--!F-
fttria or Ef.ouc C}lbm Syhmelrical components Posiliv6 sequenc€ components Negalive sequence componenrs
Zero sequ€ne conponents GeneEJ lofiutas fo. symmeric€t
696 696 696
697 6S7
Th6oretrcat power ota wabr turbine 749 SFcilic spe€d of a wat6r tulbine 750 Iu6ne generator sel up 750 Heat en6rgy @nveEbn fictoB 751 Tems used in pEdiction ofload 751
Per unit impedanc€ using a difte€nt porcr & voltage bass 697 The seougnce nelwo* Ths seqoence nslwo Singte tino to ground fautt Double lins to ground fautt
Thee-phase t.utr €tcutanons gNen
698 698 698 699 700 700 701
SIRUGIUft Planl capacity taclor TEST 26 (44 problems) SOLUTIONS TO TEST 26
752 752 753
Tt4
.l .:. TEST 27 (33 probtems) SOLUTIONS TO TEST 27
Matter
-
anything that occupi€s space and has woight
Element- a substance lhat cannol be decomposed any fudher
by chemical
765
702
Apprcximate eqlivatents belw€en a line to linefaLrtt& a 3-ph.s6 fauli ot an untoaded gene.ator 703 TEST 24 (40 probtemst 7U SOLUTIONS TO TEST 2' 715
+ .:.
OTI|IIIN
-
a combination of lwo or more elemenls
Molocule- smallest partrcle lhat s compound can be reduced to before il breaks down into its elements.
776 784
Compouncl
a
Atom - smallest part lhal an element can be reduced to and slill keeping lhe properties of lhe element.
(two dimensionat ptane)
735
(lh€6 dhensionar ptan€)
736
lrluminalion trom more than one
V.l!e
of mountrng heighrto grve ma)rmum lllumrnarron Zonal cavity or av€rage tltuninat on TEST 25 (20 probtems) SOLUTIONS TO TEST 25
9.107 x 736
1
1.672 x 10
The atomic struclure ofan atom:
737
NuJ€ls it
f3?
at"h -.?ite,.itli. il.il "li. 6r'|.Etlt !,rc,!'ns i,]J,.,.
ieutbnr .:md..l
738
743
<-a
_irr. el
x..!s, where:
N = iotal number of electrcns on a given shell
n:
nrh
shellofthe atom
i".cls
jii o,bG
!J!t!J!!!!!!!!e!!!4J!_!!!!!!t!!!Et!3e!ins
.l n $
b!,
R
Roras
rt.
Valence electrons
-
Atomic Number
represents the number ofelectrons or protons of an alom
Atomic
Mass
-
electrons found in the outermost
Electri.ity: Basic Principles 3
she or orbit ofan atom.
Volt (V)
- unit of polenlial differencs, which is equalto onejoule of work done p.r ono coulomb ofchargs. Named afierthe ltaljan physicist, Atessand.o C. Voltr (1754 - 1827) who invented the firct electric battery.
represents the sum ofprotons and neulrons ofan atom.
TITCTIIC GUIITXT Wh€n a polential difference between lwo charges forces a third charge to mov€, the charuo in motion is called an el€ctric current.
tttclntcG[
$
A body is said to be charge, if il has eirher an excess or deficit of etectrorc from its normalvalues due to sharing.
{.
Ampere (A) - unil of charge flow equalto one coulomb of charge past a given point in ons s€cond. Naned affer lhe French physicist and mathematician, Andr6 M. Ampers (1775-1836).
Coulomb (C)- unit ofetectric charge, which is equivatentto G.2S x t0r3 electrons or protons. Named afrerlhe French physicist, Charles A. Coulomb (1736
-
1806).
ms$r cI
?OTIXIIII IIITfIRTTSI
Ihe fact that a wire carrying a currenl can become hot, it is evident lhat the work done by lh€ applied force in producing the current must b€ accomptished against
Potential- the capability of doing work Any charge has lhe capabitily of doing work of movjng another charge either by attraclion or repulsion.
Bome opposition or r€slstance.
+
l\\ _ 6+ F\ (€+ betflohs* tbc:elcedbr'the,_ cpaea".r I \\ l\\ rer{rw,l b.ar,4drd b},--. cpL.e *_o'O-- l:) -". L,^n l',.1 I
o* \l \J -oo-*\
b. No+ r \\ o+ I 11 aj!+
frcll o;o+ \ \ill o-+
3 3
6rllant hol,oi of6eecr
-
praclical unit of r€sistsnce_ Named afi€r the ceman physicist,
oh5 qo nq
rohad. rhe.r c Phte
The net num ber of electrons moved in the diredion olthe positive charge ptate depends upon the porentjat difference between the two charges.
-
1854).
Ri*', where:
da1rcn5 M lbe itt,a.te,j b\ ihe,l c ptarc anJ elR:tons drllb€ Ep.led irhc -rc;hr.
mal,Dg r
Ohm (O)
Georg S. Ohm (1787
Example,A$ume1 cofcbarge.in movedi€lectron5.
R'
R = €sistance (ohm) A = cross-sectional area (square meter) p = resislivity (ohm-meter) L = Ienglh (meter) V = volume cubic met€r)
.l
Specific resistance (!$istivity)
.!
Circularmil(CM) - area ofa circle having
-
resistance offered by a unil cube of the a diarneter (d) of ono rnit.
-r---1 l00l Solrtd Ptublen$ in Elect cal Ensineeins
cM-il.. IEIGI 0I
mil= I inch McM = 1,000 CM
1,000
I
Elect citv:
by R. Rot.ts Jr.
It Pmlrufit n$t$ tc[
Experiments have shown lhal the resistance of all wires generally used in praclice in elecl cal syslems, increases as the temperature increases.
Basic
Ptihcible.\
r, = radi'rs otthe cable r, = radius 1o the ouler surface oflhe insulation
G0x0tfitxcI Conductance is a measure ofthe malerial's abilitv lo conducl electric cuffenl lt rs equalto lhe reciprocalof reaistance. Slomons (fomerly mho) - unil of conductance. Named aflerthe Geman engineer, Em6t Weme. von Siemens (181G1892)
R1= initial resistance (ohm) R, = linal resistance (ohm) T = inferred absolute temperalure = temperature when resistance of a given materialis zero lr initial tempe€ture 12 = final lemperaturc Ai = change in temperature o = temperature coefficienl of resistance
:
Tempe.ature coetficient of resislance (o)- ohmic change per degree per ohm at some specified temperature.
lffiultfl0r llst$txct
@" p = resisiivily of the insulating I = length of cable (melet
0r
flflrs
rt:*'"f male al (ohm-mete0
5
6 = conductivity (siemens per meleo L = lensth (meler)
A
-
cross-sectional area (square meler)
p = specilic resislance (ohm meter) G = conduclance (siemens) R = resislance (ohm)
Ubs{ Problcnl
t.
t!
Aietoctdc has a chsrge of 2 coutombs. tf 12.S x 1O1s tree -.- TS,ua19qg"9 eredrons are added to it, whatwillbe the net charge on lhe said dielectric?
I
A_ 4C a. -2c
c.
D.
't
0 joutes ofenergy to move 5 coutombs ofcharge. Whal rs
ofti6
ba(6ry?
D.'5V
v
1Or0
elecfrons move pasl a given point ev€ry 2 seconds. How
much is the intensity of the €lectmn
c.
D.
t!e?
A.2.121x1O{ohm 8.4.312x1O'ohm C. 3-431 r 1o6ohh 1.078 x 1Or ohm
Nk$rome ribbon r$lstor elemenls 6ach ha6 a 63i3tenco of 1 ohm, The €lement made from sh€et of nichrome alloy, 0.m5 crn thick. tf ths width of th€ ribbon is 0.3 's what longth iB requlr€d p€r cm. elcm6nt? Assurne specific €3istance of nidrrcme alloy to b€ 1m 4-cm.
C. 70.7 cn D. 67.4 cm
A
, The curenl in €n 6lectdc tamp ls S amperes. towards the litament in 6 minulesi
D.
13.8 x 106 ohm
C.5.185x105ohm D.2.96x106ohm
B. . 62.1 cm
2.5 A
Problcltt 4:
A. B. c.
B.
A.68.8crn
2A 1.5
A.7.21x1Osohm
P.aobL!A
ProblGrn tt A clorid of 2.5 x
a.
Or EE 8o.rd X|rch t99a The 3ubstation bus bar is mad€ up of 2-iches round gopper bars 20 fl. long. What is lh€ resislance of each bar if resistivlty iB 1.724 x 10'ohln-cm
Ploblcdr
D.
50V 0.5
'12 s
|esislivity.
Aiatt€ry.c€n detiver
A.2V B.
1na
D.
Problcr!
lhe potentiat difference betwe€n the termin;s
c.
32s
c.
r9y, 7r EE !o.!ll bbcr D€lemine lhe rasislanc€ ot s bus bar made of copper if th6 l€ngth is j O meters long and the crGs-section is a 4 x 4 cln2. Use f.i241 micro ;hm"ctn as the
8c
Zeto
Problqn ar
.,
2s B.
PlobLnl
72C
c
coutomb; on the ptates?
dI
from soildrawn copper having a resistivily at 2O'C of l.732x1orohm-dn. The washer is 0.125 inch thick and has inside diamele. and outside diameter ol 1 inch and g ind|eB r$pectivety. C6lculete the exacl .€6islance betwoen the two en& of th€ tum to direct curent; taking into account lhe non-uniiom qrnenl dislribution. Assume lhe contacl along the ends of lhe turn to b€ perfect ovff the 6ntire cross€eclion_
3600 C
Eroblclo 5r A constant current ot 4 A charges a capacitor. How tong wilt it lake accumulale a tolalcharge of g
lo.r.d Oetota. t99t
and spreading the end6. The washer is
30C 1800
9r EE
One lum o{ a copper bar is produced by cutting a copp€r wash€r along a mdius
Whet quanlily of etoctricity fows
to
A. B. C.
12.74 x 10 6ohm 15.53 x
10{
ohm
17.22 x 1Or ohm
t00l
O.
Prcbl.nt tn Et.ct cat 'olv.d 14.83x 10{ onln
Teet
6!.. taio ! condicior o.1o_ m tors. with . uniform drrm.ror 1.0 c"',no t.o #mf,*f::"*lr-mor rro"rne ;i;iffi:i Jffi;"fH"*ff "'."l*"ny-"irfr";;il; on€ 6nd otfrr! condqdor accordiag lo fl. foffula: p
-
O.OOf + f
O
otric,n
A.0.0852ohm 0.0015ofim
C.0.0806ohm D.0.0902otrn
htr
""iffi A. g. C. D.
C, D,
8,H,,l;tdil?T; Hff :i,1ff"H"i*"#ilJ.#:"Ji
O.01ohm
A oo.ducto{ wttos€ diameter is 0.175 inch hAs a rosislsnce ot 0.5 ohm. Th€ wire
h dnwfl lhroogh a sedes of dies udit its dianr€ter is r€duced to 0.OS ind,. A;;; lta apaclfic rrristance ot the materiat r€main3 con"t nt, *t ij ff," ,""i"L-J Itneih.ncd conducior? "f
a, c.
','Drllta},
0.e3 ohm
tiaDLrr tt
A.
ur trtlc.r{
D.
l l.,l5
o
10.22
A
12.75A 10.82
()
tr.ll.rr t& EP !o.rd
ln e. 'R 2 c. 2R o. 1n 1
D. 6n
A, 8.
A givan wir6 ha6 a r6sbtance of 17.5 ohms. It its t€ngth b 560 m, how much r|i€ in order to reduce its reiistancs to ti.s otrms.-
langth mwt b€ dd-off tom the
5rr
fo$d Octob., taca
,",r"i'gfiH,H"*ft.i"1;:H'c'
r!;rewn
untir
ft! resbt nce is 1oo tirn€" rhe
c.
12.5 m
A B.
5m
!o.rd Alstt tro:]
.#i':ffi :y#fi :?HH#'","Hl*';;lilT;:k ffl:ff s #,0#'*'.
A.
0.85ohm
t45 m 155 m
tA
EE
lo.|r! oglotcr t''6
Whet is th€ stze jn squaB miltim€l€r (mmr) is the cabte of 260 MCM 6ize?
t00 m
P.loblc.r 14, tE ,"
C. D.
160 m 170 m
PlcoblCltr
10n D.
rgtj
A,
trou.ttt t?r
8.
Apfll
lha ll..l except for havlng twico its diemet€r is
A. 4A B. 3rr
Pr!t- lc- r!|t EA
&
A corloin wir€ has a ro3ldiance R, The r$iltance of another wire fttenticel wilh
1896 ohr|! 1825 ohmt 17gZ c'vn t8O5 ohrnr
#
c.
9
1.0.?8ohm
PaotL- tcr ED loaaa
8.
I
C. D.
118.656 mm'? 126.675 ll,m2 112.509 mm'z 132.348 mm2
Pr.obl.tn l9r A 500 MCM ACSR cabt6 has 37 strands. Dete.mine the diameter in mils of each
l00l
TO
Solvcd prcblens in Etectticat Test
116.25 . B,
120.24
c.
118 34
D.
110.35
Dalalmlno lh€ l€mp€rature coefficient ol copper at 0 "C.
