Chapter # 32
1. Sol.
Electric Current in Conductors
SOLVED EXAMPLES
An electron beam has an aperture 1.0 mm2. A total of 6.0 × 1016 electrons go through any perpendicular cross section per second. Find (a) the current and (b) the current density in the beam. (a) The total charge crossing a perpendicular cross–section in one second is q = ne = 6.0 × 1016 × 1.6 × 10–19 C = 9.6 × 10–3 C. The current is q t
i=
9.6 10 3 C = 9.6 × 10–3 A. 1s (b) The current density is =
j=
9.6 10 3 A i = = 9.6 × 103 A/m2 . S 1.0 mm 2
Ans.
(a) 9.6 × 10–3 A.
2.
Calculate the drift speed of the electrons when 1 A of current exists in a copper wire of cross-section 2 mm2. The number of free electrons in 1 cm3 of copper is 8.5 × 1022. We have j = nevd
Sol.
or,
vd =
(b) 9.6 × 103 A/m2 .
j j = ne A ne
1A =
(2 10 m )(8.5 10 10 6 m -3 )(1.6 10 -19 C) = 0.036 mm/s. We see that the drift speed is indeed small. -6
2
22
Ans.
0.036 mm/s.
3.
Calculate the resistance of an aluminium wire of length 50 cm and cross-sectional area 2.0 mm2. The resistivity of sluminium is r = 2.6 × 10–8 - m.
Sol.
The resistance is R =
4. Sol.
Ans.
(2.6 10 8 m) (0.50m)
= 0.0065 . 2 10 10 m 2 We arrived at Ohm's (equation 32.6 or 32.8) by making several assumptions about the existance and behaviour of the free electrons. These assumptions are not valid for semiconductors, insulators,solutions etc. Ohm's law cannot be applied in such cases. 0.0065 =
Ans.
A
A resistor develops 400 J of thermal energy in 10 s when a current of 2 A is passed through if. (a) Find its resitance. (b) If the current is increased to 4 A, what will be the energy developed in 20 s. (a) Using U = i2Rt, 400 J = (2A)2 R(10 s) or R = 10 . (b) The thermal energy developed, when the current is 4A, is U = i2Rt = (4A)2 × (10 ) × (10 s) = 1600 J. (a) 10 . (b) 1600 J.
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Chapter # 32 Electric Current in Conductors 5. A battery of emf 2.0 V and interial resistance 0.50 supplies a current of 100 mA. Find (a) the potential difference across the terminals of the battery and (b) the thermal energy developed in the battery in 10 s. Sol. The situation is the same as that shown in figure. (a) The potential difference across the terminals is VA – VB = (VA – VC) – (VB – VC) = – ir = 2.0 V – (0.100 A) (0.50 ) = 1.95 V. (b) The thermal energy developed in the battery is U = i2rt = (0.100 A)2 (0.50 ) (10 s) = 0.50 J, Ans. (a) 1.95 V (b) 0.05 J 6.
Find the equivalent resistance of the network shown in figure between the points A and B.
Sol.
The 10 resistor and the 30 resistor are connected in parallel. The equivalent resistance between A and C is
Ans.
(10 )(30 ) = 7.5 . 10 30 This is connected with 2.5 W in series. The equivalent resistance between A and B is 7.5 + 2.5 = 10 . 10
7.
Find the value of R in figure so that there is no current in the 50 resistor.
Sol.
This is a Wheatstone bridge with the galvanometer replaced by the 50 resistor. There will be no current in the 50 resistor if the bridge is balanced. In this case,
R1 =
Ans.
10 R = 20 40 or R = 20 . R = 20
8.
The ammeter shown in figure consists of a 480 coil connected in parallel to a 20 shunt. Find the reading of the ammeter.
Sol.
The equivalent resistance of the ammeter is ( 480 ) (20 ) = 19.2 . 480 20 The equivalent resistance of the circuit is 140.8 + 19.2 = 160 . 20 V = 0.125 A. 160 This current goes through the ammeter and hence the reading of the ammeter is 0.125 A. 0.125 A.
The current is i = Ans.
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Chapter # 32
1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13.
14. 15. 16. 17.
Electric Current in Conductors
QUESTIONS
FOR
SHORT
ANSWER
Suppose you have three resistors each of value 30 . List all the different resistances you can obtain using them. A proton beam is going from east to west. Is there an electric current ? If yes, in what direction ? In an electrolyte, the positive ions move from left to right and the negative ions from right to left. Is there a net current ? If yes, in what direction ? In a TV tube, the electrons are accelerated from the rear to the front. What is the direction of the current ? The drift speed is defined as vd = l / t where l is the distance drifted in a long time t. Why don't we define the drift speed as the limit of l / t as t 0 ? One of your friends argues that he has read in previous chapters that there can be no electric field inside a conductor. And hence there can be no current through it. What is the fallacy in this argument ? When a current is established in a wire, the free electrons drift in the direction opposite to the current. Does the number of free electrons in the wire continuously decrease ? A fan with copper winding in its motor consumes less power as compared to an otherwise similar fan having aluminium winding. Explain. The thermal energy developed in a current–carrying resistor is given by U = i2Rt and also by U = Vit. Should we say that U is proportional to i2 or to i ? Consider a circuit containing an ideal battery connected to a resistor. Do "work done by the battery" and "the thermal energy developed" represent two names of the same physical quantity ? Is work done by a battery always equal to the thermal energy developed in electrical circuits ? What happens if a capacitor is connected in the circuit ? A nonideal battery is connected to a resistor. Is work done by the battery equal to the thermal energy developed in the resistor ? Does your answer change if the battery is ideal ? Sometimes it is said that "heat is developed" in a resistance when there is an electric current in it. Recall that heat is defined as the energy being transferred due to the temperature difference. Is the statement under quotes technically correct ? We often say "a current is going through the wire". What goes through the wire, the charge or the current ? Would you prefer a voltmeter or a potentiometer to measure the emf of a battery ? Does a conductor become charged when a current is passed through it ? Can the potential difference across a battery be greater than its emf ?
Objective - I 1.
A metallic resistor is connected across a battery. If the number of collisions of the free electrons with the lattice is somehow decreased in the resistor (for example, by cooling it), the current will (A*) increase (B) decrease (C) remain constant (D) become zero
,d /kkfRod izfrjks/kd dks ,d cSVjh ls tksMk+ tkrk gSA ;fn eqDr bysDVªkuW ksa dh tkyd ls VDdjksa dh la[;k izfrjks/kd esa fdlh rjg ?kVk nh tkrh gS] ¼mnkgj.k ds fy, bls BaMk djds½ rks /kkjk (A*) c<+sxh (B) ?kVsxh (C) fu;r jgsxh (D) 'kwU; gks tk,xh 2.
Two resistor A and B have resistance RA and RB respectively with RA < RB. The resistivites of their materials A and = B. (A) A < B (B) A = B (C) A > B (D*) The information is not sufficient to find the relation between A and B nks izfrjks/kdksa A o B ds izfrjks/k Øe'k% RA o RB gS] tgk¡ RA < RB A muds inkFkks± dh izfrjks/kdrk A o = B gS (A) A < B (B) A = B (C) A > B (D*) A o B ds e/; lac/a k Kkr djus ds fy, lwpuk vi;kZIr gSA
3.
The product of resistivity and conductivity of a cylindrical conductor depends on (A) temperature (B) material (C) area of cross-section (D*) none of these ,d csyukdkj pkyd dh pkydrk ,oa izfrjks/kdrk dk xq.kuQy fuHkZj djrk gS (A) rki ij (B) inkFkZ ij (C) vuqiLz Fk dkVk {ks=k ij (D*) buesa ls dksbZ ugha
4.
As the temperature of a metallic resistor is increased, the product of its resistivity and conductivity (A) increases (B) decreases (C*) remains constant (D) may increase or decrease ,d /kkfRod izfrjks/k dk rki c<+k;s tkus ij bldh pkydrk ,oa izfrjks/kdrk dk xq.kuQy (A) c<+rk gS (B) ?kVrk gS (C*) fu;r jgrk gS (D) ?kV vFkok c<+ ldrk gS manishkumarphysics.in
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Chapter # 32 Electric Current in Conductors 5. In an electric circuit containing a bettery, the charge (assumed positive) inside the battery (A) always goes from the positive terminal to the negative terminal (B*) may go from the positive terminal to the negative terminal (C) always goes from the negative terminal to the positive terminal (D) does not move. ,d cSVjh okys fo|qr ifjiFk es]a cSVjh ds Hkhrj vkos'k - ¼/kukRed ekuk gqvk½ (A) ges'kk /ku VfeZuy ls _.k VfeZuy ij tkrk gSA (B*) /ku VfeZuy ls _.k VfeZuy ij tk ldrk gSA (C) ges'kk _.k VfeZuy ls /ku VfeZuy ij tk ldrk gSA (D) xfr ugha djrk gSA 6.
A resistor of resistance R is connected to an ideal battery. If the value of R is decreased, the power dissipated in the resistor will (A*) increase (B) decrease (C) remain unchanged R izfrjks/k dk ,d izfrjks/kd ,d vkn'kZ cSVjh ls tksMk+ x;k gSA ;fn R dk eku ?kVk;k tkrk gS] rks izfrjks/k esa O;f;r 'kfDr(A*) c<+sxh (B) ?kVsxh (C) vifjofrZr jgsxh
7.
A current passes through a resistor. Let K1 and K2 represent the average kinetic energy of the conduction electrons and the metal ions respectively. (A) K1 < K2 (B) K1 = K2 (C*) K1 > K2 (D) Any of these three may occur ,d izfrjks/kd esa /kkjk izokfgr gks jgh gSA ;fn pkyu bysDVªkuW ksa ,oa /kkfRod vk;uksa dh vkSlr xfrt ÅtkZ Øe'k% K1 o K2 gks rks (A) K1 < K2 (B) K1 = K2 (C*) K1 > K2 (D) bu rhuksa esa ls dqN Hkh gks ldrk
8.
Two resistance R and 2R are connected in series in an electric circuit.The thermal energy developed in R and 2R are in the ratio nks izfrjks/kd R o 2R ,d fo|qr ifjiFk esa Js.khØe esa tksM+s tkrs gSaA R o 2R esa mRiUu rkih; ÅtkZ dk vuqikr gS (A*) 1 : 2 (B) 2 : 1 (C) 1 : 4 (D) 4 : 1
9.
Two resistance R and 2R are connected in parallel in an electric circuit. The thermal energy developed in R and 2R are in the ratio nks izfrjks/k R o 2R ,d fo|qr ifjiFk esa lekUrj Øe esa tksM+s tkrs gSaA R o 2R esa mRiUu rkih; ÅtkZ dk vuqikr gS (A) 1 : 2 (B*) 2 : 1 (C) 1 : 4 (D) 4 : 1
10.
A uniform wire of resistance 50 is cut into 5 equal parts. These parts are now connected in parallel.The equivalent resistance of the combination is 50 izfrjks/k ds ,d le:i rkj dks 5 leku Hkkxksa esa dkVk x;k gSA ;s Hkkx vc lekUrj Øe esa tksMs+ tkrs gSAa la;kstu dk rqY; izfrjks/k gS (A*) 2 (B) 10 (C) 250 (D) 6250
11.
Consider the following two statements : (a) Kirchhoff’s junction law follows from conservation of charge. (b) Kirchhoff’s loop law follows from consevative neature of electric field. (A*) Both A and B are correct (B) A is correct but B is wrong (C) B is correct but A is wrong (D) Both A and B are wrong fuEu nks dFkuksa ij fopkj dhft, : (a) fdjpkWQ dk laf/k fu;e vkos'k ds laj{k.k dks n'kkZrk gSA (b) fdjpkWQ dk ik'k fu;e fo|qr {ks=k dh lajf{kr izÑfr dks n'kkZrk gSA (A*) a o b nksuksa lgh gS (B) a lgh gS] ysfdu b xyr gS (C) b lgh gS] ysfdu a xyr gS (D) a o b nksuksa xyr gS
12.
