0.25 %oectio %oectio  < L6L his motor is 500 hp! -6 phase! 2200 volt! 06 cycle! 6 pole iductio motor "ith additioal core loss of 2000 "atts ad the frictio ad "idage losses total is 10!000 "atts. A. 1200 rpm &. 11 rpm %. 110 rpm (. 1188 rpm Problem 1-: A 506 hp! $$0 ,! ,! -6 phase! 0 /B! 6 pole suirrel cage iductio motor is operatig at full load ad 0.8 pf. he full load efficiecy is 85J ad the percetage slip is 5J. (etermie full load torue. A. 21$.55 6m &. 20.2 6m %. -12.$ 6m (. -2-.2$ 6m Problem 1$ A -06hp! $6pole! -6phase! 2-0 volts! 0 /B suirrel cage iductio motor operatig at 0J efficiecy! 85J po"er factor ad 2.5J slip slip drives a "ater pump for a reservoir. reservoir. Eid Eid the curret ad speed at "hich "hich the motor is operatig. A. -.$ -.$ A! 155 rpm &. 2.- A! A! 155 rpm %. -.$ A! A! 1800 rpm rpm (. 2.2.-  A! A! 1800 rpm Problem 15

A 506hp! 506hp! $6pole! -6phase! 2-0 volts! volts! 0 /B suirrel cage iductio iductio motor has a efficiecy of 0J! po"er factor 85J ad a slip of 2J at full load. Eid the torue i lb6ft at full load. A. 152.$5 &. 1$8.8 %. 121.-$ (. 150.5 Problem 1 he losses i a -6phase! 256/B! $6pole iductio motor at full load are as follo"s: core ' -JT Erictio ad "idage ' 2JT Dtator copper ' $JT ad @otor copper ' 2J. At 0J of full load of the machie! determie its efficiecy. A. 0.52J &. 88.15J %. 8.2$J (. 8.81J Problem 1 A 100 hp! $$0 volt! three6phase! 0 /B iductio motor operates "ith 2 percet efficiecy! efficiecy! 0.85 po"er factor! at rated load. +hat is the curret of the motor# A. 125 &. 115 %. 8 (. 55 Problem 18 A six6pole! three6phase! $$06volt! 06/B iductio motor develops 106hp at 1150 rpm! the po"er factor beig 80J laggig. laggig. Dtator ad core losses amout amout to $00 + ad ad -50 + respectively respectively.. Erictioal losses amout amout to 0.56hp. %alculate motor lie curret. A. 12.2$ A &. 11.$2 A %. 1-.$1 A (. 1$.$ A Problem 1 A three6 phase! .5 hp! 2206 ,! 6 pole! 06 /B! L6 coected iductio motors "as tested ad give the follo"ig results: o6 load test: l < .5 A P < 10 +  r < < 110 rpm ;oad test: l < 18.$ A P < 00 +  r < 11$ rpm @esistace test: test: 0.5$ ohm bet"ee ay t"o stator stator termials (etermie the efficiecy of the motor durig the load test. A. 8-.$5J &. 8$.J %. 85.28J (. 8.55J Problem 20: A three6 phase "oud iductio motor! $$0 ,! 0 cycle! $6 pole is coected to a pump that delivers 1000 cfm of "ater agaist a effective head of 8 ft. 9der this load the motor! the motor dra"s 15 3+ at a po"er factor of 0.0. +he operated "ithout load! the motor dra"s 800 "atts. he stator resistace per phase is 0.21 ohm. he turs ratio bet"ee stator ad rotor is $:1. (etermie the efficiecy of the pump. A. 8-.2J. &. 8$.5J %. 8-.1J (. 8$.2$J Problem 21: A three6 phase! 6 pole! 06 /B! L6 coected coected suirrel cage iductio motor operates "ith a slip of 2J ad dra"s 10 A ad -!500 + at full load. +he ruig light! the motor dra"s $.2 A ad -00 +. +he the rotor of this motor is bloc3ed! the motor dra"s $50 +! 15 A at 50 ,. %alculate the torue exerted "he drivig its full load. A. 22.82 6m &. 2$.5 6m %. 20.- 6m (. 2.$ 6m Problem 22: he follo"ig data "ere obtaied from a 56 hp! 1106,! 0 /)! 86 pole! -6 phase! L6 coected iductio motor: o6 load test: l < 10 A+1 < 25 + +2 < 6 $25 + ;oad test: l < 28 A+ 1 < -1$0 + +2 < 1580 r < 80 rpm (% resi resist sta ace ce test test:: 0.1 0.1 ohm ohm bet" bet"ee ee  ay ay t"o t"o stat stator or term termi ial alss %alculate the torue i 6m exerted by the motor durig the load test. Assume the A% resistace is 1.25 times the (% resistace. A. $$.$ &. 52.$2 %. $.10 (. $-.8$ Problem 2-:

A resistor assembly assembly for a "oud rotor cosist of 20 elemets of cast iro grid elemets. he ratig of each is 0.10 ohm. +he *oied i series ad tested across 220 volts! the measured curret "as 100 A. /o" much po"er "as lost i the *oits# A. 1. 3+ &. 1.8 3+ %. 2.0 3+ (. 2.1 3+ Problem 2$: A "oud rotor motor! motor! .5 hp! 2-0 volts! -6 phase ta3es a lie curret of 18.$ amperes! "he operatig at rated output at a efficiecy of 88J. %alculate the idicatios o "attmeters "he these are iserted to measure po"er by the t"o6"attmeter t"o6"attmeter method. A. $.0-5 3+! 2.-2- 3+ &. $.2-2 3+! 2.12 3+ %. -.02 3+! 2.$5 3+ (. -.855 3+! 2.50- 3+ Problem 25 "o6"attmeter method is used to test a 25 /P! 2-0 volt! 1800 rpm! 06cycle! -6phase iductio motor. +he the lie voltages are 2-0 volts! oe "attmeter reads = 1-!$00 "atts ad the other = !$00 "atts. (etermie motor lie curret. A. 55. A &. 58.$ A %. 52.- A (. 50.8 A Problem 2 A iductio motor of -0 hp! 220 volts -6phase dra"s $50 percet of the rated curret "ith the rated voltage ad delivers durig the startig period 1-0 percet of the ormal torue. he full load efficiecy ad full load pf of this motor is 80J ad 0J respectively. respectively. If a autotrasformer is used as a startig uit ad startig torue of the load is oly 50 percet of the rated torue of the motor! fid the startig voltage. A. 1-.$$ , &. 1-0.5 , %. 1-2.- , (. 1-$.20 Problem 2 A 106/P! 550 ,! 0 /B! -6phase iductio motor has a startig torue of 10J of full load torue ad a startig curret of $25J of full load curret. If the motor is used o a $$0 ,! ,! 0 /B system! "hat "ill be the startig torue expressed i percet of full load value# A. 102.$J &. 101.8J %. 10$.2J (. 10-.5J Problem 28 A 106/P! 550 ,! 0 /B! -6phase iductio motor has a startig torue of 10J of full load torue ad a startig curret of $25J of full load curret. +hat "ill the voltage i order that the startig curret be limited to the full load value# A. 1-2.$ , &. 12.$1 , %. 125.52 , (. 1-.5 , Problem 2 A 256hp! 2-0 , three phase motor "ith 85J po"er factor has a startig curret of 5.5 times rated curret. o reduce the startig curret! a "ye6delta starter is istalled. +hat shall be the e" startig curret# A. 15 A &. -02 A %. 1-5 A (. 15 A Problem -0 A suirrel cage motor is started at 50J of its rated voltage. +hat is the startig torue relative to its rated voltage startig torue# A. 50J &. 100J %. 25J (. 5J Problem -1 A delta6coected delta6coected load iductio motor dra"s 100 A from each lie of a three6phase source ad develops $0 lb6ft torue at stadstill. Eid the torue "he coected to the same service if the motor is re6coected i "ye. A. 1-.- lb6ft &. 120 lb6ft %. $0 lb6ft (. oe of these Problem -2

A iductio motor of -06/P! 220 ,! -6phase dra"s $50 percet of the rated curret "ith the rated voltage ad delivers durig the startig period of 1-0J of the ormal torue. he full load efficiecy ad full load pf of this motor is 80 percet ad 0 percet respectively. respectively. If a autotrasformer is used as a startig uit ad the startig torue of the load is oly 50 percet of the rated torue of the motor! fid the startig curret o the lie. A. -01.$ A &. 22. A %. 288. A (. 22.- A Problem -A .5 hp! 500 , three6phase motor iductio motor is developig its full load output at 0 rpm. he startig torue is eual to the full load torue. +hat "ill be the startig torue if the supply voltage falls to $00 volts# A. -2.-$ 6m &. -5.2 6m %. 28.2 6m (. -8.82 6m Problem -$ he deep "ell pump motor is 50 hp! -6phase! $6poles! 2-0 volts! 0 /B iductio motor! operatig operatig at 0.0 efficiecy! efficiecy! 0.85 po"er factor ad -J slip. (etermie the discharge of the pump i cubic meters per miute! assumig that the total head agaist "hich the pump is "or3ig is 50 meters ad the desity of "ater is 1000 3g per cubic meter. A. $.10&. $.22$ %. $.5(. $.205 Problem -5 A three6phase! 2-0 ,! ,! 6pole! 0 /B iductio motor is loaded by meas of a poy bra3e. he legth ad dead "eight of "hich are 2.5 ft ad 2 lbs respectively. respectively. +ith 2-06, impressed o this motor! the total f orce delivered by the poy bra3e is -0 -0 lbs. the po"er iput to to the meter is measured by t"o t"o "attmeters usig the the t"o6"attmeter method. hese "attmeters register $ ad 8 3+ respectively. respectively. If the motor has a slip of 55 uder this coditio! determie its efficiecy. A. -.2J &. 2.$$J %. 0.82J (. $.$-J Problem - A 5 3,A! -6phase trasformer supplies the follo"ig iitial loads: %apacitor ba3 dra"ig 25 3,A@ ad a iductive load dra"ig - 3+! 0.15 pf. (etermie the maximum hp of iductio motor at 0.8 pf ad 0J efficiecy that ca be added to the load "ithout overloadig the trasformer. A. $0.2 hp &. -8.$ hp %. -5.5 hp (. -.5 hp Problem - he feeder of a idustrial plat is has a load of $00 3,A at a po"er factor of 55 laggig. A iductio motor load dra"ig 100 3+ is added to the feeder. (etermie the operatig po"er factor this motor such that the total feeder load "ill have a laggig pf of 0J. A. 0.$2 &. 0.8$ %. 0.02 (. 0.818

TEST 2. – S5NC0N4S %T0S Problem -8: A three6phase L6coected sychroous motor "ith a lie6to6lie voltage of $$0 , ad a sychroous speed of 00 rpm operates "ith a po"er of  3+ ad a laggig po"er factor of 0.8. he sychroous reactace per phase is 10 ohms. (etermie (etermie the torue agle i electrical electrical degrees. A. -.--° -.--° &. --.51° --.51° %. -5.51° -5.51° (. -8.$° -8.$° Problem -: A 106hp! 2-06,! three6phase L6coected sychroous motor has a reactace of - ohms per phase ad a egligible resistace. o o "hat voltage must the motor is excited i order to deliver full load at uity po"er factor ad 0J efficiecy# A. 22 , &. 25$ , %. 2$8 , (. 2$2 ,

Problem $0: A 1000 3,A! 11 3,! -6phase L6coected sychroous motor has a resistace of - ohms ad a reactace of 25 ohms per phase. (etermie the iduced emf "he fully loaded at 0.8 po"er factor laggig. A. 5521 , &. 5-2 , %. 5$5 , (. 5582 , Problem $1: A 5006hp! 2200 ,! ,! -6phase star coected sychroous motor has a sychroous impedace of 0.- = * - Ω per phase. (etermie the iduced iduced emf per phase if the motor motor "or3s o full load "ith a efficiecy of $J ad ad a po"er factor of 0.8 leadig. leadig. A. 1-0 , &. 1-5$ , %. 1$02 , (. 1522 , Problem $2: A three6phase! star coected sychroous motor ta3es 50 3+ at 0 ,! the po"er factor beig 0.8 laggig. his machie has a sychroous reactace of 1 ohm per phase "ith a egligible resistace. If the emf is icreased by -0J! the po"er ta3e remaiig remaiig the same! fid the the e" leadig pf curret. A. $.1 A &. 85.2 A %. 0.- A (. 88. A Problem $-: A . 3,! L6coected sychroous motor has a sychroous reactace per phase of 10 ohms ad a egligible resistace. Eor a certai load! the motor dra"s 0 3+ at ormal voltage. At this load! the iduced emf per phase is 510 ,. ,. (etermie (etermie the operatig pf. A. 0.$8 &. 0.502 %. 0.$2 (. 0.518 Problem $$: A 56hp! 006volt! 10006rpm! three6phase L6coected L6coected sychroous motor has a armature resistace of 0.05 per phase ad a lea3age reactace of 0.$5 ohm per phase. he efficiecy at rated load! 80J pf leadig is 2J. (etermie mechaical po"er developed "ithi the armature. A. $ 3+ &. 2 3+ %. 0 3+ (.  3+ Problem $5: A 2-0 ,! 06/B! -6phase! L6coected sychroous motor dra"s a curret of 20 A ad delivers 8 bhp. Armature resistace per phase is 0.5 Ω. If the iro ad frictio losses amout to -00 +! determie determie the operatig po"er factor of the motor. motor. A. 0.80 &. 0.80 %. 0.8 (. 0.82 Problem $: A 2200 ,! ,! 200 hp! -6phase mesh coected sychroous motor is "or3ig o full load at a efficiecy 0.88 ad 0.8 pf leadig. he armature reactace per phase is 5 ohms. (etermie the iduced emf per phase. eglect resistace. A. 2- , &. 2-$5 , %. 2-80 , (. 2-0 , Problem $: A 156hp! 2-0 , sigle6phase sychroous motor is ta3ig a curret of 50 A at a po"er factor of 0.85 leadig. 7ffective armature resistace is 0.2 ohm. If this motor drives a mechaical load of 12 hp! determie its rotatioal losses. A. -2- + &. -$0 + %. - + (. -1$ + Problem $8: A $$06 ,! sigle6 phase sychroous motor is dra"ig a lie curret of 50 A at 0. pf laggig. his machie has a armature resistace ad reactace of 0.8 ad 10 ohms respectively. If stray po"er ad excitatio losses amout to 500 + ad 800 + respectively! determie the bhp output of the motor. A. 2-.2 & 20.$ %. 25.8 (. 18.5