A,
0.00427 0.00615
;#i#f #fi ,::#rx$;"1a,:"Jiff ';Hj"1r,T:"["j,ilHri;rT, jt: A. 800 ttllcM 8.820.t\rcM c. 850 MCM
trablod r3r
MCM
ll#!.nitrffilfiffi fr'#iic"gr,""g;,g;y"r".,".,,. A.72.26ohms
3
rt9!
t{l
I c 0
9OO
l1
EE Eo.rd L.!Gh Th€ r68blenc6 of a wire is 126.48 ohms at 100 'C and 100 ohms at 30'C.
PtobLo
Prioblcd 2ot
D,
I
Two wire8 A & B made trom two diffgr9nl materials have temperature coeflicient Cl r.rlslenc€ 6quel to 0.0025 and 0.0005 ohm per "C, respectively. tt is deGircd to makc s coil of wire having a aesistenc€ of 1200 ohms \Nith a tomperature coefficient Ot 0.001, u8ing suitabl€ length6 of the tvo given wires connecled in series. oalarmlne lhe required length of rvir6 A?
A, 5.5 m g. 8.2 m
C
D.
3i3i3i$i
0.0258 0 0356
8.0 m 5.0 m
D.58.15ohms
Ftobld!
P|!obLa!ta,
Ts,o heating €lem€nls which is 500 ohrns and 25O ohms are connected in sedes wllh l€mp€ratu€ co€ffci€nls of 0.001 and 0.m3 ohms per 'C, resp€dively at 20 .C. Calcul5le lh€ effediva lomp€rature coefficient of lhe combination.
fl rfli"l:H*:#$ffi A.
88.89a
D.
92.81fi
B.90.12o c.85.22()
Problcnt
'ft1ir6r:q,f
{,:r{!ffirll,1##
A[
16r EE
c.
0.00215 0.00626 0.00712
D,
000167
tt t
lnlulation thickness of0.25 inch. lfthe specific resistance of rubber is dclermine lhe insulalion Gsistance per 1000-fi length of the cabte.
A.624MO
A. 60.4.C B. 59.2.C c. 58.4 "C
D.
53.7.C
OctoDar
ProbIod !7t A cylindrical rubber insulaled cable has a diameter of 0.18 tnch and
s:*l,!";li$"fl ..*;[i'$*ur;:"f:si#Hrrli*iiJlr""#Jix D.
lo.rd
B.
c.
1014
an
ohm-cm.
682 MO 678 MO 694 Mc)
Problelr tar The diameter of a given bare conduclor is 0.50 inch. A thermoptastic insulation with lhickness of 0.1 inch is Mapped around lo jnsulate lhe conductor. Determine the ansulation resistance of thjs conductor per meler. Assume specific resistance of lhermoplastic insulation to b€ 2 )( 1 014 ohm - cm .
I2
AB.
l00l
Solved
'
Solutons to Tesl
i.tulrsbL$l
107xroP{)
l(}5x'toeo
C- llOxl0eo
D
109x lOeA
PFoblc|! t9. tE
to.!d
Oclo0cs
tgtt
Or
.
o?
- -12.5 xlor'€
+2C
x;2i*1% = -2c
Ttte insulalion resistance of a kilo.neter of the c€bl6 haling a diameler of 2 cm and an inG{dation tickoes ot 2.rn b 600 otms. It the thic*n€;s of the ansutation ig inc.eas€d to 3 crn, find the in€ulatio,r r€6i6bnce otthe c€ble.
o,n - o, +O, -2+G2)
A. 8.
Notc volt-opulci ps@ul@b
C.
725 ohms
Q*
I
850 ohms 757 olms
D.828ohms
Pr.DLd n€
B.
tiot
'o
o6
8.2 voll|
lE fc.a Oettb.r riao
a€si6k1ce ot 1m rneie{s otBire is 12 ohms. Wrat is its condl€ ance?
E::'ff:;,H"';"'
0.0521 mno 0.0€i] mho 6 mho6 !2n lo3
l
E
- 2dfipgratt
o-n-'(ur"'+#; a -18@c
o8 t4
l-2d€c
2 Tl' lincfi / 'lrir,"r,r t ^-€-a 1.D7. O !3_A t9_ A 25. C 2.A 8. A t4_C 20. a 26. D 3.8 9. B 15. A 21.C 27.O 4.O 10.C 16.O n.A28.A 5.411_A 17. A ?'3. B 29. C 6. C 12. C 1A. a 2.t- A 30_ B
??-30 - Iopnotch€.
1= 2ex r
^ :!5-e0 - Condition.l
O-I{ - F€iled
R
t
4
1319 r 2 9{ 9n ln 1n l!
= 649 6
(1.724 x 1O3Xm96)
25238
-5.t85x105drt
-m2f8rrr? o','
I
13
H l00l Solved Prcblens L = 1om x
^ 'L A
1991!
in Elechical Enpineetine by
R
Ro,as
Jt
Sofutio^ to
= looo cm
1m
n
f,
Jo
aL =
f
Jro
oor.ro 'r:r
".
L=!l=(Iqg!-9o L = 68.8 cm
*=oi
R
I
i[.-".'.
1'; ]]l' = i [.
a difiarential
elenrot
15
**'.''{S)]
- 0.0806()
*-'$ Nota Sinccthe coll ir roound with thc
same weight
orwke, thusthe vollnres
th. wir. us.d in both coDditionr rre con,Qnt. Ard witl lhc volumc
Comid€f
I
+
(4x4)
R =107.75x 103 ohm
E
.
{1-7241x 104 )(1ooo)
Test
of
coDttaDt,
rcrttance lqfleJ drFctly a. the qua'e ofthe length.
A=tdr L=2nt an
=
R=kL2;
=
d4I) ldr
R-
fbnorh o€r tumxtL 380 ( bngfi p€r tum r t{ Rz = 1895 35O
lt* \
Notc: Asruming thclc ar. no losscs in the prcc€si, tb€ volum€ ofthe mat€dal rcmains constaDt And wlth the volume
o.125inch x12
2a11.724x1041
G =6?902.25mho
I.
-1
1
G
U102.25
R = 15.527 x 103
()
"nd'?0?r, L=0.1m=10cm .
_L
ofthc lcngth.
R = k L2; k = proportionality constant
R, &
L1
TL, I, -|.r.,]
l!.J'
"^'
=*.
Rz
=54
lL,/
sr,
- o.o1z
1'
I L, .l
c::::33:::::5 L,= 2.5L1
R = k L, ; k = pfoportionality constant
R, IL, ], =l\ R.
R1
J
,
lE;=. Foo R; "="'1& I & L,
I
a: ihe squa'e
s4cm
) t l inch l (tnl.s- tno.s)
I
lr
\t ttsaool' J \ 6000./
tJl2't
c=-t 2olrfdt =frn'1* =f.tr"',-r"1r ^
proportionslity conBtant
n" tr"i' R' lL,J
^1R ce
k=
= 10m
1:::::::..:.-) &=100R)
16
l00l
Solved Problens in
n=oY= a'
Elect cal
Sofutio,"t to Te.r
I
Pv -=rr'Y
(1,\' 14)
R=*;
R:
o
l-anClh cu(
rr'zf
]'
*
*; =proporlbnality &=f d,)'
R,
14
k - propodionality
Lamfi crrl -off
26O
c.mlant
-
of
=
L -1,
=
\,,
560- 400
= 160 m
^ d14 ann"
t(b tlCM
= 500,000
Cil{
-!Q!99=1*rr.u,* cu. d
L
d-Jet=Jfu5135 d-116-25mils
of{'c
two wkcj ir botb cordltlors to be conrtaDt And wrtl thc lcngth cor5tant, tLc rerstance oftlc wrie latres rnveisely ar the square oi
R=-:i
t
Dgth 'D
p. - !.!.;0.
both @nditron! are tle eme
=
tzo-cup.rn
f = proportona$y con {ar R.
R,=l4l ld,l' n:
\E--._1,
n.^=n.l4l'=^ll')' '
\d,I
",
Notc
I
=?o
l2d, )
=*ie.
d/
.1,=2d'
R. =Rc
p.L.
_ p.L"
rz 2,,,n'
J7-ttrand ACSR c6U.
r.-
R2-1145a
th< lalgth
cut-of
---ii$,2ll"
A.l26.67mrn'?
*=o[#)'=,'(#i
Note.Arlrm.
l
Length
MCM = 25O,@O CM
, soo-1", a. JEu " Jzsoooo l0o0rn13 linch -
con3tant
,J
c=olAltdz = 4 =r
-rJ!_r L,
c -d
-__.\ R=
dott 5
' -!&.!9!l?9=1oo'' R, 17.5
ldzl-oo.r1\5,
= o.€3
-tti
ca
r.
f R' t d,.J
*,' =".f
thc
R, tR' L'
!e=f4
4,
N'n. Alrrm. R
k =proporlionatity constanl
17
u'"1 (n^r k1tr.r)nl
Notc: A5romiDg th.rc arc ro I65e5 jn thc proce$, thc voloD. of rlr rcrrainr .onstart And wtth tbc volume onstant, the rcsirtan.e ofthe *tre:nrcncly ar thc (ou,tfi powcrorthe d amerer.
I
= 10.37 r! - clil per
ff
t-
;
Solutioti; to Test
E
-="#=ffi="#
Notc Anumc thc doss s.ctional area oFth. wrrc to be constnrt And wjfh tbe rkrs 5..tronilarca conrtant, retisiaD.r o{the wire v?do dfectlyasthe lenqth
R-kLi
R. R,
_k r.(=-, t = proportionatity constanl
]'-
\5,
I
250 MCM = 250,000
d=
ris
1.-,0
4 -r, L n=ol 'A = aA2 d,
cM=d d =
4
d =
Note,.Ar5urrc th€ largtl ofthc two wr.s ir leth cor,ljtron, to be con5tant Ard wiih tlc lcngil, corrtant. tlc rerrsrarccoftlc nrcvariu: inu.oely as th"sq,a," oi
E
p.
R,= fd, l' td,j
1!r
'1=:lo
R-
lind ,54tt , l(mmtu litrn
ACSR caue
=!4@=rrra_srcu
\2d,j
d, =2d'
ir boil ondltronr
are
=!&;o. = rzo-cLrp.rt !t!r;0.
/___-t,
l' =Rl-!Ll
'7-strand
= Ji 3s13,sl 1t6 25 mils
=
-12.7.n.,'
d12n2
Note, L$gth'
p"
n,
Ci'
Jct
- propondna&y corl3tafil
Length
cut-olf
500 McM = 500,0m ca4
Rz''11.45n
=
\____rl__,J L,
A = 126.67mm2
f - = R.l.dtl'=o5lo \d,) \ 0.06 l
n,
-
. soo -r" JcM - "65oJoo '
. 'd 41
R,
R,,
= a66
d
cM=
; k = nrolortionalig conatant
k
17.5
Longth crrl o{f = lr - L" = 560 - 40O Longth cul - off = 160 m
R, Id,)' R, ldr.l
R=+;
$o(12 5)
-
^.I(dr/ R:=093o
R=f
L,
' =lEa= R,
!.=ld' i R' ld, J *,' =
k = propodionality conslant
L.
1^
rr'z o 0.,1
17
_L
Note: Aisumirg tlerc arc no lorse5 ir t6. procets, rhc volumc oF the rrateral ieirairs constart. Ar.l with tlc voJum..oDnrn{ tbe resrsence ofthe materral lzrr.5 inv.6cly atthc Foudh power ofthe,ljamerd
a,
I
p,L. _ P"L.
=
ro.:zn- cup.,a
tle
5ame
It
1001 Solved Ptublens in
Electical
49.!1) a' = &e. p" = 10.37
ars ezz
Solutloru to Test
so . do
=
i-
zgl:e6
0.00427
A. :820 cM R,
E
R,
""=+ r = --a- =zu.tgz,c 0,00427
!e_ R1
T+12
-
63.24
o
=1=-1-= do O.o(X
R.(1+0.0025^r)
Rr
0.0005^,) = (R.
+
RbXl +0 001ai)
= 3Ru
R. +Rb
= 1200
= 300
256"s
o
300
L.=6m
T +t1
so(250 * so) ffjlzl = j t 250+20
R- = R. 'lT*
'
-
R,
+ Rb(1+
Rr +3R. = 1200
Rr_T+t,
Rj
-O
Equato Eq.1toEq.2:
R.
r
)
0.0015R.Ar =0.0005RoAr
T+t1
-..=\tfaj= - /r*.-) 5o(234.1s2+1oo) ,3.rrr.3o Rz
- R.(1+ 0.0025ar) - Rb(1+ 0.0005^, - (R. + RbXl+ 0.001a,) =O
88
R, = 5OO(1 +O.0o1l') + 250(1 +0.003ar) R, = (500 + 250X1+ a* ai )
=O
=A
89o Equate Eq.1toEq.2:
1=1=-1-=zso"c ao 0.004
500 (1+ 0.001^r)+ 250(1+ 0.003a,) = (750X1+
T+t, R1 T+t!
dji
R,
_
r^ =
ft*r.)rla)-r ' \R,l
500 +0.5ar + 250 +0.75A( = 750 + 750o.nar.
