Chapter # 32
1.
Sol.
2.
Sol.
Electric Current in Conductors
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 tksM+k tkrk gSA ;fn eqDr bysDVªkWuksa 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 A The current will increase.
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/; laca/k Kkr djus ds fy, lwpuk vi;kZIr gSA D
A Resistance is depend on Material, length & Area. So RA < RB is information is not sufficient to ding. The relation between A and B. R
3.
Sol.
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) vuqizLFk dkVk {ks=k ij (D*) buesa ls dksbZ ugha D conductivity
1
Where is resistivily.
Product of conductity and resistivity = 1 4.
Sol.
5.
Sol.
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 C × = constant 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 esa] 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 B The charge (Positive) inside the bottery may go from the positive terminal to the negative terminal.
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Chapter # 32 Electric Current in Conductors [2] 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 tksM+k 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 Sol. A
Power
v2 R
, R thanpower
Because Power 7.
Sol.
1 . R
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 bysDVªkWuksa ,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 C vd drift velocity =
1 eE 2m
1 1 1 e2E2 2 2 mv m d K.E. = 2 2 4 m2 K.E. =
1 e2E2 2 8 m
1 m Mass of electron < mass of metalions. K.E. of electron > K.E. of metalions. K1 > K2
8.
Sol.
K.E.
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 A Thermal Energy developed = I2 Rt (Because in series, current is same)
R
2R
I
+ v ThermalEnergy developedin"R " I2Rt 1 2 ThermalEnergy developedin"2R " I 2R t 2
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Chapter # 32 Electric Current in Conductors [3] 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 Sol. B Thermal Energy developed =
v2 t (Because in Parallel, voltage is same) R 2R
R
+ v v2 t ThermalEnergy developedin"R " R2 2 : 1 ThermalEnergy developedin" 2R " v t 2R 10.
Sol.
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 tksM+s tkrs gSaA la;kstu dk rqY; izfrjks/k gS (A*) 2 (B) 10 (C) 250 (D) 6250 A
R
l 50 A
resistance of all ‘5’ equal parts are same. l / 5 50 10 A 5 all ‘5’ equal parts connect in parallel :-
5
5
R'
5
5
1 1 1 1 1 1 5 R eq 10 10 10 10 10 10 Req = 2 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
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5
Chapter # 32
Sol.
12.
Sol.
Electric Current in Conductors
[4]
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 A
Kirchhoff’s Junction Law follows from conservation of charge. Kirchhoff’s loop law fallows from conservation nature of electric field.
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 (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 B +
r1
-
+
-
r2
(i)
equivalent emf = 1 +2 Req = r1 + r2 +
-
r1
- +
r2
(ii) equivalent emf = 1 +2 {1 >2} Req = r1 + r2 The equivalent emf is may be larger than either of the two emfs. The equivalent internal resistance is mustbe larger than either of the two internal resistance. 13.
Sol.
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 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 C Equivalent emf 0
1 r1 2 r2 r1 r2
Equivalent resistance = r0
r1r2 r1 r2
+ -
+ -
r1 r2
The quivalent emf is larger than either of the two emfs. The quivalent internal resistane is smaller than either of the two internal resistance.
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Chapter # 32 Electric Current in Conductors [5] 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 Sol. D The net resistance of an ammeter should be small to ensure that it does not oppreciably change the current to be measured. 15.
Sol.
16.
Sol.