Problem $: 77 &oard April 1A sigle6 phase sychroous motor gives a output of .$ 3+ at 0. pf laggig "ith 00 ,. If stray po"er loss is 00 + ad field losses is 00 +! "hile effective armature resistace is 0. ohm! fid the efficiecy of the sychroous motor. A. 82.-5J &. 81.2$J %. 85.25J (. 8$.$J Problem 50: A 2-06 ,! 0 /B! sigle6 phase sychroous motor is ta3ig a curret of 20 A at 0.8 pf laggig. 7ffective armature resistace is 0.- ohm. Iro ad frictio losses amout to $50 +! determie the efficiecy of the motor at this load. eglect losses i the excitatio. A. 8.15J &. 85.-0J %. 88.52J (. 8-.20J Problem 51: A three6 phase star6 coected sychroous motor dra"s 0 A at 0.8 pf leadig "he coected from a . 3,! 3,! 06 /B! -6 phase source. At this load the machie operates "ith a efficiecy of 0J. If the sychroous impedace of the machie is 2 = *15 ohms! determie the stray po"er losses. A. $0.5 3+ &. 28. 3+ %. -2.8 3+ (. -$. 3+ Problem 52: A 20 /P! $$06 ,! 06 /B! -6 phase star6 coected sychroous motor has a armature resistace per phase of 0.$ ohm. It is givig its full load output at a po"er factor of 0. leadig. If the stray po"er losses amout to 500 +! fid the armature curret. A. 2-.$$ A &. 25.28 A %. 22.5- A (. 20. A Problem 5-: A 5006 ,! ,! 06 /B sigle6 phase sychroous motor develops 20 hp ad is operatig at a pf of 0.8 laggig. 7ffective resistace of armature is 1 ohm. Iro ad frictio losses amout to 500 +. (etermie (etermie po"er dra" by the motor from the source. A. 18! 220 + &. 21! 5$ + %. 1! 288 + (. 20! 05 + Problem 5$: 77 &oard April 11 A three6 phase sychroous motor is measured by t"o A ad & "attmeters. It reads 28. 3+ ad $.5 3+! respectively. respectively. he iput po"er of the motor is uder excited ad assumed costat. +hat are the readigs of A ad & "attmeters if the po"er factor is 85J ad the motor is over6 excited# A. $8. 3+! 2.5 3+ &. $2.5 3+! -2. 3+ %. 52. 3+! 52. 3+ (. 51.10 3+! 2$.1 3+ Problem 55: he iput po"er to a 006 , sychroous motor is measured by meas of a t"o6 "attmeter method. e "attmeter reads $ 3+ ad the other read -.5 3+. he motor is 3o" to be operatig at leadig pf. (etermie the lie curret dra". A. 10$ A &. 110 A c. 10 A (. 112 A Problem 5: A group of iductio motors rated 22006 ,! ,! -6 phase! 06 /B delivers a total of 00 hp ad is operatig at a average po"er factor of 0.8 laggig. If half of this is replaced by a group of sychroous motors operatig at a average pf of 0.8 leadig! determie the percetage reductio of the lie curret. Assume all motors operate at a average efficiecy of 0J. A. 2-J &. 25J %. 20J (. 21J Problem 5: A three6 phase trasmissio lie havig a capacity of 10! 000 3,A serves a balaced load that dra"s 5!000 3+ at a laggig po"er factor such that the trasmissio lie is loaded to its capacity. capacity. A additioal load of iductio motors operatig at a average pf of 0.8 is coected to the lie. I order to ma3e use of the lie to its full

capacity i active po"er! a sychroous codeser havig losses of -00 3+ "ill be coected to the lie. +hat is the 3,A ratig of the sychroous motor reuired# A. 11!158 &. 12!18 %. 1-!205 (. 10! 202

Problem 58: A three6 phase L6 L6 coected sychroous motor "ith a lie6to6lie voltage of $$0 , ad a sychroous speed of 00 rpm ad operates "ith a po"er of  3+ ad a laggig po"er factor of 0.8. he sychroous reactace per phase is 10 ohms. ohms. he machie is operatig operatig "ith rotor curret of 5 A. it is desired desired to cotiue carryig carryig the same load but to provide 5 3,A@ 3,A@ of po"er factor correctio to the lie. (etermie the reuired rotor curret to do this. A. .88 A &. 8.0 A %. 8.2 A (. 8.28 A Problem 5: 77 &oard ctober 11 A -!00 volts three6 phase star6 coected turbo sychroous geerator of sychroous reactace of 0.5 ohm per phase ad egligible resistace is supplyig -0! 000 3,A 3,A at 0.8 pf laggig to a large po"er system. If the steam supply is cut6 off! calculate the amout of curret that the geerator "ill the carry assumig egligible losses. A. -01 A &. -82- A %. -0$ A (. -80 A Problem 0: 77 &oard ctober 1$ A "ye6 coected turbo alterator havig a sychroous sychroous reactace of 0.5 ohm ad egligible resistace is coected to a large po"er system havig a busbar voltage of 1-.8 3, supplyig a load of 15!000 3,A at 0.80 laggig po"er factor. If the steam supply is cut6 off! the armature curret "ill the machie carry assumig egligible losses# A. 25.55 A &. 525.2 A %. -8$.2 A (. 2.55 A Problem 1: 77 &oard ctober 1$ A 1!000 3,A! !00 , "ye6 coected three6 phase alterator havig a reactace of 8.8 Ω ad a egligible resistace is supplyig po"er to a costat freuecy bus bar. he ope circuit emf at this istace is $!-11 , per phase. If steam supply is suddely suddely cut6 off! off! the armature curret i amperes is earest earest to A. 5 &. 0 %. 5 (. 5

7D 216%,7@7@D 216%,7@7@D M @7%IEI7@D Problem 2: A sigle6 phase rotary coverter has a output voltage of 110 volts. (etermie the lie curret i A% side "he the machie delivers a (% output of 50 A. assume the machie is operatig "ith a pf of 0.85 ad efficiecy of 5J. A. 0.1 A &. 0.-$ A %. 8.5 A (. 82.2- A Problem -: 77 &oard April 1$ I a three6 phase sychroous coverter! coverter! the ratio of the A% voltage to (% voltage is A. 0.12 &. 1.0 %. 0.50 (. 0.0 Problem $: A three6 phase rotary coverter delivers 150 A at 500 ,(%. (etermie the curret o the A% side if the operatig po"er factor of the machie is 0.8 ad the efficiecy is 0J. A. 1 A &. 15 A %. 18 A (. 1 A Problem 5:

hree sigle6 phase trasformers coected delta6 delta fed a three6 phase rotary coverted at 2!-00 ,. his coverter supplies po"er to a (% load that dra"s 100 3+ at 250 ,. if the rotary coverter operates at a efficiecy of 5J at uity pf! calculate the curret i the secodary "idig of the trasformer ba3. A. - A &. -55 A %. 22 A (. 2$0 A Problem : A ope delta ba3 is servig a three6 phase rotary coverter. he coverter supplies 15 A of curret to a 1.$ ohm dc resistive load. If the machie dra"s a A% curret of 100 A at 0.8 pf laggig! determie miimum siBe of trasformer eeded. A. 15.$ 3,A &. 2$. 3,A %. 25. 3,A (. 1.$ 3,A Problem : A $6 phase star6 coected rotary coverter operatig at 0.0 efficiecy ad 0.85 po"er factor is coected across a 220 ,! 0 /B mais ad dra"s 50 A. determie the (% load curret. A. 208. A &. 220.5$ A %. 21.- A (. 212.25 A Problem 8: he dc output of a six6phase star6coected star6coected coverter is 500 3+ at $00 ,. If the pf ad efficiecy of the machie is 0.8 ad 0.2 respectively! determie the ac curret dra" from the supply# A. 2 A &. $0 A %. -2 A (. 2 A Problem 0:77 &oard April 12 A 500 3+! 006, 006, (%! 126phase sychroous coverter operates as a direct coverter at a full load efficiecy of 2J ad a po"er factor of 0.-. %alculate the A% voltage bet"ee slip rigs ad the A% curret dra" from a 126phase trasformer fed supply. A. 112 ,! 22 A &. 112 ,! 215 A %. 110 ,! 22 A (. 110 ,! 215 A Problem 1: A $6pole 2-06,! 2-06,! 0 /B! three6phase iductio motor directly drives a 6pole alterator. (etermie the output freuecy. Assume a slip of -J i the motor. A. 0.0 /B &. 88.$ /B %. 8. /B (. 8.- /B Problem 2: A $6pole! three6phase sychroous motor coected across a 0 /B supply drives a 6pole alterator through a belt. If the diameter of the driver ad drive pulleys are  ad 10 iches respectively! respectively! determie the operatig freuecy of the alterator. alterator. A. 5- /B &. 5$ /B %. 55 /B (. 52 /B Problem -:77 &oard April 12 A geerator is rated 00 3,A! 2$0 ,! 0 cycles! -6phase! 6poles ad "ye6coected. +hat "ill be the speed of the drivig pulley if the drive ad drivig pulleys are 1 ft ad 2 ft i diameter respectively# A. 2$00 rpm &. 1200 rpm %. 00 rpm (. 00 rpm Problem $:77 &oard ctober 18 he pulley of a old ge6set has a diameter of 20 iches. he belt exerts a pull of -5- pouds o the pulley. he ge6set rus at 00 rpm. +hat is the approximate 3+ ratig of the ge6set# A. 5 &. 250 %. 200 (. -.5 Problem 5: A 2-06,! -6phase motor belt drives a 50 3+! 250 , (% geerator. he motor! belt ad geerator efficiecies are 0.1! 0.5 ad 0.8 respectively. If the motor operates at a pf of 0.8 laggig! determie the curret dra". A. 18 A &. 15 A %. 12 A (. oe of these

Problem : A sigle6phase sigle6phase motor is coected across a 2-06,! 0 /B supply ad dra"s -0 A at 0.85 pf laggig. his motor drives a 1206, (% geerator through direct couplig. If the motor! couplig ad geerator efficiecies are 0.85! 0. ad 0.88 respectively determie the curret delivered by the geerator usig the above coditio. A. -5 A &. -2 A %. -8 A (. $8 A Problem : A 2-06,! 0 /B! -6phase motor drives a 25 3+! 250 , (% geerator through belt ad pulley. he motor geerator efficiecies are 0.0 ad 0.88! respectively. respectively. If the motor dra"s a lie curret of 8 A at 0.85 pf laggig! ho" much is the combied efficiecy ad pulley# A. 5.12J &. .0$J %. 5.--J (. $.81J Problem 8:77 &oard ctober 11 A movig coil ammeter! a resistace of 1500 ohms ad a diode are coected i series across a 125 , siusoidal A% A% supply. he diode diode has a liear curret  voltage characteristics over the rage l < 0! 7 < 0 , to l < 200 mA ! 7 < 200 ,. %alculate the po"er dissipated i the 15006ohm resistor. A. 0. + &. 0.8- + %. 0.1 + (. 0.8 + Problem : A half6"ave rectifier dra"s po"er from a po"er trasformer at 2$ volts ad delivers 200 mA to a load. (etermie po"er dra" by load. A. 0.$8 + &. 1.25 + %. 2.1 + (2.05 + Problem 80: A A% ammeter is coected i series across the output of a full "ave rectifier. his meter reads 8 A. If a movig coil ammeter is replaced istead of the A% ammeter! "hat "ill be its expected readig# A. 8. 8.0 A &. . A %. .2 A (. 5. A Problem 81:77 &oard ctober 18 I the electroic laboratory! laboratory! a diode rectifier tube has a plate resistace of 00 ohms >assumed costat? ad is coected as a half6"ave rectifier to a load resistace of 1!$00 ohms. he eergiBig supply voltage is $$0 volts effective! 0 /B alteratig curret. Eid the average load voltage. A. 1-8.5 , &. 1-2.8 , %. 1--. , (. 150.2 , Problem 82: hree idetical trasformers are used to supply po"er to a iodiBig plat reuirig 1000 A at 28 , (% through a -6phase half "ave rectifier. he trasformers are coected delta6"ye. +hat are the secodary curret ad voltage of each trasformer# A. 10$2 A! 28.$ , &. 101 A! 1.$$ , %. 10$2 A! 1.$$ , (. 101 A! 28.$ , Problem 8-:77 &oard ctober 18 I the electrical laboratory! a diode rectifier tube has a plate resistace of 00 ohms >assumed costat? ad is coected as a half6"ave rectifier to a load resistace of 1!$00 ohms. he eergiBig supply voltage is $$0 volts effective! 0 /B alteratig curret. Eid the rms rectified curret. A. 0.12 A &. 0.155 A %. 0.208 A (. 0.-11 A Problem 8$:77 &oard April 1 A resistor is coected across a A% supply of 220 volts. he po"er dra" is 1!000 "atts. If a diode "ere coected i series "ith the resistor! "hat "ould be the po"er absorbed by the resistor# A. 850 + &. 250 + %. 50 + (. 500 + Problem 85:77 &oard ctober 1