Two non-ideal batteries are connected in series. Consider the following statements: (a) The equivalent emf is larger either of the two emfs. (b) The equivalent internal resistance is smaller than either of the two internal resistance. (A) Each of A and B is correct (B*) A is correct but B is wrong (C) B is correct but A is wrong (D) Each of A and B is wrong. nks vukn'kZ (non-ideal) cSVfj;k¡ Js.kh Øe esa tksM+h tkrh gSA fuEu dFkuksa ij fopkj dhft, (a) rqY; fo|qr okgd cy nksuksa esa ls fdlh Hkh fo|qr okgd cy ds T;knk gSA (b) rqY; vkarfjd izfrjks/k nksuksa esa ls fdlh Hkh vkarfjd izfrjks/k ls de gSA manishkumarphysics.in
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Chapter # 32 (A) a o b nksuksa lgh gS (C) b lgh gS] ysfdu a xyr
Electric Current in Conductors
gS
(B*) a lgh gS] ysfdu b xyr (D) a o b nksuksa xyr gS
gS
13.
Two non-ideal batteries are connected in parallel. Consider the following statements (a) The equivalent emf is smaller than either of the two emfs. (b) The equivalent internal resistance is smaller than either of the two internal resistance. (A) Both a and b are correct (B) a is correct but b is wrong (C*) b is correct but a is wrong (D) Each of a and b is wrong [HCV_Chp_32_Obj-1_13] nks vukn'kZ (non-ideal) cSVfj;k¡ lekUrj Øe esa tksM+h tkrh gSA fuEu dFkuksa ij fopkj dhft, (a) rqY; fo|qr okgd cy nksuksa esa ls fdlh Hkh fo|qr okgd cy ls de gSA (b) rqY; vkarfjd izfrjks/k nksuksa esa ls fdlh Hkh vkarfjd izfrjks/k ls de gSA (A) a o b nksuksa lgh gS (B) a lgh gS] ysfdu b xyr gS (C*) b lgh gS] ysfdu a xyr gS (D) a o b nksuksa xyr gS
14.
The net resistnace of an ammeter should be small to ensure that (A) it does not get overheated (B) it does not draw excessive current (C) it can measure large currents (D*) it does not appreciably change the current to be measured. ,d vehVj dk dqy izfrjks/k vYi gksuk pkfg, rkfd (A) ;g vR;f/kd xeZ uk gksA (B) ;g vR;f/kd /kkjk u ysAa (C) ;g cM+h /kkjkvksa dks eki ldsA (D*) ;g ekih tkus okyh /kkjk esa dksbZ fo'ks"k ifjorZu u djsA
15.
The net resistance of a voltmeter should be large to ensure that (A) it does not get overheated (B) it does not draw excessive current (C) it can measure large potential differences (D*) it does not appreciably change the potential difference to be measured. ,d oksYVehVj dk dqy izfrjks/k vf/kd gksuk pkfg rkfd (A) ;g vR;f/kd xeZ uk gks (B) ;g vR;f/kd /kkjk uk ysA (C) ;g cM+s foHkokUrjksa dks eki ldsA (D*) ;g ekis tkus okys foHkokUrj esa dksbZ mYys[kuh;
16.
ifjorZu uk djsA
Consider a capacitor-charging circuit. Let Q1 be the charge given to the capacitor in a time interval of 10 ms and Q2 be the charge given in the next time interval of 10 ms. Let 10 C charge be deposited in a time interval t1 and the next 10mC charge is deposited in the next time interval t2. ,d la/kkfj=k ds vkos'ku ifjiFk ij fopkj dhft,A eku yhft, fd 10 feyh lsd.M ds ,d le;kUrjky esa la/kkfj=k dks Q1 vkos'k fn;k tkrk gS rFkk 10 feyh lsd.M ds vxys le;kUrjky esa Q2 vkos'k fn;k tkrk gSA eku yhft, fd t1 le;kUrjky esa 10 ekbØks dwykWe vkos'k laxfz gr gksrk gS rFkk vxys le;kUrjky t2 esa vxyk 10 ekbØks dwykWe vkos'k laxfz gr gksrk gS (A) Q1 > Q2, t1 > t2. (B*) Q1 > Q2, t1 < t2. (C) Q1 > Q2, t1 > t2. (D) Q1 < Q2, t1 < t2.
Objective - II 1.
Electrons are emitted by a hot filament and are accelerated by an elecrtic field as shown in fig. The two stops at the left ensure that the electron beam has a uniform cross-section.
,d xeZ fQykesUV ls bysDVªkWu mRlftZr gksrs gS ,oa fp=kkuqlkj ,d fo|qr {ks=k }kjk Rofjr gksrs gSaA ck;ha rjQ ds nks vojks/k ;g lqfuf'pr djrs gSa fd bysDVªkWu iqt a dk dkV {ks=k leku jgs -
(A*) The speed of the electron is more at B than at A. (B) The electric current is from left to right (C) The magnitude of the current is larger at B than at A. (D) The current density is more at B than at A. (A*) The speed of the electron is more at B than at A. (B) The electric current is from left to right (C) The magnitude of the current is larger at B than at A. (D) The current density is more at B than at A. k 2.
A capacitor with no dielectric is connected to a battery at t = 0. Condiser a point A in the connecting wires and a point B in between the plates. (A) There is no current through A manishkumarphysics.in
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Chapter # 32 Electric Current in Conductors (B*) There is no current through b (C*) There is a current through A as long as the charging is not complete. (D) There is a current through B as long the charging is not complete. ,d la/kkfj=k ftlesa dksbZ ijkoS|rq inkFkZ ugha gS] t = 0 ij ,d cSVjh ls tksM+k tkrk gSA ds e/; ,d fcUnq B ij fopkj dhft, (A) A ls dksbZ /kkjk ugha xqtjrhA (B*) B ls dksbZ /kkjk rHkh rd xqtjrh gS tc rd vkos'ku iw.kZ ugha gksrkA (C*) A ls /kkjk rHkh rd xqtjrh gS tc rd vkos'ku iw.kZ ugha gksrkA (D) B ls /kkjk rHkh rd xqtjrh gS tc rd vkos'ku iw.kZ ugha gksrkA
la;kstu rkjksa esa ,d fcUnq A rFkk IysVksa
3.
When no current is passed through a conductor (A) the free electrons do not move (B) the average speed of a free electron over a large period of time is zero (C*) the average velocity of a free electron over a large period of time is zero (D*) the average of the velocities of all the free electrons at an instant is zero tc ,d pkyd ls dksbZ /kkjk izokfgr ugha gksrh (A) eqDr bysDVªkWu xfr ugha djrs gSA (B) eqDr bysDVªkWu dk cM+h le;kof/k ij vkSlr pky 'kwU; gSA (C*) eqDr bysDVªkWu dk cM+h le;kof/k ij vkSlr osx 'kwU; gSA (D*) lHkh eqDr bysDVªkWuksa ds osxksa dk fdlh {k.k ij vkSlr 'kwU; gSA
4.
Which of the following quanitites do not change when a resistor connected to a battery is heated due to the current ? (A) drift speed (B) resistivity (C) resistance (D*) number of free electrons ,d cSVjh ls tqM+k ,d izfrjks/k tc /kkjk ds dkj.k xeZ gksrk gS rks fuEu esa ls dkSulh jkf'k;k¡ ifjofrZr ugha gksrh gS (A) viokg osx (B) izfrjks/kdrk (C) izfrjks/k (D*) eqDr bysDVªkWuksa dh la[;k
5.
As the temperature of a conductor increases, its resistivity and conductivity change. The ratio of resistivity to conductivity (A*) increases (B) decreases (C) remains constant (D) may increase or decrease depending on the actual temperature.
,d pkyd ds rki esa tSls&tSls o`f) dh tkrh gSA bldh pkydrk ,oa izfrjks/kdrk ifjofrZr gksrh gSA bldh pkydrk ,oa izfrjks/kdrk ifjofrZr gksrh gSA izfrjks/kdrk ,oa pkydrk dk vuqikr (A*) c<+rk gSA (B) ?kVrk gSA (C) fu;r jgrk gS (D) okLrfod rki ds vuqlkj c<+ vFkok ?kV ldrk gSA 6.
A current passes through a wire of nonuniform cross-section. Which of the following quantities are independent of the cross-section? (A*) the charge corssing in a given time interval (B) drift speed (C) current density (D*) free-electron density. vleku vuqiLz Fk dkV ds ,d rkj esa /kkjk izokfgr gks jgh gSA fuEu esa ls dkSulh jkf'k;k¡ vuqiLz Fk dkV ij fuHkZj ugha djrh(A*) ,d fn;s ,d le;kUrjky ls xqtjus okyk vkos'k (B) viokg osx (C) /kkjk ?kuRo (D*) eqDr bysDVªkWu ?kuRo
7.
Mark out the correct options. (A*) An ammeter should have small resistance (C) A voltmeter should have small rsistance lgh fodYi pqfu;s (A*) ,d vehVj dk izfrjks/k vYi gksuk pkfg,A (C) ,d oksYVehVj dk izfrjks/k mPp gksuk pkfg,A
(B) An ammeter should have large resistance (D*) A voltmeter should have large resistance (B) ,d vehVj dk izfrjks/k mPp gksuk pkfg,A (D*) ,d oksYVehVj dk izfrjks/k mPp gksuk pkfg,A
8.
A capacitor of capacitance 500 F is connected to a battery through a 10 k resistor. The charge stored on the capacitor in the first 5 s is larger than the charge stored in the next 500 F /kkfjrk dk ,d la/kkfj=k ,d cSVjh ls 10 k izfrjks/k }kjk tksMk+ tkrk gSA izFke 5 lsd.M esa la/kkfj=k ij laxfz gr vkos'k T;knk gksxk (A*) 5 s (B*) 50 s (C*) 500 s (D*) 500
9.
A capacitor C1 of capacitance 1F and a capacitor C2 of capacitance 2F are separately charged by a common battery for a long time. The two capacitors are then separately discharged through equal resistors. Both the discharge circuits are connected at t = 0. manishkumarphysics.in
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Chapter # 32 Electric Current in Conductors (A) The current in each of the two discharging circuits is zero at t = 0 (B*) The currents in the two discharging circuits at t = 0 are equal but not zero. (C) The currents in the two discharging circuits at t = 0 are unequal (D*) C1 loses 50% of its initial charge sooner than C2 loses 50% of its initial charge. 1F /kkfjrk ds ,d la/kkfj=k C1 ,oa 2F /kkfjrk ds ,d la/kkfj=k C2 dks ,d mHk;fu"B cSVjh }kjk yacs
le; rd vyx&vyx vkosf'kr fd;k tkrk gSA nksuksa la/kkfj=kksa dks fQj leku izfrjks/kdksa }kjk vyx&vyx fujkosf'kr fd;k tkrk gSA nksuksa fujkosf'kr ifjiFkksa dks t = 0 ij tksM+k tkrk gS (A) t = 0 ij nksuksa fujkosf'kr ifjiFkksa esa /kkjk 'kwU; gSA (B*) t = 0 ij nksuksa fujkosf'kr ifjiFkksa esa /kkjk vleku gSA (C) t = 0 ij nksuksa fujkosf'kr ifjiFkksa esa /kkjk vleku gSA (D*) C1 vius izkjafHkd vkos'k dk 50% , C2 ds izkjafHkd vkos'k ds 50% [kksus ls tYnh [kksrk gSA
WORKED OUT EXAMPLES 1. Sol.
An electron moves in a circle of radius 10 cm with a constant speed of 4.0 × 106 m/s. Find the electric current at a point on the circle. Consider a point A on the circle. The electron crosses this point once in every revolution. In one revolution, the electron travels 2 × (10 cm) distance. Hence, the number of revolutions made by the electron in one second is 40.0 10 6 m 2
2 10 7
20 10 m The charge crossing the point A per second is 3 .2 2 ×107 × 1.6 × 10–19 C = × 10–12 C. Thus, the electric current at this point is 3 .2 ×10–12 A = 1.0 × 10–12 A.
