Direct Current Circuits (QB)

7/30/2019 Direct Current Circuits (QB)

1/48

DIRECT CURRENT CIRCUIT Question Bank

EASY QUESTIONS

1. If resistance of each wire in the network shown is r,

the equivalent resistance betweenA & Cis equal to

(a) r (b)2

r

(c)3

2r(d)

2

3r

C

A

E D

B

Sol.: For pointA & C, loopBCD shorted

HenceRAC= rr

rr

3

2

3

2

(c)E

2. In the circuit shown each capacitor has capacitance C.

The emf of the battery is and the Sw is closed. Thetotal heat generated in the wire once the switch Sw isopened is

(a) C2 (b)6

2C

(c)12

2C(d) No heat will be dissipated

Sw

C

CC

Sol.: As the charge distribution remains same on opening the switch, no charge will flow in the

circuit. So heat dissipated is zero.

(d)

E

3. In the circuit shown in figure, equivalent resistance between

A andB is

(a) 8 (b) 15

(c)

2

3 (d) 2

B

42

1

A

2

2

4

Sol.: Equivalent circuit diagram of the circuit is

B

2

A2

4

1

2 4

AB

6

43

2


2/48


So 2

3eqR

(c)

E

4. The resistance of hexagon circuit between A andB

represented in figure is

(a) r (b) 0.5 r

(c) 2r (d) 3r

r

r r

r

rr

r

r

rr

A B

Sol.: From figure (i) it is evident that the potential difference between points a, b andc is zero.

The equivalent circuit is as shown in figure (ii).

rrr

rrrr degf

22

22

222

22 r

rr

rrrAB

(b)

r

r r

r

rr

r

r

rr

A B

a

b

c

d

g f

e

g f

r r

r ra

b

d e

r r

r rb

c

A B

E

5. In the given circuit, each resistor has resistance R. Theequivalent resistance betweenA andB is

(a)4

R (b) R4

(c)4

3R(d)

3

4R

A B

Sol.: (a)

E

6. A heater coil is cut into two equal parts and only one part is now used in the heater. The heat

generated will now be (Assuming potential difference is same in both cases)

(a) one fourth (b) halved (c) doubled (d) four times

Sol.: RH1

R becomes half so heat generate will be doubled.

(c)E

7. In the circuit shown the potential difference between

points CandB will be

(a) (8/9) volt (b) (4/3) volt

(c) (2/3) volt (d) 4 volt

5

B

555

55C

AD

+

2V


3/48


Sol.:15

2

eR

VI A (Iis current in each branch)

V3

4BC VV

(b)E

8. The current through 2 resistor is(a) zero (b) 1 amp

(c) 2 amp (d) 4 amp

2

5 1010V 20V

Sol.: (a)

E

9. The equivalent resistance between pointsA and

B in the circuit shown is

(a) 4 (b) 6 (c) 10 (d) 8

4A

B

8

8

4

6

4

8

E

10. There are n similar resistors each of resistanceR. The equivalent resistance comes out to be xwhen connected in parallel. If they are connected in series, the resistance comes out to be (a)x /n

2 (b)n2x (c)x/n (d) nx

Sol.: In parallelx

1=

R

nand seriesReff= nR = n

2x

(b)E

11. In the balanced wheatstone bridge circuit as shown

in the figure, when the key is pressed, what will be

the change in the reading of the galvanometer?(a) no change (b) increased

(c) decreased (d) zero

GR

R

RR

Sol.: Under balanced conditionS

R

Q

P

Here resistances are in same proportion

Hence, there will not be any deflection in galvanometer on pressing the key. It remain same.

(a)


4/48


E

12. In the circuit shown in figure, the reading of

voltmeter will be

(a) 0.8 V (b) 1.33 V

(c) 1.6 V (d) 2.00 V

20

V

80

80

2V, r= 0

Sol.: (b)

E

13. In the circuit shown in figure

(a) current in wireAFis 1A

(b) current in wire CD is 1A

(c) current in wireBEis 2A

(d) none of the above

4 4 4

2V

A B C

DF E2V 2V

Sol.: By KVL in loop 1 02842 ii

i = 0

(d)4 4 4

2V

E2V 2V

i i

2i

1

E

14. A battery of internal resistance 4 is connected to thenetwork of resistance as shown. In order to give the

maximum power to the network, the value ofR should

be

(a) 9

4(b)

9

8

(c) 2 (d) 18

4

R R

R6RRR 4R

E

Sol.: Given circuit is balance wheat stone bridge hence no current will flow through 6 resistance.So equivalent resistance will be 2R.

For maximum power 2R = 4 R = 2

(c)

E

15. A cell of emfE is connected across a resistance R. The potential difference between the

terminals of the cell is found to be V. The internal resistance of the cell must be

(a)R

VVE )(2 (b)E

RVE )(2 (c)V

RVE )( (d) (EV)R


5/48


Sol.: As V=EI.r &I=rR

E

r=

V

RVE )(

(c)E

16. The resistance acrossAB is

(a)8

5R (b)

8

7R

(c) 1R (d) 2R

R

R

RRR

DC

A B

Sol.: The circuit can be rearranged

Now 2R andR are parallel

RRR

RAB

1

3

2

11

=

RR

1

5

3 RAB =

8

5R

(a)

2R

R

A BR

R

C

E

17. The equivalent resistance of the network shown in the

figure between the base terminals is

(a) 3 (b) 321

(c)3

22 (d) 2

1 1

1 1 1

Sol.: Req

=3

81

3

211

21

211

(c)E

18. n identical cells, each of emf and internal

resistance r, are joined in series to form a closedcircuit as shown. The potential difference across

any one cell is

(a) zero (b)

(c)n

(d)

n

n 1

r r r

Sol.:rnr

nI

, V = Ir = 0

(a)


6/48


E

19. In the given circuit it is observed that the currentI

is independent of the value of the resistanceR6.

Then the resistance value must satisfy

(a)R1R2R5 =R3R4R6

(b))(

1

)(

111

432165 RRRRRR

(c)R1R4 =R2R3

(d)R1R3 =R2R4 =R5R6

R2

R1

R5

R3

R4

R6

I

Sol.: This is condition for balance wheatstone bridge

3241

4

3

2

1 RRRRRR

RR

(c)E

20. The resistances 500 and 1000 are connected in serieswith a battery of 1.5 volt. The voltage across the 1000 resistance is measured by a voltmeter having a resistance of

1000 . The reading in the voltmeter would be(a) 1.5 volt (b) 1.0 volt

(c) 0.75 volt (d) 0.5 voltSol.: (c)

500

V

1000

1.5 V 1000

E

21. A set of n identical resistors, each of resistance R ohm when connected in series has an

effective resistance ofx ohm. When the resistors are connected in parallel, the effective

resistance is y ohm. What is the relation betweenR,x andy?

(a)R =)( yx

xy

(b)R = (y x) (c)R = xy (d)R = (x +y)

Sol.: For series connection x = nR.

For parallel connectiony =n

R.

Thereforexy = nR n

R=R

2.