A circuit cosists cosists of a resistor that is coected i series "ith a perfect diode. +he coected across a 206 , battery the curret measured is 10 A. /o"ever! "he coected across a -06, battery but "ith reversed polarity! "hat "ould be the e" curret# A. 15 A &. )ero )ero cur curret%. ret%. 20 A (. oe oe of the the as" as"er ered ed listed sted Problem 8:77 &oard April 1 A germaium diode! "hich has a offset voltage of 0.2 , ad a icremetal resistace of 20 ohms! is coected i series "ith a 10!0006ohm resistor across a 100 , (% source. +hat is the curret# A. .8 mA &. 10 mA %. .8 mA (. . mA

TEST 22# T0*NS%(SS(N L(NES Problem 8: A sigle solid roud copper "ire has a diameter of 0.50 ich. (etermie its self OC(. A. 0. 0.-8$ ich &. 0. 0.1$ ich %. 0. 0.1825 ich (. 0. 0.25 ich Problem 88: (etermie the mutual OC( of a three6phase sigle6circuit trasmissio trasmissio lie "hose coductors arraged i a triagular formatio so that the t"o distaces bet"ee coductors are 20 ft ad the third is -8 ft! respectively# respectively# A. 2$. ft &. 25.-5 ft %. 2-. ft (. 2.$- ft Problem 8: A three6phase three6phase double circuit lie has a horiBotal spacig of $0 ft ad a vertical spacig bet"ee ad*acet levels of 25 ft. (etermie the mutual OC( of the lie. A. $-.08 ft &. $2.$ ft %. $1.$2 ft (. $0.5 ft Problem 0:77 &oard Carch 18 A 2-0 3, trasmissio lie is 100 miles log. he coductor is A%D@ of 1!11-!000 %C. he coductors are horiBotally arraged "ith 20 ft spacig. he resistace per mile is 0.0 ohm ad its OC@ is 0.0$-5 ft. +hat is the impedace of the lie# A. . = * 5.25 ohms &. -.2- = * 5.25 ohms %. . = * .1 ohms (. -.2- = * 25.- ohms Problem 1:77 &oard ctober 1 A -$.5 3, feeder lie is 5 miles log. he coductors are space $ feet horiBotally. horiBotally. he coductor is $0 copper "ith OC@ of 0.018 ft. +hat is the reactace# A. 2.- ohms &. .2 ohms %. -.$ ohms (. 5.1 ohms Problem 2: A sigle6phase sigle6phase trasmissio lie 15 3m log is usig copper coductors of diameter 0.8 cm. If the distace bet"ee coductors coductors is 1.25 m! determie the self6iductace self6iductace of the trasmissio trasmissio lies. A. 0.0- / &. 0.02$ / %. 0.02 / (. 0.055 / Problem -:77 &oard ctober 18 A 5 3m log! three6phase lie has a horiBotal cofiguratio of $ ft spacig. he coductor is --.$ C%C A%D@ "ith OC@ of 0.02$$ ft ad a resistace of 0.-0 ohm per mile. +hat is the impedace# A. 2.22 cis 5 &. 1.1 cis $ %. . cis 5 (. -.5 cis $ Problem $: A 106 mile! 2-06 3,! 06 /B! three6 phase sigle6 circuit trasmissio lie uses a triagular arragemet "ith 20 ft! 220 ft ad - ft spacig respectively. If the lie coductors have a OC@ of 0.021 ft! determie the capacitive susceptace to eutral per mile. A. $.18 x 106 mh mho per mi &. 12.$ x 106 mho per mi

%.

.2 x 106 mh mho per mi

(.

2.$ x 106 mho per mi

Problem 5: A sigle6 circuit! -6 phase! 0 /B trasmissio is usig a horiBotal arragemet "ith a spacig distace of $ m bet"ee ad*acet coductors. If each coductor has a outside diameter of 0.8 cm! determie the capacitace to eutral per meter. A. 8.-52 pEm &. .- pEm %. .$2 pEm (. oe of these Problem : A sigle6 phase trasmissio trasmissio lie has total impedace of 1.5 = *0.- ohms. It delivers - C+ at 11 3, ad at uity po"er factor. (etermie the sedig ed voltage of the trasmissio lies. A. 11.$1 3, &. 11.2$ 3, %. 12.02 3, (. 11.8 3, Problem : A sigle6 phase trasmissio lie has total impedace of 5 ∠0o ohms ad supplies a total load of 120 A! -.3, ad at 0.8 po"er factor laggig. %alculate the sedig ed voltage. A. -.- 3, &. -.2 3, %. -.5$ 3, (. -.8 3, Problem 8: 77 &oard April 181 A three6 phase! -6 "ire short trasmissio lie has a resistace of - ohms ad a reactace of 8 ohms per "ire. At the receivig ed! a balaced -6 phase load dra"s a lie curret of 0 A! at 1-!500 volts lie6to6lie! 0.0 po"er factor laggig. laggig. (etermie the voltage at the the sedig ed. A. 1$!15 , &. 1$!1$- , %. 1$!2-0 , (. 15!055 , Problem : 77 &oard ctober 1! 77 &oard April 1 A load of 12 C,A! C,A! 0.8 pf laggig! 22 3, is served by a trasmissio lie! "hich has a lie resistace of ohms ad a lie reactace of 10 ohms. Dolve for the sedig ed voltage. A. 2$!-$5 , &. 2!5 , %. 2-!--5 , (. 25!$22 ,

Problem 800: 77 &oard April 182 A short -6 phase! -6"ire lie has a resistace of 5 ohms ad reactace of 12 ohms per "ire ad trasmit po"er to a three6 phase load load dra"ig 1000 3+ at 1-!120 volts lie6 lie6 to6 lie! 0.8 pf laggig! laggig! 0 /B. Dolve for the sedig ed voltage. A. 1$!200 , &. 1$!--0 , %. 1$!1-0 , (. 1$!2$0 , Problem 801: 77 &oard April 11 A -6 phase lie has a reactace of $.-2 ohms ad a resistace of $.15 ohms. he load at receivig ed is -800 3+ at 5J po"er factor laggig ad the voltage at the sedig ed is - 3,. (etermie (etermie the voltage at the receivig ed. A. -$.2$ 3, &. -$.8 3, %. -5.1$ 3, (. -5.$5 3, Problem 802: 77 &oard Carch 18 A short 2-06 3, trasmissio trasmissio lie has impedace of 5 cis 8 ohms. he sedig ed po"er is 100 C+ at 2-0 3, ad 85J po"er factor. +hat is the voltage at the other ed# A. 225.$ 3, &. 22.- 3, %. 22-.2 3, (. 228.2 3, Problem 80-: 77 &oard ctober 1 A three6 phase trasmissio lie! 15 3m log serves a substatio rated 15 C,A at -$.5 3,! 0 /B. If the lie resistace is 0.12 ohm per 3ilometer ad the lie reactace is 0.$5 ohm per 3ilometer! "hat should should be the sedig ed voltage be so that the trasformer ca be fully loaded at its rated voltage ad uity pf. A. -!200 , &. -!500 , %. -5!$08 , (. -$! 0 ,

Problem 80$: A three6 phase trasmissio trasmissio lie! 10 3m log delivers a load of 5000 3+ at 11 3, ad 0.8 pf laggig at the receivig ed. he resistace of the lie per 3m per phase is 0.1 ohm ad the reactace per phase per 3m is 0.2 ohm. Dolve for the regulatio of the lie. A. 12.$$J &. 11.05J %. 10.$5J (. 10.12J Problem 805: 77 &oard April 180 A -6 phase! -6"ire short trasmissio lie havig a impedace of -. = *1 ohms per "ire is used to supply a iductive load of 100 A at 0.0 pf ad a capacitive load of 50 A at 0.50 pf. he receivig ed voltage is $10 volts per phase. Eid percet regulatio. A. 25.$J &. 20.--J %. 2.-$J (. 2-.1J Problem 80: It is desired to deliver $000 3+! three6 phase at distace of 25 miles. he load beig -- 3,! 0 /B ad the po"er factor of the load beig beig 85J laggig. he he reactace of the lie lie per phase is 0.2 ohm per mile. mile. he lie loss is 10J of the po"er delivered. (etermie voltage regulatio of the lie. A. 11.2J &. 12.-J %. 11-.2J (. 1$.8J Problem 80: 77 &oard ctober 18 A short 2-0 3, trasmissio trasmissio lie has a impedace of 5 cis 8 ohms. he sedig ed po"er is 100 C+ at 2-0 3, ad 85J po"er factor. +hat is the percet regulatio of the lie# A. -.J &. 1.5J %. 2.2J (. 0.J Problem 808: 77 &oard ctober 18A short subtrasmissio lie serves at its ed a iductio motor rated 500 /P! 0.88 pf! 0.0 efficiecy! at 2!$00 volts phase to eutral. If the trasmissio has a resistace of 1.5 ohms ad a reactace of 2.$ ohms per phase! solve for the voltage regulatio of the lie. A. .J &. .02J %. .$8J (. .25J Problem 80: A short -6 phase! -6 "ire trasmissio lie has a impedace of 2 = * ohms per "ire! at the receivig ed are coected a three6 phase iductive load dra"ig 5000 3+ at 0.8 pf! lie6to6lie voltage of 1-!800 , ad a capacitor ba3 dra"ig 200 A lie curret. Dolve for the po"er at the sedig ed. A. 5!52 3+ &. 5!1$ 3+ %. 5!-1 3+ (. 5!2$ 3+ Problem 810: A short -6 "ire trasmissio lie has a impedace of 2 = *5 ohms per "ire. At the receivig ed! a balaced -6 phase load ad ad capacitor ba3 dra" -000 -000 3,A 3,A! 0.1 pf laggig ad 00 3,A 3,A@ respectively at 8000 volts per phase to eutral. (etermie the po"er loss of the trasmissio lie. A. 81.-2 3+ &. .5 3+ %. 0.$ 3+ (. 1.0 3+ Problem 811: A 106 3m! three6 phase trasmissio lie delivers po"er to a load rated 200 3+! 3+! . 3, ad at 80J laggig po"er factor. he resistace ad reactace reactace of each lie are 0.8 ad 1.2 ohms! ohms! respectively. respectively. +hat percetage of the the po"er geerated is lost lost i the trasmissio trasmissio lie# A. .5-J &. 8.05J %. 8.22J (. .$J Problem 812: 77 &oard ctober 17ach coductor of a three6 phase! -6 "ire trasmissio lie has impedace of 15 = *20 ohms at 0 /B. he voltage bet"ee lie coductors at the sedig ed is 1-!200 volts. he load coected to this lie is balaced ad ta3es 1000 3+ at a laggig po"er factor. he curret per coductor is 0 A. +hat is the load po"er factor# A. 0.80- laggig &. 0.82$ laggig %. 0.850 laggig (. 0.815 laggig Problem 81-:

A 15 C+! 1-2 3,! 80J pf three6 phase load is to be serve by a trasmissio lie havig coductors "hose resistace is 0.5 ohm3m. If the losses o the lie shall ot exceed .5J! ho" log must this lie be# A. 8 3m &. 0 3m %.  3m (. 5 3m Problem 81$: A three6 phase! three6 "ire trasmissio lie supplies at the receivig ed a iductive load of 5000 3+ at 0.8 pf ad the lie6to6lie voltage 1-!250 , ad i additio a capacitor ba3 that dra"s a lie curret of 0 A. he effective impedace of the trasmissio lie is 2 =*- ohms per "ire. (etermie the sedig po"er. A. 5-1 3+ &. 55-0 3+ %. 52555 3+ (. 5$10 3+ Problem 815: 77 &oard April 182 A short three6 phase! three6 "ire lie has a resistace of 5 ohms ad reactace of 12 ohms per "ire ad trasmit a po"er to a -6 phase load dra"ig 1000 3+ at 1-!120 volts lie to lie! 0.8 pf laggig! 0 /B. Dolve for the sedig ed po"er factor. A. 0.55 laggig &. 0.$5 laggig &. 0.5$ laggig (. 0.- laggig Problem 81: 77 &oard ctober 182! 77 &oard ctober 18 A short three6 phase trasmissio trasmissio lie has a resistace of - ohms ad a iductive reactace of $ ohms. At the receivig ed the voltage is1-! 250 volts lie6to6lie! "ith a load "hich dra"s 200 A at 0.0 laggig po"er factor ad capacitor ba3 dra"ig  A. solve for the po"er factor at the sedig ed. A. 0.5 laggig &. 0.02 laggig %. 0.28 laggig (. 0.-5 laggig Problem 81: 77 &oard April 185 he sedig ed voltage of a balaced -6 phase trasmissio lie is 8$10 volts! phase to eutral. he lie curret is 200 amperes "ith a sedig ed laggig po"er factor of 0.80-. he receivig ed voltage is 00 volts! phase to eutral! "ith "ith a lie curret havig a laggig laggig po"er factor of 0.85. 0.85. Dolve for the impedace of the the trasmissio lie. A. 1.$ = * 5.15 Ω &. 1.$2 = * 5.15 Ω %. 1.$ = * 5.0$ 5.0$ Ω (. 1.$2 = * 5.0$ Ω Problem 818:77 &oard April 18 A short! -6phase! -6"ire trasmissio trasmissio lie has a receivig ed voltage of $10 , phase to eutral ad servig a balaced -6phase load of 8!$00 volt6amperes at 0.82 pf laggig. At the sedig ed the voltage is $00 ,! phase to eutral ad the pf is 0. laggig. Dolve the resistace of the lie. A. 1.-$5 Ω &. 1.$2 Ω %. 1.$5 Ω (. 1.-5 Ω Problem 81:77 &oard April 18$! 77 &oard April 18 At the sedig ed of a -6phase trasmissio lie! the voltage is measured to be 2!5$0 volts phase to eutral ad the lie curret to be 0 A at a laggig po"er factor of 0.5. At the receivig ed! the voltage measured is 2!200 volts phase to eutral ad the po"er factor is 0.80! li3e"ise laggig curret. Dolve for the resistace ad reactace of the lie. A. 2.2 = * $ &. 2.$ = *  %. 2.0 = * 5 (. 2.1 = * Problem 820: A three6phase trasmissio lie 5 3m log delivers 2 C,A at a po"er factor of 0.80 laggig. he resistace ad reactace per 3m of each coductor are 0.- ohm ad 0. ohm! respectively. respectively. %alculate the voltage at the sedig ed. Assume Assume a trasmissio lie loss of .2J of the po"er factor delivered to the load. A. $$05.$5 , &. $1. , %. $810.50 , (. $52.85 , Problem 821:77 &oard April 1A certai statio i ;uBo has - 3,! -6phase! is to supply12 C+ load at -0 3, ad 0.8 po"er factor laggig over a -6phase trasmissio lie. Eid the resistace of the lie if the legth of the trasmissio lie is $!000 meters ad the efficiecy is 0J. A. 5.-- ohms &. $.$5 ohms %. .-1 ohms (. -.5 ohms