2. Sol.
A current of 2.0 A exists in a wire of cross-sectional area 1.0 mm2. If each cubic metre of the wire contains 6.0 × 1028 free electrons, find the drift speed. The current density in the wire is i 2 .0 A = 2.0 × 108 A/m2 A 1mm 2 The drift speed is
j=
= 3.
Sol.
i 2.0 10 6 A / m 2 ne 6.0 10 28 m 3 1.6 10 19 C
=
2.1 × 10–4 m/s.
Find the resistance of a copper coil of total wire-length 10 m and area of cross-section 1.0 mm2. What would be the resistance of a similar coil of aluminium? The resitivity of copper = 1.7×10–8 -m and that of aluminium = 2.6 × 10–8 -m. The resistance of the copper coil is
(1.7 10 8 m) 10m = 0.17 -m A 1.0 10 6 m 2
The resistance of the similar aluminum coil will be (2.6 10 8 ( m) 10m 1.0 10 6 m 2
4.
Sol.
= 0.26 .
A parallel-area capacitor has plates of area 10 cm2 separated by a distance of 1mm. It is filled with the dielectric mica and connected to a battery of emf 6 volts. Find the leakage current through the capacitor. Resistivity of mica = 1 × 1013 -m. The resistance of the mica between the two faces is
manishkumarphysics.in
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Chapter # 32
Electric Current in Conductors
(1 1013 m) 10 3 m A 10.0 10 4 m 2
The leakage current = 5. Sol.
Sol.
6v 1 1013
1 ×1013 .
= 6 × 10–13 A.
Find the resistance of a hollow cylindrical conductor of length 1.0 m and inner and outer radii 1.0 mm and 2.0 mm respectively. The resistivity of the material is 2.0 × 10–8 -m. The area of cross-section of the conductor through which the charges will flow is A = (2.0 mm)2 – (1.0 mm)2 = 3.0 × mm2. The resistance of the wire is, therefore, R=
6.
=
( 2.0 10 8 m) 1.0m = 2.1 5 × 10–3 . A 3.0 10 6 m 2
A battery of emf 2 V and initial resistance 0.5 is connected across a resistance 9.5 . How many electrons cross through a cross-section of the resistance in 1 second ? The current in the circuit is 2V = 0.2 A. 9 . 5 0 .5 Thus, a net transfer of 0.2 C per second takes place across any cross-section in the circuit. The number of electrons crossing the section in 1 second is therefore,
i=
0.2C 1.6 10 19 C 7.
= 0.125 × 1019 = 1.25 × 1018.
A battery of emf 2.0 volts and internal resistance 0.10 is being charged with a current of 5.0 A. What is the potential difference between the terminals of the battery.
Sol. As the battery is being charged, the current goes into the positive terminals as shown in figure. The potential drop across the internal resistance is 5.0 A × 0.10 = 0.150 V. Hence the potential drop across the terminals will be 2.0 V + 0.50 V = 2.5 V. 8.
Figure shows n batteries connected to form a circuit. The resistances denote the internal resistance of the batteries which are related to the emf’s as r1 = ki where k is a constant. The solid dots represent the terminals of the batteries. Find (a) the current through the circuit and (b) the potential difference between the terminals of the ith battery.
Sol.
(a) Suppose the current is i in the indicated direction. Applying Kirchoff’s loop law. 1 – ir1 + 2 – ir2 + 3 – ir3 + ............+n – irn = 0 or
i=
1 2 3 ...... n r1 r2 r3 ...... rn
1 2 3 ...... n 1 = k( ...... ) = . k 1 2 3 n (b) The potential difference between the terminals of the ith battery is i = iri 1 (ki) = 0 k A copper rod of length 20 cm and cross-sectional area 2 mm2 is joined with a similar aluminum rod as shown in figure. Find the resistance of the combination between the ends. Resistivity of copper = 1.7×108 W-m and that of aluminum = 2.6 × 108 -m. = 9.
i
manishkumarphysics.in
Page # 8
Chapter # 32
Sol.
Electric Current in Conductors
The resistance of the copper rod =
(1.7 10 8 m) (20 10 2 m) A 2.0 10 6 m 2
= 1.7 × 10–3
Similarly, the resistance of the aluminium and = 2.6 × 10–3 These rods are joined in parallel so that the equivalent resistance R between the ends is given by =
1 1 1 3 R 1.7 10 2.6 10 3
or,
R=
1 .7 2 . 6 × 10–3 = 1.0 m. 4 .3
10.
A wire of reistance 10 W is bent to form a complete circle. Find its resistance between two diametrically opposite points.
Sol.
Let ABCDA be the resistance 10 . We have to calculate the resistance of this loop between the diametrically opposite point A and C. The wires ADC and ABC will have resistance 5 each. These two are joined in parallel between A and C. The equivalent resistance R between A and C is, therefore, given by R=
5 5 = 2.5 . 5 5
11.
Find the currents in the different resistors shown in figure.
Sol.
The two 2 resistors are in series so that their equivalent resistance is 4. The two 8 resistors are in parallel and their equivalent resistance is also 4 . The circuit may be redrawn as in figure. Suppose the middle 4 resistor is removed. The remaining circuit in figure. It is easy to see that no current will go through any resistor. If we take the potential at b to be zero, the potential at
d will be 2 V. The potential at a dn c will also be 2V. As there is no current in the 4 resistors, the potential at e will also be 2V. Thus, there is no potential difference between d and e. When a 4 resistor is added between d and e, no current will be drawn into it and hence no change will occur in the remaining part of the circuit. This circuit is then the same as the given circuit. Thus, the current in all the resistors in the given circuit is zero.
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Page # 9
Chapter # 32 Electric Current in Conductors 12. Find the current supplied by the battery in the circuit shown in figure.
Sol.
All the resistors shown in the figure are connected in parallel between the terminals of the battery. The equivalent resistance R between the terminals is, therefore given by
1 1 1 1 1 R 12 12 12 12 or, R = 3. The current supplied by the battery is V 3V = 1 A. R 3 Find the equivalent resistance between the points a and b of the network shown in figure.
i=
13.
Sol.
The two resistor 4 and 2 at the right end are joined in series and may be replaced by a single resistor of 6. This 6 is connected with the adjacent 3 resistor in parallel. The equivalent resistance of these two is 6 3 = 2. 6 3 This is connected in series with the adjacent 4 resistor giving an equivalent resistance of 6 which is connected i parallel with the 3 resistor. Their equivalent resistance is 2 which is connected in series with the first 4 resistor from left. Thus, the equivalent reistance between and b is 6 .
14.
Find the effective resistors between the points A and B in figure.
Sol.
The resistors AF and FE are in series. Their equivalent is 3 + 3 = 6 . This is connected in parallel with 6 3 = 3 . 6 3 This 3W resistance between A and E is the series with ED and the combination is in parallel with AD. Their equivalent between A and D is again 3. Similarly, the equivalent of this 3, DC and AC is 3. This 3 is in series with CB and the combination is in parallel with AB. The equivalent resistance between A and B is, therefore.
AE. Their equivalent between A and E is, therefore.
15.
6 3 = 2. 6 3 Find the equivalent resistance of the network shown in figure between the points a and b when (a) the switch S is open (b) the switch S is closed.
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Page # 10
Chapter # 32 Electric Current in Conductors Sol. (a) When the switch is open, 6 and 12 resistors on the upper line are in series giving an equivalent of 18. Similarly, the resistors on the lower line have equivalent resistance 18. These two 18 resistances are connected in parallel between a an db so that the equivalent resistance is 9. (b) When the switch is closed, the 6 and 12 resistors on the left are in parallel giving an equivalent resistance of 4. Similarly the two resistors on the right half are equivalent to 4. These two are connected in series between a and b so that the equivalent resistance is 8. 16.
Each resistor shown in figure has a resistance of 10 and the battery has an emf of 6V. Find the current supplied by the battery.
Sol.
Suppose a current i starts from the position terminal of the battery. By symmetry, it divides equally in the resistors ab and fe, so that each of these carries a current i/2. The current going into the negative terminal is also i and by symmetry, equal currents should come from ed and bc. Thus, the current in ed is also i/2 and hence there will be no current in eb. Va – VC = (Va – Vb) + (Vb – Vc) or, giving
i i ×10 + ×10 2 2 i = 0.6 A. This is balanced wheat stone bridge. 6V =
17.
Find the equivalent resistance of the network shown in figure between the points A and B.
Sol.
Suppose an ideal battery of emf is connected across the point A and B. The circuit is a Wheatstone bridge with the galvanometer replaced by a 50 resistance. As the bridge is balanced (R1/R2 = R3/R4), there will be no current through the 50 resistance. We can just remove the 50 resistance without changing any other current. The circuit is then equivalent to two resistances 30 and 60 connected in parallel. The equivalent resistance is R=
(30) (60) 20 (30 ) (60)
18.
In the circuit shown in figure E,F,G and H are cells of emf 2,1,3 and 1V respectively. The resistances 2,1,3 and 1 are their respective internal resistances. Calculate (a) the potential difference between B and D and (b) the difference across the terminals of each of the cells G and H.
Sol.
Suppose a current i1 goes in the branch BAD and a current i2 in the branch DCB. The current in DB will be i1 – i2 from the junction law. The circuit with the currents shown is redrawn in figure. Applying the loop law to BADB we get, manishkumarphysics.in
Page # 11
Chapter # 32
Electric Current in Conductors (2)i1 – 2V + 1V + (1)i1 + (2) (i1 – i2) = 0 or, (5)i1 – (2)i2 = 1V. ............(i) Applying the same law to the loop DCBD, we get, – 3V + (3)i2 + (1)i1 + (2) (i1 – i2) = 0 or, – (2)i1 + (6)i2 = 2V ............(ii) From (i) and (ii) i1 =
5 6 A, i2 = A 13 13
1 A. 13 The current in BD is from B to D.
so that i1 – i2 = –
(a) (b)
1 2 VB – VD = (2) A = V.. 13 13 The potential difference across the cell G is VC – VD = – (3)i2 + 3 V
18 21 V = 3V V.. 13 13 The potential difference across the cell H is
=
19. Sol.
6 19 VC – VB = (1)i2 + 1V = (1) A + 1V = V.. 13 13 Find the equivalent resistance between the points a and bof the circuit shown in figure. Suppose a current i enters the circuit at the point a,a part i1 goes through the 10 resistor and the rest i – i1 through the 5 resistor. By symmetry, the current i coming out from the point b will be composed of a part i1 from the 10 resistor and i –i1 from the 5 resistor. Applying Kirchoff’s junction law, we can find the current through the middle 5 resistor. The current distribution is shown in figure.
We have Va – Vb = (Va – Vb) + (Vc – Vb) = (10)i1 + (5) (i – i1) = (5)i + (5)i1 ..................(i) Also, Va – Vb = (Va – Vc) + (Vc – Vd) + (Vd – Vb) = (10 )i1 + (5) (2i1 – i) + (10)i1 Multiplying (i) by 6 and subtracting (ii) from it,we eliminate i1 and get, 5(Va – Vb) = (35)i Va Vb = 7 . Thus, the equivalent resistance between the points a and b is 7.
or,
20.
Find the currents going through the resistors R1,R2 and R3 in the circuit of figure.
Sol.
Let us take the potential of the points A to be zero. The potential at C will be S1 and that at D will be S2. Let the potential at B be V. The currents through the three resistors are i1,i2 and i1+ i3 as shown in figure. Note that manishkumarphysics.in
Page # 12
Chapter # 32 Electric Current in Conductors the current directed towards B equals the currents directed away from B. Applying Ohhm’s law to the three resistors R1,R2 and R3 we get 1 – V = R1i1 ..................(i) 2 – V = R2i2 .................(ii) and V – 0 = R3(i1 + i2) .................(iii) Adding (i) and (ii) 1 = R1i1 + R3(i1 + i2) = (R1 + R3)i1 + R3i2 ,,,,,,,,,,,,,,,,,,(iv) and adding (ii) and (iii), 2 = R2i2 + R3(i1 + i2) = (R2 + R3)i2 + R3i1. ...................(v) Equations (iv) and (v) may be directly written from Kirchoff’s loop law applied to the left and the right half of the circuit. Multiply (iv) by (R2 + R3), (v) by R3 and subtract to eliminate i2. This gives i=
1(R 2 R 3 ) 2R 3 (R1 R 3 )(R 3 R 3 ) R 32
1(R 2 R 3 ) 2R 3 = R R R R R 1 2 2 3 1 Similarly, eliminating i1 from (iv) and (v) we obtain,
2 (R1 R 3 ) 1R 3 i = R R R R R R 1 2 2 3 3 1 And so,
1R 2 2R1 i1 + i2 = R R R R R R 1 2 2 3 3 1
21.