(c)


7/48


E

22. In the circuit shown in figure, the current

through

(a) the 3 resistor is 0.50 A(b) the 3 resistor is 0.25 A(c) the 4 resistor is 0.50 A(d)the 4 resistor is 0.25 A

3 A 2 C 2

8 8 49V

2 B 2 D 2

Sol.: The equivalent resistance between pointsA andB to the right ofAB is 4 . Therefore, totalresistance = 3 + 4 + 2 = 9 . CurrentI= 9 V/9 = 1 A. This current is equally divided in the8 resistor between A and B and the remainder 8 resistor. Hence current inAC= 0.5 A.This current is equally divided between the 8 resistor in CD and the circuit to the right ofCD. Therefore, current in the 4 resistor = 0.25 A.

(d)E

23. In the arrangement of resistances shown in the figure, the

potential difference between the points B andD will be zero

when the unknown resistanceXis

(a) 4(b) 2(c) 3

(d) e.m.f. of the cell is needed to find outX

A C

B

D

3

1

12

4X

1

1

Sol.:2/1

416

X,X= 2

(b)E

24. The currentIdrawn from the 5 V source will be

(a) 0.33 A (b) 0.5 A

(c) 0.67 A (d) 0.17 A

5V

I10

20105

10

Sol.:10

5

20

10 . So it is a balance wheat stone bridge.

1045

1530e

R ,2

1

10

5I A

(b)E

25. Five cells, each of e.m.f. E and internal resistance rare connected in series. If due to over

sight, one cell is connected wrongly, then the equivalent e.m.f. and internal resistance of thecombination, is

(a) 5Eand 5r (b) 3Eand 3r (c) 3Eand 5r (d) 5Eand 3r


8/48


Sol.: EMF = EE 3)14( .Internal resistance = 5r (c)

E26. Five equal resistors, each equal toR are connected as shown in the

following figure; then the equivalent resistance between points A

andB is:

A B

(a) R (b) 5R (c) R/5 (d) 2R/3

Sol.: It is a case of wheat stone bridge.

(a)E

27. A wire has resistance 12 is bent in the form of a circle. The effective resistance betweenthe two points on any diameter of the circle is

(a) 12 (b) 24 (c) 6 (d) 3Sol.: Rtotal = 12

RAB upper= 6 =RAB lower

Combination21

111

RRReff

6

1

6

11

effR 3effR

(d)

A B

E28. When cells are connected in series

(a) the emf increases (b) the potential difference decreases

(c) the current capacity increases (d) the current capacity decreases

Sol.: (a)

E

29. Which of the following has the maximum resistance?

(a) voltmeter (b) milivoltmeter (c) ammeter (d) miliammeter

Sol.: (a)

E

30. A conductor with rectangular cross-section has

dimensions (a 2a 4a) as shown in figure. ResistanceacrossAB isx, across CD isy and acrossEFisz. Then

(a) zyx (b) zyx

(c) xzy (d) yzx

C

4aA B

2a

F

E D

a

Sol.:A

lR

,

aaa

ax

2

2

4,

aaa

ay

824

then yzx


9/48


aaa

az

24

2

(d)

E

31. A wire l = 8m long of uniform cross-sectional area A = 8 mm2, has a conductance of

G = 2.45 1. The resistivity of material of the wire will be(a) 2.1 107m (b) 3.1 107m (c) 4.1 107m (d) 5.1 107m

Sol.:Gl

A

l

RA

845.2

108 64.1 107 meter

(c)

E

32. A galvanometer of resistance 400 can measure a current of 1mA. To convert it into avoltmeter of range 8V the required resistance is

(a) 4600 (b) 5600 (c) 6600 (d) 7600

Sol.: VRGig , 8400103 R , R = 7600

(d)

E33. An ammeter reads upto 1A. Its internal resistance is 0.81 . To increase the range to 10A,

the value of the required shunt is

(a) 0.03 (b) 0.3 (c) 0.9 (d) 0.09

Sol.: 110

81.01

g

g

II

GIS 0.09

(d)

E

34. The resistance of the series combination of two resistances is S. When they are joined inparallel, the total resistance is P. IfS= nP, then the minimum possible value ofn is

(a) 4 (b) 3 (c) 2 (d) 1

Sol.: For two resistancesR1 andR2

21 RRS (in series),21

11

RRP (in parallel)

According to S= nP,

21

2121

RR

RRnRR

Ifn is minimum RRR 21

then n = 4

(a)


10/48


E

35. A wire of resistance 4 is stretched to twice its original length. What is the resistance ofthe wire now?

(a) 1 (b) 14 (c) 8 (d) 16 Sol.: Volume of wire remains constant 2211 lAlA , 1211 2lAlA

So, 21 2AA ,1

11

A

lR ,

2

22

A

lR ,

2

1

R

R

1

2

2

1

A

A

l

l , 2R 16

(d)E

36. The net resistance between points P andQ in the

circuit shown in the figure is(a)R/2 (b) 2R/5

(c) 3R/5 (d)R/3

RR

R

RQP

Sol.: (b)

E

37. The equivalent resistance between pointsMandNis

(a) 2 (b) 3 (c) 2/3 (d) none of the above

1

11

1 1

1M N

Sol.: When a battery is connected between pointsMandN. NO current is found is PQO. Hence

this section may be removed from the circuit.

3

2

12

12effR

Q

M N

P

I

I

O

M N

O

s

(c)


11/48


E

38. The potentiometer wire AB is 600 cm long. At what

distance fromA should the jockey Jtouch the wire to get

zero deflection in the galvanometer?

(a) 320 cm (b) 120 cm

(c) 20 cm (d) 450 cm

E r

A BJ

r

R=15r

E/2G

Sol.: In case of zero deflection in galvanometer

2

EV

AJ ,

2

EiR

AJ ,

2600

15

15

EAJ

r

rr

E

AJ= 320 cm

(a)E

39. The emf of the battery shown in the figure is

(a) 6 V (b) 12 V

(c) 18 V (d) 8 V6 2 1

122

I=1.5A

Sol.: According to KVL 0 irE (ris effective resistance in circuit)05.14 E

E= 6 volt

(a)

E

40. In the figure, the steady state current in 2resistance is

(a) 1.5 A (b) 0.9 A

(c) 0.6 A (d) zero

2

3A B

6 V

2.8

4

C= 0.2 F

Sol.: In steady state, current through batteryI=2.18.2

6

= 1.5 A

I2 = 5.132

3

= 0.9 A

(b)


12/48


E

41. The charge on the capacitor in the figure is

(a) 2 C (b) 2/3 C

(c) 4/3 C (d) zero1F

4

1

2V/0.5

Sol.: Inet =

Ar 3

4

2/3

2

1

2

. VVT

3

45.0

2

42 . C

3

4

3

4F1 VCVQ

(c)

E

42. Each of the resistance in the network shown in

the figure below is equal to R. The resistancebetween the terminalsA andB is

(a)R (b) 5R

(c) 3R (d) 5/3R

M

RR R

R

RA

B

ON

Sol.: Resistance between M and N can be removed (Balanced whetstone bridge)

Reff=R

(a)

E

43. Kirchoffs second law is based on the law of conservation of

(a) momentum (b) charge

(c) energy (d) sum of mass and energy

Sol.: A charge if taken around a closed loop work done is zero

(c)E

44. The current i in the figure below is

(a) 1/5 A (b) 1/10 A

(c) 1/15 A (d) 1/45 A

30

30

30

i

+2V

Sol.: Reff= 20 , AR

Ei

eff 10

1

(b)


13/48


E

45. The time constant of anRCcircuit shown in the

figure is(a) 3RC (b) 2/3RC

(c) 6RC/5 (d) 2RC

R

R

3R

C

Sol.: RRR

RRR

eff5

6

32

32

,5

6.