Problem 822: A three6phase! three6phase! -6"ire trasmissio lie has a impedace per "ire of - = * ohms! the receivig ed load is 150 3+! 0.5 pf laggig "ith the lie voltage of 1-!200 ,. (etermie the efficiecy of the lie. A. 2.$J &. 5.21J %. 0.2$J (. -.05J Problem 82-:77 &oard ctober 180 A short lie lie has a resistace of $ ohms ad a reactace of 12 ohms per "ire ad trasmits po"er to a cocetrated load of 1000 3,A! 3,A! 1-!200 volts! -6phase! 0 /B! 0.8 pf iductive load. Eid its efficiecy. A. -.-0J &. .5J %. 5.5J (. .21J Problem 82$:77 &oard ctober 18 A 2-0 3, trasmissio lie is sedig 100 C+ po"er at 2-0 3, ad 0J po"er factor. he impedace is 5 = * 20 ad its capacitive reactace is 2500 ohms. (etermie the receivig ed voltage. A. 221.2 3, &. 222.8- 3, %. 22.15 3, (. 22$.28 3, Problem 825:77 &oard ctober 1 A -6phase! 0 /B trasmissio lie delivers 20 C,A to a load at  3, at 80J pf laggig. he total series impedace of each lie is 15 = * 5 ohms. If a omial GpiH circuit is used! "hat "ould be the trasmissio efficiecy if the admittace is *0.000 mhos# A. 0.8J &. 1.J %. -.5J (. 2.J Problem 82:77 &oard Carch 18 A 2-0 3, trasmissio trasmissio lie has a impedace of 50 cis 8 ohms ad a capacitive reactace of 1200 ohms. It trasmits the po"er of a base load plat.  a certai dry seaso the sedig ed po"er is 100 C+ at 2-5 3, ad 5J po"er factor cotiuously for a period of oe moth. If cost of geeratio is 1.-0 pesos per 3+6hr! "hat is the cost of the lie losses for the oe moth period# A. P 55!000 &. P 12.2 millio %. P 5. millio (. P 2.- millio

TEST 23# P6E0 *CT0 C00ECT(NS Problem 82:77 &oard ctober 1 A 1-23, lie three6phase system delivers 0. C,A of a balaced delta load of po"er factor 0.J laggig. (etermie the reactace ecessary to attai uity po"er factor. A. Kc < 1!02 Ω &. Kc < 5 Ω %. Kc < 1!1$2 Ω (. Kc < 1!0$5 Ω Problem 828:77 &oard ctober 10 A sigle6phase iductive load ta3es 50 3,A at 0.0 po"er factor laggig. Dolve for the 3,A@ of a capacitor reuired to improve the po"er factor to 1.0. A. -0 3,A@ &. 20 3,A@ %. 22.5 3,A@ (. $0 3,A@ Problem 82: A sigle6phase load o 220 , ta3es 5 3+ at 0. laggig po"er factor. Eid the 3,A@ siBe of capacitor! "hich maybe coected i parallel "ith this motor to brig the resultat po"er factor to 1.0. A. . &. .-2 %. 8. (. .2 Problem 8-0: A 5 hp! 220 ,! 0 /B! sigle6phase iductio motor operates at a efficiecy ad po"er factor of 0.88 ad 0.8 respectively. respectively. +hat capacitace should be coected across the motor i order for the feeder supplyig this motor to operate at uity po"er factor# A. 201.55 µE &. 18-.52 µE %. 1$.22 µE (. 212.5 µE Problem 8-1:77 &oard ctober 18

Istalled i oe of the customer of %7PA;% are t"o sigle phase trasformers each rated 5 3,A are coected , or ope delta to serve a -6phase load of 120 3+ at 0.8 pf laggig. o prevet the overloadig of the trasformers! determie the siBe of the capacitor i 3,A@. 3,A@. A. $0 &. $1 %. - (. $2 Problem 8-2: hree trasformers each rated 100 3,A are coected delta supplyig a load of 10 3+ at 0.8 pf laggig. e of the trasformers is ta3e for repair ad the rest are coected ope delta. +hat 3,A 3,A@ >miimum? of capacitor must be coected "ith the load so that the load of the remaiig trasformers "ill be 5J of their combied rated capacities. A. 85.-2 &. 81.2 %. 8.58 (. 80.2 Problem 8--:77 &oard ctober 1hree sigle6 phase trasformers each rated 5 3,A are ba3ed i delta ad supplyig a -6 phase load dra"ig 10 3,A at 0.8 laggig po"er factor. If oe trasformer is removed for repairs! solve for the miimum amout i 3,A@ of capacitor eeded to prevet overloadig of the remaiig uits. A. 0.-2 &. -.$ %. 2.$ (. -.28 Problem 8-$ "o sigle6 phase trasformer each rated 150 3,A are coected ope delta supplyig a three6 phase iductio motor rated 250 hp! 0. pf ad 80J efficiecy. (etermie the miimum siBe i 3,A@ of a capacitor eeded to prevet overloadig the trasformers. A. 1-5 &. 12 %. 120 (. 12Problem 8-5: 77 &oard ctober 182 "o "o sigle6 phase trasformers are each rated 5 3,A are coected i , or ope delta to serve a -6 phase load of 120 3+ at 0.80 po"er factor laggig. (etermie the siBe i 3,A@ 3,A@ of the capacitor eeded to prevet overloadig of the trasformers. A. $0.25 &. $1.28 %. $5.2$ (. $-.50 Problem 8-: 77 &oard April 18$ A plat has a load of 20 3ilo"att "ith a average po"er factor of 0J. he o"er reuests you to correct the po"er factor to reduce its po"er cosumptio. /o" much capacitor 3,A@ 3,A@ is reuired to icrease the po"er factor to 0J# A. 152.$ &. 155.- %. 150.-$ (. 15$.58 Problem 8-: 77 &oard ctober 182 A 15063,A6trasformer ba3 "ill serve a load expected to dra" 1-5 3+ at 0.80 laggig po"er factor. Dolve for the siBe of the capacitor ba3 eeded to be added i order to prevet overloadig of the trasformer ba3. A. -2.50 3,A@ &. -5.8 3,A@ %. $0.-1 3,A@ (. 28.2 3,A@ Problem 8-8: 77 &oard ctober 181 A -6 phase geerator has the follo"ig -6 phase loads: a iductive load dra"ig $00 3,A at 0.0 po"er factor ad a resistive load dra"ig 80 3,A at 1.00 po"er factor. Dolve for the siBe i 3,A@ of the capacitor ba3 eeded to improve the po"er factor of the combied loads to 0.85 laggig. A. 120.58 3,A@ &. 121.8 3,A@ %. 12$.5$ 3,A@ (. 122.82 3,A@ Problem 8-: 77 &oard April 1 A load of 10!000 3,A 3,A!! 80J pf laggig is coected to a 1-!200 volts lie. /o" much capacitive reactive po"er is eeded to correct correct the po"er factor to 0. laggig# laggig# A. 5!15 3,A@ &. -!1-8 3,A@ %. 2!5$ 3,A@ (. $!-5 3,A@ Problem 8$0: 77 &oard April 18

A short! -6 phase! -6 "ire trasmissio lie has a receivig ed voltage of $10 , phase to eutral a servig a balaced -6 phase load of 8!$00 volt amperes at 0.82 pf laggig. At the receivig ed the voltage is $00 , phase to eutral ad the pf is 0. laggig. Dolve for the siBe i 3,A 3,A@ of the capacitor eeded to improve the receivig ed pf to 0. laggig maitaiig $10 ,. A. 181 &. 15 %. 12 (. 18 Problem 8$1: A short! -6 phase! -6 "ire trasmissio lie has a impedace per "ire of - = * ohms! the receivig ed load is 150 3+! 0.5 pf laggig "ith the lie voltage of 1-!200 ,. determie the 3,A@ of the capacitor to be coected at the receivig ed to ma3e the pf at that ed to 0.8 laggig. A. 82$.5 &. 81.$5%. 80-.88 (. 81.-2 Problem 8$2: 77 &oard ctober 180 A balaced! balaced! 500 3,A! 3,A! -6 phase! $$0 volt! 0 /B iductive load operates at a pf of 5J. (etermie the total capacitor 3,A@ reuired improvig the pf to 5J. A. 20.$ &. 1.$2 %. 210.5 (. 12.21 Problem 8$-: 77 &oard Carch 18 A sigle6 phase iductio iductio motor is rated 5 hp! 5J po"er factor ad 220 volts. +hat approximate siBe of the capacitor siBe is ecessary to raise the po"er factor to about 5J# A. - 3,A@ &. 2 3,A@ %. 2.5 3,A@ (. -.5 3,A@ Problem 8$$: 77 &oard ctober 18$ A balaced -6phase load dra"s 150 A phase curret at .5 3, phase to eutral! 0.81 po"er factor laggig. It is desired to raise the po"er factor to 0. leadig. Dolve for the amout of capacitor 3,A@ 3,A@ eeded to achieve such pf. A. 22- 3,A@ &. 2$0 3,A@ %. 250 3,A@ (. 2-5 3A,@ Problem 8$5:77 &oard April 181 A -6phase! -6"ire! short trasmissio trasmissio lie has a resistace of - ohms ad a reactace of 8 ohms per "ire. At the receivig ed! a balaced -6phase load dra"s a lie curret of 0 A! at 1-!500 ,! solve the siBe i 3,A@ of capacitors eeded to raise the po"er factor at the receivig ed to 0.5 leadig. A. 10$-.5 &. 115$.2 %. 102.5 (. 12-2.2 Problem 8$: he iput to a iductio motor from a 1- 3,! 0 /B lie is 1000 3,A! at 0.8 pf laggig. A capacitor is placed i parallel "ith "ith the motor to improve improve the pf. %alculate the capacitace capacitace reuired raisig the po"er po"er factor to 0. leadig. A. 2 µE &. 50 µE %. $- µE (. 5 µE Problem 8$:77 &oard April 181 A three6phase balaced load dra"s a lie curret of 80 A at 0.0 laggig po"er factor. Dolve the miimum siBe i 3,A@ 3,A@ of the capacitor ba3 eeded to raise the po"er factor to 0.5 leadig! if the lie to lie voltage is 1-!200 volts. A. 1-10.15 &. 1--8.25 %. 12$.5$ (. 1$-0.12 Problem 8$8:77 &oard ctober 10 "o L6coected! L6coected! 50° 50 ° rise iductio motors are fed by a $10 ,! lie to lie! -6phase 0 /B motor6cotrol ceter 20 feet a"ay. Cotor V1 drives a 006hp compressor. he efficiecy of this motor is 0J ad its po"er factor is 0.5. Istrumets of motor V2 idicate 1-0 3+! 2 amperes. (etermie the capacity i microfarads per phase of a "ye6coected ba3 that is reuired to correct the po"er factor of the total load to 0. laggig. A. 12.$ µE &. 1-.8 µE %. 1.2 µE (. 182.1 µE Problem 8$:

A 2.- 3, three6phase system supplies a 12063+! 12063+! 0. pf laggig balaced load. (etermie the capacitace i each phase of a "ye6coected capacitor ba3 to ad*ust the po"er to 0.0 laggig. A. 51.1 µE &. 52.2 µE %. 2.5 µE (. 8.2 µE Problem 850:77 &oard ctober 1 A sigle6phase! 0 /B! 5 hp suirrel cage iductio motor dra"s a curret of 5- A at 11 ,. If it has a 8.5J electrical to mechaical coversio efficiecy! "hat capacitace should be coected at the termials of the motor i order to icrease the po"er factor of the load combiatio to 2J# A. $80 µE &. -80 µE %. -20 µE (. $20 µE Problem 851:77 &oard April 18 A three6phase! 0 /B! 2200 volts iductio motor develops 500 /P! 0.8 laggig pf ad efficiecy of $J. he po"er factor is raised to 0.0 laggig by coectig a ba3 of codesers i delta across the lies. If each of the capacitace uit is built up of four similar 550 , codesers! calculate the capacitace of each codeser. A. .0$ µE &. 5.$2 µE %. .12 µE (. 2.-0 µE Problem 852:77 &oard April 1A star6coected star6coected $00 /p >metric?! 2000 ,! 50 cs motor "or3s at a po"er factor of 0. laggig. A ba3 of mesh6coected codesers are used to raise the po"er factor to 0.- laggig. %alculate the capacitace of each uit reuired reuire d if each is rated 500 ,! ,! 50 cs. he motor efficiecy is 85J. A. 1$ µE &. 225 µE %. -02 µE (. 2-- µE Problem 85-: A factory dra"s a laggig load of 2000 3+ at a po"er factor of 0J from a  3, bus lie. A sychroous sychroous capacitor is istalled to raise the overall po"er factor to uity. Assumig Assumig the sychroous capacitor losses are 25 3+! 3+! calculate the operatig po"er factor of the sychroous capacitor. capacitor. A. 0.20 0.201 1 lead leadi ig g &. 0.10 0.10- lead leadi ig g %. 0.-0 0.-02 2 lead leadi ig g (. 0.11 0.11- lead leadi ig g Problem 85$: A $006, three6phase balaced load ta3es $0 A at a laggig po"er of 0.8. A over6 excited sychroou s ychroouss motor is coected to raise the po"er factor to uity. uity. If the mechaical output of the motor is 15 hp ad its efficiecy is 85J! fid the 3,A iput to the motor. A. 2$.2 3,A &. 18. 3,A %. 20.- 3,A (. 21.2 3,A Problem 855: A three6phase load of 100 3,A operates at 0.06pf laggig. Dolve for the operatig po"er factor of a sychroous codeser ecessary to raise the system po"er factor to 0.8 laggig! ad so arrage that the system apparet po"er is the same as before the correctio. A. 0. 0.-2 leadig &. 0. 0.882 leadig %. 0. 0.05 leadig (. 0. 0.0 leadig Problem 85: A idustrial plat has a electrical load of 5000 3+ at 80J pf laggig. It has bee decided to replace a 5006hp iductio that drives a pump. his motor operates at a efficiecy of 0J ad a pf of 0. laggig. If a sychroous motor is purchased as a replacemet! "hich is capable of operatig at 0.80 pf leadig. +hat "ill be the e" plat pf. Assume the sychroous motor to have the same efficiecy. A. 0.8&. 0.82%. 0.8- (. 0.852 Problem 85: A sychroous sychroous motor is dra"ig 0 3+ is coected i parallel "ith a load dra"ig 250 3+ at a laggig po"er of 0.8. If the combied combied load has a po"er factor of 0.! at "hat "hat po"er factor is the sychroous sychroous motor operatig# A. 0. 0.82 leadig &. 0. 0.8$ leadig %. 0. 0.88 leadig (. 0. 0.8$ leadig Problem 858:77 &oard ctober 185