Find the equivalent resistance between the points a and c of the network shown in figure. Each resistance is equal to r.
Sol.
Suppose a potential difference V is applied between a and c so that a current i enters at a and the same current leaves at c. The current i divides in three parts at a. By symmetry the part in ad and in ab will be equal. Let each of these current be i1. The currents through a is i – 2i1. Similarly, currents from dc, bc and of combine at c to given the total current i. Since the situation at c is equivalents to that at a, by symmetry, the currents in dc and bc will be i1 and that in oc will be i – 2i1. Applying Kichhoff’s junction law at d, we see that the current in do is zero. Similarly, the current in ob is zero. We can remove do and ob for further analysis. It is then equivalent to three resistance, each of value 2r, in parallel. The equivalent resistance is, therefore, 2r/3.
22.
Twelve wires, each having resistance r,a re joined to form a cube as shown in figure. Find the equivalent resistance between the ends of a face diagonal such as a and c.
Sol.
Suppose a potential difference V is applied between the points a and c so that a current i enters at a and the same current leaves at c. The current distribution is shown in figure.
manishkumarphysics.in
Page # 13
Chapter # 32
Electric Current in Conductors
By symmetry, the paths ad and ab are equivalent and hence will carry the same current i1. The path a will carry the remaining current i – 2i1 (using Kichhoff’s junction law). Similarly at junction c, currents coming from dc and bc will be i1 each and from fc will be i – 2i1. Kirchhoff’s junction law at b and d shows that currents through be and dg will be zero and hence we may be igonires for further analysis. Omitting these two wires, the circuit is redrawn in figure. The wire hef and hgf are joined in parallel and have equivalent resistance
(2r )(2r ) = r between h and f. this (2r ) (2r )
is joined in series with an and fc giving equivalent resistance r+r+r = 3r. This 3r is joined in parallel with adc (2r) and abc (2r) between a and c. therefore, given by
1 1 1 1 , R 3r 2r 2r
giving
R=
3 r. 4
23.
Find the equivalent resistance of the circuit of the previous problem between the ends of an edge such as a and b in figure.
Sol.
Suppose a current i enters the circuit at the point a and the same current leaves the circuit at the point b. The current distribution is shown in figure. The paths through ad and ah are equivalent and carry equal current i1. The current through ab is then i – 2i1. The same distribution holds at the junction b. Currents in eb and cb are i1 each. The current i1 in ah is divided into a part i2 in he and i1 – i2 in hg. Similar is the division of current i1 in ad into dc and dg. The rest of the currents may be written easily using Kirchhoff’s junction law. The potential difference V between a and b may be written from the paths ab, aheb and ahgfcb as V = (i – 2i1)r V = (i1 + i2 + i1)r and V = [i1 + (i1 – i2) + 2(i1 – i2) + (i1 – i2) + i1]r which may be written as V = (i – 2i1)r V = (2i1 – i2)r and V = (6i1 – 4i2)r. Eliminating i1 and i2 from these equations,
V 7 r i 12 which is the equivalent resistance. 24.
Find the equivalent resistance between the points a and b of the infinite ladder shown in figure
. manishkumarphysics.in
Page # 14
Chapter # 32 Electric Current in Conductors Sol. Let the equivalent resistance between a and b be R. As the ladder is infinite, R is also the equivalent resistance of the ladder to the right of the point c and d. Thus, we can replace the part to the right of cd by a resistance R and redraw the circuit as in figure
This gives
or, or,
rR r R rR + R2 + r2 + 2rR R2 – rR – r2 = 0
or,
R
R=r+
r r 2 4r 2 1 5 r. 2 2
25.
Find the equivalent resistance of the network shown in figure, between the points a and b.
Sol.
Suppose a current i enters the network at point a and the same current leaves it at point b. Suppose, the currents in ac, ad and ae are i1, i2 and i3 respectively. Similar will be the distribution of current at b. The current i leaving at b is composed of i1 from db, i2 from cb and i3 from eb. The situation is shown in figure.
As the current in ae is equal to that in eb, the current in ce will be equal to the current in ed from the junction law. If we assume that the branches ced and aeb do not physically touch at e, nothing will changed in the current distribution. We can then represent the branch aeb by a single resistance of 10 connected between a and b. Similarly, the branch ced may be replaced a single 5 resistor between c and d. The circuit is redrawn in figure. Thus us same as the circuit in figure connected in parallel with a resistance of 10. So the network is equivalent to a parallel combination of 7 and 10 resistor. The equivalent resistance of the whole network is, therefore, R
26.
(7 ) (10 ) 4.1 7 10
(A) Find the current i supplied by the battery in the network shown in figure in steady state. (B) Find the charge on the capacitor.
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Page # 15
Chapter # 32 Electric Current in Conductors Sol. (A) Once the capacitor is charged, no current will go through it and hence the current through the middle branch of the circuit is zero in steady state. The 4 resistor will have an current in it and may be omitted for current analysis. The 2 and 6 resistors are, therefore, connected in series and hence 2V 0.25 A 2 6 (B) The potential drop across the 6 resistor is 6 × 0.25 A = 1.5 V. As there is no current in the 4 resistor, there is no potential drop across it. The potential difference across the capacitor is, therefore, 1.5 V. The charge on this capacitor is Q = CV = 2 F × 1.5 V = 3 C i
27.
A part of a circuit in steady state along with the currents flowing in the branches, the values of resistancess etc, is shown in figure. Calculate the energy stored in the capacitors.
Sol.
To get the energy stored the capacitor, we shall calculate the potential difference between the points P and Q. In study state, there is no current in the capacitor branch. Applying Kirchhoff’s junction law at P, the current i the 5 – 1 branch will be 3A and hence VP – VS = 6 × 3A = 18V, Applying the same theorem at Q, the current in the 2 resistor will be 1A towards Q so that VS – VQ = 2Q × 1A = 2V Thus, VP – VQ = (VP – VS) + (VS – VQ) = 20 V The energy stored in the capacitors
=
1 1 CV2 = ×4F × 400 V2 2 2
= 800 J. 28.
(A) Find the potential drops across the two resistors shown in figure. (B) A voltmeter of resistance 600 is used to measure the potential drop across the 300 resistor. What will be the measured potential drop ?
Sol.
The current in the circuit is
100 V = 0.2 A. The potential drop across the 300 resistor is 300 × 300 200
0.2 A = 60 V. Similarly, the drop across the 2 resistor is 40 V. (b) The equivalent resistance, when the voltmeter is connected across 300 , is 600 300 = 400 600 300 Thus, the main current from the battery is
R = 200
100 V = 0.25 A 400 The potential drop across the 200 resistor is therefore, 200 × 0.25 A = 50V and that across 300 is also 50V. This is also the potential drop across the voltmeter and hence the reading of the voltmeter is 50 V.
i =
29.
A galvanometer has a coil of resistance 100 showing a full-scale deflection at 50A. What resistance should be added to use it as (a) a voltmeter of range 50V (b) an ammeter of range 10 mA ? manishkumarphysics.in
Page # 16
Chapter # 32 Electric Current in Conductors Sol. (a) When a potential difference of 50 V is applied across the voltmeter, full-scale deflection should take place. Thus, 50A should go through the coil. We add a resistance R in series with the given coil to achieve this
We have, 50 V 100 R or, R = 106 – 100 = 106 . (b) When a current of 10 mA is passed through the ammeter, 50 A should go through the coil. We add a resistance r in parallel to the cell to achieve this (figure). The curernt through the coil is :
50A =
50 A – (10 mA) or, 30. Sol.
r r 100
r = 0.5 .
The electric field between the plates of a parallel–plate capacitance 2.0 F drops to one third of its initial value in 4.4 s when the plates are connected by a thin wire. Find the reistance of the wire. The electric field between the plates is
Q0 Q E = A = A e–t/RC 0 0 or,
E = E0e–t/RC
In the given problem, E =
31. Sol.
1 E at t = 4.4 s. 3 0
4.4s RC
Thus,
1 e 3
or,
4.4s = In 3 = 1.1 RC
or,
R=
4.4s = 2.0 . 1.1 2.0F
A capacitor is connected to a 12 V battery through a resistance of 10. It is found that the potential difference across the capacitor rises to 4.0 V in 1s. Find the capacitances of the capacitor. The charge on the capacitor during charging is given by Q = Q0(1 – e–t/RC). Hence, the potential difference across the capacitor is V = Q/C = Q0/C (1 – e–t/RC). Here, at t = 1 s, the potential difference is 4V whereas the steady potential difference is Q0/C = 12V. So, 4V = 12V(1 – e–t/RC) 1 3
or,
1 – e–t/RC =
or,
e–t/RC =
or,
t 3 n = 0.405 RC 2
or,
RC =
or,
C=
2 3
t 1s = = 2.469 s 0.405 0.45
2.469 s = 0.25 F.. 10
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Page # 17
Chapter # 32 Electric Current in Conductors 32. A capacitor charged to 50 V is discharged by connecting the two plates at t = 0. If the potential difference across the plates drops to 1.0 V at t = 10 ms, what will be the potential difference at t = 20 ms ? Sol. The potential difference at time t is given by V = Q/C = (Q0/C)e–t/RC or, V = V0e–t/RC According to the given data, 1V = (50 V) e–10 ms/RC 1 50 The potential difference at t = 20 ms is V = V0e–t/RC
33. Sol.
or,
e–10 ms/RC =
=
(50 V) e–20 ms/RC = (50 V) e 10ms / RC
=
0.02 V
2
A 5.0 F capacitor having a charge of 20 C is discharge through a wire of resistance 5.0 . Find the that dissipated in the wire between 25 to 50 s after the connections are made. The charge on the capacitors at time p after the connection are made is Q = Q0e–t/RC dQ = –(Q/RC) e–t/RC , dt Heat dissipated during the time t1 to t2 is
or,
i=
t2
i Rdt
t2
2
U=
t1
=
Q0
2
RC
2
e
2 t / RC
t1
2
dt
Q0 = 2C
2t 2 2 t1 e RC e RC .
...............(i)
The time constant RC is 5 × 5F = 25 s. Putting t1 = 25 s, t2 = 50 s and other values in (i), U=
( 20C) 2 (e–2 – e–4) = 4.7 J. 2 5.0F
EXERCISE 1.
The amount of charge passed in time t through a cross-section of a wire is Q(t) = At2 + Bt + C (a) Write the dimensional formulae for A, B and C (b) If the numerical values of A, B and C are 5, 3 and 1 respectively in SI units, find the value of the current at t = 5 s. ,d rkj ds dkVj {ks=k ls t le; esa fuEu vkos'k izokfgr gksrk gS % Q(t) = At2 + Bt + C (a) A, B o C ds foeh; lw=k fyf[k,A (b) ;fn A, B o C ds vkafdd eku 5, 3 o 1 (S ek=kdksa esa½ gS] rks t = 5 lsd.M ij /kkjk dk eku Kkr dhft,A
2.
An electron gun emits 2.0 × 1016 electrons per second. What electric current does this correspond to? ,d bysDVªkWu xu 2.0 × 1016 bysDVªkWu izfr lsd.M mRlftZr djrh gSA blds laxr fo|qr /kkjk dk eku fdruk gS\ 3.2 × 10–3 A
Ans. 3.
The electric current existing in a discharge tube is 2.0 µA. How much charge is transferred across a crosssection of the tube in 5 minutes? ,d foltZu ufydk esa 2.0 µA fo|qr /kkjk izokfgr gks jgh gSA ufydk ds dkV {ks=k ls 5 fefuV esa fdruk vkos'k LFkkukarfjr
gksrk gS\ 4.