=0.00107
9 "=*"7 't
'
= tzso
r,
* r stl.119) 't120, - zso
= 59.2 "C
R=
,J,*" \
x
1O1a
t,,
lQ:gt 3.281ff1
R = 694.05 MO
R, _ T+1, Rr T+ lr 100= T+30 126,48 T-1oo T+30
= 0.7906 T + 79.06
T = 234-288 0C
@ ^=*^? 1oi'
2x I0.35l 2r(100cln) \0.25,
R=107.'102x1Oen
hfg-!r) \0.09./
od'^,)
I
19
L)C
E
Not : spdifi c5irt neofth. n
cabl€ a'rd
DCthffik
&=s R" nC
R)=757o
^1
Cba.itt 2t
l.ngth are @nrt?rt iD both conditiors
==!1nL
-,--.lH] .-[#,J
Electi.
0[r8ut Ohm'. Lrw alates lhct lhe cunenl florving ln 6n eloctric circuil ls dlr.cty iilJrtr"ii ii tt'L imprersed enrf applied lo the circuit snd lnv'r'6lv lo lhe
5iui".r.t t .".i.r.n"" & lh€ said orqiil l- Ohn {1787 - 1854).
ilamed afier th€ 6€rman physrqst Gcorg
WWmm
1
R12
G = 0.0833 mho
ri\6ro
E = impr€3s€d voltdge (voft) I = currenl drswn (ampere) R = r*iEtsnce (ohm)
lltgnltll?affn Electrlcal powor Watt
- un
-
rale of Using or con6uining the olectrical onergy
ol electdcal energy equal to one loula of enctgy consumed in one Bri-iish engineer and inventor Jam€ Watt (1 736
iecona. ltarneU aner Oe 1819).
,s$ue where:
P = electrical Power ($,att) E = vohage (volt) I=
dfient (amperg
R = re3istance (ohm)
-
DC
IUG| {.
C
Energy- the capacily to
where
r ltHI
IXtXCy
toulllErls
tlltlttm$unloi
do work
E]
tulD
Ele.tic Cn.uits 23
It[ tlsltflns G0rrts Ilr8I
t8
E.
W = etectncatene.gy (oute) Q : heal energy (calone) P = etectflcatpower (wait time (second).
t:
m = mass ofthe materiat (gram) c = sp€cific heat of the matenat (catone per gram .C) dr = change in temperature of the mate at (.C)
Note
N.n. llth.i..rcthree
l.ihl,r!
lD
rn.t
or more reenances In serier, rcduce b.fore applying the VDr
l calode:4186 ioales BII z 252 .alo:jes
1
.l
Kilowat-hour (kw-hr) _ unil
Note
1,lay
=
two 'nto
?rillu+onlcTm ftsBr0ns in which etectricat energy is sotd to a customer.
a
Prnllal clrcult - th€ resistancos are connecled across each olher.
24 boql5
l"
l Donth = 30,lays 1
6l'tthecrrc!ii
yerr = 365 day5 = 8760 hou6
snl$_G0tIIGII0 trsFTtBS
.i
Series circuit- the resislances are connecled end to end.
+Et-
+E2-
L
t
+ E3
,t I3
N,n. 1,,
'l
lniependen{ oF
h. t hc
tota
th. .ircuit .oDn€.lion
either serl::, parallel or corDbination
of
power drawn by tbe circu it i5 equiE lent to the powels d'awn by each lord
DC
cufin r Dilt$t3r
llfittrl
IGm ron
?illl1fl
ffstsnis E0xxrsrrl tr
EleciicCircuits 25
0[trt r wYt-G0rltcllll Rlsl ols
lE, \o(e .rth.,"a'e.hreF ,e<
ror.
rn
or no,e.ern4nce,.n para,lel, reduce t-l)r (he -tr.u pdallFlbeforc applyinq tre lDT
r ntot*o
U
Wya to delta trenlformation3
U
D.lli
sHllt-pttflfit G0tIIGIMISI$I0[8 t
Series-parallel
circuit-a
into a series circuat.
conbinationat circuit which when simpl
iedwj
resutt
to wye transformationr
_gc
-
?m|lut$Inl[$ c0xrlcltD [IStSr0ns Parallel-series circuit- a combinational circuil which when simptifjed wiltresutt
into a parallel circuil.
u
tf
A=B=c=Rrandx=Y=z=Ry
-A.+
BtC
26 l00l Solyed Prcblens
in Etecbicat
T?tt
trotl.d ttr
EE
2
27
f,o.rd Apdl a98a
A lrd'trrUay power stalion supplies 60 kW to a load over 2,500 fi, 100 mm'?, rwo rNnrlrr,brr coppor losder lhe resistance ofwhich is 0.078 ohm per 1000 tt The bus vollr{,o rB maintained constanl at 600 V. Delermine lhe load curenl
lfl F
ti
ttr EE Bo.6d Oetobcs r99t load of 10 ohms was connected to a 12_voll battery. The curGnt .1.18 Aamp€res. drawn What is
Problc|[
the intemat resistance of the
B. c_ D.
0.35 0.20 0.25 O 30
batei?
D.
ohm
ohm
D
t
r EE
Board Aprit
tt9?
127 05 V
.
l2g 32 V
tralLd
l7r EE Bo.rd Ostobc. r9a6 A LRT csr, 5 km distance from the Tayuman, takes 100 A over a 100 mrn hard alrawn copp€r trolley wile having a €sistance of 0.270 ohm per ftm. The rail and gtound rolum has a €sistance of 0,06 ohm per km. lf the slation voltage is 750 V, ihrl l. tho vollage of the cal? c D
685 690 685 690
V V V V
trobl.n lar 00
ohms ohms ohms ohms
Th6 hol resistance of an incandescenl lamp is 10 ohms and the rated voltage is Find the series resislance required to operate the lamp lrom an 80 V supply.
V
8() 4O c. 6C) o 10 C) B
A battery is formed of fve ce s Joined in series When the e{erna| resrstance rs . 4 ohms. tne cunenl is I 5 A and wh;n the enernar ,"r,"t"n"u i" S o1.", lrii"ni talls to 0. 75 A. F rnd the intemat ressrance of each ce .
;"
Problca t9r A rcsislive coil draws 2 A at 110 V aier
0.5 ohm
operating for a long time. lf lhe lcmp€ralure rise is 55 'C above the ambient lemperature of 20 "C, calculale lhe oxlemal rcsislance which mustbe initially connected in series with thecoillo limitthe
1.0 ohm
currenl to 2 A. The lempeBture coeflicienl of the materialof the coil {s 0 0043
c. 0.2 ohm
o
l)
ll
ProbleE !,4t
L
125 32 V 130 24 V
c
trr
ora standard ce is measured with a potenliometer that _...^^t!"_-"l11rotng!'1"_lT9e gMes a readrng oI1.3562 V When a I 0 megaohm r€"i"t", i" it" slandard ce termrnals, the potentiomete. ;ding arop" "onir*r.a ""i""" to t.S56O V. Wh;ii"inlernal resisiance ofthe standard cel? " B,
I
20
174.5 145.7 147.5 157.4
16r
cuffent of 100 A and is located 1000 fl from the supply ll lh6 diameler of lhe copper transmission line is 0.45 inch whal must be lhe ll,1146 ol lhe 6upply?
5cl 4Q 3cl
ProbL|[
c.
ttalLm to0r|.l
The potentiat at lhe terminats of lhe banery fa s from 9 V on open circuit to 6 vo[s when a resistor ol 10 ohms is connecled across s terminals. What is lhe rnternal resislance ot the banery?
c
lt0A
lr 120 V DC motor draws
ohm ohin
Problc|n
B,
t08 A
il20'C.
0.3 ohm
I
10.12 ohms 10-52 ohms
per'C
20 l00l Solyed Prcblens in Electical
C. D.
Tesr
A carbon resislor dissipates 60 W of power from a 12d V source et 20.C H much power will be dissipal6d in th6 resi3tor al 120.C jf connectsd across the source? Assume th6 temperalure coefficionl of carbon at 20 'C is -O,0OO5 per .C,
B-
c
61.50 W 65.21 W
6234W
lo.rd Aprlt rgatt EE,Eo.rd
Octobcr
rt34
An arc lamp takes 10 A al 50 volts. A resistanco R is to b€ ptace in serios so lhat tho lamp may bum correclly from a 110 V suppty. Find lhe power wasted in this resislor.
w'fli
900 wstt6
trctLo 4tt
Conduclor'x" ofs cedain materialand a givon crcss seclion has a resistanc€ of 0.1 ohm p€l met€r and a lomp€rature co6fici6nl of O.OO5 p6r .C. Conduclor ,y" of anothor maledal and a given cross soclion has a resistance of 0,5 ohms p€r nioter and e t6mp€ralure coemcient of 0.001 pd 'C. ls desired to make s coil havno a rc5i6lance of 50O ohm6 and a tompe€Ure coeficiont of O OO2 by u3ing suttait. longths of thc lwo wires conn€cted in geries. CalqJlato lhe requhed length of wiro
'l'
L
c. D.
m m
ln m
610 5
!06
2
tV kv
607.7 kV 603.8 kV
EE
lo.rd Aprll lt3t
'16r ll lr rcquh€d lhat a loeding of 3 kW bc mainlain6d ln e h€eling al€menl at an lntlal lanpareturo of 20'C, a voltago of 220 V b nao€.3gry for he puryo.e. Afigr tho aLmani fu! 3.ttbd down to rloady atato, ll Ir iound lhrl a voLta of 240 voll6 ia t!.o...rry to malnlain tho 3 kW loading, Tha olamont reildinca l€np€ralu€ o. dcnl l. 0.0008 p.r do9r.o c.itigmd! at 20'C, Caloulato th. tn l l€mp€raluro ol lh. h.allng.l6m6nl.
A 4 c
B.600watts C.700watts
1225 1250 1240 1210
I
tr.tlin
0.48 A
Prcb|...lo 4Ct
D.
A
D.
o.0.57A
800
Aaauma lhc loop reristanco of th€ lino to b6 1 mO p€r km,
0.
4.0.52A B.0.64A
A.
A 0 30 ohm l020ohm 0 0.23 ohm 0 0 03 ohm A hlgh vollage OC tran6mission lin6 deliverc 1000 MW al 500 kV to 6n {er.Orta load ov6r a dblance of 900 km. Determine lhe voltago at the sending 6nd.
Two (2) 11$V incandesc€nl lamps A and B are conneclod in sedes acro6s 23GV sourco. lf lampA b mt6d 75 watb and lamp B is rated SOwa s, det€rmineth curlent drawn bylhe sedes conneclion.
c.
a4r EE to.rd Aprll 1997 ll a rcrlrlor mtod al 5 welts and 6 volts are connected across a battery with an cran ckcult voltago of I volts. What 13 th6 inlemal rosistance of the baltery if lhe raaulllng cuflcnl i3 0.6 A?
tratldn 4tr
D.63.16W lrroblcln 4rt EE
29
tr.t!.o
11.45 ohms 12.05 ohms
Ptoblcltr 4Ot
A.
2
345.43
'C
328.12',C 33r| 84
'C
o. 318.48'C taol!.r atr
r
r
A 200'W, 11GV lncandaacont lrmp haa flhmonl hay{no biprnlu.r @afidcnt of ruBlrllnca aqual lo 0.006 al 0 'C. lf tha nollnal gprr-te ad,l9araturc ol tha bulb ii 2600 'C, how mloh cuarlnt rvrll lhr bulb drrw .t lha hda.( i ta lllliod on. Arsume N rcom tcnrparalurc ot 20 'C.
A.
29.42 A
o.
22.31,' 24.214
8. c.
18.37 A
hl-ra.t
d..*nbtrr*-a--g-alt tOV..d h ll'e G ly t cj .l 1tt V- 0 'C. Ih. coeilnc,i!.i oli-Fd rO 'C b O.OrOl.
Th. po$qr
10 'C. Calcuht! lamparetuE
t
,-t!F* J0
c.
D.
l00l
Solred Prcblems in Eled,ical
thsr
2
3I
H
183 W 225 W
c (.)
Nona ol
Probleltl 49!
trobLd
A. 2.43 mils B. 2.52 mils C 3 21 mils D. 1.35 mils
t
(l
10 12 13
t,
11o
An eleclric wals healer has a rating of 1 kW.230 V lhe coit used as heating elenent is 10 m long and has a resistivity of 1.724 x 1O$ ohm-cm. Detemi the required dianeter of lhe Mre in mils?
J4r
Ihroc rssistors of 10, 12 and "/ ohms, tespectiv€ly are connected in pardLl .iro.r a constant current source of8 A Determine "x'ifthis resislor draws 2.5A ()
t) Q
trobl.E
ProblqN! 5or
lh6ie
55t
lhe resislance Rato resislance Rb. Assume Ra < Rb.
fwo rosislors A and B made of differenl maleriali have tefiperature coefficignla ol rol6tance at 20 'C of 0.004 and 0.006 respectively. When connected acrola e vollrg6 source al 20 'C, lhey draw cunent equallY What percenlage of th€ lol.l o rranl al I 00 'C does r€sistor A c€rry?
A B. c.
s c
a7.14% 62 a6% a1 34%
D
30.86%
When two reslstors A and B are connected in series, the lotal resistance is ohms. When connected in parallel, the totatrcsistance is I ohms_ What jslhe ratio
D.
0.5 0.4 0.8 0.6
Problqn
tt:
EE
lo.rd X.rch r99a
Ihree resislors of 10, lhe equivalent resislance?