A iductive load cosumes 10 3+ at 0.5 pf laggig. A sychroous sychroous motor "ith a pf of 0. leadig is coected i parallel "ith the iductive load. +hat is the miimum reuired 3+ siBe of the sychroous motor so that the combied load "ill have a pf of 0.8 laggig. A. 1.08 3+ &. 1.20- 3+ %. 1.05 3+ (. 1.10 3+ Problem 85:77 &oard April 180! 77 &oard April 12 A idustrial plat dra"s 500 3+ at 0. po"er factor from a three6phase system. I order to raise the po"er factor to 0.8 laggig laggig ad to supply eeded additioal po"er! a sychroous sychroous motor is added . his motor dra"s -00 3+! brigig the e" total plat load to 800 3+. eglectig the losses of the sychroous motor! calculate the exact reuired 3,A ratig. A. --.2$ Wva &. -$5.-$ 3,A %. --.-$ 3,A (. -0.-5 3,A Problem 80: A iductio motor load of 1500 3+ cosists of several uits i parallel! operatig at a average po"er factor of 0.80 laggig. I order to improve the po"er factor! a portio of the iductio motor loads is to be replaced by a sychroous motor! motor! operatig operatig at the same efficiecy as the the iductio motors motors ad at a po"er factor of 0.0 leadig. Eid the 3,A ratig of the sychroous motor reuired brigig the po"er factor of the total load to 0.0 laggig. Assume Assume the iductio motor load! "hich has to be replaced! operates at the same pf as the iductio motor group. A. --2 3,A &. -25 3,A %. -00 3,A (. -22 3,A Problem 81: A alterator rated 5-063,A! is operatig at 0.0 pf laggig. A sychroous codeser is i parallel to improve the pf to 0J laggig. (etermie the operatig po"er factor of the sychroous codeser. Assume Assume the alterator is ot supposed to be overloaded. A. 0. 0.5$ leadig &. 0. 0.$$ leadig %. 0. 0.$- leadig (. 0. 0.5- leadig Problem 82: A over6excited sychroous motor is coected across a 150 3,A! 3,A! 0.0 laggig po"er factor load. he motor ta3es 12 3+ "hile ruig o o6load. %alculate the 3,A 3,A ratig of the motor reuired i order to brig the overall pf of the motor6iductive load combiatio to uity. A. 10.22 &. 10.8 %. 110.20 (. 105.58 Problem 8-: A idustrial load ta3es 250 3+ at 0J pf from a 00 ,! 0 /B! -6phase feeder. It is desired to raise the pf of the etire feeder to 0J by meas of a sychroous motor! motor! "hich at the same time is to drive a dc compoud geerator! reuirig reuirig that the sychroous motor ta3es 0 3+ from the lie. (etermie the reuired curret ratig of the sychroous motor. A. 15.5- A &. 18$.- A %. 12.5 A (. 188.$$ A Problem 8$:77 &oard April 18 A existig idustrial plat has a average load of 00 3+ at 0. pf laggig. o raise the overall po"er factor to 0.2! a sychroous motor drivig a (% geerator is to be istalled. If the iput of the sychroous motor "ill be 250 3+! calculate its 3,A iput ratig. A. 5- 3,A &. 88 3,A %. 2- 3,A (. 15 3,A

TEST 2" – *4LTS N ELECT0(C*L S5STE% Problem 85: I symmetrical compoets! "hat is the vector sum of 1 = a = a 2# A. 1 &. 0 %. '1 (. Ifiity Problem 8:77 &oard ctober 18

If the load of a "ye6coected trasformer are: la< 10 10 cis cis >6-0 >6-0?? l b< 12 cis 215 l c< 15 cis 82 +hat is the phase b positive seuece compoet# A. 1-.$ 1-.$ cis >6>6-2.2? .2? &. 10 10.2 cis cis 2$0 %. 12 12.2 .2 cis 208. 208.$ $

(. 12.2 12.2  cis >6>6-1.? .?

Problem 8: Oive three ubalaced three6phase voltages: ,a < 150 = *0 , , b < 60 ' *120 , ,c < 6120 = *0 , (etermie ,a1. A. 1$2.$- = *12.-5 &. 1-5.-2 ' *21.-$ %. 1$5.2 = *1-. (. 1$0.2- ' *.-2 Problem 88: 77 &oard Carch 18 he three6 phase ubalaced currets are: la < 10 cis >6-0? >6-0? l b < 0 lc < 10 cis 150 Eid the egative seuece curret of phase a. A.

8. cis -0

&.

5. cis >60?

%.

Problem 8: 77 &oard ctober 18$ Oive the follo"ig currets: la < 0 = *0 A l b < 6-6 *$8 A Dolve for the egative seuece compoet of l a. A. 8. 8 .$1 ' *1.5$-

&. . .$- = *1.-$

65.

(.

5.

lc < 6$8 = *- A

%. . .51 ' *1.$$

Problem 80: 77 &oard ctober 18 he three6 phase ubalace currets are: la < 10 cis >6-0? >6-0? l b < 0 lc < 10 cis 150 Eid the Bero seuece curret. A. -.-- cis -0 &. 0 %. 5.

(.

Problem 81: 77 &oard ctober 18$ Oive the follo"ig currets: la < 0 = *0 A l b < 6- 6- ' *$8 *$8 A lc < 6$8 = *- A Dolve for the Bero seuece compoet of l a. A. 10 = *$ &. 8 ' * %. 68 ' *$

(. 8. 8.-5$ = *1.0-$

-.--

(.

12 ' *

Problem 82: 77 &oard ctober 1 he seuece curret of phase a curret are as follo"s: )ero seuece curret < 1$.1-∠ 1$.1- ∠1.-$o Positive seuece curret < 2.8∠ 2.8 ∠10.0o egative seuece curret curret < 08.2∠ 08.2 ∠6-1o (etermie phase a curret. A. 20∠6-0o &. -0∠615.2o %. 10∠88o

(.

5∠15.2o

Problem 8-: 77 &oard ctober 18 A seuece compoet compoet of phase a curret are: )ero seuece compoet < 0.$ = *1.$ Positive seuece compoet < 18.$ cis >6-1.? egative seuece compoet compoet < -.2- cis 18.2 (etermie the phase b curret. A. 18 cis 215 &. 15 cis 2$0 %. 1 cis 220

(.

20 cis 225

Problem 8$:77 &oard Carch 18 he seuece compoets of phase a curret are: )ero seuece < 0.$ = *1.$ Positive seuece < 18.$ cis >6-1.? egative seuece < -.2- cis 18.2 (etermie phase c curret. A. 1.5 cis 1 &. 18 cis 215

%.

22.5 cis 82

(.

15 cis 100

Problem 85: 77 &oard April 12 (etermie the symmetrical compoets of the lie curret i lie \a if oe of the i phases impedace of its delta6 coected load coected across lies \ca is removed. he delta load "ith impedace of 10 ∠0o ohms per phase is supplied from from a 220 volts! 0 cycle! cycle! -6 phase. Assume Assume a phase seuece of a6b6c. A. la1 < 11 A! la2 < 11 A! la0 < 0 A &. la1 < .-- A! 1 a2 < .-- A! l a0 < .-- A %. la1 < 22 A! la2 < 22 A! la0 < 22 A (. la1 < 25.$ A! l a2 < 12. A! l a0 < 0 A Problem 8: 77 &oard April 11 A star6 coected balaced load ta3es 5 A from a balaced -6 phase! $6 "ire supply. supply. If the t"o supply lies of the fuses are removed determie the symmetrical compoets of the lie currets after the fuses are removed. A. l1 < 25 A! l2 < 25 A! l0 < 25 A &. l1 < 25 A! l2 < 50 A! l0 < 0 A %. l1 < 5 A! l2 < 5 A! l0 < 5 A (. l1 < 5 A! l2 < 0 A! l 0 < 0 A Problem 8: 77 &oard Carch 18 A 50 C,A! -- 3,11 3,! three6 phase! "ye6 delta coected trasformer has a -J impedace. +hat is the percet impedace at 100 100 C,A base ad -$.5 3, base# base# A. 1.-J &. 5.$J %. 5.$J (. .5J Problem 88: At a certai locatio of a electrical system! the available short circuit C,A is10 at 110 3, "hile its heveis heveis euivalet reactace is 0.05 pu. (etermie the per uit reactace uit of this poit usig a base 20 C,A ad 115 3,. A. 0.08 &. 0.0 %. 0.0 (. 0.0 Problem 8: A 1-.8 3,$$0 ,! 50 3,A sigle6 phase trasformer has a lea3age reactace of -00 ohms referred to the 1-.8 3, side. (etermie the per uit value of the lea3age reactace for the voltage base. A. 0.0$ &. 0.082 %. 0.08(. 0.0 Problem 880: 77 &oard April 180 A 5 3,A! 2$0061202$0 volt distributio trasformer "he give a short circuit test had $.2 volts applied "ith rated curret flo"ig i the short6 circuited "irig. +hat is the per uit impedace of the trasformer# A. 0.0-2 ohms &. 0.0-8 ohms %. 0.0$15 ohms (. 0.0$-5 ohms Problem 881: A -6 phase! -5 3,A! $80 ,! 50 /B! "ye6 coected alterator has a eual positive ad egative seuece reactace of 10J. Eid the symmetrical fault curret! if a -6 phase fault occurs at the alterator termials. A. $510.5 A &. $20.- A %. $255.2 A (. $8.1 A Problem 882: 77 &oard April 12 A geerator rated 00 3,A! 2$00 ,! 0 cycles! -6 phase! 6 poles ad "ye6 coected has 10J sychroous reactace. If a -6 phase fault occurs! "hat "ill be the short circuit curret#

A.

1$$- A

&.

15-2 A

%.

1$-5 A

(.

1$28 A

Problem 88-: A 20 C,A! 1-.8 3,! 0 /B! three6 phase sychroous turbo alterator has a positive ad egative Bero seuece reactaces of 0.25! 0.-5! 0.15 respectively. If lies b ad c are short6 circuited! determie the subtrasiet fault curret i lie b. Assume the alterator is operatig at o load ad at rated voltage "ith the eutral of the alterator solidly grouded. A. 21$ A &. 288 A %. 2$15 A (. 2510 A Problem 88$: 77 &oard Carch 18 At a certai locatio i a electric system! the available fault C,A is $00 C,A. A 15 C,A! -$.5 3,.2$ 3,! 2.5J impedace! "ye6 "ye grouded trasformer is istalled at that locatio. (etermie the short circuit C,A at the secodary side of the trasformer. trasformer. A. 15 C,A &. 150 C,A %. 0 C,A (. 2$0 C,A Problem 885: "o "o three6 phase trasformers are coected i parallel at the primary as "ell as at the secodary sides. e is rated 10 C,A! -$.51-.8 3, ad $J impedace "hile the other is rated .5 C,A! -$.51-.8 3, ad 5J impedace. he primary tappig poit has a -6 phase short circuit C,A of 1000 C,A. (etermie the fault curret delivered to a -6 phase fault at the secodary side bus bars. A. 1-!501 A &. 10!$52 A %. 12!5-2 A (. 11!5- A Problem 88: A 100 C,A! C,A! 22 3, sychroous turbo geerator has positive ad egative reactaces of 0.20 p.u. ad a Bero seuece reactace of 0.05 p.u. he eutral of the geerator is grouded through a reactor of 0.2$2 ohm. (etermie the ratio of the subtrasiet curret for a sigle lie to groud fault to the subtrasiet curret for a three6phase fault. Assume Assume the geerator is operated "ithout a load ad at rated voltage. A. 1.50 &. 1.25 %. 1.00 (. 0.80 Problem 88: 77 &oard ctober 12 A geerator is rated 100 C,A! 15 3,! it is L6 coected! solid grouded ad is operated at rated voltage at o6 load ad is discoected from the rest of the system. Its positive ad egative seuece reactaces are each 0.10 p.u. ad its Bero seuece reactace reactace is 0.05 p.u. %alculate i ohms of iductive reactace reactace to be iserted i the the eutral coectio of the geerator to limit the fault curret for a sigle lie to groud fault to the fault curret of a symmetrical three6 phase fault. A. 0.02-- Ω &. 0.0225 Ω %. 0.0- Ω (. 0.0-25 Ω Problem 888: 77 &oard ctober 182 he reactace of a three6 phase alterator is 8 percet. If the alterator is rated 25 C,A at 1-.25 3, output voltage! lie to phase! solve for the magitude of the fault curret geerated "he a short circuit occurs bet"ee 2 phases at the termials. termials. A. 0$ A &. 1-5 A %. 808 A (. $58 A Problem 88: A -6 phase! -5 3,A! $80 ,! 50 /B! "ye6 coected alterator has a eual positive ad egative seuece reactace of 10J. Eid the usymmetrical fault curret! if a 26 phase fault >lie6to6lie? occurs at the alterator termials. A. -80 A &. -0 A %. $510 A &. $180 A Problem 80: 77 &oard Carch 18 A 5 C,A! 1-.8 3,$80 ,! 5J impedace trasformer is tapped at 1-.8 3, lie "here the heveis euivalet impedace is 4 ohm. (etermie the fault curret at the primary for a three6 phase fault at the secodary. A. 10!500 A &. -!-00 A %. $!200 A (. !050 A