The current through a wire depends on time as i = i0 + t, where i0 = 10 A and = 4 A/s. Find the charge crossed through a section of the wire in 10 seconds.
,d rkj esa /kkjk le; ij fuEukuqlkj fuHkZj djrh gS i = i0 + t, tgk¡ i0 = 10 A o = 4 A/s gS rkj ds dkV {ks=k ls 10 lsd.M esa izokfgr vkos'k Kkr dhft,A Ans.
300 C
manishkumarphysics.in
Page # 18
Chapter # 32 Electric Current in Conductors 5. A current of 1.0 A exists in a copper wire of cross-section 1.0 mm2. Assuming one free electron per atom calculate the drift speed of the free electrons in the wire. The density of copper is 9000 km/m3. 1.0 feeh2 dkVj {ks=k ds ,d rkacs ds rkj esa 1.0 ,Eih;j dh /kkjk izokfgr gks jgh gSA izfr ijek.kq ,d eqDr bysDVªkWu ekurs gq, rkj esa eqDr bysDVªkWuksa ds viokg osx dh x.kuk dhft,A rkacs dk ?kuRo 9000 fdxzk/eh3 gSA Ans. 0.074 mm/s 6. Ans. 7.
A wire of length 1 m and radius 0.1 mm has a resistance of 100 . Find the resistivity of the material. 0.1 feeh f=kT;k ,oa 1 eh- yEckbZ ds ,d rkj dk izfrjks/k 100 gSA rkj ds inkFkZ dh izfrjks/kdrk Kkr dhft,A × 10–6 -m. A uniform wire of resistance 100 is melted and recast in a wire of length double that of the original. What would be the resistance of the wire ? 100 izfrjks/k ds ,dleku rkj dks fi?kykdj blls nqxuh yEckbZ dk ,d rkj cuk;k tkrk gSA bl rkj dk izfrjks/k fdruk
gksxkA 8.
Ans. 9.
Ans. 10.
Consider a wire of length 4m and cross-sectional area 1 mm2 carrying a current of 2A. If each cubic metre of the material contains 1029 free electrons, find the average time taken by an electrons to cross the length of wire. 4eh- yEckbZ ,oa 1 feeh2 dkV {ks=k ds ,d rkj esa ,d ,Eih;j /kkjk izokfgr gks jgh gSa ;fn blds inkFkZ ds izfr ?ku ehVj esa 1029 eqDr bysDVªkWu gks] rks ,d bysDVªkWu dks rkj dh yEckbZ ikj djus esa yxk vkSlr le; Kkr dhft,A 3.2 × 104 s 8.9 hours. What length of a copper wire of cross-sectional area 0.01 mm2 will be needed to prepare a resistance of 1k? Resistivity of copper = 1.7 × 10–8 -m. 1 fdyks vkse ds izfrjks/k fuekZ.k gsrq 0.01 feeh2 dkV {ks=k ds ,d rkacs ds rkj dh yEckbZ fdruh gksuh pkfg,A rkacs dk fof'k"V izfrjks/k = 1.7 × 10–8 -m 0.6 km. Figure shows a conductor of length having a circular cross-section. The radius of cross-section varies linearly from a to b. The resistivity of the material is . Assuming that b – a << l. find the resistance of the conductor. fp=k esa ,d o`Ùkkdkj dkV {ks=k okys yEckbZ ds ,d pkyd dks n'kkZ;k x;k gSA dkV {ks=k dh f=kT;k a ls b rd jsf[kd :i ls cnyrh gSA inkFkZ dh izfrjks/kdrk gSA b – a << l ekurs gq,] pkyd dk izfrjks/k Kkr dhft,A
b
a 11.
Ans.
A copper wire of radius 0.1 mm and resistance 1 k is connected across a power supply of 20 V. (a) How many electrons area transferred per second between the supply and the wire at one end? (b) Write down the current density in the wire. 0.1 feeh f=kT;k ,oa 1 k izfrjks/k ds ,d rkacs ds rkj dks 20 V ds 'kfDr lzkr s ls tksMk+ tkrk gSA (a) rkj ds ,d fljs ,oa lzksr ds e/; izfr lsd.M fdrus bysDVªkWu LFkkukarfjr gksrs gSaA (b) rkj esa /kkjk ?kuRo Kkr dhft,A (a) 1.25 × 1017 (b) 6.37 × 108 A/m2 .
12.
Calculate the electric field in a copper wire of cross-sectional area 2.0 mm2 carrying a current of 1A. The resistivity of copper = 1.7 × 10–5 -m. 2.0 feeh2 dkV {ks=k ds ,d rkacs ds rkj esa fo|qr {ks=k dh x.kuk dhft,] tcfd rkj esa 1 ,Eih;j /kkjk izokfgr gks jgh gSA rkacs dh izfrjksd/kdrk = 1.7 × 10–5 -m A
13.
A wire has a length of 2.0 m and a resistance of 5.0 . Find the electric field existing inside the wire if it carries a current of 10 A. ,d rkj dh yEckbZ 2.0 eh- ,oa izfrjks/k 5.0 gSA ;fn rkj esa 10 ,Eih;j /kkjk izokfgr gksrh gS rks rkj esa mifLFkr fo|qr
{ks=k Kkr dhft,A Ans.
25 V/m.
14.
The resistance of an iron wire and a copper wire at 200C are 3.9 and 4.1 respectively. At what temperature will the resistances be equal ? Temperature coefficient of resistivity for iron is 5.0 × 10–3 K–1 and for copper it is 4.0 × 10–3 K–1 and for copper it is 4.0 × 10–3 K–1 and for copper it is 4.0 × 10– 3 K–1 . Neglect any thermal expansion. 200C ij yksgs ds ,d rkj ,oa rkacs ds ,d rkj ds izfrjks/k Øe'k% 3.9 o 4.1 gSA fdl rki ij nksuksa ds izfrjks/k cjkcj gksxsa\ izfrjks/kdrk dk rki xq.kkad cjkcj gksxsa\ izfrjks/kdrk dk rki xq.kkad yksgs ds fy, 5.0 × 10–3 K–1 o rkacs ds fy, manishkumarphysics.in
Page # 19
Chapter # 32
Electric Current in Conductors –3
–1
4.0 × 10 K 15.
gSA m"eh; izlkj dks ux.; eku yhft,A
The current in a conductor and the potential difference across its ends are measures by an ammeter and a voltmeter. The meters draw negligible currents. The ammeter is accurate but the voltmeter has a zero error (that is , it does read zero when no potential difference is applied). Calculate the zero error if the readings for two different conditions are 1.75 A, 14.4 V and 2.75 A, 22.4 V.
,d pkyd esa /kkjk ,oa blds fljksa ds e/; foHkokUrj ,d vehVj ,oa ,d oksYVehVj }kjk ekik tkrk gSA ehVj ux.; /kkjk ysrs gSA vehVj fcYdqy lgh gS] ysfdu oksYVehVj esa 'kwU;kadh =kqfV gSA ¼vFkkZr~ tc dksbZ foHkokUrj ugha yxk;k tk, rks ;g 'kwU; ikB~;kad ugha nsrk gSA 'kwU;kadh =kqfV dh x.kuk dhft,A ;fn nks fHkUu fLFkfr;ksa esa ikB~;kad 1.75 A, 14.4 V rFkk 2.75 A, 22.4 V. gSA Ans. 16.
0.4 volt Figure shows an arrangement to measure the emf and internal resistance r of a battery. The voltmeter has a very high resistance and the ammeter also has some resistances. The voltmeter reads 1.52 V when the switch S is open. When the switch is closed the voltmeter reading drops is to 1.45 V and the ammeter reads 1.0 A. Find the emf and the internal resistance of the battery ? fp=k esa ,d cSVjh ds fo-ok-cy rFkk vkarfjd izfrjks/k r ekius dh ,d O;oLFkk n'kkZbZ xbZ gSA oksYVehVj dk izfrjks/k dkQh vf/kd gS rFkk vehVj dk Hkh dqN izfrjks/k gSA tc fLop S [kqyk gSA rks oksYVehVj dk ikB~;kad 1.52 V gksrk gSA tc fLop S can fd;k tkrk gS rks oksYVehVj dk ikB~;kad 1.45 V rd de gks tkrk gS rFkk vehVj dk ikB~;kad 1.0 A gksrk gSA cSVjh
dk fo-ok-cy rFkk vkarfjd izfrjks/k Kkr dhft,A
Ans.
1.52 V, 0.07
17.
The potential difference between the terminals of a battery of emf 6.0 V and internal resistance 1 drops to 5.8 V when connected across an external resistor. Find the residences of the external resistor. tc ,d 6.0 oksYV fo-ok-cy rFkk 1 vkarfjd izfrjks/k okyh cSVjh dks ,d cká izfrjks/k ls tksMk+ tkrk gS rks bldh VfeZuy oksYVrk 5.8 oksYV jg tkrh gSA cká izfrjks/k dk eku Kkr dhft;sA 29
Ans. 18.
Ans. 19.
The potential difference between the terminals of a 6.0 V battery is 7.2 V when it is being charged by a current of 2.0 A. What is the internal resistance of the battery ? tc ,d 6.0 oksYV okyh cSVjh dks 2.0 A /kkjk izokfgr djds vkosf'kr fd;k tkrk gS rks blds VfeZuyksa ds e/; foHkokUrj 7.2 V gks tkrk gSA cSVjh dk vkarfjd izfrjks/k fdruk gS\ 0.6 The internal resistance of an accumulator battery of emf 6V is 10 when it is fully discharged. As the battery gets charged up, its internal resistance decreases to 1. The battery in its completely discharged state is connected to a charger which maintains a constant potential difference of 9V. Find the current through the battery (a) just after the connections are made and (b) after a long time when it is competitively charged. 6V fo-ok-cy okyh lapk;d cSVjh dk iw.kZr;k fujkos'ku dh fLFkfr esa vkarfjd izfrjks/k 10 gSA tc cSVjh vkosf'kr gks tkrh gS rks vkarfjd izfrjks/k de gksdj 1jg tkrk gSA iw.kZr;k fujkosf'kr cSVjh dks ,d pktZj ls tksM+k tkrk gS tks 9V foHkokUrj fu;r j[krk gSA cSVjh ls izokfgr /kkjk Kkr dhft;sA (a) rqjra tksMr+ s gh rFkk (b) yEcs le; i'pkr~ tc ;g iw.kZr;k vkosf'kr
gks tkrh gSA
3 A = 0.3 A 10
Ans.
(a)
(b) 3A
20.
Find the value of i1/i2 in figure if (a) R = 0.1 , (b) R = 1 (c) R = 10 . Note from your answer that in order to get more current from a combination of two batteries they should be joined in parallel if the external resistance is small and in series if the external resistance is large as compared to the internal resistances. fp=k esa iznf'kZr ifjiFk ds fy;s i1 / i2 dk eku Kkr dhft;s] ;fn (a) R = 0.1 , (b) R = 1 (c) R = 10 .
vius mÙkj }kjk bl ij /;ku nhft;s fd nks cSVfj;ksa ds la;kstu ls vf/kd /kkjk izkIr djus ds fy;s ;fn vkarfjd izfrjks/k dh rqyuk esa cká izfrjks/k de gS rks budks lekukarj la;ksftr djuk gksxk ,oa ;fn vkarfjd izfrjks/k dh rqyuk esa cká manishkumarphysics.in
Page # 20
Chapter # 32
Electric Current in Conductors
izfrjks/k vf/kd gS rks budks Js.kh la;ksftr djuk pkfg;sA
1.2 = 0.57 (b) 1 2.1
(c)
10.5 = 1.75 6
Ans.
(a)
21.