RCRC

eff

(c)E

46. What is the current through the resistorR in the circuitshown below? The emf of each cell is Em and internal

resistance is r

(a)rR

Em

2(b)

Rr

Em

2

(c)rR

Em

2

2

(d)

rR

Em

22

r

rR

+

+

Sol.:2

2

2

rRrRR

eff

, meff EE ,

rR

E

R

EI m

eff

eff

2

2

(d)E

47. CurrentI3 in the given circuit shown in the figure is

(a) A11

5(b) A

11

7

(c) A11

2(d) none of these

R

3V

3

I3

2

2V

I2I1

1V

1

Sol.: Applying Kirchoffs law AI 11

53

(a)E

48. Six resistors each of resistance R are

connected as shown in figure. What is

the effective resistance between pointsAandB?

R

A BRR

R R

R

(a)3

R(b)R (c) 3R (d) 6R


14/48


Sol.: (a)

E

49. The current at which a fuse wire melts does not depend on

(a) cross-sectional area (b) length

(c) resistivity (d) density

Sol.: (b)

E

50. In the circuit shown in figure the heat

produced in the 5 resistor due to acurrent flowing in it is 10 calories per

second. The heat produced in the 4resistor is

4 6

5

(a) 1 cal s1

(b) 2 cal s1

(c) 3 cal s1

(d) 4 cal s1

Sol.: LetIbe current through 5

1052 I (i)

current through 4 will be2

I

Heat produced in 4 resistance 44

2I

2

(b)E

51. In the circuit shown in the figure, thecurrent through

(a) the 3 resistor is 0.50 A

(b) the 3 resistor is 0.25 A

(c) the 4 resistor is 0.50 A(d) the 4 resistor is 0.25 A

3 2 2

2 2 2

8 8 4

A C

B D

9V

Sol.: (a)

E

52. Figure shows currents in a part of an electricalcircuit. The current i is

(a) 1 A (b) 1.3 A

(c) 1.7 A (d) 3.7 A

2A

2AP Q

R

1A

1.3A

i

Sol.: (c)


15/48


E

53. The meter bridge circuit shown in figure is

balanced when jockeyJdivides wireAB in twoparts AJ andBJ in the ratio of 1: 2. Theunknown resistance Q has value

(a) 1 (b) 3

(c) 4 (d) 7 A B

J

P Q

G

1.5

Sol.:JB

Q

AJ R

R

R

5.1,

1

2

AJ

JB

QR

RR 1.5, 3QR

(b)

E

54. n identical cells, each of emf and internal resistance r, are joined in series to form a closedcircuit. The potential difference across any one cell is

(a) zero (b) (c)n

(d)

n

n 1

Sol.: Current in circuit i =rnr

n

The equivalent circuit of one cell is shown in thefigure p.d. across the cell

= VA - VB = + ir= + 0. rr

(a)

r BA

+

E

55. In the circuit shown, P R, the reading of thegalvanometer is same with switch Sopen or closed. Then

(a)GR

II (b)GP

II

(c)GQ

II (d)RQ

II

P Q

S

R G

V

Sol.: As P R and reading of galvanometer is same so wheat bridge must be balanced and in thatcaseIR =IG

(a)


16/48


E

56. The currentIdrawn from the 5 V source will be

(a) 0.33 A (b) 0.5 A

(c) 0.67 A (d) 0.17 A

5V

I10

20

10

5

10

Sol.:10

5

20

10 .

So it is a balance wheat stone bridge.

1045

1530eR ,

2

1

10

5I A

(b)E

57. In the steady state in the circuit shown

(a) potential difference across C1 is 4 V

(b) potential difference across 10 is 2V

(c) potential difference across C2 is 4 V

(d) charge on C1 orC2 is 0 C

10F 10 4F

14V

C1 C2

Sol.: 14

410

qq

q = 40 C.

Potential difference across10

401C = 4 V

(a)

E

58. Find the current supplied by the battery as shown in thefigure.

(a) 1.5 amp (b) 5 amp

(c) 1.2 amp (d) 2.4 amp

4

6

4

6

24V

Sol.: Circuit become simple, then

i20 =24 i 1.2 amp4.824V

(c)


17/48


E

59. What is the equivalent resistance between A andB?

(Each resistor has resistanceR)

(a)3

4R(b)

3

5R

(c)5

4R(d)

4

3R

BA

Sol.: In the figure5

4eff

RR

(c)BA

2R/3 2R/3

R R

E

60. The ammeter will read the value of current

(a) 3A (b) A3

10

(c) 30 A (d)3

100A

5

5

5

55

5

B

A

5

A10V

Sol.:

3

10

ABR (use wheat stone bridge)

3/10

10I = 3A

(a)E

61. Each cell has emf and internal resistance r in thefigure. Find the current through resistanceR

(a)r

4(b)

r

3

(c)r

(d) zero

R

A

B

Sol.: Potentialdifference betweenA andB is zero the current throughR is zero.

(d)

E

62. If emf in a thermocouple is 2TT then the neutral temperature of the thermocoupleis

(a) 2/ (b) /2 (c) 2/ (d) /2


18/48


Sol.: For neutral temperature 0

dT

d 02 T

Then ,2

T

(c)E

63. The charge flowing through a resistanceR varies with time tas 2btatQ . The total heatproduced inR from t= 0 to the time when value ofQ becomes again zero is

(a)b

Ra

6

3

(b)b

Ra

3

3

(c)b

Ra

2

3

(d)b

Ra3

Sol.: dtbtadHHba

22/

0

(a)E

64. In the steady state in the circuit shown

(a) potential difference across C1 is 4 V

(b) potential difference across 10 is 2V

(c) potential difference across C2 is 4 V

(d) charge on C1 orC2 is 0 C

10F 10 4F

14V

C1 C2

Sol.: 14410

qq

q = 40 C

Potential difference across10

401 C = 4 V

(a)E

65. The charge flowing through a resistanceR varies with time tas 2btatQ . The total heatproduced in R from t = 0 to the time when value of Q becomes again zero is

60

10

15 51A1A

i

(a)b

Ra

6

3

(b)b

Ra

3

3

(c)b

Ra

2

3

(d)b

Ra3

Sol.: b

RadtRbtadtRIH

baba

32

32

/

0

2

/

0

(b)


19/48


E

66. The currentvoltage (I-V) graphs for a given metallic

wire at two different temperatures T1 andT2 are shown inthe figure. It follows from the graphs that

(a) T1 > T2

(b) T1 < T2

(c) T1 = T2

V

T1T2

I

(d) T1 is greater or less than T2 depending on whether the resistanceR of the wire is greater

or less than the ratio V/I.