15 and 20 ohms each are connected in para et. What is
tlcDl.a
55t
Two re6blors A afld B made of different maledals have temperature co€fficiontl ol rcrislance of 0,003 and 0.005, resPctively. When connected in parallel asosa
I
45 ohms
consumes equal power, What is the ratio of the pow€l drawn by rcsistor B to that in resistor A wh€n lemperatute rises to 60 'C? Assurn!
B.
17-2 ohms
aupply vottage is constant.
c.
0.22 ohm 4.62 ohms
D
P.oblqn!
5l
The equivalent resisiance of three resistors A, B and C connecled in parallel is 1.714 ohms. lf A is twice of B and C is hatf as muctr as I, find the equivatent resisiance when lhe three of them are connected in se es.
A.
17.5 ohms
D.
28-0 ohms
volhgs source al 15 'C,
A B c 0
3!
Three resislors of 10, 12 and 15 ohms are connected in parallel. Evaluate the value of curent lo the parattel system thal will make the cufient in the 10 ohm resislor €qual to 2 A.
0.829 0.926 o.s@r
0.882
Itroblcln
ttr
EE
Eo.rd
Lrsft t9te
Three resistoB of 10, 15 and 20 ohrns each are conflecled in parallel. whal lhe lolal conductance?
B.21.0ohms C.24.5ohms
Problcn
at
B
c D
0.217 mho 3.41 mhos
4.62 mhos O 52 mho
ll
.
12
!t)!!,1!!!!!t'tohtens
ih Eteuticat En
Plobtcor jAr
Test
,3ij,,-i;1,.;1-$i$:tHiyn$:}"?,.,"ki
jrf::":*:#:q{sfi
33
s 0
llJ ohrn.
0
615.32 W 506.58 W
B.
2
tttll.l.
lrln
c. 582.45 W D 604.38 W
atr lho ctrcull as shown, determine the rosislance between terminals a & b.
Problqn 59!
:"9nru;'"m":i*"rl""il";git i';iii,:i'""':"Jii:g;U:;:*i",il
A
A. B.
c.
D.
8.10 r) 8.52 rl 7.84 n 9.22 rl
votta, what is ths current through th€ 1O_ohm re3i6lanca?
c. 200 ohrns
arr EE
a
3t33iilt
D.
2'1.2
C.
I21 A
a 271A c a02A 0 672A t otl.n| 6tit rc.o..a l20vsource
lcad Ostotrr roor
f ""*:J*;iijttfi"*.xfl{*".,"r,#."J"1:iTfii
j:i:ffi ""
*t":iil
1a.i ohm6 ohmi
"#*ffi{",:;5';
Find the minimum veru€ of
lo lho pow6r lak6n by th66-ohm rcslslor
A B c o
10.35 12.24
itf"
a"r,Inl."ii.r"i i"o i_..iii ih; il;;il;;';;;l;'fii,"":
rl
A 10.21A 1146(r
lc.fit lrlch rrra r wo r€srgtanca3 of i0 and 15 ohms, sach respoctivoty 616 conn6cted _, Ihs rwo are then connrded in s6ri€s Pr.oblco Car EE
Pr.ll- ht ff a-.a l-!t &_
tr
A
*jprrl"ir""" iJ'ii
A 30 ohm reststor is @nn6cted ln psre €l wilh e variabte resistanc€ R. .lh6 prr.tlot com_brnation is th6n connocted in s6n€swith
180 ohm!
;* ;f,:Hf
2362(l
C4t EE Xro'.rd Octob.r t99? 10.ohm and s 2Gohm r€sistanc€ ar6 connected n.r.r.ncc or s-ohm is cDnnecrod in 6efi6s w h lhe t"". rr r," in paraltet Another
150 ohms 100 ohms
h..ll.r
t,
A
,H:"::#t{tlifl iirstrni{{.{#ifi $;ffi :"ji{lxr#trJ D,
c
I(rcon e23 n ,oE30 t
;Lu.d
Ptobtqt| 60l
B.
I
"*i"".sf:t*;I
icross a 12-V ba ary, rhel are
ffi
trro,ifi :i?j"f:
1.2
a
c r)
A,17.28W
096A. t152W 1.00 A 13.1 1 5 A. 20.2s
W
w
li€
rn pare gl with a s_ohm reeisianc€. tt le than coi;.e-;Ja crlrr6nl and ,0116r?
J J4 l00l Soltcd Ptoblems in Elechicdl
Test
P.obLn| 6t!
(;
An 8 ohm resislor is connected in series with a parallel combinatior of resistors, R and 24 ohms. Determine R if the power consumed by the pa connected resistors js equalto the power consumed by the 8-ohm resistor.
tlallar
c.
10 ohms 16 ohms 12 ohms
D
20 ohms
B.
Probt€lf 6ar A multi'tap resistor R is connecled aqoss a 220-V supply. A voltmeler intemal resislance is 15-kO is connecled across the cenler tap and one end of supply terminals. lfthe voltmeter regislors 100 V, what is lhe value of resistor R.
2 Jt
lr
?rr Ef, ao.rd octoDct t997
A procc.! 6quipm6nl contains 100 gallons of waler at 25 'C. lt i3 rsquired to Llrt0 ll lo bolllng in 10 minules. The heat loss is estimaled to be 5%. what as lhe kW
ltll
I
e ol
lhr hcalofl
t:6 kw
,!2 kw 0 t06kw o t07 iw
itltat. ttr 3E Eo.!d octobc! ltto placed l[n.
in a l"kw elect c ket e. How long a lol.l ot 0.8 kg ofwater at 20 'C as lll mlnulol! nesded lo raise the temperature ofthe water to 100'C?
I
4 a6 filh 6 32 mln
t:
0Umin
Problclr 6l't
l)
a 60 mln
A potontial divider of resistance of 50 ohms is connected across a 100 V DC sourc€, A load resislance of 10 ohms as connected across a tap an the potential divider afld the negative teminal of lhe source. lf a current of 4 A flows towards th6 load, what is the curent supplied bythe source?
tiallitr
A
B.
c. D
5.5 kO 5.0 ko 6.0 ko 6.5 kO
5.32 A
5.05A c. 5_21 A D. 5.48 A B.
Ploblclr 7or Two €sistors A and B are connectod in series across a 220 V DC source, Wh6n a vollm6ter wilh an intemal resistance of 10 kO ohms, is connectod across resistor A, th6 instrument reads 100 V and $hen connected across resislor B, il reads 80 volls. Find the resislance of resislor A.
A
I
c
0
l0 1000 a2 25
tta$.a lwh
I
0
It
lo.rd oc@b.. t9t7
roquired to 6ise the temperature of water in a pool is 1000 lhr hoat loss€s are 25%, the heating energy rsquned will be
1266 1750 1333
Ptobldr 7tr EE lo.td &rll 1992
36',F?
79.1% 75.30k
75r EB
lhc clectric snergy
t)
B.
Bo.!d ogtobcr l9glt
?3 88
4ka B. 3ko c. 5ko D. 6ka An electric kelde was ma*ed 500 W, 230 V found to take 15 minutes to bring 1 kilogram of water at 1 5 "C to boiling point. Delemin€ the heat effciency of the kettte.
74l EE
Ho{r meny calories does an electdc h€at€r ot 10o watls generate per second?
-.
kwh kwh kwh
Problqr ?6!
EE
8o.rd April
l9'r
An olecldc healer carries 12 A at 110 V, is subfieQed in 22.5 lbs of water for 30
nrlnulss. What will be lhe linal lemp€ralurc of lhe waler if ils initial lemperature ia
A lt (; t)
'F 42'F 133 56'F 135 43 125
12a 33
'F
36
l00l
T?-rt
Soh'ed Prcblens in Elecbical
lrroblqli 7'l
EE
ao.rd Octotcr r99o
ln an electric hoater the inlel temperatur€ is '15'C. Water is tlowing st the rate 300 grams per minute. The voltmeler measuring voltage adoss the heating ele reads 120 volts and an ammeter m6asuring cunent tak6n rceds lO ampercs. sleady slate is linally reached, what islhe finalreading ofthe oultel thermometer?
A. 57.6 "C B. 68.4'C c. 72.4', C D.
EE
lo.rd
Octob.s
lR I
0
!
0lR ttall.ir !
ttgt
Four cubic meters of waler as to b€ healed by m6ans of four 1.5 kW, immersion heating elemenls. Assuming the offciency of lhe h6al6r es determine lhe time required boiling lhe water ifthe initiat temperature is 2O.C and allfour el€ments are connect€d ifl parallel.
B. C. D.
att
l I
0
ootnorr 2 ohm.
I ohm! a ohmr
0
63 hrs 69 hrs 66 hls
Ploblcnt 79 EE Bo.rd Octobcr r99r
Four cubic meters of waler is to be heated by means of four 1.5 kW, 230, imn€r6ion heating elemenls. Assuming the efficiency of the heater as determine the time required boiling lhe water iflhe initiat temperaturc is 2O"C and the elemenls are connocled lwo in series in parallelwilh two in series.
A. 275.6 hrs B. 295.3 hrs C.252.2hts D. 264.4 hrs Probhltl aor EE 8o.rd Aprit t99? A carcuil consisting ofthree resistors rated: 10 ohms, 15 ohms and 20 ohms connecled jn delta. Whal would be the resislances ofthe equivalent wye load?
A. B.
C. D.
0.30, 0.23 and 0.15 ohm 3-0, 4.0 and 5.0 ohms 3.33, 4.44 and 6.66 ohms
5.77.8.66and 11.55ohms
Probl.rlt
ttr
37
lhraa r6sislors of Aohm resistance are connected in della lnside the delta lmlll.l thrac &ohm resistors aae connected in wye Find ils resistance betwe€n any
42.6.C
Probl.ll ?t!
I
2
EE
Bo.rd Octobcr 1994
The equivalent wye element in delta is
of3 equal resistors each equallo
R and con
B 42. B 53. A 64. B 75. D 32A43.854.865.D76.4 33C44.A55.B66.C77.C 34C45.D56.867.C78.C 35. B 46. C 57. A 68. C 79. A 36 B 47. C 58.8 69.8 80. C 37A48.C59.A70.C81.C 38. C 49. B 60. C 71. A 82. B 39. B 50. A 61. A 72. O 40D51.D62.A73.4 41 A52.8 63.8 74.8 31.
qg-5?
-
AE-35
- Conditional
Top
notcher
tt
SolutiorLt to Test
100) Solved Ptublens in Elect cal
r ;.1'#::*:i:fi'"-"-'"
Sllutlonrtolts[2
E=* ,=E-s=-11-1e I 1.18 =
t
R:t0t)
0.169o
I
600l
r=&=9=oer R 10
tr
=9-6 =3V
=E-vR
R:
I
t00
0_6
-
1538 46+153846= o
L.,.lng lh6 quadratic fomula
---
,' vR 1.3560 R trlo6
N,,t.
l=1.356x10rA
I
=1.3562-1.3560 = 0.0002
v. |
t=
1
147.5
R=lMO
A
,E
5r+RL Substitute condilionsl
15=-l
5r+4
E=7.5r+6-O 075= l 5r+9
E = 3.75r +6.75
+O
EquateEq.ltoEq.2: 7.5r +6 = 3.75r + 6.75
1538.46 +1323.433
= 1s7 513
x
lq 1O8A Let
A
- 4{1X153846)
n
2
v
356x10 "
:
Vre negativ€ rigD folminimum curent
- 1!!!39:.1!39-1!9
0 0002
Lel: r = internal resistance per cell.
t = O2r)
80,000 + 0.391'
.,-1538 461fi538.46r
r=5O
v
+lRi
ooo-4!99*rro.gsr
:0.2O
v
I. - '
2
-------->
R = tolal resistance of the line
. d, =(0.45x1000)'?=
Ft00A
202,500 cM
R-"! n
-]9!21?49) 202,fio
= o roz+
o
E. = VR + lR = 120+ 100(0.1024) E. = 130.24volts
Lr-+-1000
t
Assume iR = equivalent resistance of the circuit Vn = E' -lR
v.
= 750
VR
=585V
-1oo ft0.27+ o.o6x5t
v50 E,,= R,
10
=E-r:lL=99:!9=eo
I
39
10
l00l
Solved Prcblen: in Elecbical
a.=f R, R1
=f
n,f
+o
=so a,)
*Rr= t+od,
0.0043(55) Rr = 55 - 44.48 'l
+
= 44.48
o
R = Rr R - 10.52fi Ar
20€:
n.'P60 = l4 = 13{ =zoo., At1200c:
R,
= R1(1+ aa, ) = 2ao
- E2 R, P
!-
o.ooo50 20"
- 20.)l=
12o' 228
-63.16W
Ei = ll![ = 176 333 o = "475 R- = Ei = l!l![ = ,* ,,, 'Pb50 R-
I
rl
Rr =R. +Rb = 176.333 + 264.5 = :140.833 O
, E,
230
""&=410'a33 '052A
lr
E
*.=f=#=u. n, =f -ff=rro R-R( -RL = 11'5
=
6O
Pbs=rr'?R=(10)r(6)
R, = Rr ('l +0,^,)+ Ry(1+ arlr) R, = R,(1+0.00s^r) +Ry(1+ 0.0014,) R, = (R" R, = (R,
+
RyXl+ 0rar)
+
RyXl+0.002^,)
+O
+O
22s
n
Solutio,te to Test
11
+R, . lQg
i,
. !00-R, =1p
lqu.[. Eq.I io Eq.2 : ir(l+0.m64) + Ry(i
ir
+R,
+O,OO6
+o
+
0ol4) = (Ri +RyXl+0.0o24r)
0.00larRy
=
'R,
-O.mlR' =0=@
C,O3R,
h lubdtlt Eq.3in Eq.4
:
0,00!Rt -0.001(500-R,) - 0
;
0,OaR' '0 6 ir.126ohme
t'
.r.