Consider N = n1n2 identical cells, each of emf and internal resistance r. Suppose n1 cells ar joined in series to form a line and n2 such lines are connected in parallel. The combination drives a current in an external resistance R. (a) Find the current in the external resistance. (b) Assuming that n1 and n2 can be continuously varies, find the relation between n1,n2 R and r for which the current in R is maximum. N = n1n2 ,d tSls lsyksa esa izR;sd dk fo-ok-cy rFkk vkarfjd izfrjks/k r gSA ekukfd n1 lsyksa dks Js.kh Øe esa tksMk+ tkrk gS rFkk ,slh n2 Js.kh;k¡ lekUrj Øe esa la;ksftr dh tkrh gSA ;g la;kstu ,d cká izfrjks/k R esa /kkjk izokfgr djrk gSA (a) cká izfrjks/k esa izokfgr /kkjk Kkr dhft;sA (b) ;g ekurs gq, fd n1 rFkk n2 fujUrj ifjofrZr fd;s tk ldrs gSAa R vf/kdre /kkjk izokfgr djus ds fy;s n1,n2 , R rFkk r ds e/; laca/k Kkr dhft;sA
22.
A battery of emf 100 V and a resistor of resistance 10k are joined in series. This system is used as a source to supply current to an external resistance R if R not greater than 100 then the current through it is constant upto two significant digits. Find its value. This is the basic principle of a constant-current source. 100 V fo-ok-cy dh ,d cSVjh rFkk 10k eku dk izfrjks/k Js.kh Øe esa tksM+s x;s gSA ;g fudk; ,d cká izfrjks/k R esa /kkjk izokfgr djus ds fy, iz;D q r fd;k tkrk gSA ;fn R dk eku 100 ls vf/kd ugha gS] rks blls izokfgr /kkjk nks lkFkZd
vadksa rd fu;r jgrh gSA bldk eku Kkr dhft;sA ;g fu;r /kkjk lzkrs dk ewyHkwr fl)kUr gSA 23.
If the reading of ammeter A1 in figure is 2.4 A, what will the ammeters A2 and A3 read ? Neglect the resistance of the ammeters. ;fn fp=k esa iznf'kZr vehVj A1 dk ikB~;kad 2.4 A gS] rks vehVj A2 rFkk A3 ds ikB~;kad D;k gksx\as vehVjksa ds izfrjks/k ux.;
eku yhft;sA
Ans.
1.6 A, 4.0 A.
24.
The resistance of the rheostat shown in figure is 30 . Neglecting the meter resistance, find the minimum and maximum currents through the ammeter as the rheostat is varies. fp=k esa iznf'kZr /kkjk fu;a=kd dk izfrjks/k 30 gSA vehVj dk izfrjks/k ux.; ekurs gq,] /kkjk fu;a=kd esa ifjorZu ds lkFk
vehVj ls izokfgr U;wure rFkk vf/kdre /kkjk,¡ Kkr dhft;sA
Ans.
0.15 A, 0.83 A
25.
Three bulbs, each having a resistance of 180 , are connected in a parallel to an ideal battery of emf 60 V. Find the current delivered by the battery when (a) all the bulbs are switched on, (b) two of the bulbs are switched on and (c) only one bulb is switched on. rhu cYcksa esa izR;sd dk izfrjks/k 180 gS] budks ,d60 V fo-ok-cy dh vkn'kZ cSVjh ds lekUrj la;ksftr fd;k x;k gSA cSVjh ls izokfgr /kkjk Kkr dhft;s tcfd (a) lkjs cYc pkyw gksA (b) nks cYc pkyw gksA (c) dsoy ,d cYc pkyw gksA manishkumarphysics.in
Page # 21
Chapter # 32 Electric Current in Conductors 26. Suppose you have three resistors of 20 , 50 and 100 . What minimum and maximum resistances can you obtain from these resistors ? ekuk fd vkids ikl 20 , 50 rFkk 100 eku okys rhu izfrjks/k gSA bu izfrjks/kksa ls vki U;wure rFkk vf/kdre fdruk
izfrjks/k izkIr dj ldrs gSa\
Ans.
12.5 , 170
27.
A bulb is made using two filaments. A switch selects whether the filaments are used individually or in parallel. When used with a 15 V battery, the bulb can be operated at 5W, 10W or 15W. What should be the resitances of the filaments ? ,d cYc esa nks rUrq gSA ,d fLop ;g fu/kkZfjr djrk gS fd rarq vyx&vyx iz;D q r gksxas ;k lekUrj Øe esa 15 V dh cSVjh ds lkFk iz;D q r djus ij cy dk mi;ksx 5W, 10W ;k 15W ij fd;k tk ldrk gSA rUrqvksa ds izfrjks/kksa dk eku fdruk gSA
Ans. 28.
45 , 22.5 Figure shows a part of a circuit. If a current of 12 mA exists in the 5k resistor, find the currents in the other three resistors. What is the potential difference between the points A and B ? fp=k esa fdlh ifjiFk dk ,d Hkkx iznf'kZr fd;k x;k gSA ;fn 5kizfrjks/k ls 12 mA /kkjk izokfgr gks jgh gSA rks vU; izfrjks/kksa ls izokfgr /kkjk,¡ Kkr dhft;sA fcUnqvksa A rFkk B ds e/; foHkokUrj fdruk gS ?
Ans.
4 mA in 20 k resistor, 8 mA in 10 kW resistor and 12 mA in 100 k resistor, 1340 V
29.
An ideal battery sends a current of 5A in a resistor. When another resistor of value 10 is connected in parallel, the current through the battery is increased to 6A. Find the resistance of the first resistor. ,d vkn'kZ cSVjh] ,d izfrjks/k esa 5A /kkjk izokfgr djrh gSA tc blds lekUrj esa 10 eku dk ,d vU; izfrjks/k yxk;k tkrk gS rks cSVjh ls izokfgr /kkjk c<+dj 6A gks tkrh gSA igys okys izfrjks/k dk eku Kkr dhft;sA
30.
Find the equivalent resistance of the network shown in figure between the points a and b. fp=k esa iznf'kZr ifjiFk tky dk fcUnqvksa a rFkk b ds e/; rqy; izfrjks/k Kkr dhft;sA
Ans.
(a) r/3
31.
A wire of rsistance of 15 is bent to form a regular hexagon ABCDEFA. Find the equivalent resistance of the loop between the points (a) A and B, (b) A and C and (c) A and D. ,d rkj dk izfrjks/k 15 gS] bldks eksM+dj ,d fu;fer "kV~Hkqt ABCDEFA cuk;k x;k gSA ywi dk izfrjks/k fcUnqvks (a) A rFkk B, (b) A rFkk C (c) A rFkk D ds e/; Kkr dhft;sA (a) 2.08 (b) 3.33 (c) 3.75
Ans. 32.
Consider the circuit shown in figure. Find the current through the 10 resistor when the switch S is (a) open (b) closed. fp=k esa iznf'kZr ifjiFk ds fy;s 10 izfrjks/k ls izokfgr /kkjk Kkr dhft;s] tcfd fLop S is (a) [kqyk gqvk gSA (b) can gSA
Ans.
(a) 0.1 A
33.
Find the currents through the three resistors shown in figure
(b) 0.3 A
fp=k esa iznf'kZr rhuksa izfrjks/kksa ls izokfgr /kkjk,¡ Kkr dhft;sA
manishkumarphysics.in
Page # 22
Chapter # 32
34.
Electric Current in Conductors
Figure shows a part of an electric circuit. The potentials at the points a, b and c are 30V, 12 V and 2V respectively. Find the currents through the three resistors. fp=k esa fdlh fo|qr ifjiFk dk ,d Hkkx iznf'kZr fd;k x;k gSA fcUnqvksa a, b rFkk c ds foHko Øe'k% 30V, 12 V ,oa 2V gSA
rhuksa izfrjks/kksa ls izokfgr /kkjk,¡ Kkr dhft;sA
35.
Each of the resistors shown in figure has a resistance of 10 and each of the batteries has an emf of 10 V. Find the currents through the resistors a and b in the two circuits fp=k esa iznf'kZr izR;sd izfrjks/k dk eku 10 rFkk izR;sd cSVjh dk fo-ok-cy 10 V gSA nksuksa ifjiFkksa esa izfrjks/k a rFkk b ls
izokfgr /kkjk,¡ Kkr dhft;sA
36.
Ans.
37.
Find the potential difference Va – Vb in the circuits shown in figure. iznf'kZr ifjiFkksa esa foHkokUrj Va – Vb Kkr dhft;sA
1 2 R1 R 2 1 1 (a) 1 R1 R 2 R 3
1 2 R1 R 2 1 1 (b) 1 R1 R 2 R 3
In the circuit shown in figure, 1 = 3V. 2 = 2V, 3 = 1V and r1 = r2 = r3 = 1. Find the potential difference between the points A and B and the current through each branch. fp=k esa iznf'kZr ifjiFk esa 1 = 3V. 2 = 2V, 3 = 1V rFkk r1 = r2 = r3 = 1gSA fcUnqvksa A rFkk B ds e/; foHkokUrj rFkk
izR;sd 'kk[kk ls izokfgr /kkjk Kkr dhft;sA
38.
Find the current through the 10 resistor shown in figure fp=k esa iznf'kZr 10 izfrjks/k ls izokfgr /kkjk Kkr dhft;sA
manishkumarphysics.in
Page # 23
Chapter # 32 Ans. zero 39.
Electric Current in Conductors
Find the current in the three resistors shown in figure
fp=k esa iznf'kZr rhuksa izfrjks/kksa ls izokfgr /kkjk Kkr dhft;sA
40.
What should be the value of R in figure for which the current in it is zero? fp=k esa iznf'kZr izfrjks/k R dk eku fdruk gks fd blls izokfgr /kkjk 'kwU; gks\
41.
Find the equivalent resistance of the circuits shown in figure between the points a and b. Each resistor has a resistance r. fp=k esa iznf'kZr ifjiFk dk fcUnqvksa a rFkk b ds e/; rqY; izfrjks/k Kkr dhft;sA izR;sd izfrjks/k dk eku r gS -
42.
Find the current measured by the ammeter in the circuit shown in figure.
fp=k esa iznf'kZr vehVj }kjk ekih x;h /kkjk Kkr dhft;sA
43.
Consider the circuit shown in figure (a). Find (a) the current in the circuit, (b) the potential drop across the 5 resistor, (c) the potential drop across the 10 resistor, (d) Assume the parts (a),(b) and (c) with reference to figure. fp=k (a) esa iznf'kZr ifjiFk ij fopkj dhft;sA Kkr dhft;s % (a) ifjiFk esa /kkjk (b) 5 izfrjks/k ds fljksa ij foHkokUrj (c) 10 izfrjks/k ds fljksa ij foHkokUrj] (d) fp=k (b) ds fy, Hkkx (a), (b) rFkk (c) ds mÙkj nhft;sA
manishkumarphysics.in
Page # 24
Chapter # 32 Electric Current in Conductors 44. Twelve wires, each having equal resistance r, are joined to form a cube as shown in figure. Find the equivalent resistance between the diagonally opposite points a and f. izfrjks/k r okys ckjg rkjksa dks ?kukÑfr cukus ds fy;s fp=kkuqlkj tksMk+ x;k gSA fod.kZr% foifjr fcUnqvksa a rFkk f ds e/; rqY;
izfrjks/k Kkr dhft;sA
45.
Find the equivalent resistance of the networks shown in figure between the points a and b. fp=k esa iznf'kZr ifjiFk tkyksa dk fcUnqvksa a rFkk b ds e/; rqY; izfrjks/kksa dks eku Kkr dhft;sA
Ans.
(a)
46.
5 4 1 r (b) r (c) r (d) r (e) r 8 3 4 An infinite ladder is constructed with 1 and 2 resistors as shown in figure. (a) Find the effective resistance between the points A and B. (b) Find the current that passes through the 2 resistor nearest to the battery. tSlk fd fp=k esa iznf'kZr fd;k x;k gSA 1 rFkk 2 eku okys izfrjks/kksa ls ,d vuar lh<+h cuk;h x;h gSA (a) fcUnqvksa A rFkk B ds e/; izHkkoh izfrjks/k Kkr dhft;sA (b) cSVjh ds lehi fLFkr 2 izfrjks/k ls izokfgr /kkjk Kkr dhft;sA
47.