Sol.: (a)

E

67. The potential difference between pointsA andB

in the following circuit diagram will be

(a) 8V (b) 6V

(c) 4V (d) 2V

5 5

5

55

5 2V

A

B

Sol.: (c)

E

68. The current in the arm CD in the circuit shown in

the figure will be

(a) 21 ii (b) 32 ii

(c) 31 ii (d) 321 iii

i1i2

Ai3

B

C

D

Sol.: (b)

E

69. In the given circuit resistance of voltmeter is 400 and its reading is 20V. Find the value of emf of battery

(a) 130/3 volt (b) 65 volt

(c) 40 volt (d) 33.6 voltE

V

200 300

Sol.; Current in voltmeter, .amp20

1

400

20I

Current in .amp15

1300

Current in60

7

300

35

15

1

20

1200

V3

13020

60

7200 E

(a)


20/48


E

D70. In the given circuit, find the equivalent resistance

between pointA andB.(a) 18 (b) 12

(c) 20 (d) 27

3 6 9

4 5A B

6 12 18

Sol.: Equivalent circuit is

12eq

R

(b)

3 6 9

A B

6 12 18

E71. In the given circuit diagram, current through the

battery isR

V

2

3, if and only if

(a)R1 =R2 =R (b)R1 >R2

(c)R1


21/48


Sol.: Equivalent ciruit is balanced Wheat- stone

bridge as shown

RAB =3

10

(a)

2 2

10/3 A B

1010

E

80. The current in branch CD of given circuit is,

(a) zero (b) 1 A

(c) 2 A (d) 3 A

A B

C D

E F

8V

4

3

4V

2 12V

Sol.: The equivalent emf of 12V and 8V battery =3/12/1

3

8

2

12

=

23

1636

= 4 V

5

6

32

32eqr

The equivalent circuit is

4V 6/5

4V 4C D

I= 0

(a)E

81. Two sources of emf 6V and internal resistance 3and 2 are connected to an external resistance R asshown. If potential difference across sourceA is zero,then value ofR is

A B

R

6V,3 6V,2

(a) 1 (b) 2 (c) 3 (d) 4

Sol.:R

I

5

12and 0

5

1236

R

R

5

366 R = 1 (a)

E


22/48


82. The equivalent resistance between pointsA andB is

(a) 2R (b) R4

3

(c) R3

4(d) R

5

3

R

RR

RA B

Sol.: Circuit can be rearranged as follows

5

3

2

3

2

3

R

RR

RR

Req

A BR R

R

R

(d)E

83. In the circuit shown, current through 3 resistance is(a) 1 amp (b) 2 amp

(c) 3 amp (d) 4 amp

6 6V

3

Sol.: Current through battery

36

36

6i 3 amp

Current through 3 is i

63

63 2 amp

(b)E84. The circuit as shown in figure. The ratio of

current21 / ii is

(a) 2 (b) 8(c) 0.5 (d) 4

8 2 4 3

2 8 8

1 V16

V8 2i 1i

Sol.: The simplified circuit can be drawn as

4 8 V16

V8

A4

A2 A2

AI

AI

2

1

4

2

1

(b)


23/48


E

85. In the circuit shown in the figure, reading of voltmeter is

V1 when only S1 is closed, reading of voltmeter is V2

when only S2 is closed and reading of voltmeter is V3

when both S1 andS2 are closed. Then:

(a) 123 VVV (b) 312 VVV

(c) 213 VVV (d) 321 VVV

S1

S2

3R

6R

R

VE

Sol.: In series P.D. R

When only S1 is closed, EEV 75.04

31


62

And when S1 andS2 are closed, combined resistance of 6R and 3R is 2R.

EEV 67.03

23

312 VVV

(b)E

86. The resistance of a wire is 10. Its length is increased by 10% by stretching. The newresistance will now be nearly

(a) 12 (b) 1.2 (c) 13 (d) 11

Sol.: SinceR I2 If length is increased by 10% resistance is increases by almost 20%

Hence new resistanceR = 10 + 20% of 10 = 10 + 10100

2012

(a)E

87. The same mass of copper is drawn into two wires 1 mm and 2 mm thick. Two wires are

connected in series and current is passed through them. Heat produced in the wire is in theratio

(a) 2 : 1 (b) 1 : 16 (c) 4 : 1 (d) 16 : 1

Sol.:2

222

A

Vtit

A

liRTiH

(V= volume)

4

1

rH

4

1

2

2

1

r

r

H

H=

1

16

1

24

(d)


24/48


E

88. In the circuit shown, a meter bridge is in its balanced

state. The meter bridge wire has a resistance 0.1 ohm/cm.

The value of unknown resistanceXand the current drawn

from the battery of negligible resistance is

X 6

5V

G

40cm 60cmA B

C

(a) 6, 5 amp (b) 4, 0.1 amp (c) 4, 1.0 amp (d) 12, 0.5 amp

Sol.: Resistance of the partAC

RAC= 0.1 40 = 4 andRCB = 0.1 60 = 6

In balanced condition

6

4

6

X

X= 4

Equivalent resistanceReq= 5 so current drawn from battery5

5i = 1A

(c)E

89. Find the equivalent resistance acrossAB

(a) 1 (b) 2(c) 3 (d) 4

2

2

2

2

2

A

B

Sol.:

2

2

2

2

2

A

B

22

A

B

22

22

ABR = 1

(a)

E

90. The reading of the ammeter in the figure shown is

(a) A8

1(b)

4

3A

(c) A2

1(d) 2A

2VA

2

2

2

2

Sol.: (b)


25/48


E

91. The total current supplied to the circuit by the battery is

(a) 1A (b)2A(c) 4A (d) 6A

26

1.5

36V

Sol.: Net resistance =2

3

Then by Kirchoff law 6 =2

3i, i = 4 amp

(c)E92. The magnitude ofi in ampere unit is

(a) 0.1 (b)0.3

(c) 0.6 (d) 0.4

Sol.: (a)

E

93. AB is a wire of uniform resistance. The galvanometerG

shows zero current when the length AC= 20 cm and

CB = 80 cm. The resistanceR is equal to (a) 2 (b)8

(c) 20 (d) 40

80R

AC

BG

Sol.: By Balanced wheat stone bridge80

80

20

RR = 20

(c)

MODERATE QUESTIONS

M

94. In the shown arrangement of the experiment of the

meter bridge ifACcorresponding to null deflection

of galvanometer isx, what would be its value if the

radius of the wireAB is doubled?

B

G

xC

R1

R2

A

(a) x (b)x/4 (c) 4x (d) 2x

Sol.: (a)


26/48


M

95. Two cells with the same emf Eand different internal resistances r1 andr2 are connected in

series to an external resistanceR. The value ofR so that the potential difference across thefirst cell be zero is

(a) 21rr (b) 21 rr (c) 21 rr (d)2

21 rr

Sol.: Current in the circuit is21

2

rrR

EI

, 011 IrEV

02

21

1

rrR

ErE 21 rrR

(c)M

96. A battery of internal resistance 4 is connected to thenetwork of resistances as shown in the figure. In order

that maximum power can be delivered to the network, the

value ofR in ohm should be

(a)9

4(b) 2

(c)3

8(d) 18

R R

R

4RR

R 6R

Sol.: (b)

M

97. In the adjoining circuit, when the key Kis pressed at

time t= 0, which of the following statements about

currentIin the resistorAB is true?