Problem 81: 77 &oard ctober 180 A 10 3,A! 110 ,! -6 phase! $6 "ire! 0 /B alterator geerates 0 volts per phase "he excited to give rated voltage at full load. he armature sychroous impedace per phase is 0.2 ohm. +hat is the curret i each phase if the geerator geerator termials are short6 circuited# circuited# A. -50 A &. 85 A %. 20 A (. 5-$ A Problem 82: 77 &oard ctober 18 At a certai poit of the et"or3 the positive! egative ad Bero seuece impedaces are 0.25 p.u.! 0.25 p.u. ad 0.- p.u. respectively. respectively. he base C,A C,A is 100. he voltage voltage level at that poit poit is -$.5 3,. 3,. (etermie (etermie the Bero seuece curret for a oe lie to groud fault. A. 2255 A &. 81-2 A %. 51 A (. 201 A Problem 8-: I a short circuit aalysis! the positive! egative ad Bero i mpedaces are 0.20 pu! 0.20 pu ad 0.25 pu! respectively! respectively! usig a base C,A of 50. 7stimate the fault curret o the faulted lies if the double lie to groud fault at the 1-8 3, level occurs. A. 120$ A &. 1011 A %. 100 A (. 110$ A Problem 8$: A 15 C,A! !00 ,! 0 /B! L6 coected sychroous alterator has a positive! egative ad Bero seuece per uit reactace of 0.20! 0.20! 0.20 ad 0.10 respectively. respectively. he eutral of the the geerator is grouded through through a reactor "ith a per uit reactace of 0.05 based o the geerator ratig. If a double lie to groud fault occurs at the alterator termials! estimate the curret that flo"s i the reactor. A. 520 A &. 52- A %. 125 A (. -$2 A Problem 85: 77 &oard Carch 18 I a short circuit study! the positive! egative ad Bero seuece impedace are 0.25 pu ad 0.25 pu ad 0. pu! respectively. respectively. he base C,A C,A is 100. (etermie the fault curret curret for a three6 phase fault at at the 115 3, level. A. -000 A &. 2000 A %. 2500 A (. -500 A Problem 8: A three6 phase 2206 ,! L6 coected alterator has a sychroous impedace of 0.15 = *2 ohms per phase. he alterator is coected through a short trasmissio lie "hose impedace per "ire is 2 = *1 ohms to a L6 L coected trasformer ba3 "hose total euivalet impedace referred to the secodary >high side? is 50 = *8. ohms. A three6 phase load is coected from the secodary side of the ba3 through a trasmissio lie "hose impedace is 50 = *5 ohms per "ire. If a symmetrical three6 phase fault occurs at the load side! ho" much is the curret "ill flo" to the alterator "idigs. Assume Assume the ba3 has a trasformatio ratio of 1:10. A. 22.- A &. 2-. A %. 21.5$ A (. 2$.15 A Problem 8: 77 &oard April 18$! 77 &oard April 18 hree sigle6 phase trasformers each rated 50 3,A! 2$00 volts primary! 2 volts secodary! four >$? percet impedace have their their primaries coected i delta ad secodaries i i "ye. %alculate the fault curret curret dra" if a short circuit occurs bet"ee t"o phases at the secodary termials. A. 220 A &. 2585 A %. 20- A (. 225 A Problem 88: 77 &oard ctober 18 ;a ode ña (istillery Ic.! located i %alubag! %alamba! ;agua istalled a emergecy -6 phase geerator! -6 "ire OC geerator is rated -50 3,A! 3,A! $0 volts! 0 /B "ith a reactace of 8J. Dolve for the symmetrical -6 phase short circuit. A. 5!-21 A &. 5!0- A %. 5!28$ A (. 5!$1 A Problem 8: 77 &oard ctober 1 A 15 C,A! C,A! -$.5 3,.2$ 3, trasformer is coected at a ifiite bus. he percet impedace of the trasformer is 2.5J. +hat is the curret at the -$.5 3, side for a three phase short at the .2$ 3, side# A. 55!500 A &. 10!000 A %. 5!000 A (. 25!000 A

Problem 00: 77 &oard April 185 A three6 phase! -6 "ire geerator is rated -25 3,A! 3,A! $80 volts! 0 /B "ith a reactace of eight >8? percet. Dolve for the fault curret delivered y the geerator durig a 26 phase short circuit at the termials. A. $2-2 A &. $$2 A %. $012 A (. $15 A Problem 01: he secodary side of 5 C,A! -$.51-.8 3,! -6 phase trasformer is coected to a bus bar served by a 20 C,A! 1-.8 3,! -6 phase alterator. he primary side of the trasformer is tapped from a ifiite bus. 7stimate the fault curret delivered to a -6 phase fault ear the commo bus bar. he trasformer ad alterator have $J ad .5J reactace! respectively. respectively. A. 12!0 A &. 18!00 A %. 15!-$0 A (. 1!$00 A Problem 02: (etermie the miimum rupturig capacity of the circuit brea3er as sho"! if a three6 phase symmetrical fault occurs at poit P. A. -0 C,A &. 20 C,A %. 10 C,A (. 15 C,A Problem 0-: Oive the oe lie diagram as sho"! determie the curret delivered by geerator A "he a three6 phase fault occurs at poit P. A. 12.5 A &. 1.1 A %. 1$5.5 A (. 18.$ A Problem 0$: Oive the oe lie diagram as sho"! determie the reactace of the limitig reactor x i order to limit the three6 phase symmetrical fault po"er at poit P to $00 C,A. A. &. %.

0.5  0.52  0.$ 

&.

0

TEST 2$ # (LL4%(N*T(N Problem 05: 77 &oard April 12 A piece of paper lies o the table 2m a"ay from a poit directly belo" a bulb of 100 cd ad is $ m above the table. %alculate the illumiatio o the ceter of the paper i lux. A. 5.2 &. . %. $.5 (. -.$ Problem 0: A ushaded lamp is  m above a table. It is lo"ered do" by 2 m. &y ho" much is the illumiatio o the table icreased# A. 2.0 times &. 2.25 times %. 2.50 times (. oe of these Problem 0: 77 &oard April 18 A medium ushaded lamp hags 8 m directly above the table. o "hat distace should it be lo"ered to icrease the illumiatio to $.$5 times to its former value# A. $.02 m &. -.8 m %. -. m (. $.21 m Problem 08: A ushaded lamp is placed 5 cm from the scree of a photometer. If a glass is placed bet"ee the scree ad the lamp! the lamp must be moved 5 cm closer to the scree to produce the same illumiatio as before. he glass bloc3s "hat percet of the light produced by the lamp#

A.

11.5J

&.

10.2J

%.

12.$J

(.

.08J

Problem 0: 77 &oard April 1$ A floodlight floodlight emittig 25!000 cadelas i the ceter of its beam is aimed at 0 o to a poit o the groud 20 meters a"ay. he illumiatio of the poit i lux is A. 5$.12 &. 2.5 %. -1.25 (. 25 Problem 10: A u3o" lamp placed  m from a photometer scree provides the same illumiatio as a 06cd lamp placed $ m from the scree. +hat +hat is the cadlepo"er cadlepo"er of the u3o" lamp# lamp# A. 202.5 cd &. 215.2 cd %. 211. cd (. 208.$ cd Problem 11: "o "o lamps A ad & havig itesities of -00 cp ad 500 cp respectively are situated 12 ft apart. A scree is placed bet"ee them i order order that the illumiatios illumiatios o ay sides are eual. eual. /o" much is this illumiatio# illumiatio# A. 11.$5 fc &. 10.2 fc %. 12.$5 fc (. 1-.15 fc Problem 12: 77 &oard April 185 A light is to be placed o a "all i order to obtai maximum brightess of illumiatio. o o achieve this brightess! ho" high high o the "all should should this light bulb bulb be placed to a poit o the floor that is -. m from the the "all# A. 2.5$ m &. 2.25 m %. 2.08 m (. 2.8 m Problem 1-: "o "o lamps x ad y are haged at a distace of 20 ft bet"ee their ceters ad 10 ft above the "or3ig surface. ;amp x gives 1$0 cp "hile lamp y gives 2-0 cp. At "hat distace from a poit directly belo" lamp A o the lie *oiig their ceters "ill the t"o lamps give eual a mout of illumiatio# A. 8.22 ft &. 8.-$ ft %. 8.05 ft (. 8.2 ft Problem 1$: A 1006 cd lamp! "hich emits light uiformly i all directios! is suspeded 2.5 m above the ceter of the "or3ig table! "hich is - m suare. %alculate the illumiatio at each corer of the table. A. 2.5 lux &. 2.- lux %. 2$.$ lux (. 28.-8 lux Problem 15: 77 &oard ctober 11 A lamp lamp of 500 cd is placed at the ceter of a room! 20 m x 10 m x 5 m. calculate the illumiatio illumiatio i each corer of the floor. A. 0.8- lux &. 0.81 lux %. 0.8 lux (. 0.8 lux Problem 1: Eour lamps are suspeded  m above the groud at the corer of the la" $ m o each side. If each lamp emits 250 cd! calculate the illumiatio at the ceter of the la". A. 20.5 lux &. 22.-$ lux %. 21.2 lux (. 1.85 lux Problem 1: 77 &oard ctober 11 A lamp lamp of 500 cd po"er is placed at the ceter of a room! 20 m x 10m x 5 m. %alculate the illumiatio i a poit i the middle middle of a 10 m "all at a height of of 2 m from the floor. A. $.5 lux &. $.8 lux %. $. lux (. $.88 lux

Problem 18: 77 &oard ctober 12 ;umiaries of 2000 lumes output "ill be istalled i a room. he desired illumiatio is 200 foot6cadle. he coefficiet of utiliBatio is 0.$5 ad maiteace factor is 0J. (etermie the umber of lumiaries eeded if the siBe of the room is 120 m 2. A. -0 &. 28 %. -2 (. - Problem 1:77 &oard ctober 1A hall -5 x 20 m is to be provided "ith a geeral illumiatio of 1-0 flux. If the depreciatio factor is 1.$ ad the coefficiet of utiliBatio is 0.50! determie the umber of fluorescet lamp reuired. Assume Assume lumious efficiecy of the fluorescet tube as 50 lumes per "att for 806"att lamp. A. 2 lamps &. 5 lamps %. - lamps (. $ lamps Problem 20:77 &oard April 10 A hall -0 m log ad 12 m "ide is to illumiated ad the illumiatio reuired is 50 meter6cadles. %alculate the umber of lamps reuired ta3ig a depreciatio factor of 1.- ad utiliBatio coefficiet of 0.5 give that the output of the lamp to be used is rated 100 +! 115 lumes. A. -0 &. 2 %. 28 (. 2 Problem 21:77 &oard April 1A room 20 m x 10 is to be illumiated by eight lamps ad the average illumiatio is to be 50 lumes per suare meter. If the utiliBatio factor is 0.$8 ad depreciatio factor 1.2! calculate the mea spherical po"er per lamp. A. 2$ cp &. 255 cp %. 2-5 cp (. 22$ cp Problem 22:77 &oard ctober 11 It is reuired to provide a illumiatio of 150 meter6cadle i a factory hall $5 m x 15 m. Assume that the depreciatio factor is 0.80! coefficiet of utiliBatio is 0.$0 ad efficiecy of the lamp is 20 lumes per "att. %alculate the umber of 250 + lamps. A. 0 &. 2 %. (. 1 Problem 2-:77 &oard Carch 18 he illumiatio of a room is beig desiged. he data as follo"s: @oom legth < -0 ft @oom "idth < 20 ft %oefficie %oefficiett of utili utiliBatio Batio  < 0. 0. Caiteac Caiteacee factor factor < 0J @euired foot6cadle < 50 ;umes per lamp < --00 +atts per lamp < 5,oltage < 220 Po"er factor < 5J +hat is the total curret curret of the lamp reuired to satisfy satisfy the reuiremet# A. 21 A &. 15 A %.  A (. 25 A Problem 2$:77 &oard ctober 18 he floor area of a room is 10 meters by 10 meters. he desired illumiatio level is 150 lux. Assume Assume the coefficiet of utiliBatio is 0.51! the maiteace factor is 0.5 ad i the mar3et a fluorescet lumiaire provide 1200 lumes output of light. (etermie the umber of lumiaires eeded for the room. A. 2$ lamps &. 28 lamps %. 25 lamps (. 2 lamps

TEST 2' – ELECT0(C*L P6E0 P6E0 PL*NT Problem 25:77 &oard Carch 18 A deep "ell pumps "ater ito a ope reservoir at the rate of 100 gpm. he "ater table is 200 ft belo" the output pipe. he system has a overall efficiecy of 0J. +hat shall be the miimum hp of the pump# A. 5.- hp &. 8. hp %. 12.$ hp (. .2 hp Problem 2:77 &oard Carch 18

