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ANDHERI / BORIVALI / DADAR / CHEMBUR / THANE / MULUND/ NERUL / POWAI

TIME: 1 HRS TW TEST MARKS: 65

TOPICS: ELECTRIC CURRENT AND CIRFCUITS

SECTION-I (Single Choice Questions)

This section contains10 multiple choice questions. Each question has 4 choices (A),(B ),(C ) and(D ) for its answ er, out w hich ONLY ONE is correct. (+ 3, - 1)

1. Equivalent resistance between A and B in the figure is

a. 8r/15 b. 7r/15 c. 15r/8 d. 15r/7

2. Each of the resistors shown in the figure has resistance R. Find the equivalent resistance between

A and B.

a.7R

4 b.

5R

4 c.

9R

4 d.

11R

4

3. Find the equivalent resistance between A and E (Resistance of each resistor is R).

4. The circuit shown has resistors of equal resistors of equal resistance R. Find the equivalent

resistance between A and B, when the key is closed:

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a.11R

12 b.

13R

12 c.

R

5 d.

15R

12

5. In the circuit shown, current through 25 V cell is

a. 7.2 A b. 10 A c. 12 A d. 14.2 A

6. Figure represents a load consisting of three identical resistances connected to an electric energy

source of e. m. f. 12 V and internal resistance 0.6 . The ammeter reads 2A. The magnitude of each

resistance is

a. 3.6 b. 7.5 c. 16.2 d. 10.8

7. Figure represents a part of closed circuit. The potential difference (A B

V V ) is

a. 24 V b. 0 V c. 6 V d. 18 V

8. Two cells A and B, each of e .m. f 2 V, are connected in series to an external resistance R 1 . If

the internal resistance of cell A is 1.9 , what is the potential difference between the terminals of cellA ?

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9. Two resistors of resistances 200 k and 1 M , respectively, form a potential divider with outer

junctions maintained at potentials of +3 V and -15V.

What is the potential at the junction X between the resistors?

a. +1 V b. 0 V c. -0.6 V d. -12 V

10. The equivalent resistance between A and B in the network in Fig. is

a.4

3 b.

3

2 c. 3 d. 2

SECTION-II (Multiple Choice Questions)

This section contains7 multiple choice questions.Each question has 4 choices (A ), (B), (C ) and

(D ) for its answ er, out w hich ONE OR MORE is/are correct. (+ 3, 0)

11. In the network shown in Fig. points A, B and C are at potentials of 70 V, 0 and 10 V,

a. Point D is at a potential of 40 V.

b. The currents in the sections AD, DB, DC are in the ratio 3:2:1.

c. The currents in the sections AD, DB, DC are in the ratio 1:2:3.

d. The network draws a total power of 200 W.

12. In the circuit shown in Fig. the cell has e. m. f. = 10 V and internal resistance = 1 .

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a. The current through the 3 resistor is 1 A

b. The current through the 3 resistor is 0.5 A

c. The current through the 4 resistor is 0.5 A

d. The current through the 4 resistor is 0.25 A

13. In the circuit shown in Fig. , some potential difference is applied between A and B. The equivalent

resistance between A and B is R.

a. No current flows through the 4 resistor. b. R 15

c. R 12.5 d.18

R5

14. In the given circuit Fig.

a. the current through the battery is 5 A b. P and Q are the same potential

c. P is 2 V higher than Q d. Q is 2 V higher than P

15. In the circuit shown in Fig., mark the correct options.

a. Potential drop across R1is 3.2 V. b. Potential drop across R

2is 5.4 V.

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c. Potential drop across R1is 7.2 V d. Potential drop across R

2is 4.8 V.

16. The capacitor C is initially without charge. X is now joined to Y for a long time, during which H1

heat is produced in the resistance R. X is now joined to Z for a long time, during which2

H heat is

produced in R

a. 1 2H H b. 1 21

H H2

c. 1 2H 2H d. the maximum energy stored in C at any time is 1H .

17. In the circuit shown in Fig., the cell is ideal, with e.m.f. = 2 V. The resistance of the coil of the

galvanometer G is 1 . Then in steady state

a. no current flows in G b. 0.2 A current flows in G

c. potential difference across1

C is 1 V d. potential difference across2

C 1.2 V

SECTION-II I (Paragraph Type)

This section contains2 groups of questions. Each question has 4 choices (A ), (B ), (C ) and (D ) for itsansw er, out of w hich ONLY ONE is correct. (+ 3, -1)

Figure shows two ideal voltmeters and an ammeter which are connected across the various circuit

elements. If the voltmeter connected across 9 resistance reads 4.5 V, then answer the following problems.

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18. The current through 12 resistance is

a. 0.1 A b. 0.75 A c. 0.5 A d. 1.25 A

19. The reading of the voltmeter connected across 20 resistance

a. 15 V b. 2.25 A c. 1.5 A d. 0.1 A

SECTION-IV (Integer Answer Type)This section contains 2questions. The answ er to each of the questions is a single digit integer,

ranging from 0 to 9. The correct digit below the question num ber in the O RS is be bubbled. (+ 4, 0)

20. Find the potential difference (in V ) between points A and B shown in the circuit.

21. In the circuit shown, each resistance is 2 The potential1

V is as indicated in the circuit. What is

the magnitude of V1in volt?