(a)I= 2 mA at all t

(b)Ioscillates between 1 mA and 2 mA

(c)I= 1 mA at all t

(d) At t= 0,I= 2 mA and with time it goes to 1 mA

AK2V 1000 B

10001F

Sol.: (d)

M

98. A,B andCare voltmeters of resistancesR, 1.5R and 3R

respectively. When some potential difference is appliedbetweenXandY, the voltmeter readings are VA, VB and

VCrespectively.

AX

B

C

Y

(a) VA = VB = VC (b) VAVB = VC

(c) VA = VBVC (d) VBVA = VC


27/48


Sol.: VA = iR

iRR

i

VB

5.13

2

iRRi

VC

33

A

B

C

i

Ri/3

2i/3

3R

1.5R

(a)M

99. A nucleus with mass number 220 initially at rest emits an -particle. If the Q value of thereaction is 5.5 MeV, calculate the kinetic energy of the -particle.(a) 4.4 MeV (b) 5.4 MeV (c) 3.4 MeV (d) 5.6 MeV

Sol.: By conservation of linear momentum

mvMV

54

vV (i)

By energy conservation

J106.15.52

1

2

1 1322 MVmv (ii)

From (i) and (ii)

K.E. of-particle = 22

1mv 5.4 MeV

(b)

M

100. A galvanometer of resistance 19.5 gives full scale deflection when a current of 0.5ampere is passed through it. It is desired to convert it into an ammeter of full scale current

20 ampere. Value of shunt is

(a) 0.5 (b) 1 (c) 1.5 (d) 2

Sol.: 19.5 0.5 = S(20 0.5)

S= 0.5

(a)M

101. A galvanometer of coil resistance 1 is convertedinto voltmeter by using a resistance of 5 in seriesand same galvanometer is converted into ammeter

by using a shunt of 1. Now ammeter and voltmeterconnected in circuit as shown, find the reading of

voltmeter and ammeter.

15

30 V

A

V

12

15 64.5

(a) 3 Volt, 3 amp (b) 2 volt, 2 amp (c) 4 Volt, 3 amp (d) 3 volt, 4 amp


28/48


Sol.: RVoltmeter= 6, Rammeter= 0.5Req= 10

AI 310

30

Reading of voltmeter = 1 3 = 3 volt. (a)

M

102. In the arrangement shown, the magnitude of each

resistance is .2 The equivalent resistance between OandA is given by

(a)

15

14(b)

15

7

(c) 3

4(d)

6

5 A

B

C

DO

Sol.: From symmetry.B andD are points having same

potential so, redrawing the network as

15

14OAR

(a)B, D O

A

C

M

103. A F4 capacitor is given C20 charge and is connected with an uncharged capacitor of

capacitance F2 as shown in figure. When switch Sis closed.

(a) charged flown through the battery is C3

40

(b) charge flown through the battery is C3

20

(c) work done by the battery is J3

200

(d) work done by the battery is J3

100

10V

2F

20C

C

S

++++++

4F

Sol.: Using Kirchhoffs loop law and conservation of charge,

final distribution of charge on the capacitors will be as

shown in the figure.

Charge q flown through the battery = charge on F2

capacitor and work done by the battery qV

(b) (c)10V

C3

80++++++

2F

C3

20+

+++++

4F


29/48


M

104. In the circuit shown if point O is earthed, the

potential of pointXis equal to

(a) 10 V (b) 15 V

(c) 25 V (d) 12.5 V

2

215V

5 10 V10 V

O5 V

5 V

X

5

Sol.: V0 + 10 5 + 10 = Vx Vx = 15 V (b)

M

105. Figure shows a network of a capacitor andresistors. The charge on capacitor in steady state

is

(a) 4 C (b) 6 C(c) 10 C (d) 16 C 6V

24

4

1F 10 V

8 V

4 V

8

Sol.: Let the potential of the junction be

V. Then

04

8

4

4

2

6

VVV

084212 VVV

V424

6V volt

6V2

4

4

1F 10 V

8 V

4 V

8i1

i3

i2

Potential drop across capacitor

106

V16

Charge on capacitor = 16 C (d)


30/48


M

106. A parallel plate capacitor is connected with a resistanceR and

a cell of emf as shown in figure. The capacitor is fullycharged. Keeping the right plate fixed, the left plate is moved

slowly towards further left with a variable velocity v such that

the current flowing through the circuit is constant. Then the

variation of v with separation x between the plates isrepresented by curve

v

R

x

v(a)

x

v(b)

x

v(c)

x

v(d)

Sol.: IRx

AIRCq

0 (i)

q

IRAx

0 (ii)

On differentiation of equation (ii) and from (i)

IRA

Ixv

0

2

(b)

x

I

R

M

107. The electric potential variation around a single closed

loop containing an ideal battery and one or more resistors

as shown in figure. If current of A1 flows in the circuit,

the circuit can not have

(a) two resistors and two batteries

(b) one resistor and three batteries

(c) maximum net emf of 6 volt

(d) three resistor and one battery

4

68

10

V

Sol.: The possible circuit of close loop corresponding to graph are

(i)

1R 2R

V2 V4

(ii)

R V2 V4 V2


31/48


(iii)

V4 V2 1R 2R

(iv)

R V4 V2 V4

(d)

M

108. An ammeter is obtained by shunting a 30 galvanometer with a 30 resistance. Whatadditional shunt should be connected across it to double the range?

(a) 15 (b) 10 (c) 5 (d) none of these

Sol.: For ammeter, GII

IS

g

g

1

gI

I

S

G gII 2

New range is doubled, i.e. 4Ig

Now shunt required, GII

IS

gg

g

4

= 10

This can be obtained by shunting the earlier shunt of 30 with an additional shunt of 15. (a)

M

109. An ideal ammeter and an ideal voltmeter are connected

as shown. The ammeter and voltmeter reading for

R1 = 5,R2 = 15,R3 = 1.25 andE= 20V are given as(a) 6.25 A, 3.75 V (b) 3.00 A, 5 V

(c) 3.75 A, 3.75 V (d) 3.75 A; 6.25 V20V

15

5A V

1.25

R1R2 R3

Sol.: eqR of the circuit = 321

21 RRR

RR

=

4

5

20

75

100

125

155

1555

5

20

eqR

EI = 4 A

20V

15

5A V

1.25

R1R2 R3

Current in ARR

IRR 3

21

21

P.D. across V533 IRR

(b)


32/48


M

110. A moving coil voltmeter is generally used in the laboratory to measure the potential

difference across a conductor of resistance r, and carrying a currentI. The voltameter has a

resistanceR and will measure the potential difference more accurately as

(a)R approaches r(b)R becomes larger than r

(c)R becomes smaller than r

(d)R equals to zero

Sol.: A voltmeter should have high resistance.