'5;67
Lr
.
r ;'
t26n
t260m
r.f "P5
.{=z.zn
.E ' r+R
rl
,.E-n=9-z.z '| 0.8 r.
T
O,3o
r. (0.001)(2 x 900) - 1.8 o t-l.1mox10.6 = 2ooo A
V,
Er . V
E.
'
500
x10'
k = 500,000+2000(0 18) 503 6 kV +
R.-:L=:::=18133O '
P,
3000
''' P, 3000 "^=Ei=4=rn2n R, - R1[+ o (1, - t1)] 19.2 = 16.133 h + 0.0006 (12
t2 = 336 84"C
2
- 2o')l
R" +0 0024rR.
+ Ry + 0
002d,Ry
1I
[(.oz
-
zr)
3.t8 occ . ll sooo o+r€o! sr. z 6r tt D+qru - tu [(,}-
oom zd uzst=TA=-E="8 ome zd Ueet Sl=-;;=
!:='U
Argeoe-13 (8! o)oooz + ooo'ms .0!x v0o0z = u s.!
=
=
h-'3
rt+
009 h =
^m (006 x
d.l
zX!00 0)
-I
UC0=I
80 Z/- 9 =U--=r I
u+l
I "l
u
11-;9d= -:;- - u ur
oez! =
ui/ut
0
''l ,_
fi;zr=rU|
gurqo
g?t-
I o = 'utoo o ('u-oo9)loo o= o-'uooo o :tbf ute blqnuaqns ru'vzoo
'
@=o= ^ulooo-'ucooo
rU+ o+ 'u'vzoo o* 'a = ^f 'vi,oo o+ ^u + "Urv900 O+ 'U ('vzoo o+l)(ru+ 'u) - ('vroo o+ t)ir +('vgoo o +!)'u' :Z b3ol!'b3 elenbf
G)- 'u - ooe = 'u 009 = ^a+ 'u
It
a
,tal
o, sltolt vos
a2
E
1001 Solved
tuoblens in Elect ical
'
*,=f,=ig=*' 1
'T L=zoo.c r=-o.oo5 R, _ T+t,
R1 T+tj
*, = *,
ffi ,E110 R,
I
)
=
*.
(r...,.*_ry-)
=
.,.
"
4.93
=22 31A
E *,=#=#=*' n,* = n.! + ca,]=
E?=zz:e 11s' -F,oo=nq* = ieaw
ss[r+0.003s3(100 *20j = 72.29o
.=f,=ffi=u,nn n=o! A=
e
e!=
1.724x1:0::(0x100) =
32.sg9x1oacm,
=ra' 1
Ei
106) ^ ^^--lo0omits 6dnx_
4{32.589 x
la
n
2-S4cm
d = 2.52 mils When aeri€s connecl€d Ra +Rb
:
=36+O
When parallol connecled
El*;='-o
:
Solutiotll to
lub.tltut. Eq.I
T?^tt
to Eq.2 :
R.(ra-R.)-8(36)
i.'.30R.+288-o UdnC thc qusdretic fo.mula
-
ae
r
:
'6rlfoxzs8) 36ii2
36-12
Rr.36-12*24o nrlo Rrtlo
l!21
- 0.5
1111 Rt Rl R2 R3 ttl1 Rr 1o 15 20
n,
tqa
Rr-4615o
1111
R, R. Rb Ro 1111 1714 2Rb Rb
0.5Rb
R! =6O R. =12o
R"=3o wh6n connoct€d in series
:
Rr=R"+Rb+R.=12+6+3 Rr = 21o
I
n=ffi=ffi=uu'n ByCDT:
,
I,R
"=R,+E ,\= | (R, +R)=2(10+6.67)
R
\ =5A
n,t
6's7
t
Rr*
15r,1
Rr
200
2
13
12
1001 Solvcd
Problens in Ele.tical
=4=]l{=*..,
n,'P200
-T
1
T=---1-=2oo.c 0.m5
R,
Rj
_
T+t, T+ir
*, = *, (Jf!.u - * J
u(ffifr) =.'.'
.E110 R, 4.93 l=231A
E
*,"=g=g=*'' n,* = n.! + ca,l= ss[r +o.ooseqroo -eo] = zz.ze o
...
E2 n.=C=rrfi 1,1.s2
P,- =ioew q = E1=
P
n
3g!i
10m
= sz
e.,
=.!
A=
!!
=
1.724 x 1:0-6(0 x 109)
= 32.s89 x 1o€cm2
'o=Io"
'=',F= d= 2.52 mils
4(32.589r10€) -
r
When sedes connected
R. +Rb
=
10O0 mits
2.54crn
:
3s+O
Wh€n paral[6]@nnected
El*='-o
^----
:
Fr' Solutiora to Te$
trb
llul.Eq.1toEq.2:
i.(16-R.)-q36) i.'|-36R.+288=o UtlnC lhr quadrallc formula
:
t.- o i .,..6t- a6es8)
36r12 3a-12
22
2(1)
_-
Rr.36-12*24o rrlo - 11 21
R.tb.0.5
't111
Rr Rr R2
R3
&
1111 R, 10 15 20 Rr
.4.615O
1111
R, R. Rb
R.
1111
't.714 2Rb Rb
0.5Rb
R! =8O R. = t2o R. =3O Wh€n @nnecl€d in s€ri6s
Rr
:
=R"+Rr+Ro =12+6+3
Rr = 21o
R,R, i3M n=R,+Rr = 12+15 By CDT
,= "
.
I
I,R
R;;E lj (R,
+R) 2(10+6.67) =-- a-?
R "= =5A \
= o.oz
o
n,{ t00t
Rr{ tsat
R,
200
2
13
aa
I00l
!
ued ptublens in
Etect cat
Solutions to Test
n."ffi=1wa=s.*n
I',
ByCOT:
lt, t Pt = 2280)O
l,R ,''-R,-R
=f-n=!.(!!9
2P,
t'. l"
=12()
760
w
2(760) = 1520 W
=Re!+o.ooa(loo 2ot=1.32Rla = R1b11+ 0.006(1oo
- 2o)l= 1.a8R$
Not<_\hethev.lrawedudr .ur..,rhat20.C. r\en r.,=
p,!,
.r. =
R,.
AyCDT
L tl.4aRl
p
ll
P, -
=0,5286rr
-;R;-;R
E-'-
P,= 760 = 13.15 e l{ _E1=lq=658,, Pr '1520 E' 1oo2 ,, '"' R-- - 1316+6.58 |k.. 506.58W -
R.
= 52-86% of lr
Pl
At 100.c:
1200 W
1.i3sR
=1 'Rl 'l l1l R, R1 R' ^111 10 15
12O
r,
,t,
E':
E=F;="t*
750+250+200
Pr=f?99=ror
- 120
Note, Since they daw equal powe6 .C. ar j5 rheD Rr= &b_ R
P. R,,
=
1,-Pr=I99=75y
Rr. -R1.(1-oa^r) - Rr! * 0.003(60 - 1st t.135R," R2b ' Rlb(l * ob^r) - Rrlr , o.oosloo tstj. r zzsn,o E, . -11.
1991
+P, +P.
' - r't
r.
o
:
-----"> I, -2.5 A
At 100.c:
l.
15
+O
lirb8tilule Eq.1 in Eq.2 l', | 2P, = 22AO
5a5
2
10
75 =
45V
4s' -E; P, - 25O
l"
nJ
lr, -81cr
x--E1=!{=zoo., 'P,50
=on'u
n"-E1=194=roon ' Py 100
G-
N,n. !J thrt th. R3
20 Gr =O.2167mho
R,l
tqal
R,
rsn
NJ
200
hmp5 may opeiate prcpefy when connecte,l in se;ie5 too5s a ,'(){) V ntuilc, thc votage a*oss each amp must be equalto lOO V. rlus, the ,Arn,rL.acos: thc paralcL bEnch irust bc equalto thc rejijtanc. oFlamp y.
u -L
aa lql
Solwd Pmbf.nt tn Etectricdt Ekoi
1oo.
!.egl R+20()
R+20O=2R
R=200o
E'-#*i.* R'
-678o
toa 10a
^,'#**u -11 o
Rr
R."
-
=
1+
Rn
=
1.e23o
2(3X4)
2(3) + 3(a) + a(2)
80
t Solurior,.t to Tert
=ffi+s=rr.ezo 1. -Er=-IL-rrrae ' R, 11.67 nr
EI By CDT :
48t/
, f, R, 4.113t2O\ "-R,*&= 10-m
\,2.74 A
s"cor r" =]i4l ' 3O+R trr(6)
=
r.'(R)
'*--(#ffi) i
(ao + n)'! =
T
900+00R+R'?=1a,oR Rt -goR+9oo
-
o
!y quadr.tc fumula
:
t Ji6F - a{teoo) -"._ln co
Tala
I i
T
-
G)
for minimum value
g0
i
87.08
ofR:
00-870a 2
R.lt.46 n
R,..lgM*5=11o ' 10+ 15
r,.f,L-lf
=r.oor
. Eil' .l2C.0s) A .13 08w Pr
E,
t2v
1,,
,2
2
17
lE
I00l
Solved prcbtens in Etectticat
Solutions to Test
\ote 5rn." power. d%wF are -qua' rhc equralent ..s.
U.lng lh. quodretlc lormub
R"q=8
8=l1L 24 +R
2(1)
'2 R
24R = 192 +8R
:
Jzsi,lttx sool zs*st.za
zr ,
., -
2
2
26 r 61.23 38.12 A
R=12rr &rbtlllulc R inEq.2: lj = l, +1, 120_ 100-
R/2 R/2
100
120
Ra-1qooo=
Ro
=20 v
Rb
i;
t.R
R+R, =
qnJ to)
100, +
:
100
10,000(120) 120R.
="it.";=o
Uring the second condiLon:
=O
v -Vr V = 100-a(10) =60V = Er
100 1m 1-
by CDT:
r.(Rf R.)
50- 38.12
5.05 A
Urlng lho lirst condition
Er
r, =
-
60
100
15.OOO
n=ao1#99=oeoo R=6ko
_ '
80 '' 50 R t,
15.000
240_2OO-
R R 40 _ 100 R 15,000
t00 y
t20I/
""""> It
---'>L
t#=*m.#=u
= Vr -|LRL
,'=#=effi-a
substitute Eq.l inEq.2:
r4o so. ^^l t - s) = rpoo 6R" R,
Equale Eq.land Eq.2: 4(R+10) _ 60 R sO_R (R + 10)(50- R) = 15R
50R,R, +500
1OR = 15R
R, -25R_500 =o
r
oooo
"'[
r
140= 1t 400 R" ?50.6R"
t14o=t*-u*"J* tt mol^ l*" 140 = 0.0'146677R. + 66.6667
R"
= 4999.88
R.:5kO
O
]
h
=220
v
19
!! ! AJ!!!!!!!!!!2!!r, O = (lOoO
,r
i, Et""t,i""t
g)(4 186 J/g ,C)(
=l5g!=-
Solutions to Test 1Oo _
1
5) = 356.810 .t
3558r0
w"* 5oof;;*]=o?eo6
fir
-
.J/
r,
-
!1r
|I () 0(,
. r00s"r,,ffi"ffi,ff_a,are
L-
X:-
il1'.itj:
!T.T[',?']lo.
-
283s0 kca,
= o.24
{J
Pt
P-2o7.24kw
t
"
!9lgg l
o
Q= (80osX4.lss J/g.CX100 - 20) = 2679o4 J
,w2674 = -i6o'-
=,ut n*s.c E 4.46hjn
x
ou sec =!!L
Note,l atore = 4.j86./
10-.(1)+1.667=5746.C e(slrot
*.,
f("a,)
p9!,-1rL \ min
57 6oC
l,
= ar
\
=72.6"C
-!1
rle
111g,r 60 sec J s ,c I
l,^ ./,-'
.