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; ifjorZu uk djsA D The net resistance of a voltmeter should be large to ensure that it does not apperciably change the potential difference to be measured. 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 laxzfgr gksrk gS rFkk vxys le;kUrjky t2 esa vxyk 10 ekbØks dwykWe vkos'k laxzfgr gksrk gS (A) Q1 > Q2, t1 > t2. (B*) Q1 > Q2, t1 < t2. (C) Q1 > Q2, t1 > t2. (D) Q1 < Q2, t1 < t2. B Condition for charging capacitor :Q = Q0 (1-e-t/Rc) Q = Q0 (1-e-10m/Rc) ....(i) Q1 + Q2 = Q0 (1-e-(10m+10m)/Rc) Q1 + Q2 = Q0 (1-e-20m/Rc) ....(ii) from eq. (i) & (ii) we get :Q1 > Q2 Given Q = Q0 (1-e-t/RC) 10c = Q0 (1-e-t/Rc) ....(iii)
10c 10c Q0 1 e(t1 t2 )/Rc
20c Q0 1 e(t1 t2 )/Rc
....(iv)
from eq. (iii) & (iv) we get t2 > t1
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 iqat dk dkV {ks=k leku jgs -
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Chapter # 32 Electric Current in Conductors [6] (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. Sol. A Electric field goes higher potential to Lower potential. The drift velocity fo the electron at higher potentail is greater than the lower potential. So the speed of the electron is more at B that at A. 2.
Sol.
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 (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|qr inkFkZ ugha gS] t = 0 ij ,d cSVjh ls tksM+k tkrk gSA la;kstu rkjksa esa ,d fcUnq A rFkk IysVksa 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 BC
B C
A
+ -
I 3.
Sol.
4.
Sol.
V
Ther is no current through B There is a current through A as long as the charging is not complete.
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 CD No current is passed through a conductor means. That the average velocity of a free elecron over a large period of time is zero or the average of the velocity of all the free electrons at an instant is zero. 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 D When a resistor connected to a battery is heated due to the current that causes drift speed, resistivity & resistance may e change But number of free electrons remains same.
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Chapter # 32 Electric Current in Conductors [7] 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.
Sol.
6.
Sol.
7.
Sol.
8.
Sol.
,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
A Temperature of a cunductor increases that causes resistivity () is increases & due conductivity () is decrease. 1 resistivity ratio of 2 is increase conductivity 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 vuqizLFk dkV ds ,d rkj esa /kkjk izokfgr gks jgh gSA fuEu esa ls dkSulh jkf'k;k¡ vuqizLFk 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 AD e i vf E qm A ne i j A Vd drift speed j current density i current A cross-section Area Mark out the correct options. (A*) An ammeter should have small resistance (B) An ammeter should have large resistance (C) A voltmeter should have small rsistance (D*) A voltmeter should have large resistance lgh fodYi pqfu;s (A*) ,d vehVj dk izfrjks/k vYi gksuk pkfg,A (B) ,d vehVj dk izfrjks/k mPp gksuk pkfg,A (C) ,d oksYVehVj dk izfrjks/k mPp gksuk pkfg,A (D*) ,d oksYVehVj dk izfrjks/k mPp gksuk pkfg,A AD An ammeter should have small resistane. To measure the accurate reading of current in the circuit by Ammeter. A voltmeter should have large resistance. To measure the accurate reading of voltage across voltmeter. 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 tksM+k tkrk gSA izFke 5 lsd.M esa la/kkfj=k ij laxzfgr vkos'k T;knk gksxk (A*) 5 s (B*) 50 s (C*) 500 s (D*) 500 ABCD Q = CE (1-e-t/Rc) C - 500 ×10-6 F R = 104 = Rc = 104 × 500 × 10-6 = 5 = 5 sec.
1 Q = c 1 e t / c 1 0.63c e
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Chapter # 32 Electric Current in Conductors Thus, 63% of the maximum charge is deposited in one time constant. Q
[8]
t with the help of the figure we can say that the capacitor in the first 5s is larger than the charge stored in the next any second.
at
t
Q Q0 1 e Q0 c
t t5 c .63 .37
after t = 5 sec., maximum charge is deposited is only 37%. 9.
Sol.
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. (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 BD Charging
(i) Discharging
(ii)
C1
C2
R
R
(iii) Q C1 e i1
(iv) t /Rc1
Q C2 e t /Rc2
dQ C1 t /RC1 e dt RC1
t /RC1 e R at t 0
i2
dQ t /RC 2 e dt R
at
t 0
i1
i1
R
i2
R
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