(b)M

111. A potentiometer has a driving cell of negligible internal resistance. The balancing length ofa Daniel cell is 5 m. If the driving cell has internal resistance, the balancing length of the

same Daniel cell would have been(a) more (b) less

(c) same (d) cannot be said from the data

47. (a)

M

112. A simple potentiometer circuit is shown in the figure. The

internal resistance of the 4V battery is negligible. AB is a

uniform wire of length 100 cm and resistance 2. Whatwould be the lengthACfor zero galvanometer deflection?

(a) 78.5 cm (b) 84.5 cm

(c) 82.5 cm (d) 80.5 cm

G

2.4

C BA

1.5 V

4 V

28.11

10

1.1

1

4.4

4

24.2

4

I

IfAC=x

Then 50

xRAC

5.15011

10

x 5.821155.1 x cm

(c)

M

113. As the switch S is closed in the circuit shown in figure

current passed through it is

(a) 4.5 A (b) 6.0 A(c) 3.0 A (d) zero

S

2

42BA

20V 5V


33/48


Sol.: Let Vbe the potential of the junction as

shown in the figure, applying junctionlaw.

VVV 25240 orV= 9Volt

2

3

Vi 4.5 Amp

S

2

42BA

20V 5V

(a)M

114. A potentiometer wire AB is 100cm long and has

total resistance of 10. Find the value of unknownresistance R so that null point is obtained at adistance 40 cm fromA.

(a) 1 (b) 2

(c) 3 (d) 4

10V

40cm

AC B

R

r= 1E= 5VG

Sol.: 11

5540

100

10

R R = 4

(d)M

115. A miliammeter of range 10mA and resistance 9 is joined in a circuit as shown. The meter gives full

scale deflection for current I when A andB areused as its terminals, i.e. current enters at A and

leaves atB (Cis left isolated). The value ofIis

(a) 100 mA (b) 900 mA

(c) 1 A (d) 1.1 A

9 ,10mA

0.1 0.9

A B C

Sol.: 33 101011.09.91010

or I= 1A

(c)

9 10mA

0.1 0.9

A B C


34/48


M

116. A resistanceR carries a current i. The power lost to the surroundings is ( - 0). Here is a

constant, is temperature of the resistance and0 is the temperature of the atmosphere. If thecoefficient of linear expansion is . The strain in the resistance is

(a) Ri 2

(b) iR

(c)

2

2Ri

(d) proportional to the length of the resistance wire

Sol.: Under steady state condition power developed = power loss

or 0

2 Ri

Ri2

0

Now, strain = =Ri 2

(a)

M

117. Two identical batteries, each having emf of 1.8V

and of equal internal resistances are connected asshown in the figure. Potential difference between

A andB will be equal to : (Ignore the resistance oflead wires)

(a) 3.6 V (b) 1.8 V

(c) zero (d) none of these

A B

Sol.:r

I2

6.3 , 08.1 IrVAB

(c)M

118. A milliammeter of range 10 mA has a coil of resistance 1. To use it is an ammeter ofrange 1A, the required shunt must have a resistance of

(a) 101

1(b)

100

1(c)

99

1(d)

9

1

Sol.: ig = 10 mA = 0.01 A r= 1 I= 1AVA VB = igr= (Iig)S

S=

991

01.01

101.0

)( g

g

iI

ri

(c)S

BI A ig

(I = ig)


35/48


M

119. A 100 W bulb B1, and two 60 W bulbs B2 andB3, are

connected to a 250 V source, as shown in the figure. Now

W1, W2 andW3 are the output powers of the bulbs B1,B2

andB3, respectively. Then.

(a) W1 > W2 = W3

(b) W1 > W2 > W3

(c) W1 < W2 = W3

(d) W1 < W2 < W3

B1 B2

B3

250 V

Sol.: Resistances of bulbsB2 andB3 are equal but that ofB1 is smaller than their resistance, hence

resistance of path of bulb B3 is less than that of the series combination of bulbs B2 andB3,therefore power consumed byB3 will be maximum possible.

Since B1 andB2 are in series, therefore current through them will be the same. Since

resistance ofB2 is greater, therefore W2 > W1.

(d)

M

120. In the circuits shown in the figure, the heat produced in

the 5- resistor due to the current flowing through it, is10 cal s1. The heat generated in the 4- resistor is

(a) 1 cal s1 (b) 2 cal s1(c) 3 cal s1 (d) 4 cal s1

4 6

5

Sol.: Heat produce in 5 resistance is ti 215 i1 = 2i/3

where i is the total current again heat produce through 4 is ti 224 i2 = i/3

given that 105 21 ti

therefore 109

20 2

ti

2

92 ti

Heat produce in 4 is equal to2

9

9

4

9

4 2 ti = 2

(b)


36/48


M

121. For what value ofR, power developed across 6

resistor is equal to the power developed across 24 resistor

(a) 12 (b) 6

(c) 24 (d) 8

6

R

24

Sol.: We have,I1 =24R

RI

I2 =24

24

R

I

P1 = 24)24( 2

22

RIR

P2 = I2.6

P2 = P1

R = 24 (c)

I

I2R

24

I1

6

M

122. If a cell produces the same amount of heat in two resistors R1 andR2 in the same time

separately, the internal resistance of the cell is

(a) (R1 +R2)/2 (b) 21 RR (c) 2/21 RR (d) (R1 R2)/2

Sol.: Let ris the internal resistance of cell

Case I: current in loopI1 =rR

1

H1 =I12R1 = 12

1

2

)(R

rR

Case II: Current in loopI2 =rR

2

H2 =I22R2

222

2

)(R

rR

i.e.,2

2

22

21

12

)()( rR

R

rR

R

rR

rR

R

R

2

1

2

1

r= 21RR (b)


37/48


M

123. Eight identical resistances r, each are connected along edges

of a pyramid having square base ABCD as shown in figure.

The equivalent resistance betweenA andD is

(a)15

2r(b)

15

r

(c)15

4r(d)

15

8r

A D

CB

O

Sol.: The circuit can be represented as C1OC2

So we can arrange the circuit in following way

rrrrq

1

2

1

3

22

1

Re

1

=rrr

1

2

1

8

3

Req =15

8 r

(d)

B C

DA C1

C2

O

r

2r

2r

r

r r

M

124. What is the potential difference between points

CandD in the circuit shown in figure in steady

state?