,
= 57.8_15
iating
ofrh. hearo
is
rlways the power drawn from rhe suppty
Pd,M = 4(1.5) 6kW = P*a* = Pon I = 6(0.9) 5 4kW = o O = mcdl
Q = 23.98 calories
+O o=Pt+o
Wn = 10oo+o.25wn
=4
m.rlqg&
l= 4ooo
Wn = 1333.333kW-hr
'n
o
equate eq.t roeq.Z:
= 0.24Pr =0.24Ett
mci! min),
!9!99
=
szo.zr
r""r
ri
J.
f'c,::=
zler=
Note,
E
loo{1)
Q = 0.24(1 10X12X30
r.mz"c
570.24
I, =
o zo1'zoy1rel =
4Js6 =;.i86
E
szy =
- n,c^, rO 0 24pr = 0.24Etr+O
A
Pt n--
:zy-
10.22kS
Ltriats Eq 1lo Eq.2l 0 24Etr = mcA,
tmin J
Q = mcAr
E
=
(),
zesrz. t os = o.zap f1o ,nin
I=
2
l, - tJ5.43.F
cl,n = 29842. 105 kc€t
o
orbt
f,1:sl] -, r: r lnc(1, tr)
\=79.1o/o
25)
1h
i"
= Pt
mcA, _ 4000 ks
= 248059.259
sec
{4.1s6l{IqI).(!g:!qlg
2
Sl
32
l00l
Solyed Prcblens in Electrical
I = 248059.259
l:69
hrs
=;
=
R
t 2 2
Sol
x3600thrsec = 68 9h.g
e- t0 1o(2ol--lllc, 20 + '16 +
Jo(16) .rsso r. - lO+16+20 tsoo
= 35.266
c- lO l6(?Q= s.eeo , 16 +20
o
rs rlway,tlr powcr,l.awn 6or the suppli Vlheh two Scaterecncnt. are.on.ected In e!. {he'ort,:9.a ros. eacr element wll be equalto l,/2 oFrhetotalvo hse 5,1c. rct spe.l'icq. a*Lhe (he. rLir -onnF.l,ol qoe. no, a4a trc e#iciency o(the heato.
dre 'atrng ofrne heare.
€
lr-rR
R,
i,-:
Convcn lhe Y resislors lo a R^
""* =.[+] =.[-q*] =,uoo * =, .** P6ftdtu = PdM n = 1.5(0,9) = 1.35 kW Q = mc^l
+O
.
=+
t
mc^r _ 4000
kg (4.186
P
kJ/kg oCX100
1.35kW
=992237.037secx
t = 275.6 hrs
Ry - 4.5+ 4.5 = 9O
thi
3600sec
R,R, 4.5(9) 'R. -& = 45-,
Rrt =3rt
ro
EquateEq.l to eq.2:
t_
18+6
"*
," fl!99-9) = rooo t 1m' ./
3Ry = 3(6) = 18O
p' - 1!{9I=a5e
,.
o=Pt+O
-
-20)'C
:
tio8
to Test
2
53
-----------xF51
1001 Solved Problens m
Electrccl
Netvatk Ldvs &
',1,,1
1
,
thetens
SS
i...lr!l lr.ric cvr uate.lby tDultarco!t trbstjt!irons ofthe
, I' rt,r r l,rtrrLrl.rtc.l,A n9 KCLand
KVL
lllxwltls lsll flIl0[ lhll
trncitoF$ uw Named afrer the Geman physi;$, cusrav Rob€.t
,,
lnvolves a set of independent loop
curent3
assagned to as many
.r rl oxisl in lhe circuil and lhese curents are employed in connection wilh itath'r.lhod lftxot
rrln rosislances when the KVL equations ale written
Kirchhotf(1g24- 1887)
CunEnt L.w (KCL) - the atgebraic sum ofthe cur.ents al any junction or of an €lectric circuit is zero.
t
n/
Voltags Layv (KVL) - the atgebraic sum of th€ emfs and the resistance drops in any closed loop of an electdc
ciroit
6^r,
is zero.
Sign Convenlions for Kirchhoffs Law:
O Currenl towards lhe node, positivecunsnt. o Cufi€nt away frcm the node, negative curent. g ln a vollage source, ifloop enters on mrnus and goes oul on plus, positive Cl ln a voltage source, if loop enters on plus and goes out on miirus, negative O ln a resislance, if the loop direclion is the same as the cur€nl direction. n€getlve rrabtance volta06 drop O ln a resistance, if the loop directioh is opposite to the clrent direction, reriatance voltage drop_
R)
Itlvl ,
@
''t'A Ei - lA(Rr++ R,) - lBRl = o 't'lJ l::- lr(Ri R6) - lAk = O
,.
M.\h.!rent5la
and Lua'ee€luated by 5im u ltaneou, 5u brtitutions oFthc -1 ,.rt ons lotmulate,l6om each oopormerh using KVL
q;lqi
q"]rl
s|PtlP0sm0x [oRfii of resistors, the currenl in any resistot is equal to lhe algebraic aum ot lha curronts delivered by each independent sources assuming that each source is tolhr0 alone or independently wilh respect lo lhe olhers.
h r r$lwofk By KCL By KVL
:
E1-liRr-lrRr=O E:
-
trR.1, lsRs
E1- lRj +
Ir
=
l,&-
O Er= O
56 I00l Solved Prcblens
in Electti@l
Neteork Ltus & Theorens 57
tottlrou flllo! lhl. m.lhod, r clrcult wth "n" nodo!, has a solulion with only
lill.llon.
Note,
5TEP 1
:
nacd.d.
lFa 5ource (eithda cutrentoia voltagesourcc) rs acting alone, ihe other cureDt sources a'e open circujted whileth€ other voltage sources aie rhod
L
lfEr ii acting alon.
Connon node
\rr.
tl)crc
aicthre (n = 3) node5
rn thrs crrcuit,
therefore on ly two equitions
.'r. r.al.d toiolvcthr5 PDblem
I sTEP
2:
lfE,
i5
acting alone,
| {l i!rl. b, lr = l!
+
15
Nd,lc voltage5 v.and Vbare daluated by sim ultaneous substitutiont oFthe .quitioni f;'mualtcd using KCL an,l otespondrn!!, currctts flowiDg througb tr.h rcrstancescan be solvcd.
Note
,
ltre cunentr ll', l:', L', l1', l:' & 13'arc cqluate,i ushg
basic electrc
circd(
If,Iwilt$flI0nil ll a taalalor of R ohms be connected between any lwo terminals of a linear network
ra.ulling steady state current through the resi6lor is the ratio of the potential E. (betw€€n lhe two points prior to the conn€ction) and the sum of tho resistance R. (resistance b€tween lh6lwo points) and lhe connect€d ol lhe Yalu.. lltla{anco R. Named affer the French telegraph engineer, Chatlea Leon Thevenin
ll.
allLr.nc. (1667
'
1926)
St l00l
Solled Ptublens in Electical
Thevenin s equivalent circuil:
Example, solve for tr.
0
sTEP 1
|
Note,
Open circuit &, and:olve for the voltage acbsr the op.n crrcu(ed t€lminals.
Eo rs
computcd uring any rhetbo,l5 (Ktrchhoffs, Ma&e I, Nodal, etc..)
aDalyziog oetwo* Problcmt.
of
Netwotk
Conprin
t
hll.Fn{.nt
LNs & 1'heorcns 59
(rhort circutall iodcpcndcnt voltage source5 and open clrcuit all curcDt tourc.t).
R,
RJ
+
h
lahcornputcd
w
using the basic priDciple5of iindinq thc tota I rcrista nce
ofa
llwn clrcu(.
(on t,
u.l thc Thev.nin': equiv?lent ciEuit and solvethe resultiog cunent RO
R3
rcnmffilf,t0ltx l! analogous to Thevenin's theorem except thet instead of lhe open theorgm uses the Bhort drcuil te6t and tho 6quivalent circuit is a lhla , Cmllt. Nomed afrer lhe Amedcan engineer, E. L. Norton (1898 - ) aqulv6l€nl circuil:
l" I
N.t\|o
50 lo0l Sobed Pmhlens in ElecEical
Laws &
Theot.ns
61
ruxl $fltaltr E,
numbat ol voliag6 aourc€s of arbrtrary geneBled vohage and finhe iiinoo ddhrcnt trim zero are connecl€d in parallel the resulting voltage '.'dh__-'_toomutnatton ia ttre ralio of lhe algebraic sum of lhe currenls that shorl circuited to the algobratc sum of the Siidirioir. v 0.r,""r.
lIltlolanc.
tQulvll.nl clrcuil sTEp
1,
Note
short crrcuii Rj ?n,1 soh€ tbe sholt .irc! it currcnt that llowstowardsthe
l'.,5 computed using?nymethods (Kirchhoffs, Ma"well, Nodal, et..) oF analyzng octwork poblems
5TtP2,
Not.,
Compute R.
R.6 computed uting the given
5TEp
basic principle,
of Fnding thetotalresistaDc. ofa
ciruit.
3, CoDstiuctthe Norton's equi!4lent circ! it and rolve For I'
l^,
"ten
62
l00l
5TEP
1
Solved Prcblens in Electlical Enpineehns br R. R
Drawthe Millman sequN?lentcirco( and solve for Vrb
5TTP2 5olve for
lJ.
b and l:
stuncr
rmrs[0[ttll0I ilHIt0!
Thls method srmplifies lhe numb€r ofmeshes rn lhe nelwork and thus simplilies the number ofequations needed.
*
Voltags source to curentsource
tre .,
Cunent source lo voltage source
Es Rs
It./l,'or[ LaN &Thzon
tblw
(or
ll
thc rcltagc and cuncnt sourcet to
th.lr
cqu
i€lcnts'
branch andtransform bacl
ltmplltthc Ptallel
-------->
llP
!
By
It
K\iL (utng thc aivcD looP d irc€tioD ), tt crn thcn bc solv'd
9&-1"(p.
'p"
"
p.)-
1"p.
-s
s
63
61
I 001 Solved
Ptubtens in Electrical Test
3
65
xt ru t0xftfiIr $Ennr0nl Maximurn power transferred to 6load rss,stor occurs ontywhen the sad resistor
1-"j.tu.,_"q*t
to the resistance (R.) of the networi ro"t,ng u""k
Ex?mple Find
iFit 6to absotbe,l maximum pryer.
R
f._-t;"'
aL-
alr
EE
Bo.rd Oc3ot
r99C
A 12 V baflory ol o.os-ohm resFtance " and another battery of 12 ta.r.r.nc€ supply power to a 2_ohm resistor. What is lhj curent V and 0.075
R3-
lhrough the
Is
llnA 103 A
tf2
A
lt 91 A
solltion: o?enrrcu( a
rcsistor R, shortcrcqit ail independeDt volhge, 5ource5 aDd lI in,lepen,lent .uneDr sou rces.
04r
p'ip.:itn1["str":,fi3"f,ffi:1"}[,s";j;H:x,lr"J?"",ilffi Rt
RJ
R,
A, zrv
|
0 0
.Ro
2sv 25v 3ov
trobls!
a5! EE Bocrd
At dt 1989
Tho LRT lroll€y syslem 10 mites tond ir fed by two substalrcns that generate 100 volls and 560 votrs, rospectivety. The ;esistanc€ or lhe lroley wire and rait ler n tl o 3 ohm per mite tr the car is toiated 4 mite" r.o. U," OOo*ilr lrom the line. How much is the cunent supptied by eactr "i"iii"-oi*J)"riii"
station?
^
No(e,
ifrhe @aesuo'rl "s .Lr.c1, or po*e. drah') by,c!s,o, p 5 to elhe the\en|n ror \o,lon s thco,em ro.o.ve tre p.ob.em
be
ro .ed ..e
I
c D
_ -
133.33 A, 66.67 A 123.67 A, ?6.33 A 1 17 .44 A, 82.56 A
125.54A.63.05A
Problqn 36t Ten cells each ol emf 1.5 volls and inlemal resistance of 0,2 ohm are ioined in pamrer and connected to an exremar circuit resisran* o ir,rn!.
i
A 0.45 A 8.0.65A c 0.50A D 0.48 A
#r,"iii,i#iiiili
i
66
lOAt S.)lr!Lr 1,tu1)k"8 u-Etetunol t hgtneeting
b
TPtl3
R
Problem a7: Two battedes A and A are connected across each other with lerminats of
same polarity together. The open ci.cuit emf and internal resistance of each b;i js respeclively 24 V and 1.5 O. Oetemine the resistance ot a healjng toad conn;i across lhe parallel combinalion of baneies so lhal tne power mnsimeO in tne t is 100 watts.
I
0
4.12o-
Problcm 8a: EE f,oard Octobe! rtx)o The lead storage bafleies'A" and "8" are connected in para et. .A,
,r.u.rn 9tr - '-a has
open circujt voltage of 12 V and an inlernat resistance of 0.2 ohm. Ba(ery,,B,, h;; open c'rcuil voltage of .12.2 V and an internat resistance of 0 3 ohm: tf the b€ efles together detivers power to a 0 5 ohm power resistor. Negtecting effecls
i
temperature how much cLirrent is contibuled bi baflery.A"? 16.00
c.
12.85 A 25.24 A
D
t "lr" . respeclrvely. i t """rii"i,n
l. / lllit
I
ll
A
Three resistors of 2 ohm resistance arc connecled in della. lnside the d( 2 ohm resisrors a.e connected tn wye. six oatrertes oi neirigi lTlti ]l':: inle.nal resistance and ot different emf are inse.b; i"6 .";;;;; U"Y;;; ammeler. lhe current rn one ot tne defla Dranch was tound oul ro be 3 A. tt a 4_;h
wi
be the new cunent?
A.2.00A
D.
1.75 A
1.50 A 1.25 A
Problclr
9Or EE Bo.rd Apltt r99" ln Man a, the LRT runs between Grt puyat Statron and layuman Stallon wh. _ v^ott.ees of 420 votrs and 410 v;rs respecrivery ::_-.^l(Il-?:l_i"9 resrslance ofgo andTlllta,ns retum rs 0.05 ohm per km. The lra," d,"; a;;;;i;;i;ijr or ruo A whrre in motion. whal arc the currents supphed by the two statrons rf
i
train ls al lhe distance of minimum polentiat?