A pump reuires 100 hp to move "ater "ith a specific gravity of 1.0 at a certai flo" rate to a give elevatio. +hat horsepo"er does the pump reuire if the flo" rate ad elevatio coditios are the same but the fluid pumped has a specific gravity of 0.8# A. 80 hp &. 100 hp %. 0 hp (. 120 hp Problem 2:77 &oard April 1A electric motor is used to pump "ater from a "ell $0 ft deep. he amout of "ater pumped per hour to the groud level is 2-!000 gallos. +hat is the hp of the motor assumig there is o po"er loss# A. $- hp &. $2 hp %. $1 hp (. $0 hp Problem 28: A pump! drive by $$0 ,! -6phase iductio iductio motor lifts 1!100 cubic feet of "ater per miute agaist a total head of 100 ft of "ater. he he efficiecy of the pump ad that of its motor are 0.5 ad 0.2 respectively. respectively. he po"er factor of the motor is 0.. %alculate the cost of operatig the pump f or 2$ hours a day duty "he po"er costs per P 0.-0 per 3+6hr. A. P 1!2&. P 2! 0-5 %. P 1!1 (. P 1! 05 Problem 2:77 &oard ctober 18-! 77 &oard &oard ctober 10 A atural "aterfalls! 50 meters high! cosistetly discharges discharges 1.- m - per secod. A mii hydroelectric plat is to be costructed at the bottom of the "aterfalls. %alculate the electric geerator 3+ ratig! assumig 0J mechaical to electrical coversio efficiecy ad "ater turbie desig efficiecy of 0J. A. $01.2 &. $10.55 %. $20.$2 (. $25.10 Problem -0:77 &oard Carch 18 A po"er plat gets "ater from a dam from a height of 122.$5 m at the rate of 1000 cubic meters per miute. If the output of the plat is 15!000 3+! "hat is the plat efficiecy# A. 80J &. 0J %. 5J (. 5J Problem -1:77 &oard April 15 %alculate the curret ta3e by a motor at $$0 , havig a efficiecy of 0J. Coreover the motor is used to drive a pump! "hich lifts 2$0!000 3g of "ater per hour to a height of -0 meters. he pump is give to be 80J efficiet. A. 58 A &. 5 A %. 8- A (. 2 A Problem -2:77 &oard April 1$ A three6phase three6phase motor drive pump at 1!800 rpm is dischargig 500 gpm of "ater at a head of 25 meters. he pump efficiecy is is 5J. (etermie the hp of the the motor. A. 15 hp &. 12 hp %. 1$ hp (. 10 hp Problem --:77 &oard April 1A hydroelectric hydroelectric statio has a turbie efficiecy of 80J ad a geerator efficiecy of 85J. (etermie the volume of "ater reuired "he deliverig a load of $0 C+ for  hours if its effective head is 120 m. +ater "eighs 1200 3gm -. A. 8!$2$ m- &. $5!-52 m- %. 8$5!0$1 m(. 05!$12 mProblem -$: A hydroelectric hydroelectric po"er plat is to be costructed. he place has a aual raifall of 10 cm. he available catchmet area is 250 s. 3m "ith a available head of -0 meters.  the average! oly 85J of the raifall ca be collected i ad 0J of the impouded "ater is available for po"er. Assumig Assumig turbie ad geerator efficiecies of 80J ad 2J respectively! respectively! "hat "ill be the average po"er geerated. A. 2!-.5 2!-.5 3+ &. 1!-. 1!-. 3+ %. 1!$1 1!$1.2 .2 3+ (. 2!0- 3+ Problem -5:77 &oard Sue 10! 77 &oard April 11

A 100 C+ hydroelectric statio is supplyig full load for 10 hrs. i a day. %alculate the volume of "ater! "hich has bee used. Assume effect effective head of the "ater as 200 m ad the overall efficiecy of the statio as 80J. Assume 1 m - of "ater < 1000 3g. A. 2!2-2!$-5 m&. 2!2-!50 m%. 2!-$5!$5 m(. 2!-1!$02 m Problem -: A farmer has a small stream o his property! property! "hich he thi3s might supply eough po"er to light his farm load. he stream "as foud to be flo"ig at 0.05 m -sec. he available fall is 5 m. a small hydraulic turbie ad geerator "ill have a combied efficiecy of 5J. Assumig a trasmissio lie efficiecy of 82J! ho" may 506 + lamps ca be served by the geerator simultaeously# simultaeously# A. 28 28 lamps &. -2 -2 lamps %. - - lamps (. -0 -0 lamps Problem -:77 &oard April 1$ A "aterfall is 0 meters high. It discharges at a costat rate of 1.0 cubic meter per secod. A mii6hydro mii6hydro electric plat is to be costructed belo" the "aterfalls. he turbie efficiecy is 80J ad the geerator efficiecy is 5J. %alculate the 3+ output of the geerator. A. $$.-$ 3+ &. $$.22 3+ %. $5.2 3+ (. $1.-0 3+ Problem -8:77 &oard April 18 Caria %ristia falls located at south has a height of 100 meters! it discharges at costat rate 2.5 m -sec. atioal Po"er corporatio corporatio has costruct costruct a hydroelectric plat belo" belo" the "ater falls. If the turbie turbie efficiecy is 0J ad geerator efficiecy is 0J! calculate the aual eergy i 3+6hr! if the discharge rate is costat throughout the year. Assume specific gravity costat. A. 1-!5-$!85 &. 1-!$!5$ %. 1-!55!5$ (. 1-!50$!2-1 Problem -:77 &oard ctober 18-! 77 &oard ctober 10 A atural "aterfall! 50 meters high! cosistetly discharges discharges 1.- m - per secod. A mii hydroelectric plat is to be costructed at the bottom of the "aterfalls. %alculate the aual eergy i 3+6hr 3+6hr that ca be geerated by the po"er plat. Assume Assume the turbie ad ad geerator efficiecies efficiecies eual to 0J ad 0J! 0J! respectively. respectively. A. -!51!0 &. -!2!-5 %. -!12!212 (. -!-$5!815 Problem $0: A "ater turbie operatig at 85J efficiecy is uder a head of 100 m. If the costat discharge is 1000 galsec! evaluate the operatig speed of the turbie. A. $0 rpm &. 5$ rpm %. - rpm (. 08 rpm Problem $1: A mii6 hydroelectric hydroelectric po"er plat is uder a head of 25 feet. he turbie efficiecy is 0.85 ad the geerator efficiecy is 0.80. If the turbie speed is limited to $50 rpm! calculate the geerator 3+ output. A. -$02 &. -$8 %. $02$ (. -02 Problem $2: 77 &oard ctober 1 A po"er plat gets "ater from a dam from a height of 100 meters at the rate of 1000 cubic meters per miute. If the overall efficiecy of the plat is 5J! "hat is the 3+ output of the plat# A. !550 3+ &. 15!250 3+ %. 12!250 3+ (. 5!250 3+ Problem $-: 77 &oard ctober 1 A po"er plat cosumes cosumes 100!000 pouds of coal per hour. he heatig value of the coal is 12!000 &9 per poud. he overall overall plat efficiecy efficiecy is -0J. +hat is the the 3+ output of the the plat# A. 105!500 3+ &. 205!000 3+

%.

1$2!500 3+

(.

15!000 3+

Problem $$: A po"er plat has a fuel cosumptio of 1 poud per 3+6hr geerated. (etermie the heatig value of the fuel used i &9 per poud. he overall efficiecy of the plat is -J. A. 10!000 &. !500 %. !800 (. 10!050 Problem $5: 77 &oard ctober 18 A certai coal6 fired po"er plat has a heat rate of 2.88 x 10  calories per 3+6 hr. %oal costs P 2!500 per to. /o" much is the fuel cost compoet of producig 1 3+6hr# Assume Assume the heatig value of coal used eual to 1-!000 &9lb. A. P 2.50 &. P 1.5 %. P 1.00 (. P 1.25 Problem $: A 100 C+ thermal plat has a overall efficiecy of -$J. If the coal used has a heatig value of 10!800 &9 per poud! calculate the coal cosumptio of the plat per 3+6hr output. A. 0.8$ lb &. 0. lb %. 0.8 lb (. 0.- lb Problem $: 77 &oard ctober 1 A coal po"er plat has a overall plat efficiecy of 28J. %oal "ith a heatig value of 12!000 &9 per poud cost P 1.50 per poud. +hat +hat is the fuel cost pf producig producig oe 3+6hr# 3+6hr# A. P 1.0 &. P 1.52 %. P 1.5 (. P 1.2 Problem $8: A po"er plat has a overall efficiecy of -0J. If this plat ca cosumed $200 3ilograms of coal per hour! estimate the total electric eergy produced i oe day. Assume Assume the calorific value of the coal beig used is 8000 3cal per 3ilogram. A. 18- C+6hr &. 18 C+6hr %. 155 C+6hr ( 1 C+6hr Problem $: 77 &oard April 1$ A plat has a total operatig capacity of 800 3+. he coal cosumptio is 100 lbs per hour. he heatig value of the coal is 500 &9 per lb. +hat approximate percet of the heat i the coal is coverted ito useful eergy# A. 12.&. 15.0 %. 8. (. 1.5 Problem 50: 77 &oard ctober 1 A 100 C+ coal6 fired po"er plat has a average heat rate of 500 &9 per 3+6hr. 3+6hr. the plat load factor is 5J. he heatig value of the coal is 12!000 &9 per poud. %alculate the amout of coal usage for oe day. A. 1.$25 x 10 po pouds &. 2.2-5 x 10  pouds 5 %. 2.2-5 x 10 po pouds (. 1.82 x 10  pouds Problem 51: 77 &oard ctober 18 A diesel geerator set burs diesel "ith a heatig value of 18!000 &9 per poud. he diesel egie has a efficiecy of -0J ad the alterator has a efficiecy of 5J. (etermie the fuel cost compoet of producig oe 3+6hr if diesel costs P 2.80 per poud. A. P 0.15 &. P -.28 %. P 2.15 (. P 1.8 Problem 52: A diesel electric geeratig uit supplies a load at 0 3+. he heatig value of the oil used is 12!000 3cal per 3g. If the overall efficiecy efficiecy of the uit is $0J $0J determie the mass of oil reuired per hour. hour. A. 11.8 3g &. &. 10.- 3g %. %. 12.5 3g (. (. . 3g Problem 5-: 77 &oard April 1-

A 500 3+ electric lightig cost P 1000 per 3+ istalled. Eixed charges are 1$J! operatig cost is P 1.50 per 3+6hr. 3+6hr. he he plat averages 150 150 3+ for 5000 hours hours of the year! $20 3+ for 1000 hrs ad 20 3+ for the remaider. +hat +hat is the uit cost productio of electric eergy# A. P 1. &. P 1.8$ %. 2.15 (. 2.05 Problem 5$:77 &oard ctober 18 A coil fired Po"er Plat "ith -000 3+ rated capacity costs P 15!000 per 3+ istalled. Aual Aual operatig cost p 12 millioT aual maiteace maiteace cost P 8 millioT aual depreciatio depreciatio P 15 millioT iterest iterest o ivestmet per year 8J ad cost of coal P 800 per to. If oe poud of coal is eeded to geerate 1 3+6hr! fid the total aual cost to operate the plat. Plat capacity factor is 50J. A. P $.$5 millio &. P $-.-8 millio %. P $0.28 millio (. P $2.-2 millio

Problem 55:77 &oard ctober 18 o produce produce oe 3+6hr! a po"er plat burs 0. lb of coal "ith a heatig value of 1-!000 &9. +hat is the heat rate of the plat# A. ! !250 &93"6hr &. ! !550 &93"6hr %. 11!00 &93"6hr (. 8!00 &93+6hr Problem 5: 77 &oard Carch 18 he curret loads of four circuits are as follo"s: %ircuit o.1 < 25 amperes %ircuit o. 2 < -8 amperes %ircuit o. - < 18 amperes %ircuit o. $ < $5 amperes If the diversity factor is 1.5! "hat is the miimum ampacity of the feeder coductor# A. 12 A &. 18 A %. 8$ A (. 152 A Problem 5: 77 &oard ctober 18 he po"er costumer has four feeder circuits of 220 ,! ,! three6 phase. he circuits have maximum demad as follo"s: %ircuit o. 1 < -5 A %ircuit o. - < 2 A %ircuit o. 2 < $ A %ircuit o. $ < 5 A he diversity factor is 1.5! determie the maximum 3,A demad "ith the costumer. A. $ &. 2.5 %. 80 (. 5Problem 58: 77 &oard April 180 A substatio substatio trasformer is to servig the follo"ig loads: %lassificatio otal load (emad factor ;ightig -00 3+ 0J Po"er 1200 3+ 80J /eatig 500 3+ 0J If the diversity factor amog the load types is 1.5! determie the maximum demad o the trasformer. A. 2000 3+ &. 100 3+ %. 1225 3+ (. 1180 3+ Problem 5: 77 &oard ctober 18 he po"er costumer has four circuits of 220 ,! three6 phase. he circuits havig maximum demad as follo"s: %ircuit o. 1 < -5 A %ircuit o. 2 < $ A %ircuit o. - < 2 A %ircuit o. $ < 5 A

he diversity factor is 1.5 A! the load factor is 50J ad the po"er factor is 80J. +hat is the approximate 3+ demad of the costumer# A. 15 3+ &. $- 3+ %. 21 3+ (. 5- 3+ Problem 0: A idustrial plat has a aggregate load of 100 3+. he demad factor is 0J. If the average total eergy cosumptio of the plat i oe year is 18 C+6hr! C+6hr! "hat I the yearly load factor of the plat# A. -2.$5J &. $2.25J %. -8.J (. -5.-J Problem 1: Po"er plat has t"o 5 C+ geerator sets each operatig at 200 hours per year ad a 50 C+ geerator set operatig at -000 hours per year. If the plat output is 800 x 10  3+6hr per year! solve for the plat use factor. A. -.$J &. 5.0J %. 8.-J (. 0.J Problem 2: he average "ee3ly eergy geerated by a thermal plat is !000!000 3+6hr. 3+6hr. he pea3 load of the plat is 5!000 3+ durig the "ee3. If the plat has a istalled capacity of 100 C+! solve for the load factor of the plat. A. $8.2J &. 5.-$J %. 50.$$J (. 55.55J Problem -: 77 &oard April 10 A certai po"er plat has reserve capacity above the pea3 load of 10 C+. he he aual factors are load factor < 5J! capacity factor < $1J ad use factor < $J. (etermie the istalled capacity. capacity. A. --.$5 C+ &. -1.85 C+ %. -2. C+ (. -0.$ C+ Problem $: 77 &oard April 10 A certai po"er plat has reserve capacity above the pea3 load of 10 C+. he he aual factor are load factor < 5J! capacity factor < $1J ad use factor < $J. (etermie hours per year ot i service. A. 52 hours &. 5 hours %. 2$ hours (. - hours Problem 5: 77 &oard April 1A cetral geeratig statio has a aual factor report as follo"s: f ollo"s: Plat capacity factor < 50J ;oad factor < 0J 9se factor < $5J @eserved < 10!500 10!500 3+ (etermie the rated capacity of the statio. A. -!000 3+ &. 0!000 3+ %. 5!000 3+ (. 0!000 3+ Problem : 77 &oard ctober 10 A 55!000 3+ thermal plat of atioal Po"er %orporatio delivers a aual output of 28-!000!000 3+6hr "ith a pea3 load of $$ C+. +hat +hat is the capacity factor# A. $.$J &. $2.$J %. $8.2J (. $$.J

Problem : A thermal po"er plat has t"o geeratig sets rated 00 3+ ad 800 3+ respectively beig operated at rated capacity. capacity. he total coal cosumptio is 80 3g per hour. If the coal has a heatig value of 20!000 *oules per gram! "hat fractio of heat produced from the coal is coverted ito useful electrical eergy# A. -.0J &. $0.20J

%.