(a) 3.6 V (b) 7.2 V

(c) 10.8 V (d)12 V

12 V

3

1

6

C = 1 F1 C = 2 F2

B A

D

C

I

Sol.: A2.1163

12

VI , VAB= 12 1.2 1 = 10.8 V, VAD = 6 1.2 = 7.2 V

FFCC

CCC

eff

6

21

21 103

2

3

2

21

21

,

Q = VAB Ceff= 10.8 6

103

2 = 7.2 10-6 C

VC

QVAC 6.3

102

102.76

6

2

, VCD= VAD VAC= 7.2 3.6 = 3.6 V

(a)


38/48


M

125. A battery of emfEand internal resistance r is connected to a resistor of resistance r1 andQ

joules of heat is produced in a certain time t. When the same battery is connected to another

resistor of resistance r2, the same quantity of heat is produced in the same time t, the value of

ris

(a)2

2

1

r

r(b)

1

2

2

r

r(c)

2

1(r1 + r2) (d) 21rr

Sol.:I1 =rr

E

1, Q1 =I

2r1t=

2

1

rr

E r1t

Q2 =

rrE

2

r2t, 22

2

2

1

1

)()( rr

r

rr

r

, r= 21rr

(d)

M

126. In the network shown in the figure, each resistance is 1 ohm.The effective resistance betweenA andB is

(a) (4/3) (b) (3/2)

(c) 7 (d) (8/7)

A B

Sol.:7

8

23

8

23

8

eR

(d)A B

1

1

11

1

11

M

127. A ammeter is to be constructed which can read current upto 2.0 A. If the coil has a

resistance of 25 and takes 1 mA for full-scale deflection, what should be the resistance

of the shunt used?(a) 2.25 102 (b) 1 102

(c) 1.25 102 (d) 1.25 104

Sol.: A2I

So, 1 103 25 = 2R

R= 1.25 102

A25

R

2A

I

1mA

(c)


39/48


M

128. Figure shows a network of eight resistors numbered

1 to 8, each equal to 2, connected to a 3V batteryof negligible internal resistance. The currentIin thecircuit is

(a) 0.25 A (b) 0.5 A

(c) 0.75 A (d) 1.0 A

3V

A1 B C4 6 D

23 5

87

E F

Sol.: No current will flow through 3 and 5.

So,Req=

66

663 ,

3

3

eqR

Vi 1A

(d)M

129. Two heater wires of equal length are first connected in series and then in parallel. The ratio

of heat produced in two cases will be

(a) 1 : 2 (b) 1 : 4 (c) 2 : 1 (d) 4 : 1

Sol.: Applied potential difference is same

Rseries = 2R , Rparallel =R/2

Power

R

V2

, Pseries

R2

1

Pparallel 2/

1

R,

4

1

22

1

R

RP

P

parallel

series

(b)M

130. The effective wattage of 60 W and 40W lamps connected in series is equal to

(a) 24 W (b) 20 W (c) 100 W (d) 80 W

Sol.:21

21

PP

PPPeff

24effP W

(a)M

131. A heater boils a certain quantity of water in time t1. Another heater boils the same quantity

of water in time t2. If both heaters are connected in series, the combination will boil the

same quantity of water in time

(a) 212

1tt (b) 21 tt (c) 21

21

tt

tt

(d) 21tt

Sol.: Q = quantity of energy requiredQtP 11 , QtP 22


40/48


Pseries21

21

PP

PP

Pseriest0 = Q , QtPP

PP

02121

Solving t0 = t1 + t2

(b)M

132. The filament of an electric heater should have

(a) high resistivity and high melting point

(b) low resistivity and high melting point

(c) high resistivity and low melting point

(d) low resistivity and low melting point

Sol.: (a)

M

133. The voltage across a bulb is decreased by 2%. Assuming that the resistance of the filament

remains unchanged, the power of the bulb will

(a) decrease by 2% (b) increase by 2% (c) decrease by 4% (d) increase by 4%

Sol.:R

VP

2

2VP [R = constant]

(c)M

134. A student has connected a voltmeter, an ammeter and

a resistorR as shown. If voltmeter reads 20V andammeter reads 4A, thenR is

(a) = 5 (b) > 5

(c) < 5

(d) > or < 5 depending upon its material.

V+

A

4A

+ R

Sol.: Ri 420

54

20

iR

V

AR

(4-i)

i

(b)


41/48


M

135. In the figure, the potentiometer wire of length

l = 100 cm and resistance 9 is joined to a cell ofemf E1 = 10V and internal resistance r1 = 1.Another cell of emfE2 = 5V and internal resistance

r2 = 2 is connected as shown. The galvanometerG will show no deflection when the lengthACis

(a) 50 cm (b) 55.55 cm

(c) 52.67 cm (d) 54.33 cm

E1 = 10 V

r1 = 1

G

A B

E2 = 5 V

r1 = 2

C

Sol.: 1 5100

9

x x = 55.55 cm

(b)M

136. An electrical cable of copper has just one wire of radius 9 mm. Its resistance is 5. Thissingle copper wire of the cable is replaced by 6 different well insulated copper wires each of

radius 3 mm. The total resistance of the cable will now be equal to

(a) 7.5 (b) 45 (c) 90 (d) 270

Sol.: AsR2

1

r

93

9

5 2

2

R

R = 45

Req. =6

R= 7.5

(a)M

137. A uniform wire of resistance R is shaped into a regular n-sided polygon (n is even). Theequivalent resistance between any two corners can have

(a) the maximum value2

R(b) the maximum value

n

R

(c) the minimum value

21

nnR (d) the minimum value

nR

Sol.: Resistance between opposite corner is2

Rand

2

Rwhich is parallely connected.

Maximum value4

R

For adjacent corner two resistancen

Rand R

n

n

1are parallel connected

So minimum resistance is

2

1

n

nR

(c)


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M

138. In the circuit shown, current through 25V cell

is(a) 7.2 A (b) 10 A

(c) 12 A (d) 14.2 A

10V 5V 20V 30V 25V

5 10 5 11

Sol.: Applying KVL in loop ABCDA,

ABFEA, ABHGA and ABJIA we get

30 i1 11 = 25 ..... (i)

20 + i2 5 = 25. .... (ii)

5 i3 10 = 25 ..... (iii)10 + i4 5 = 25 ..... (iv)

i1 = 5A, i2 = 1A, i3 = 3A and i4 = 3A. (c)

I G E C A

10V 5V 20V 30V 25V

5 10 5 11

J H F D B

i4 i3 i2

i1

i1+i2+i3+i4

M

139. Seven identical lamps of resistances 2200 ohm

each are connected to 220 volt line as shown in

the figure. What will be the reading in theammeter? A

(a) (1/10) ampere (b) (3/10) ampere

(c) (4/10) ampere (d) (7/10) ampere

Sol.: Current through each resistor will be same current passing through ammeter

10

4

2200

2204

(c)M

140. In the part of a circuit shown in the figure, the

potential difference between points G and H

(VG VH) will be

(a) 0 V (b) 15 V

(c) 7 V (d) 3 V

G

2A 1A

3A

5V2

4

3V

H

1

Sol.:HG

VV 1222342

V7HG VV

(c)


43/48


M

141. In the circuit shown in the figure, the ratio ofVB as to

VC is(a) 2/5 (b) 5/2

(c) 1 (d) 1/3

1C

2A

B

D

5V10V

2V

Sol.: 5VV CA , 2VV BA ,5

2

C

B

V

V

(a)M

142. A 3 resistor as shown in the figure, is dipped into acalorimeter containing H2O. The thermal capacity of

H2O + calorimeter is 2000 J/K. If the circuit is activefor 15 minutes find the rise in temperature of H2O is

(a) 2.40

C (b) 2.90C

(c) 3.40 C (d) 1.90C

1

6

6V

3

Sol.:

12

6

1

6

ABRI 2A

2

36

36ABR

Current through 3 resistors 9/6' II 4/3 A(mC) T= RtI2

601533

42000

2

T

T 2.40C (a)

M

143. In the figure AB is 300 cm long wire havingresistance 10 per meter. Rheostat is set at 20.The balance point will be attained at

(a) 1.0 m (b) 1.25 m

(c) 1.5 m (d) cannot be determined

2V 0.5

1.5

A

G

K

6V 20

B

Sol.:

50

306ABV 3.6 V

Terminal voltage of cell

2

5.12 = 1.5 V


44/48


Using klV l300

6.35.1 orl = 125 cm

(b)M

144. In the circuit shown in the figure, reading of

voltmeter V1 when only S1 is closed, reading ofvoltmeter is V2 when only S2 is closed and reading

of voltmeter is V3 when both S1 andS2 are closed.