B,
c. D.
l) l) ll ()
ltt EE aoard octob.t t96t tr.tl.D - ,,harqer, a battery and a load are connected in patallel. The vollage acrcss lhe,
Plobleln 89!
B. c.
Deler mrne the ohmic value of R if lhe powet absor bed by R
wAlls
0 -
resislance is inserted into that branch, what
n connected across two batleries A and B connecled in patallel 'n"'"iun"e s.rs '" ana inlernal resrstances oflhe banenes are 12 V 2 ohms and
fm
A
29.62 A
a.
100 A over a 100 mm hard
of 0.270 ohm per km The rail and *rrri ,.'l'r- irottev wire having a €sisl;nce per km lf the siation vollage is 750 V 0 06 ohm n* *"i"lance df lii"-rri lrr.,'lliciency '"i''" " / oltransmrssron llnl
^ li
B. 422() c. 4.o2 0 D.
9rl EE Do.rd Octobcr t9a5 '-rr 'tFtl.m r iii i;llr, s tm oistance fiom lhe Tavuman, lakes
67
A
nlrl0r
rs
i2
6 vons anO the banery has an eml of 12 volls and rnlemal Iesistance of
ili i""' r""
load consists of a 2 ohms lesrslor Frnd lhe cLrrtenl through lhe
(il A r|254 0424 l) o50A
t(:
{r
,"obl.D _
94! EE
l {: t)
w w
lo.rd octobcr lt9a
lead ;l;rage battery is €ted al 12 volls lf lhe internal lesrslance is 0 01 ohm rllntls the maxi;um power lhal can be delivered to the load?
3.600 / 200
1800w
175 A, 125 A 183 A, 117 A
Probrer 95! EE B(|.td Attttl a995 a l2o-tiaterv havrnq an inlernal resislance
164A,136A 172 4.128 A
lh,, ballery delivet to the load
of O 5 ohm is connecled lhrough What maximum power will load r€srslor' io a vanable n h,(, resrslance of9.5 oh;s
resistof
68 B.
1001 Sotyed
pnblens in Etectical
Test
63 watts
toor EE Bolr'd Aprll t99t lw.lv. .imilar wires each of resistance 2 ohm6
C.630wafts D.360watts
;]urfr lTi"jJ,l#j"!T.:"T,:;;X?ll:r1x,:"i::iil;til,,":#itr: ;-ifffi "''":"".#'.T1"rJ T B'f5|",f '*: ,,'3;"'j:,
*l,Sgiffi 8.0
1.8
92, EE
Bo.rd l.pdt
rrsr
ll,?:*:#
022v 02av
tqr
EE
Bo.rd Augun t97C
the value of the voltage V.
L 31V fi 24V
223.94 kW
g8r
12 19V
1166V
21v
D. 142j2W
cornem
c. D
Lanp 60w
+
t2v
t9tt
stmjtar wires eech of resislance 2 ohms are connected so as uuue. rrno lhe resislan@ between lo form ^,.h-rweNe the two diagona y opposile B,
0.10
rots EE lo.rd Aprll R.lcrring to the circuit diagram below, if the chaqer voltage is 130 volts and tha
Probte& t9l EE Bo.rd Aprtl r99r
1.45 ohrns 1.66 ohms 2.01 ohms 1.28 ohms
difference d€veloped between
0riv
tlnd
._iffi,i#,,;.,: ffi:1""1""1*3""':'i"i 0.25 # ?#t?i:r:i,i!l ohm. sorve ro, the m",_,,i'i"*iiiitl,i#v,11,:f ffffJ"yf,i:,9jeli*, A. 130.20 w B. 115.52 W c. 120.21W
I
l99l
0lo v
230.77 kW
Ploblqr
EE Boord Ostobcr
ly opposite comer. Calculate the
8.220.35kW c.242.73kW
D.
I l0l l)
torr
;T';[L",!i{fffl.r.':.ffi :i{irri}$;:3l-".", A.
M6o
hrlv. idonticalwircs of resistance 6 ohms each are arrang€d to form lhe edge cubs at one corner and out at the olher I trbr A curent of 40 mA is led into thepotential
()
PlolFln
are c.nnecled so
I ll3 (l I t02 (l
rr
a. 1.5 f,r c. 2.0 cr D.
69
llfid lh€ r6sislance between the two com€ls of the sam6 edge.
P.oblqn 96t
A.
3
vollage is 120 volts, solve for the cunent
-o.215 A
0215A -o 306 A
0306A Batrery
Chatget
Ib.
]iF 70
l0Al Solw.l Ptoblens in Itlecbict
Test
by R. Rolas Jr.
Problcm ao4r EE Bo.rd August ag1z ln the ligure below Rr = 1 ohm, R2: 1 ohm, R3 = 3 ohms,
ll.i
A. B. c. D.
0trA
V. Fand Es.
182.41V 153.32 V 164.67 V
R/
loTl
iolv. lor
I uslnq Sourc€
03llA
R?
I
157.22y
Rr
+
Problern 1o5r EE Bo.rd Ostobcr r9ao, EE Board Apdt r9a4 ln the dc circuil as shown, the high resislance vottmeter gives a reading
211/
)-b-
5 ohms 3 ohms 2 ohms
c.
0.028 A 0.010 A 0.025 A
D-
0.0't4
B,
a
V ustnq Maxwells mesh melhod
I?V
l6v l4v
t2v
EE to.rd Aprll Detemine I in the figure-
Ptobl.rn 106!
roct
lolvo lor
+0.435 volt- What is the value oflhe resistance R?
B. C. D.
llansformalion method
rgto
hrDtoo ro9:
'--rr.i.*in" A
I, 0 0
254 60A 33A
tt
rn
" "un"nt
the 1-ohm resistor using Norton s theorem
3
7l
------.F!2
tl!!.!:\:t::!tn*ren.g
n
trec
-!u!1ty!!g
by
R_ggt1,
Ploblcllr ltor
a.
c. D.
Test
Detemine the curenl in the 1o_ohm resLtor using Thevenin,s theorem. 0.8334 0.667A
0.707 A 0.508 A
3t2
t0y
Plobte&
t0!)
lrlt
Oeleffnin€ Ihe;urent in the 6,ohm resislor by Superposition theorem. B.
4.2 A
c.
4.O
D.
3.8 A
hurnrur
A
I2v
lirttrm rr.lr
l,otormine lhe vottage V
A
I0
0
l2v l0v t0
v I2
Probledr
t'
u2!
D€termtne the looking back resislance belween terminals a & b.
1lo
B.
9f)
c.
10o 12o'
D
30r
D 90. A 97. A 1O4 C 1.t1. B c91 A 98 C 105. C D A 92. A 99. B 106 D 112. 113 c 93. D 100. D 107. A 114 c D 94. B 101. A 108. c 95. D 102. A 109. AC c96 B 103. A 110. B
aA-3P
Topnotcher
- (onditionat 0- 1,5 - Failed l,E-el,
3
73
74
l00l
Sobed Problens in Electical
Solutiotts Tat
Sol&BbTlrH -2\
=0
+O
I = 2a0 aoll 12- 0.075t,
+A
= 133.333
1.2
A
560
1.8
Note opeD circuit
+O
+1, =lL
l,
600- 440
lJse Thevenin s lfroolem: Solve for the looking back resistance acrcss lhe load terminals.
Apply KCL at node'b':
I
600
VL _ 560 - 440 1.8 r, = 66.667 A
I
=0
2tL
l, =i60-26.671
v, 1.2
, t" '
Apply KVL on loop fcbef
75
VL = zl40 volt6
Apply KVL on loop'dabed
12-O.O5\
3
'^n,ar.e
Subslitut€ Eq.1 and Eq.2 in Eq.3: 240-4011 +160 26.67|1 =lL
s equa I
R!. 5hort-ciicuit allthelocclls. By in'pection the lookinq back to 'l.c egur\a lcnt .esistance of 10 rdenriLqIre.ftols -onn;e,l n
1= 93 =o.ozn n^-10 = 10
67.6711 = 400
lr =5.91A Solv6lhe open circuil vollaqe across lho load lerminals:
\'tre
'=#=o Apply KVL on
,T
loop'odbacr ./J0r,
u-,lE)
I
o
l3I 'lt)= l2 ,/
v
Srnce
the cejh
cr
t'trdl e .ohbin"r.on, thui Eo = 1.5 volts
U8ing the Thev€nin's equivalent circliitl
0.167rR=0-o
Substilute Eq-1 To Eq.2:
tRr',
J
="|.t=ffi=orsoo
r,
v_o j 67fl!q)(R)
rdentrcal therc s no ctrcrd{rrg cureri in,ouqn the i. equal rorhe ope.r c,rcu,r.mfofore ceit
Eo
]L
ll :o.sA
=o
v - 25.05 V
=
=0 25 volts
N,,ta:5rncethecellsare identjca, the load cuffent
q = 0.3(4) = 1.2 O f, =0.3(6) = 1.8o
,
UseNodalnode method
ApplyKcLatnode'a: lr +1, =lL
- vr 1.2
600
560
-
1.8
vL
0r)
5
djvided equally
.r,,,Jr!thetcncel, Refeitotlecjrcuit,{ragianl, conydera loopwith
-^^
)
y r')d ihe loa,l
t =4.2O
+
6oov
ie5r5taDce.
r s.
r5
3.02tr =0
rr
. 0.4086
rr
:0 5A
"necell
IL
I
L
rr (3) = 0 ;(0.2) -
0/1o)
+
I
.F 76
l00l
Solutions Test
Solved Prcblens in Electticol
U.. Nod.lnod. mclhod:
Use Thevenin's lheorem: Solve lor looking back resistance across the load lerminals:
Note Open.,rcL i P . )Lror-.0, d i'lae lwo I dre.ies. B/
r..p+ro.'F.
Itdy KcL .l lr'|ts-lr l
ba.k rerstan.e F equalto the equ valent rcti(ance oftwo rdentical re5ittois
r "22
15
=24v
Using th6Thevenin s equivalent circuit:
a=l-54 .
'.-**;=;#"=o,,, +O IL'RL
=
Uta lhavcnin'3 lheorem.
0t
n ckcull RL and solve for Ro
is-R,,
2+R
-!-O
r,- l.;o
Thov6nan' s equation :
i-n"rh,
100
0.5625+1.5R1
+Rr'
R.'?-4.26R.
0.s625 = o
= 5.76Rr
lli, -2o lr-t' 1no *n.1= 312121
U6ing the quadratc fomula:
-'
+.20.,,{+:of -otrxosozsl
RL(.)
=412o
211)
Rrl-) =0.136o
rt:l 59
126-10.03
lr '12 E6A
Udng
57ffit
+
03 volle
sr-2'213*ut-zo
',*={ 'o I'n,' \0.75+RL, )t
l0
aUldllul. R"in Eq.li
Substitule Eq.1 to Eq.2:
ro 75*R,
.
. t2 6 V, tt'oz'T
Note, Since the batteies are identi.al, tltre j5 Do circulating cuft.nt i'r the for the two batteres, thu, the open circuit voltage is equal to ile open ci voltage oF.a.h battery.
PL =
lrn vr ,122 V1 _ Vr 0.3 0.5 0, It llv, ' 10 66 3.33V1 = 2Vl l0 vl - 103 88 Vr
Solve the open circuil voltago;crcss the oulput.
Ea
nodo 6
4.26 r3.9A7 2
lu - tZ votts
I a l) r€sislor is insert€d i -2r4=6O a
Eo R"tq-
Ir
-
ll
-15A
12
2+6
1
3
77
7t l00l Solved
t,=005x r, = 0.05(4 x)
i-027(6)-1.36n h.0.0.06(6)-03o
Apply KCL on node 'b': lr +1, = 300 11
= 300
Let
- l,
tlo
-loor, =o tlo..loq1.35)-E, -1oo(0 l' . 6tl v
+(D
VL = voltage across
th; tain
420-11(0.0sx)-VL =0 = a20- l(0.05x)
+O
V!
=
x)
4t0- lr(0.2-
l||. llodrl nod.
vL =0
0.o5x)
+ 0.05x1,
,,-
02 l, =75x-50
Subsiilute
in Eq.4: VL = 420-15x + 0.05x(75x 12
-50) Vr =420-15x+3.75x'z -2.5x
vt
=
420
17.5x+ 3.75x2
!L=6-175*75'
u. v 8-V. 2l
75(2.333)-50
300,t' = 300
t1
=
11
=1754
+
+ 0.05x1,
t.ft-1F=o lqud.
la
Eq.1 lo Eq 2: 7 656
1.5v1
=-
llvr 1.5v.' = 7.656 v,'s::v,*s.to+=o -4(1)(5.104) 9.33:8.162 = z u''I -'33,Jtg.33))o;-
x = 2.333 =
8-vr =ll .11- l.5v! +lD
a-0.0v1
|
,
Ulng lhe quad.alic fomula
0=7.5x-17.5 l, = 75x-50 lz=125p'
method: nodo '4.
l. rlr - ll
+O
Lqualo Eq.3 to Eq.4: 11O-0.22 +O.05x12 = 420 - t5x , 410 -420 + 15x
d
Arrly KcL
=)O
SubslMe Eq.1 in Eq.2: V! = 420 - (3cro - 1, X0.05x) Vr = 420 - 15x
3) = o
i. tltt
Apply KVL on loop fcbef:
alo - lr(0.05)(a -
loorr -E,
n'l'fl'*=#"*
Apply KVL on loop dabed':
vL
Solutions ?est
Ptoblens in Elecoical
125
V.(')=875volts
V|
=o
584voltg
'
t!!-
3
79
t0 l00l Solv.d Ptoble^
$lutit' Te 3
in F.le.bical
fl"l, l!t
Substitute VL = 8.75 in Eq.2
" =Z!!9=e6754 8.75
n=!'-='875
Oll.n clrcuil
1,
rL
0.875 n,,
p
,,l.lrr.t!n) Pow.r transfcr' lor,l
l,i,l ,!tl.r, !,.nd.rD.. meilre.l
,-29
Apply KVL on loop abcdefa': =
0
I2I/
= 6.5 A
=r=00'lo
N'(r 1,, n)itimom Power van56r, load rc5istance must be equN€lent h'lrnt h.rc[ rotstance measu'cd at the load teiminaG
Using Thev6fl in's equation:
=uft.=ffi"=ooon =r!'?Rr = (600)r(0.01)
E.