The emf and the internal resistance r of the battery shown in figure are 4.3 V and 1.0 respectively. The external resistance R is 50 . The resistances of the ammeter and voltmeter are 2.0 and 200 respectively. (a) Find the readings of the two meters. (b) The switch is thrown to the other side. What will be the readings of the two meters now ? HCV_Ch-32_Ex._47 fdlh cSVjh dk fo0ok0cy rFkk vkUrfjd çfrjks/k r fp=kkuqlkj Øe'k% 4.3 V vkSj 1.0 gSA cká çfrjks/k R, 50 gSA vehVj rFkk oksYVehVj dk çfrjks/k Øe'k% 2.0 rFkk 200 gSA (a) nksuksa ehVjksa dk ikB;kad crkb,A (b) vc fLop dks nwljh rjQ yxk;k tkrk gS] vc nksuksa ehVjksa dk ikB;kad crkb, ?
Ans.
: (a) 0.1 A, 4.0 V
(b) 0.08 A, 4.2 V manishkumarphysics.in
Page # 25
Chapter # 32 Electric Current in Conductors 48. A voltmeter of resistance 400 is used to measure the potential difference across the 100 resistor in the circuit shown in figure. (a) What will be the reading of the voltmeter ? (d) What was the potential differences across 100 before the voltmeter was connected ? fp=k esa iznf'kZr 100 izfrjks/k ds fljksa ij foHkokUrj ekiu ds fy;s iz;D q r fd;s x;s oksYVehVj dk izfrjks/k 400 gSA (a) oksYVehVj dk ikB~;kad D;k gS\ (d) oksYVehVj la;ksftr djus ls iwoZ 100 izfrjks/k ds fljksa ij foHkokUrj fdruk Fkk\
Ans.
(a) 24 V
(b) 28 V
49.
The voltmeter shown in figure reads 18 V across the 50 resistor. Find the resistance of the voltmeter. fp=k esa iznf'kZr 50 izfrjks/k ds fljksa ij yxs oksYVehVj dk ikB~;kad 18 V gSA oksYVehVj dk izfrjks/k Kkr dhft;sA
Ans.
130
50.
A voltmeter consists of a 25 coil connected in series with a 575 resistor. The coil takes 10mA for full scale deflection. What maximum potential difference can be measured on this voltmeter? 25 izfrjks/k okyh dq.Myh ls cuk gqvk ,d oksYVehVj] 575 izfrjks/k ds Js.khØe esa yxk gqvk gSA iwjs iSekus ij fo{ksi ds fy;s dq.Myh ls izokfgr /kkjk 10mA gksuh pkfg;sA bl oksYVehVj }kjk ekik tk ldus okyk vf/kdre foHkokUrj fdruk gks
ldrk gSA Ans.
6V
51.
An ammeter is to be constructed which can read currents upto 2.0 A. If the coil has a resistance of 25 and takes 1mA for full-scale deflection, what should be the resistance of the shunt used? ,d ,slk vehVj cukuk gS ftlls vf/kdre 2.0 A /kkjkekih tk ldsA ;fn dq.Myh dk izfrjks/k 25 rFkk iwjs iSekus ij fo{ksi ds fy;s /kkjk 1mA gS] rks iz;D q r fd;s x;s 'kaV rkj dk izfrjks/k fdruk gksxkA –2 1.25 × 10 .
Ans. 52.
A voltmeter coil has resistance 50.0 and a resistor of 1.15 k is connected in series. It can read potential deference upto 12 volts. If this same coil is used to construct an ammeter which can measure currents upto 2.0 A, what should be the resistance of the shunt used ? ,d oksYVehVj dh dq.Myh dk izfrjks/k 50.0 rFkk blds Js.khØe esa 1.15 k izfrjks/k tqM+k gqvk gSA ;g 12 oksYV ds foHkokUrj dk ekiu dj ldrk gSA ;fn ,slh gh dq.Myh iz;D q r djds ,d vehVj cuk;k tk;s tks 2.0 A rd /kkjk eki lds
53.
The potentiometer wire AB shown in figure is 40 cm long. Where should the free end of the galvanometer be connected on AB so that the galvanometer may show zero deflection ? fp=k esa iznf'kZr foHkoekih dk rkj 40 lseh yEck gSA rkj AB ij /kkjkekih dk eqDr fljk dgk¡ ij la;ksftr fd;k tk;s rkfd
rks yxk;s x;s 'kaV rkj dk izfrjks/k fdruk gksxk\
/kkjkekih esa 'kwU; fo{ksi izkIr gks\
Ans.
16 cm from A.
54.
The potentiometer wire AB shown in figure is 50 cm long. When AD = 30 cm, no deflection occurs in the galvanometer. Find R. fp=k esa iznf'kZr foHkoekih dk rkj AB, 50 lseh yEck gSA tc AD = 30 lseh gS] /kkjkekih esa 'kwU; fo{ksi izkIr gksrk gSA R dk
eku Kkr dhft;sA
manishkumarphysics.in
Page # 26
Chapter # 32
55.
Electric Current in Conductors
A 6 volt battery of negligible internal resistance is connected across a uniform wire AB of length 100 cm. The positive terminal of another battery of emf 4V and internal resistance 1 is joined to the point A as shown in figure. Take the potential at B to be zero. (a) What are the potentials at the points A and C ? (b) At which point D of the wire AB, the potential is equal to the potential at C. (c) If the point C and D are connected by a wire, what will be the current through it ? (d) If the 4V battery is replaced by 7.5 V battery, what would be the answers of parts (a) and (b) ? [HCV_Chp_32_Exer_55] 100 lseh yEcs ,d le:i rkj ds fljksa ij ux.; vkarfjd izfrjks/k okyh 6 v dh cSVjh yxh gqbZ gSA fp=kkuqlkj 4V fo-okcy rFkk 1 vkarfjd izfrjks/k okyh ,d vU; cSVjh dk /ku VfeZuy fcUnq A ij tqMk+ gqvk gSA B ij foHko 'kwU; eku yhft;s% (a) fcUnqvksa A rFkk C ij foHko ds eku D;k gS\ (b) rkj AB ds fdl fcUnq D ij foHko] C ij foHko ds cjkcj gS\ (c) ;fn fcUnqvksa C rFkk D dks ,d rkj ls tksM+ fn;k tk;s rks blls fdruh /kkjk izokfgr gksxh\ (d) ;fn 4V dh cSVjh dks 7.5 V okyh
cSVjh ls izfrLFkkfir dj fn;k tk;s rks Hkkx ¼v½ rFkk ¼c½ ds mÙkj D;k gksxa \s
Ans. 56.
(a) 6 V, 2 V (b) AD = 66.7 cm (c) zero (d) 6 V, – 1.5 V, –1.5 V, no such point D exists.
Consider the potentiometer circuit arranged as in figure. The potentiometer wire is 600 cm long. (a) At what distance from the point. A should the jockey touch the wire to get zero deflection in the galvanometer ? (b) If the jockey touches the wire at a distance of 560 cm from A, what will be the current in the galvanometer ? fp=k esa la;ksftr foHkoekih ij fopkj dhft;sA foHkoekih dk rkj 600 lseh yEck gSA (a) A ls fdruh nwjh ij folihZ dqt a h dk laidZ fd;k tk;s fd /kkjkekih esa 'kwU; fo{ksi izkIr gksA (b) ;fn folihZ dqt a h A ls 560 lseh nwjh ij Li'kZ djrh gS rks
/kkjkekih ls fdruh /kkjk izokfgr gksxh\
r
A
R = 15r
/2 Sol.
(a)
When Jockey is not connected.
E = 16 r
=
G
tc tk¡dh tqM+h gqbZ ugha gSA ................(i)
Resistance per unit length
,dkad yEckbZ dk izfrjks/k
15r /cm 600
Let be the length when we get zero deflection. ekuk 'kwU; fo{ksi dh fLFkfr esa yEckbZ gS.
E = () 2 E E 15r 2 16r 600
(b)
r
B C
= 320 cm
Let potential at A is zero ekuk A ij foHko 'kwU; gS Then apply Kirchoff’s st law fdjpkWQ ds fu;e ls
manishkumarphysics.in
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Chapter # 32
Electric Current in Conductors
x
x 0 + 14r
0 ( x 0) 14 2 + = 0 x = 2r 22 r
14 2 = 22 2 = 3 22r r r
x g = (b)
When tc = 560 cm
r’ = (560) ×
15 r 600
r’ = 14r Using KVL in loop (i) ywi (i) esa KVL ls E – 1 .14r – r – r = 0 ................(i) and in loop (2) rFkk ywi (2) esa
E =0 ...............(ii) 2 Solving equation (i) and (ii) we have lehdj.k (1) o (2) dks gy djus ij (30.1r – E) = (E – 141 r) – 1 14r + ( – 1)r +
2E 1 = 44 r E 1 = 22 r
4E and rFkk = 22 r
So current in galvanometer Branch
3E g = 22 r 57.
4E E = 22 r 22 r
vr% /kkjkekih 'kk[kk esa /kkjk = ( – 1)
Ans. (a) 320 cm (b)
3 22 r
Find the charge on the capacitor shown in figure.
fp=k esa iznf'kZr la/kkfj=k ij vkos'k Kkr dhft;sA
58.
(a) Find the current in the 20 resistor shown in figure. (b) If a capacitor of capacitance 4F is joined between the points A and B, what would be the electrostatic energy stored in it in steady state ? (a) fp=k esa iznf'kZr 20 izfrjks/k ls izokfgr /kkjk Kkr dhft;sA (b) ;fn fcUnqvksa A rFkk B ds chp 4F /kkfjrk dk la/kkfj=k
tksM+ fn;k tk;s] rks LFkk;h fLFkfr esa fdruh fLFkj oS|qr ÅtkZ lafpr gksxh\
59.
Find the charges on the four capacitor of capacitance 1F, 2F, 3F and 4F shown in figure. fp=k esa iznf'kZr pkj la/kkfj=kksa dh /kkfjrk,¡ 1F, 2F, 3F rFkk 4F gS] bu ij vkos'k Kkr dhft;sA
manishkumarphysics.in
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Chapter # 32
Electric Current in Conductors
60.
Find the potential difference between the points A and B and between the points B and C of figure in steady state. fp=k esa iznf'kZr ifjiFk esa LFkk;h fLFkfr fcUnqvksa A rFkk B ds e/; ,oa fcUnqvksa B rFkk C ds e/; foHkokUrj Kkr dhft;sA
61.
A capacitance C, a resistance R and an emf are connected in series at t = 0. What is the maximum value of (a) the potential difference across the resistor, (b) the current in the circuit, (c) the the potential difference across the capacitor, (d) the energy stored in the capacitors. (e) the power delivered by the battery and (f) the power converted into heat. le; t = 0 ij ,d la/kkfj=k C, ,d izfrjks/k R ,oa ,d fo-ok-cy Js.khØe esa tksM+s x;sA buds vf/kdre eku D;k gksxas (a) izfrjks/k ds fljksa ij foHkokUrj, (b) ifjiFk esa /kkjk , (c) la/kkfj=k ds fljksa ij foHkokUrj , (d) la/kkfj=k ds fljksa ij foHkokUrj (e) cSVjh }kjk iznku dh xbZ 'kfDr rFkk (f) m"ek esa ifjofrZr 'kfDr
62.
A parallel-plate capacitor with plate area 20 cm2 and plate separation 1.0 mm is connected to a battery. The resistance of the circuit is 10 k. Find the time constant of the circuit. ,d cSVjh ls tksM+s x;s lekukUrj iV~V la/kkfj=k dh IysV dk {ks=kQy 20 lseh2 rFkk IysVksa dk varjky 1.0 feeh gSA ifjiFk dk izfrjks/k 10 kgSA ifjiFk dk le; fLFkjkad Kkr dhft;sA
63.
A capacitor of capacitance 10 F is connected to a battery of emf 2V.It is found that it takes 50 ms for the charge on the capacitor to become 12.6 c. Find the resistance of the circuit. 10 F /kkfjrk dk la/kkfj=k] 2V fo-ok-cy dh cSVjh ls tksMk+ x;k gSA ;g izfs {kr fd;k x;k gS fd 50 feyh lsd.M esa la/kkfj=k ij vkos'k 12.6 c gks tkrk gSA ifjifk dk izfrjks/k Kkr dhft;sA
64.