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Answers Key

1. a 2. d 3. a 4. c 5. c 6. c 7. d

8. c 9. b 10. a 11. A, b, d 12. A, d 13. A, d 14. A, d

15. c, d 16. A, d 17. B, c, d 18. c 19. a 20. (6) 21. (9)

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SOLUTIONS

1. a.

The given figure can be redrawn as shown.

We have broken the circuit at C and then connected as shown the in figure. It is because the whole of

the current from DC will go to CE and that from AC will go to CB. Now solve to getAB

R 8r /15 .

2. d.

The figure can be redrawn as

AB

3R 11R R R R

4 4

3. a.

The figure can be redrawn as

eq

7RR

12

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4. c.

A careful observation will reveal that one end of each resistor is connected to A and the other end of

each resistor is connected to B. Hence , the resistors are in parallel. Soeq

R R / 5 .

5. c.

Applying KVL in loop CDBAC, EFBAF, GHBAG and IJBAI, we get

1 130 i 11 25 i 5A

2 220 i 5 25 i 1A

2 35 i 5 25 i 3A

4 40 i 5 25 i 3A 1

Current through 25 V cell is 1 2 3 4i i i i 12A

6. c.

12 R2 0.6 6 R 16.2

R 30.6

3

7. d.

A B A BV 3 3 18 V or V B 18V

8. c.

E 4V , total resistance = 1 0.9 1.9 3.8

Now,4

I A3.8

Again, the terminal potential difference across A is

42 1.9 2 2 0

3.8

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9. b.

Current I through the resistors is

3 15I mA 0.015mA

200 1000

Potential at X is thus

3 3xV 3 200 10 0.0.15 10 0

10. a.

The equivalent of the network is given Fig.,

The equivalent of the above network is a parallel combination of 3 , 4 and 6 , I.E.,

1 1 1 1 4Rr 3 4 6 3

11. a, b, d.

Let V = potential at D

1 2 1 270 V 10i ,V 0 20i , V 10 30 i i

Solve for 1 2i , i , and V.

12. a, d.

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The last three resistors 2 ,4 and 2 are in series having the equivalent resistance as

2 + 4+ 2 = 8 . This will be in series with the 8 next to then. So, their equivalent resistance

becomes 4 . In this way, net equivalent resistance of the circuit becomes eqR 9 . This will be in

series with r = 1 . So, the current through 3 is

I e / R r 10 / 9 1 1A .

Further, current will get divided at C and E into half at each point. So, finally the current reaching in

4 will be 0.25 A.

13. a, d.

Rearrangement of the circuit as shown in Fig. gives a balanced Wheatstone bridge, and no current

flows through the 5 resistor. It can thus be removed from the circuit.

14. a, d.

Resistors 2 and 2 are in series. 2 and 4 are in series, then their resultant are in parallel.

Producing net resistance R = 2.4

20 20I 5A

r R 1.6 2.4

1 2I I 5A (i)

AB 1 2 1 2V 4l 6l 21l 3l (ii)

From Eqs. (i) and (ii),1 2

I 3A,I 2A

V V 6V V V 2I 4V

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Thus,q P

V V 2V

15. c, d.

Note that point a, h, g and f have the same potential. They are connected by conducting wires

without any circuit elements between them. Similarly, points b, c, d and e have the same potential.

Hence, the potential drop across branch e and f, and a and b is the same. The two resistors ( 6 and

4 in series) are directly connected across the terminals of 12 V battery.

The current1 2

V 12I 1.2 A

R R 10

Hence, 1 2V 1.2 6 7.2V, V 1.2 4 4.8V

16. a, d.

When X is joined to Y for a long time (charging), energy stored in the capacitor = heat produced in R= H.

When X is joined to Z (discharging), the energy stored in 1C H reappears as heat 2H in R.

Thus, 1 2H H .

17. b, c, d.

Disregard the capacitors and find the current through G. The potential difference across each

capacitor is then found from the potential differences across the resistances in parallel with them.

18. c. 19. a.

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18. c.

Potential difference across AB:

1 2 2 1 24.5 3I 6I I 1A, I I I 1.5A

Equivalent resistance between C and D:1

R 4

Potential difference across1

CD R I 4 1.5 6V

Current through3

6 112 , I A

12 2

19. (a)

Potential difference across AD is

2 36I 2I 12I 6 1 2 1.5 12 0.5 15V

This will be equal to the reading of voltmeter across 20 .

20. (6)

9 V and 165 can be ignored because potential difference between A and C is fixed which is 4 V.

The circuit cab be drawn as shown below.

Equivalent e. m. f. :

E / r 10 / 2 4 / 2e 7V

1/ r 1/ 2 1/ 2

Equivalent internal resistance: 01 1

r 11/ r 1/ r 1/ 2

Now AB AB24

V 0.5 V 1 V12

21. (9)

The resulting circuit can be drawn as shown. No current will flow into the earth.

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A 1 Ai 12 5 / 7 1A, V 2i 2i 5 V but V 0

1V 9 V .