--.J

(.

-8.28J

Problem 8: 77 &oard ctober 18 he meterig of a po"er customer "as read ad gave the follo"ig data: 3+6hr < 200!000 3,A@6hr < 180!000 Caximu Caximum m (ema (emad d < -80 3+ &illi &illig g (ay (ayss < -0 -0 +hat is the load factor# A. 8J &. -J %. 80J (.

2J

%(SCELL*NE4S TP(CS – TEST 2) Problem : 77 &oard ctober 18! 77 &oard ctober 1A electric truc3 is propelled by a (% motor drive by a 1106 , storage storage battery. he truc3 is reuired to exert a tractive effort of 220 lb at a speed of 56mph. he overall efficiecy of the motor ad drive is 0 percet. +hat curret is ta3e from the battery# A. 25.2$ A &. 28.$5 A %. -0.22 A (. 2.18 A Problem 0: 77 &oard ctober 10 A electric motor drive coveyor belt 80 feet log is reuired to ta3e 2!200 lbs of dirt! roc3s! etc per miute from a excavatio site ad raise it 15 ft vertically. If all frictio i the machiery is eglected! "hat hp of motor is reuired# A. 1.5 &. 1.0 %. 2.0 (. 2.5 Problem 1: 77 &oard Sue 10 A electric motor is eeded to drive a escalator i a departmet store. he escalator is - i "ide ad ca accommodate t"o >2? riders side by side hece the loadig ca be cosidered as uiformly distributed "ith 250 lbft. ;iear speed is 80 ftmi. the escalator agle is -8 o ad its liftig height is .5 meters. Cechaical efficiecy of the system is 0J. (etermie the motor siBe i hp. A. 21.5&. 22.01 %. 1.$2 (. 20.8$ Problem 2: 77 &oard Sue 10 A electric lift is reuired to raise a load of 5 toe through a height of -0 m. oe uarter of the electrical eergy supplied to the lift is lost i the motor ad gearig. If the time reuired raisig the load is 2 miutes fid the curret ta3e by the motor! the supply voltage beig 2-0 , (%. A. $.8- A &. -.$2 A %. 5.2 A (. .05 A Problem -: 77 &oard ctober 12! 77 &oard April 1$ A coveyor coveyor belt of legth 156m is at -5 o "ith respect to the groud or horiBotal. It ca hold $!000 3g of ore over its total legth ad it travels the distace of its legth i -0 secods. %alculate the po"er ratig of the motor i 3+ that drives the coveyor belt. A. 10.50 3+ &. 11.25 3+ %. 1-.58 3+ (. 12.5 3+ Problem $: 77 &oard ctober 18 +hat is the "or3 i ergs eeded to raise a 106gram "eight 1006m up# A. $. x 105 &. 8 x 10 %. .8 x 10 (. 1. x 10 Problem 5: 77 &oard ctober 10 A electric lift ma3es120 double *oureys i a day ad a load of  toes is raised to a height of 100 m i oe ad a half miutes. I the retur *ourey! the cage of the lift is empty ad it completes the *ourey i 0

secods. he "eight of the cage is 00 3g ad the couter"eight is - toes. %alculate the horsepo"er of the motor. Assume the efficiecy of the lift to be 80J. A. $.2$ hp &. 5.5 hp %. 58.0 hp (. 0.$- hp Problem : 77 &oard ctober 18 he po"er reuired by a plae flyig at 5$5 3mhr is 1800 3+. (etermie (etermie the average resistig force i e"tos. A. 11!80 &. 12!$20 %. 11!520 (. 12!0 Problem : 77 &oard ctober 10 A trai "eighig 100 tos is pulled up a 2J grade by 006hp. rai resistace is 1200 lbs. +hat is the speed mph# A. 22 &. 25 %. -2 (. 28 Problem 8:77 &oard April 1A (% series motor operates a crae at ormal voltage ad lifts a load of 26 toe at 50 mmi. (etermie the speed at "hich it "ill lift a load of 56toe. A. -0 mmi &. 20 mmi %. 25 mmi (. 28 mmi Problem : 77 &oard April 1$ /o" log could a 10006hp motor be operated i the heat eergy liberated by 1 cubic 3ilometer of ocea "ater "he the temperature of the "ater is lo"ered by 1 o % if all its heat "ere coverted to mechaical eergy# A. 10 yrs &. 1 yrs %. 182 yrs (. 188 yrs Problem 80: 77 &oard ctober 11 %alculate the miimum rms /P reuired of a ope type electric motor "hich operates at 15 /P load for 8 miutes! 10 /P load for  miutes! 20 /P for  miutes ad the motor is at rest for 12 miutes after "hich the cycle is repeated. A. 15.00 &. 12.25 %. 10.-2 (. 1.5 Problem 81: 77 &oard ctober 1A trasformer trasformer is desiged to carry -0 times its rated curret for 1 secod. he legth of time that a curret of 20 times the rated curret ca be allo"ed to flo" is A. 0. sec &. 2 sec %. 1.5 sec (. 2.25 sec Problem 82: 77 &oard ctober 1 A three6 phase motor dra"s 50 A of curret at full load. +hat shall be the maximum iverse6time circuit brea3er that may be used for short short circuit protectio# protectio# A. 50 A &. 100 A %. 125 A (. 5 A Problem 8-: 77 &oard ctober 10 A sychroous codeser costs P 1000 per 3,A. If a cosumer is charged at P 25 per aum per 3,A of maximum demad ad the iterest ad depreciatio charges are 10J o the capital ivested! determie his most ecoomical operatig po"er factor. he po"er factor of the cosumer load is 0.80 laggig. Assume Assume o losses i the sychroous codeser. codeser. A. 0.8 &. 0.88 %. 0.8 (. 0.8 Problem 8$: 77 &oard April 182

I a group of - motors! oe dra"s 15 A! the other $0 A ad the third 52 A. +hat siBe of coductor i terms of ampacity must be used for the feeder circuit# A. 120 A &. 10 A %. 1-$ A (. 10 A Problem 85: 77 &oard ctober 12 Oive the follo"ig three6 phase iductio motors operatig at full load. 156hp! 220 ,! 0.5 eff.! 0.80 pf 256hp! 220 ,! 0.0 eff.! 0.0 pf 20 hp! 220 ,! 0.2 eff.! 0.8 pf (etermie the ampacity of the feeder circuit coductors supplyig these motors. A. 18.$ A &. 18-.- A %. 1$.0 A (. 205.0 A Problem 8: 77 &oard April 182 A sigle6 phase motor has rated full6 load curret of 28 amperes. /o" large must be the ampacity of the fuses to be used# A. -5 A &. $2 A %. 0 A (. 8$ A Problem 8: 77 &oard ctober 18 +hat surge arresters omial ratig shall you recommed to protect distributio trasformer i a -$.5 3, grouded system# A. -$.5 3, &. -0 3, %. 22 3, (. 2 3, Problem 88: 77 &oard April 11 A iductio motor is euipped "ith a fly"heel of momet of iertia of $ 3g6m 2 for drivig a rollig mill. If the speed of the iductio motor is 1000 rpm! "hat extra eergy the fly"heel is capable of supplyig "he the load comes to the motor# A. 22 3S &. 20 3S %. 28 3S (. -2 3S Problem 8: 77 &oard ctober 12 A three6 phase! 220 ,! 0 /B iductio motor drives a fa! "hich provides $0 m - sec air reuiremets to a furace. Assume Assume fa efficiecy of 0J ad 150 mm / 2 gauge pressure. (etermie the lie curret dra" by the motor if efficiecy ad po"er factor is 80J ad 0J respectively. respectively. A.

$0 A

&.

$$ A

%.

$82 A

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$$2 A

Problem 0:77 &oard ctober 18 A te6 year ivestigatio of a rivers potetial gave a average "ater flo" of 25 cm per secod at the bottom ad 0 cm per secod secod at the surface. he average average cross6 sectio at the same same locatio is 80 s. s. m. "hat is the average flo" rate i cubic meter per hour# A. 15!00&. $ %. 5!500 (. 2!0 Problem 1: 77 &oard April 1-! 77 &oard ctober 1$ A 50 3+ electric lightig costs costs P 1000 per 3+ istalled. Eixed charges charges are 1$J! operatig cost is P 1.50 per 3+6hr. 3+6hr. he he plat averages 150 150 3+ for 5000 hours hours of the year! $20 3+ for 1000 hours ad 20 3+ for the remaider. +hat +hat is the uit cost of producig electric eergy# A. P 2.10 &. P 1. %. P 2.01 (. P 1.82 Problem 2: 77 &oard ctober 10 If the po"er gai i a circuit is 20 decibels. +hat is the actual ratio of the output to iput po"er# A. 200 &. 100 %. 20 (. 10 Problem -: 77 &oard ctober 10

If 20 decibels "ere po"er losses! "hat is the po"er output to iput ratio# A. 0.05 &. 0.01 %. 10 (.

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Problem $: 77 &oard ctober 12 +hat is the output of a 26d& gai amplifier "hose iput po"er is 25R+# A. 12.5- m+ &. 11.2$ m+ %. 15.2 m+ (. 1$.05 m+ Problem 5: 77 &oard April 11 It is desired to deposit a thic3ess of 1 mm of ic3el o the surface of a "or steel pi 10 iches log ad $ iches i diameter. he available curret is 50 A. the electromechaical euivalet of ic3el is 0.000-8 gmcoulomb ad its relative desity is 8.. %alculate the time reuired. A. 11.-2 hrs &. 12.5 hrs %. 10.08 hrs (. .55 hrs Problem : 77 &oard ctober 18 he use of electricity for meltig a so" i a drive"ay 10 ft "ide by -0 ft log is beig cosidered. At At a 0.20 cets a 3ilo"att6hour! "hat "ould be the cost of meltig  iches of so"# Assume Assume the follo"ig data: specific "eight of so"! 10 lb per cu. Et! efficiecy of operatio! 50J. A. P 20.25 &. P 25.-5 %. P 28.$0 (. P -0.Problem : 77 &oard April 10 A car is beig self started agaist a resistig torue of 06m ad at each start the egie is cra3ed at 5 rpm for 8 secods. Eor each start the eergy is dra" from a lead acid accumulator. If the fully charged battery has a capacity of 100 "h per charge! calculate the umber of starts that ca be made "ith such a battery. Assume Assume a overall efficiecy of the motor ad gears as 25J. A. 2 &. 25 %. 20 (. 2Problem 8: 77 &oard ctober 10 I a electric heater! the ilet temperature is 15 o %. +ater is flo"ig at the rate of -00 grams per miute. he voltmeter measurig voltage across the heatig elemet reads 120 volts ad a ammeter measurig curret ta3e reads 10 A. +he steady state is fially reached! "hat is the fial readig of the outlet thermometer# A. 2. o% &. 81.2 o% %. 5.- o% (. 8.8 o% Problem : 77 &oard April 12 A 2-06 volt is supplied to a electric solderig machie "hich dra"s a curret of 0.8 A! fifty6five percet >55J? of the heat geerated is lost i heatig other metal parts of the machie ad i radiatio. he copper6 solderig tip "eighs 150 grams. Celtig poit of solder ' -00 o% Ambiet temperature ' 15 o% Dpecific heat of copper ' 0.0$ calg6 o% (esity of copper ' 8.0 gcc (etermie the time elapsed after s"itchig o the machie before the solderig tip is heated to the meltig poit of solder ad the the machie ready for use. A. -.05 mi &. 2.2 mi %. -.- mi (. $.2$ mi Problem 1000: 77 &oard April 10 A room measures measures  m x  m x $ m. the air i it has to be r ee"ed t"ice a hour ad is to be maitaied at o least 15 % above the icomig air temperature. Eid the capacity of the heater assumig a loss of 10J of the heat. Assume air desity as 12.8 x 10 6$ grams per cubic cm ad specific heat of air as 0.2$ calg6 o%. A. 10 + &. 15$5 + %. 1-$ + (. 121 +

Problem 1001: 77 &oard April 10! 77 &oard ctober 1A foudatio foudatio -0 ft log! 12 ft "ide ad  ft deep is to be composed of a 1:2:$ cocrete. /o" much gravel i cubic yards "ill be reuired# A. 150 &. 10 %. 110 (. 100

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