A 20 F capacitor is joined to a battery of emf 6.0 V through a resistance of 100 . Find the charge on the capacitor 2.0 ms after the connections are made. 20 F /kkfjrk okyk la/kkfj=k 100 izfrjks/k ds lkFk ,d 6.0 V fo-ok-cy okyh cSVjh ls tksM+k x;k gSA la;kstu djus ds 2.0 feyh lsd.M i'pkr~ la/kkfj=k ij vkos'k Kkr dhft;sA
65.
The plates of a capacitor of capacitance 10 F, charged to 60 C, are joined together by a wire of resistance 10 at t = 0. Find the charge on the capacitor in the circuit at (a) t = 0, (b) t = 30 s, (c) t = 120 s and (d) t = 1.0 ms. 10 F /kkfjrk okys la/kkfj=k dh IysVas 60 C vkos'k rd vkosf'kr djus ds i'pkr~ le; t = 0 ij 10W izfrjks/k okys rkj ls vkil esa tksM+ nh xbZ gSA ifjiFk esa la/kkfj=k ij vkos'k Kkr dhft;sA (a) t = 0, (b) t = 30 s ij , (c) t = 120 s ij rFkk (d) t = 1.0 ms. ij
66.
A capacitor of capacitance 8.0 F is connected to a battery of emf 6.0 V through a resistance of 24 . Find the current in the circuit (a) just after the connections are made and (b) one time constant after the connections are made. ,d 8.0 F /kkfjrk dk la/kkfj=k 24 izfrjks/k ds lkFk 6.0 V fo-ok-cy okyh cSVjh ls tksM+ fn;k x;k gSA ifjiFk esa /kkjk Kkr dhft;s % (a) la;kstu ds rqjra i'pkr~ rFkk (b) la;kstu ls ,d le; fLFkjkad ds rqY; le; ds i'pkr~
67.
A parallel-plate capacitor of plate area 40 cm2 and separation between the plates 0.10 mm is connected to a battery of emf 2.0 V through a 16 resistor. Find the electric field in the capacitor 10 ns after the connecting are made. manishkumarphysics.in
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Chapter # 32
Electric Current in Conductors
,d lekukarj iV~V la/kkfj=k dh IysV dk {ks=kQy 40 lseh2 rFkk IysV varjky 0.10 feeh gS] bldks 16 izfrjks/k }kjk ,d 2.0 V fo-ok-cy okyh cSVjh ls tksM+kx;k gSA la;kstu ls 10 ns i'pkr~ la/kkfj=k esa fo|qr {ks=k Kkr dhft;sA 68.
A parallel-plate capacitor has plate area 20 cm2, plate separation 1.0 mm and a dielectric slab of dielectric constant 5.0 filling up the space between the plates. This capacitor is joined to a battery of emf 6.0 V through a 100 k resistor. Find the energy of the capacitor 8.9 s after the connections are made. ,d lekukarj iV~V la/kkfj=k dh IysV dk {ks=kQy 20 lseh2, rFkk IysV varjky 1.0 feeh gS] bldh IysVksa ds varjky esa ijkoS|qrkad 5.0 gSA bl la/kkfj=k dks 100 k izfrjks/k }kjk ,d 6.0 V fo-ok-cy okyh cSVjh ls tksM+k x;k gSA la;kstu ds 8.9 s i'pkr~ la/kkfj=k dh ÅtkZ Kkr dhft;sA
69.
A 100 F capacitor is joined to a 24 V battery through a 1.0 M resistor. Plot qualitative graphs (a) between current and time for the first 10 minutes and (b) between charge and time for the same period. ,d 100 F /kkfjrk dk la/kkfj=k 1.0 M izfrjks/k }kjk 24 V dh cSVjh ls tksMk+ x;k gSA xq.kkRed ys[kkfp=k cukb;sA (a) izFke 10 fefuV ds fy;s /kkjk rFkk le; ds chp rFkk (b) blh le; ds fy;s vkos'k rFkk le; ds chp
70.
How many time constants will elapse before the current in a charging RC circuit drops to half of its initial value ? Answer the same question for a discharging RC circuit. RC ifjiFk esa vkos'kudky esa /kkjk dk eku izkjfEHkd eku dk vk/kk gksus rd fdrus le; fLFkjkad xqtjrs gS\a blh iz'u dk mÙkj RC ifjiFk ds foltZu dky ds fy;s Hkh nhft;sA
71.
How many time constants will elapse before the charge on a capacitor falls to 0.1% of its maximum value in a discharging RC circuit ? RC ifjiFk ds foltZu dky esa la/kkfj=k ij vkos'k vf/kdre dk 0.1% gksus rd fdrus le;&fLFkjkad xqtjrs gSa\
72.
How many time constant will elapse before the energy stored in the capacitor reaches half of its equilibrium value in a charging RC circuit ? RC ifjiFk esa vkos'ku dky esa la/kkfj=k esa lafpr ÅtkZdk eku LFkk;h fLFkfr dh ÅtkZ dk vk/kk gksus rd fdrus le; fLFkjkad
O;rhr gks tk;sxsa\ 73.
How many time constants will elapse before the power delivered by the battery drops to half of its maximum value in an RC circuit ? RC ifjiFk esa cSVjh }kjk iznku dh tkus okyh 'kfDr dk eku] vf/kdre eku dk vk/kk jg tkus rd fdrus le; fLFkjkad O;rhr
gks tk;sxsa\ 74.
A capacitor of capacitance C is connected to a battery of emf at t = 0 through a resistancsd R. Find the maximum rate at which energy is stored in the capacitor. When does the rate has this maximum value le; t = 0 ij /kkfjrk dk ,d la/kkfj=k] izfrjks/k R }kjk fo|qr okgd cy dh cSVjh ls tksM+k tkrk gSA la/kkfj=k esa ÅtkZ lap;
dh vf/kdre nj Kkr dhft;sA ;g vf/kdre nj dc gksxh\ 75.
A capacitor of capacitances 12.0 F is connected to a battery of emf 6.00 V and internal resistance 1.00 through resistanceless leads. 12.0 s after the connections are made, what will be (a) the current in the circuit, (b) the power delievered by the battery, (c) the power dissipated in heat and (d) The rate at which energy stored in the capacitor is increasing. 12.0 F /kkfjrk dk ,d la/kkfj=k] 6.00 V fo-ok-cy rFkk 1.00 vkarfjd izfrjks/k okyh cSVjh ls izfrjks/k jfgr la;kstu rkjksa dh lgk;rk ls tksMk+ tkrk gSA la;kstu djus ds 12.0 s i'pkr~] fdruh gksxhA (a) ifjiFk esa /kkjk (b) cSVjh }kjk iznÙk 'kfDr, (c) Å"ek ds :i esa O;f;r 'kfDr rFkk (d) la/kkfj=k esa c<+ jgh lafpr ÅtkZ dh nj
76.
A capacitance C charged to a potential difference V is discharged by connecting its plates through a resistance R. Find the heat dissipated in one time constant after the connections are made. Do this by calculating
i R dt at and also by finding the decrease in the energy stored in the capacitor.. 2
V foHkokUrj
rd vkosf'kr C /kkfjrk okys la/kkfj=k dh IysVksa ls izfrjks/k R dks tksM+dj folftZr fd;k tkrk gSA la;kstu djusds
,d le; fLFkjkad esa m"ek gkfu Kkr dhft;sA bldks
i R dt dh x.kuk djrs gq, dhft;s ,oa lkFk gh la/kkfj=k esa lafpr ÅtkZ 2
esa deh Kkr djds Hkh dhft;sA 77.
By evaluating
i R dt, show that when a capacitor is charged by connecting it to a battery through a resistor 2
the energy dissipated as heat equals the energy stored in the capacitor.
la/kkfj=k dks izfrjks/k }kjk cSVjh ls tksMu+ s ij
i R dt, dh x.kuk djds O;Dr dhft;s fd m"ek ds :i esa ÅtkZ gkfu] la/kkfj=k 2
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Chapter # 32
Electric Current in Conductors
esa lafpr ÅtkZ ds cjkcj gksrh gSA 78.
A parallel-plate capacitor is filled with a dielectric material having resistivity and dielectric constant K. The capacitor is charged and disconnected from the charging source. The capacitor is slowly discharged through the dielectric. Show that the time constant of the discharge is independent of all geometrical parameters like the plate area or separation between the plates. Find this time constant. ,d lekukarj iV~V la/kkfj=k esa ijkoS|rq inkFkZ Hkjk gqvk gS] bldh izfrjks/kdrk rFkk ijkoS|rq kad K gSA la/kkfj=k dks vkosf'kr
djds vkos'ku lzkrs ls i`Fkd dj fy;k tkrk gSA la/kkfj=k /khjs&/khjs ijkoS|rq kad inkFkZ }kjk folftZr gksrk gSA O;Dr dhft;s fd foltZu leLr T;kferh; jkf'k;ksa ij fuHkZj ugha djrk gS vFkkZr~ IysV {ks=kQy rFkk IysVksa ds e/; nwjh ijA bldk le; fLFkjkad Kkr dhft;sA 79.
Find the charge on each of the capacitors 0.20 ms after the switch S is closed in figure. dqt a h S dks can djus ds 0.20 ms i'pkr~ izR;sd la/kkfj=k ij vkos'k Kkr dhft;sA
80.
The switch S shown in figure is kept closed for a long time and then opened at t = 0. Find the current in the middle 10 resistor at t = 1.0 ms. fp=k esa iznf'kZr fLop S yEcs le; rd can j[kk tkrk gS] blds i'pkr~ bls t = 0 ij [kksy fn;k tkrk gSA t = 1.0 ms ij e/; okys 10 izfrjks/k esa /kkjk Kkr dhft;sA
81.
A capacitor of capacitance 100 F is connected across a battery of emf 6.0 V through a resistance of 20 k for 4.0 s. The battery is then replaced by a thick wire. What will be the charge on the capacitor 4.0 s after the battery is disconnected ? 100 F /kkfjrk dk ,d la/kkfj=k 20 k izfrjks/k }kjk ,d cSVjh ds fljksa ij 4.0 lsd.M ds fy;s tksM+ fn;k tkrk gSA blds i'pkr~ cSVjh dks ,d eksVs rkj ls izfrLFkkfir dj fn;k tkrk gSA cSVjh gVkus ds 4.0 lsd.M i'pkr~ la/kkfj=k ij vkos'k fdruk
gksxk\ 82.
Consider the situation shown in figure. The switch is closed at t = 0 when the capacitors are uncharged. Find the charge on the capacitor C1 as a function of time t. fp=k esa iznf'kZr fLFkfr ij fopkj dhft;sA tc la/kkfj=k vukosf'kr gS] t = 0 ij fLop can fd;k tkrk gSA la/kkfj=k C1 ij le;
ds Qyu ds :i esa vkos'k Kkr dhft;sA
83.
A capacitors of capacitance C is given a charge Q. At t = 0, it is connected to an uncharged capacitor of equal capacitance through a resistance R. Find the charge on the capacitor as a function of time. C /kkfjrk okys ,d la/kkfj=k dks Q vkos'k iznku fd;k tkrk gSA t = 0, ij bldks ,d izfrjks/k R dh lgk;rk ls ,d leku
ls ,d leku /kkfjrk okys la/kkfj=k ls tksM+ fn;k tkrk gSA nwljs la/kkfj=k ij le; ds Qyu ds :i esa vkos'k Kkr dhft;sA 84.
A capacitor of capacitance C is given a charge Q. At t = 0, it is connected to an ideal battery of emf through a resistance R. Find the charge on the capacitor at time t. C /kkfjrk okys ,d la/kkfj=k dks Q vkos'k iznku fd;k tkrk gSA t = 0 ij bldks ,d fo-ok-cy okyh vkn'kZ cSVjh ls ,d izfrjks/k }kjk tksM+ fn;k tkrk gSA la/kkfj=k ij le; t ij vkos'k Kkr dhft;sA manishkumarphysics.in
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