Then

3RS1

S2R

V

6R

(a) 123 VVV (b) 312 VVV (c) 213 VVV (d) 321 VVV

Sol.: When only S1 is closed . EEV 75.04

31


62

And when both S1 andS2 are closed, combined resistance of 6R and 3R is 2R.

EEV 66.03

23

312 VVV

(b)

M

145. In an experiment to measure the internal resistance of a cell by a potentiometer, it is found

that the balance point is at a length of 2 m, when the cell is shunted by a 5 resistance andat a length of 3 m, when the cell is shunted by a 10 resistance. The internal resistance ofthe cell is

(a) 1.5 (b) 10 (c) 15 (d) 1

Sol.:)(

)(

)/(

)/(

12

21

22

11

2

1

2

1

rRR

rRR

rRER

rRER

l

l

V

V

,)5(10

)10(5

3

2

r

r

, 10r

(b)M

146. A galvanometer of 10 ohm resistance gives full scale deflection with 0.01 ampere of

current. It is to be converted into an ammeter for measuring 10 ampere current. The value of

shunt resistance required will be

(a)999

10ohm (b) 0.1 ohm (c) 0.5 ohm (d) 1.0 ohm

Sol.:999

10

01.010

01.010

g

g

ii

GiS ohm

(a)


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M

147. In the given circuit the current i1 is

(a) 0.4 A (b) 0.4 A(c) 0.8 A (d) 0.8 A

30

40

40

i1

i2

80V

40V

i3

Sol.: Applying KCL at junctionA

213 iii (i)

Applying Kirchoffs voltage law for loopABCDA

0404030 31 ii

040)(4030 211 iii 447 21 ii ...(ii)

30

40

40

i1

i2

80V

40V

i3

F E

D

CB

A i3

Applying Kirchoffs voltage law for the loopADEFA.

040804040 32 ii

32 21 ii (iii)

On solving equation (ii) and (iii) A4.01 i

(b)M

148. In the diagram shown, the reading of voltmeter is 20 V and that of ammeter is 4A. Thevalue ofR should be (Consider given ammeter and voltmeter are not ideal)

(a) Equal to 5(b) Greater from 5(c) Less than 5(d) Greater or less than 5 depends on the material ofR

V

A

4A

20V

R

Sol.: If resistance of ammeter is rthen 4)(20 rR

5 rR R < 5

(c)M

149. For ensuring dissipation of same energy in all three

resistors (R1, R2, R3) connected as shown in figure,their values must be related as

R1

R2 R3Vin

(a)R1 =R2 =R3 (b)R2 =R3 andR1 = 4R2

(c) 32 RR and 214

1RR (d)R1 =R2 +R3


46/48


Sol.: As the voltage in R2 andR3 is same

therefore, according to,

tRVH

2

, 32 RR

Also the energy in all resistance is same.

tRitRi 22

11

2

R1

R2 R3

i

i1 i2

D C

Vin

Using iiRR

Ri

RR

Ri

2

1

33

3

32

3

1

Thus, tRi

tRi 2

2

1

2

4 or,

4

21

RR

(c)M

150. The measurement of voltmeter (ideal) in the following circuit

is

(a) 2.4 V (b) zero

(c) 4.0 V (d) 6.0 V

6V

V600

400

Sol.: If the voltmeter is ideal then given circuit is an open circuit, so reading of voltmeter is equal

to the e.m.f. of cell i.e., 6V

(d)M

151. A current I is passing through a wire having two sections P andQ of uniform diameters dandd/2 respectively. If the mean drift velocity of electrons in section P andQ is denoted by

vP andvQ respectively, then

(a)QP

vv (b)QP

vv2

1 (c)

QPvv

4

1 (d)

QPvv 2

Sol.: Drift velocityneA

ivd

Avd

1 or

2

1

dvd

4

12/22

d

d

d

d

v

v

P

Q

Q

P QP vv

4

1

(c)M

152. A coil of wire of resistance 50 is embedded in a block of ice. If a potential difference of210 V is applied across the coil, the amount of ice melted per second will be (Latent heat of

fusion of ice = 80 cal/gm)

(a) 4.12 gm (b) 4.12 kg (c) 3.68 kg (d) 2.625 gm

Sol.: L

t

m

R

V

t

Q.

2.4

2

; 625.2

80502.4

)210(

2.4

22

RL

V

t

mgm

(d)


47/48


M

153. A resistance of 4 and a wire of length 5 metres and

resistance 5 are joined in series and connected to a cell ofe.m.f. 10V and internal resistance 1. A parallelcombination of two identical cells is balanced across 300 cm

of the wire. The e.m.f. of each cell isE

EG

4 10V,1

5, 5m

(a) 1.5 V (b) 3.0 V (c) 0.67 V (d) 1.33 V

Sol.: lL

iR

L

lVlE E= l

L

R

rRR

E

h

V335

5

145

10

E

(b)M

154. A 500 W heating unit is designed to operate from a 115 volt line. If the line voltage drops to110 volt, the percentage drop in heat output will be

(a) 10.20% (b) 8.1% (c) 8.6% (d) 7.6%

Sol.: 46.457500115

11022

consumed

R

R

A PV

VP W

So, percentage drop in power output = %6.8100500

)46.457500(

(c)M

155. Three equal resistances each ofR are connected asshown in figure. A battery of emf 2V and internal

resistance 0.1 is connected across the circuit. Thevalue ofR for which the heat generated in the circuit

will be maximum is

(a) 0.3 (b) 0.03

(c) 0.01 (d) 0.1

R R R

2V

0.1

Sol.: (a)


48/48


DIFFICULT QUESTIONS

D

156. A galvanometer of coil resistance 1 is convertedinto voltmeter by using a resistance of 5 in seriesand same galvanometer is converted into ammeter

by using a shunt of 1. Now ammeter and voltmeterconnected in circuit as shown, find the reading of

voltmeter and ammeter.

15

30 V

A

V

12

15 64.5

(a) 3 Volt, 3 amp (b) 2 volt, 2 amp (c) 4 Volt, 3 amp (d) 3 volt, 4 amp

Sol.: RVoltmeter= 6, Rammeter= 0.5Req= 10

AI 310

30

Reading of voltmeter = 1 3 = 3 volt. (a)

Direct Current Circuits (QB)

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