Circuit Networks(11th And 12th > Physics ) Questions and answers for exam Preparation

Question 1.

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

$\frac{4}{9}$
$\frac{8}{9}$
2
18

Explanation:-

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

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

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

or

R = 2 Ω

Question 2.

When the two identical cells are connected either in series or in parallel across a 4 ohm resistor they send the same current through it. The internal resistance of the cell is

$1.2 Ω$
$2 Ω$
$4 Ω$
$4.8 Ω$

Explanation:-

Answer: Option C. ->

4 Ω

:
C
In series the current through the resistor R is

\frac{2 E}{R + 2 r}

In parallel connection, current through the resistor R is

\frac{E}{R + \frac{r}{2}}

I = \frac{2 E}{R + 2 r} = \frac{E}{R + \frac{r}{2}} 2 R + r = R + 2 r R = r r = 4 Ω

Question 3.

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

100
150
250
300

Explanation:-

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 4.

36 cells each of e.m.f. 2 volt and internal resistance 0.5 ohm are connected in mixed grouping so as to deliver maximum current to a bulb of resistance 2 ohm. How are they connected. What is the maximum current through the bulb (n=no. of bulbs in series and, m = no. of bulbs in parallel)

n=12 ; m=3; 6 A
n = 9; m=4; 5A
n=5; m=7’ 6A
n=7; m=4; 8A

Explanation:-

Answer: Option A. -> n=12 ; m=3; 6 A
:
A
To deliver maximum current, the external resistance should be equal to internal resistance.
mn = 36; Rm = r n

\frac{m}{n} = \frac{1}{4} = \frac{2}{8} = \frac{3}{12} \frac{m}{n} = \frac{r}{R} = \frac{0.5}{2} = \frac{1}{4} l_{m a x} = \frac{m n E}{2 n r} = \frac{3 \times 12 \times 2}{2 \times 12 \times 0.5} 6 A \to (1)

Question 5.

In the given circuit, if the galvanometer shows zero reading, then
x= ___ ohms

20
30
50
100

Explanation:-

Answer: Option A. -> 20
:
A

V_{x} = 2 V V_{100} = 10 V I = \frac{V_{100}}{R} = \frac{10}{100} = 0.1 A V_{x} = I X 2 = 0.1 X X = 20 Ω

Question 6.

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, 2 $Ω$
63V, 1 $Ω$
61V, 1 $Ω$
64V, 2 $Ω$

Explanation:-

Answer: Option A. -> 62V, 2

Ω

:
A
V= E - Ir
50 = E - 6r

\to

(1)
60 = E - 1r

\to

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

Ω

E = 62V

Question 7.

24 cells of each emf 4V are connected in series. Among them if 5 cells are connected wrongly, the effective emf of the combination is

18V
56V
24V
4V

Explanation:-

Answer: Option B. -> 56V
:
B

E^{'} = (N - 2 n) E = (24 - 2 \times 5) E = 14 E = 14 \times 4 E^{'} = 56 V

Question 8.

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

$10 V, 16 Ω$
$3 V, 20 Ω$
$16 V, 10 Ω$
$24 V, 32 Ω$

Explanation:-

Answer: Option C. ->

16 V, 10 Ω

:
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 9.

The P.D between A and B will be zero when the resistance R is

4.5 ohm
12 ohm
9 ohm
6 ohm

Explanation:-

Answer: Option A. -> 4.5 ohm
:
A
From the circuit

\frac{15}{4} = \frac{R + 3}{2} 15 = 2 R + 6 R = \frac{9}{2} = 4.5 Ω

Question 10.

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

Explanation:-

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