Circuit Networks(12th Grade > Physics ) Questions and answers for exam Preparation

12th Grade > Physics

CIRCUIT NETWORKS MCQs

Total Questions : 30 | Page 2 of 3 pages

Question 11. A cell is connected in the secondary circuit of a potentiometer and a balance point is obtained at 84 cm When the cell is shunted by a

5 Ω

resistor, the balance point changes to 70 cm. If the

5 Ω

resistor is replaced with

3 Ω

resistor then the balance point will be at

53 cm
63 cm
42 cm
112 cm

Discuss Question

Answer: Option B. -> 63 cm
:
B

r = R [\frac{l_{1} - l_{2}}{l_{2}}] = R^{1} [\frac{l_{1} - l_{2}^{1}}{l_{2}^{1}}] 5 [\frac{84 - 70}{70}] = 3 [\frac{84 - l_{2}^{1}}{l_{2}^{1}}] \frac{l_{2}^{1} \times 14}{14 \times 3} = 84 - l_{2}^{1} l_{2}^{1} + \frac{l_{2}^{1}}{3} = 84 4 l_{2}^{1} = 84 \times 3 l_{2}^{1} = 63 c m

Question 12. When resistance of 30

Ω

is connected to a battery, the current in the circuit is 0.4 A. if this resistance is replaced by 10

Ω

resistance, the current is 0.8 A. Then e.m.f. and the internal resistance of thebattery are

10V,16Ω
3V,20Ω
16V,10Ω
24V,32Ω

Discuss Question

Answer: Option C. -> 16V,10Ω
:
C

E = I_{1} [R_{1} + r] = I_{2} [R_{2} + r] 0.4 [30 + r] = 0.8 [10 + r] 30 + r = 20 + 2 r r = 10 E = 0.4 [30 + 10] E = 16 V

Question 13. A potentiometer wire of length 1 m and resistance

10 Ω

is connected in series with a cell of e.m.f 2 volt with internal resistance

1 Ω

and a resistance box including a resistance R. If the potential difference between the ends of the wire is 1 mV, then the value of R will be

9989Ω
10000Ω
19989Ω
20000Ω

Discuss Question

Answer: Option C. -> 19989Ω
:
C

i = \frac{V_{2}}{R_{1}} = \frac{1 \times 10^{- 3}}{10} = 10^{- 4} A E = i [R + r + R_{L}] 2 = 10^{- 4} [R + 1 + 10] 20000 = R + 11 R = 20000 - 11 = 19989 Ω

Question 14. A battery of internal resistance

4 Ω

is connected to the network of resistances as shown. In order to give the maximum power to the network, the value of R (in

Ω

) should be

A Battery Of Internal Resistance 4Ω Is Connected To The Net...

Discuss Question

Answer: Option C. -> 2
:
C
The equivalent circuit becomes a balanced wheatstone bridge

For maximum power transfer, external resistance should be equal to internal resistance of source

\Rightarrow \frac{(R + 2 R) (2 R + 4 R)}{(R + 2 R) + (2 R + 4 R)} = 4

i.e.

\frac{3 R \times 6 R}{3 R + 6 R} = 4

R = 2 Ω

Question 15. The scale of a galvanometer of resistance

100 Ω

contains 25 divisions. It gives a deflection of one division on passing a current of

4 \times 10^{- 4} A

. The resistance in ohms to be added to it, so that it may become a voltmeter of range 2.5 volt is

Discuss Question

Answer: Option B. -> 150
:
B
Current sensitivity of galvanometer

= 4 \times 10^{- 4}

Amp/div.
So full scale deflection current

(i_{g}) = C u r r e n t s e n s i t i v i t y \times T o t a l n u m b e r o f d i v i s i o n = 4 \times 10^{- 4} \times 25 = 10^{- 2} A

To convert galvanometer in to voltmeter, resistance to be put in series is

R = \frac{V}{i_{g}} - G = \frac{2.5}{10^{- 2}} - 100 = 150 Ω

Question 16. The potential difference across the terminals of a battery is 50V when 6A of current is drawn and 60V when 1A is drawn. The emf and internal resistance of the battery are

62V, 2Ω
63V, 1Ω
61V, 1Ω
64V, 2Ω

Discuss Question

Answer: Option A. -> 62V, 2Ω
:
A
V= E - Ir
50 = E - 6r

\to

(1)
60 = E - 1r

\to

(2)
(2) - (1) 10 = 5r; r = 2

Ω

E = 62V

Question 17. Two resistances X and Y are in the left and right gaps of a meter bridge. The balance point is 40 cm from left.Two resistances of

10 Ω

each are connected in series with X and Y separately. The balance point is 45 cm. The value of X and Y are

12Ω,8Ω
4Ω,6Ω
8Ω,12Ω
12Ω,16Ω

Discuss Question

Answer: Option C. -> 8Ω,12Ω
:
C

\frac{x}{y} = \frac{40}{100 - 40} = \frac{40}{60} = \frac{2}{3} \to (1) \frac{x + 10}{y + 10} = \frac{45}{55} = \frac{9}{11} \frac{\frac{2}{3} y + 10}{y + 10} = \frac{9}{11} \frac{2}{3} \times 11 y + 110 = 9 y + 90 20 = 9 y - \frac{22}{3} y 20 \times 3 = 5 y y = 12 Ω x = \frac{2}{3} \times 12 = 8 Ω

Question 18. A

1 Ω

resistance is in series with an ammeter which is balanced by 75 cm of potentiometer wire. A standard cell of 1.02 V is balanced by 50 cm. The ammeter shows a reading of 1.5A. The error in the ammeter reading

0.002 A
0.03 A
1.01 A
no error

Discuss Question

Answer: Option B. -> 0.03 A
:
B

i p = \frac{V}{l} = \frac{1.02}{50}

Voltage across ammeter =

\frac{V}{l} l_{1}

= \frac{1.02}{50} \times 75 = \frac{1.02}{2} \times 3 = 0.51 \times 3

The true current

I_{A}^{1} = \frac{V_{A}}{R} = \frac{1.53}{1} I_{A}^{1} = 1.53 A E r r o r = I_{A}^{1} - I_{A} = 1.53 - 1.5 = 0.03 A

Question 19. A wire of length L and 3 identical cells of negligible internal resistances are connected in series. Due to this current, the temperature of the wire is raised by

Δ T

in time t. A number N of similar cells is now connected in series with a wire of the same material and cross-section but of length 2L. The temperature of wire is raised by same amount