Rh* +Rr
I
600 0.39+ 0.39
E"= 600 v
.230
R1
= (769.23)'(0.39)
77
kW
Nrrt. l(r'naximump ertransFer,load rcsittance must Lr,Liir,I'i.k rcsistance measured atthe loa,l teminals
Note, For maximum pola$tianjF.r, Joad rcsittance D{]5t be equivalentiothc looking back rcristarce mcasurcd at the load tem in.ls tn th rs problem. Rr = r
th.
A t - 760.23 rR,
ft
Pr = 3600walts
to
IL
Ir
a - t,
12v
-z[ffi'zsoot) -039(l
I,
t:0.011)
PL
& r20v
r,-.,
Note, Fofmarrmum powei tranrFer, loa,l resistance must be Eoi€lenttoth€ looLing back rctrtance mearurcd atthe loadteminals. ln thrs problem, tu = r
r.
2(8) 2'8 ".^
i
Apply KCL at node'b: lc = lr +lb = 6.25+0 25
RL =Ro
and solve lh€ looking back resislance across its terminels
R:20
ls = 0.25A
L
t" th'
ri the Lo!,]terDinalt
llr -Ro - 1.5 O
r,-13r=e.zse 12.5-tb?)-12
Ry
rcs rtancc mu5t be cqu vrLcnt
EI
Otan ckcuit
Ry
be equie lcnt
to thc
and solve lhe looking back rssistance acloss its terminalg.
R,,-=9.50 RL =Ro = r
+Rlr =05+9.5=10O r=0.5n
Using Thevenin's equalion:
.t
=
E.
Rn
RL=fu
12O
+R, = 10+10 =oA
Pr =rr?Rr = (6),(10) Pr = 380 watts
120
ir -o-' = !?1]!!l = o zoa o 0-25+1.25
E2
Solutions Test
1001 Solyed Prcbtens ih Etectricat
n.=ozs'fffffi-orzeo
:q
rl
r,=-;=oi;.28.044
1
ln Eq.2
\ -0 t, -0
rl)
tfiI
By CDT:
, 'L
(1.25)
l,
28.O4t1 25\
lor tha
i25E = -E i2os = 21 04 A
=
.t
P, - t.zRL - (24.0a),(0.208) Pt =12O21W Noie,.ln,l€pendcnt.of.be cn-u rt .on n.d roD. rhe toial powe..l,awD by
equal lo the sum o{
th. powe6 td!f,D
b
Lel: R = resistance of eadt leai3ld
R
=6(*)'R.6(*)'R
rn,
=f(,n).${,n)
R. R.
I l^
!
I
6
I
Apply KVL on toop ,abcda': (r- ztj)R - tlR - (1 _ t')R - trR 0 = lR 2llR - 2lrR -trR + t,R =
511
=0
+l=O
Apply KVL on toop ,efghe':
(, l,)R irR 2trR t,R=o lrR r,R t,R 2t,R t,R = 0 st2 \
=
0-@
t-
.112' ,2, .!
I
Lel: R = resistance of each rosistor
l,
'.O
167 Q
lda {.
o
=o
.rR,
tt2
=9R=!(a 86
r-Srr+ru
t
gq 4 to Eq.3:
3
@
-['-z(1)1"=f n-6
12
+:R =:R 93A
Rr = 1.66
a,ra
lor th. tot.l curr€nt:
!
= Pr +P2 +P3 +........+p1,
,,R,
a
eaLh rcr;stqnce.
vollagc 6cros6 branch 'ab:
R
.
thc P oblen #99 ro, 'l
e'o'-)ur. 'o
/or
P
roalltence ot each resistor
.6a - 9ror=sn r1
,
O,K
-1,
II J
I
t I._,
d *-t,
I-2r l
'tn1".
68
l,.t,Ri.(4Ox10'Xs) lr - o.2o v
/,1
I
+ 21V
y'
ita t
\)
_lj,.,.
,l"'+r''
,l
4atL
60w
r2v
3
83
E4
I 00
I
SoMng for the resistanc€ of the lamp:
=:.L=\:L=21c 'Pl 80
R,
r,.f -$-...'o
,t
Arply KCL rt nod.'b'l
l,.l'
UBo Nodal node malhod: Apply KCL on node'e': ll =1, +13 +l.i
24,V V -12 V 1 0.1 4
Snlutions Test
Elect.itul
SobeA Problens in
tr
V
-
+h - 2+ a0.67
ta lo
{
Ir-24
+
a2.67 A
+
I
+
Es
-
v
3
Es
f" 30
120v
Atply KVL on loop dabed':
21.
24- V = 10V -120 + 0.25V +0.4'l6V 11.666V = 144 V ='l2.34volts
a. - 12.g7ll) -2(1, lr .164 87 volts Ndt.,
it'.
''
I
5iDcc
"r'*'t.
12O =
0
in thc voltmet€r has a very L'igh rc5itta nce .omP?red to the resistanc's no* tow.rds th; voltmeter is negligible
tt'"
-"*t
12
20+R
I
'| 1o+50 '---lL=o2A
I30v
ch
t2y
Apgly KVL on looP 'cdabc': O 435 + l,R -lr(10) = 0 Use Nodalnode melhod: Apply KCL at node'a': 130
-
VL 120-VL
3240
a3.333-0.333V! 0.858Vr
=
ocu.(jfo)n-to,x'or=o = t.ses -1?L 20+R
Vi
+60
0.5Vr
=
0.025V!
103.333
VL = 120.43 volts
|'r- 120 12O 43 2lb = -0.215 A
Notc: Tte negativ. sign denotes only that the actual curcnt is oppo5jte given dircctron (charging the batt€ry).
Apply KVL on loop ebcfe:
120+r;Rr-V=0 12o
+211)-v =0
V =122volts
12R=31.3+1.565R 10.435R = 31 3
R-3() L,so Thevenin's
Not.
theorem
Shoit circuit the 12-V suPPly' oPcn crrcuit the
,)(.l{ln9 baak resirtancc
ofthe.rr.uit
at
5o-fl
thesat€minah
resrstor and solve
tle
86
Solt uuns
I00l Sol'ed Ptuhlens ib Electti.al R^ =
FlmIllly lh€ parall€l
!!{M* lq9qr
" 20+40 10+30
Ro = 20.833 O
,. ''
Solve forthe open circuit voltage:
',
r.= 12 =o.ze 12 ' =10+30 =ogl
200
t"
l2v Apply KVL on loop'dcbad': Eo
+r,00)-
tr(20) = 0
Eo = 0.2(20) Eo = l volt
10n
-+ 10n
300
0.3(0)
Ljsing the Thevenin s equivalent circuit: Ro
+Rr
20.833+50
l=0.014A
- 4rB '16\ -zqa - fa) '' 0"
f rrllrlorm lh6 currenl sources lo their equivalent vollage sources:
t,-0(2.4)=14.4v l, - 10(1.6)= 16V
!0 ll1 6+1+24)-1a4=0 ln 5 0 l'032A R..draw the circuil:
Transform the voltage sources lolheir equivalenl current sources.
Apply KVL on loop dabcd'i
(l^ ls)2 lB(8)- (lB + lc)2 = 0
N,n.
lA = 6/4r lc =
34
Sub6litute:
(8
ls
)2- 8lB (B + 3)2
=0
12 2ts 8ls-2lB-6=o lr.0.5A
v
rB(8) = 0,5(8)
Te
3
87
It
I 00
I Solyed Prcblens in Etectricat
Solutions Test
Open-circuit lhe 1-ohm r6sistor and solve th6 looking back .esistance circlil al these terminals.
9p.fi
Irn
3
89
olroull lhc lo-ohm rosistance and solvo the op€n circuit vollage atthese
nal.
50 o
n,
(l)
toa
vt
t0r Note, T1'e cu[ent source
i5
open€d wbilethe voltage soorces aie short_ci
Afrt
v
By insp€ction, Ro = 4C)
KVL uBino the loop as shown :
(0lv,x5)-10
=0
vr 0 6vr 10=0
Shorl-circuil th€ 1-ohm resistor and solve the shorl circuil cur€nt that to lhis branch.
Vr
-
20
lr-Vr
vottr
_20volls
ahorl clrcult lho lo-ohm resistor and solve the short circuit cunenl
i r' Apply KCL at node 'a':
t)a
"l
l1_
I
h=0
IL
Norton k equival ent c itcu it Th ev enin's equ iY ol e ht c it cu
^_t
-2O-12
, the Nonon's equivatent circuit: t*Ro l2)t4, ,''-R"*&=4;
t
l! =164
Nol. lh.
volt.ge-controled.urrcnt
Alply KvL using lhe loop as shown:
l0
l.(o+
lr'tl
5) = o
r- - E-q=?9=zoo Ua. lho Thevenin's equivalenl circuit:
'|r. - R"El=-?L=6667n +R' 20+10
source
t
a' sho,n will delivei zeto cu*ent
90
1001 Solyed Prcblens tn
Notc: When
so!r.€
Sol
Elec|ical
opefating alone, the other c! reDt Fdrces mun bc while the oth€r voltage sou rces must be shod-circuited. a
D
rio6
Test
3 9I
0llrlr rtrcull lho 2-ohm resistsnco and solve lhe open circuit voltage
------->I
-----> I
across
=0
Lel, the vollage source lo opersle alone:
+
tatv, 6r/
6t)
+ tciminah 'ab' E oPcD-cir.uited (l = o), the current controlled curent sro cuifent (equi!4lenttoan oPen circuit) wrl,lclver |.rn,.
N.il. ! r!c 6+v1-t(6)+3V=0 6+4V, =61',+O
lf
V =I(1)=r+A
ApDly KcL at node
Substitule Eq.2 in Eq.1:
6+4(l')=d'
l=9=se 2
ln.p6clion,
Eo =
30 V
llrorl clrcuil the 2-ohm resistance and solve lhe shorl circuit current
'x:
l,-lr0.6l l| -
1
.8r
-to6v, -., arv. -l 4J
J
t
Lel, the currenl souce to operal€ alone: Apply KCL at node'a': 3 = 11+ f'
02v, =0.4vr V, - l0 volts
r,
-r !:=f
=zsr
..-i:=# 3 = 1.667V
1=1!-=1[9)=12a
66
l= l'+ l"=
3 + 1.2
t=42A Notei since the drection of cureots for botb I' an,l rcsultant cu"ent that will 6ow i,r the 6-ohm r6irtance,
5!r, ofthetwo cunentyeipectively.
1,, -
12
{}
U16 Nodalnode method:
"""'> 1' x -"'t'l
=
I.
92
1001 Solrc.l Ptablens ib Apply KCL at node 'a 6 = l+ li
6= % r
24
Ele.biul
R.R
:
k3:13
6 = 0.5V. +0.25% + 0.51- 3
n=orru *ool]'l
"
I
\2
)
= 0.75V, +0.25V.
G0u10
tlrrl
t=:a=:=45A 22 lJse Nodal node methodl
I2V
l.yv ol
Bsuw
€l.cttGt tic6: Like chaQes
a[r!..
rllract sach othet.
aaoonrl
l.w ot olocuostatlca:
repoleach othet and unlike
The lorce of altracion or repulslon between
charges and inversely ir]u.r 'i.jrocily proporlionalto lhe prcducl of lwolhem Fq';'t 'trlto lh€ square ofthe distance b€tween
(lr r charge in each tody (coulomb) -.I,lolule P€rmitlivity - LE54 x '10 t'zfarad psr meter l, - rolalrvo pemittivity or dielectric constant - t(ono), for free space rl - (lritdnc€ between the lwo bodies (meleo Sl unils equalloI r10r l-
rJ, l"
Apply KCL al node 'a'
l+4=li+41 4=11+31
r=Y*"ftz-v'1
",slantn
b'',.rr,,', ".1)"e. O =.hr -ou omb. d ".entmete ri.r,,,l {, I, n a I coulombofcharqe
4=0.25V+9-0.75V tr[CIf,OSTING
0.5v = 5 V = 10 volls
lo
V
potlntlal-the
.n,l
NIttI I
electdc potential resulting from the location ofcharged
- .ln(lrostalic potential (volt) at a distance ofd l!rly ol charue Q (coulomb)
(meter) from a charce
