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

Question 1. In a metre bridge experiment, null point is obtained at 20 cm from one end of the wire when resistance X is balanced against another resistance Y. If X < Y, then where will be the new position of the null point from the same end, if one decides to balance a resistance of 4X against Y?

50 cm
80 cm
40 cm
70 cm

Explanation:-

Answer: Option A. -> 50 cm
:
A

\frac{X}{Y} = \frac{20}{100 - l} = \frac{20}{80} = \frac{1}{4} Y = 4 X \frac{4 X}{Y} = \frac{l^{'}}{100 - l^{'}}; l^{'} = 50 c m

Question 2. 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 3. Eels are able to generate current with biological cells called electroplaques. The electroplaques in an eel are arranged in 100 rows, each row stretching horizontally along the body of the fish containing 5000 electroplaques. The arrangement is suggestively shown below. Each electroplaques has an emf of 0.15 V and internal resistance of 0.25

Ω

. The water surrounding the eel completes a circuit between the head and its tail. If the water surrounding it has a resistance of 500 ohm, the current an eel can produce in water is about

Eels Are Able To Generate Current With Biological Cells Call...

1.5 A
3.0 A
15 A
30 A

Explanation:-

Answer: Option A. -> 1.5 A
:
A
Given problem is the case of mixed grouping of cells
So total current produced

i = \frac{n E}{R + \frac{n r}{m}}

Here m = 100, n = 5000, R = 500

Ω

E = 0.15V and r = 0.25

Ω

\Rightarrow i = \frac{5000 \times 0.15}{500 + \frac{5000 \times 0.25}{100}} = \frac{750}{512.5} \approx 1.5 A

Question 4. 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...

49
89
2
18

Explanation:-

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

or

R = 2 Ω

Question 5. A primary cell has an emf of 1.5 volts, when short-circuited it gives a current of 3 amperes. The internal resistance of the cell is

4.5 ohm
2 ohm
0.5 ohm
14.5 ohm

Explanation:-

Answer: Option C. -> 0.5 ohm
:
C

i_{S C} = \frac{E}{r} \Rightarrow 3 = \frac{1.5}{r} \Rightarrow r = 0.5 Ω

Question 6. 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 7. 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 8. 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

1.2Ω
2Ω
4Ω
4.8Ω

Explanation:-

Answer: Option C. -> 4Ω
:
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 9. In an experiment for calibration of voltmeter, a standard cell of emf 1.5V is balanced at 375cm length of potentiometer wire. The potential difference across a resistance in the circuit is balanced at 1.25 m. If a voltmeter is connected across the same resistance, it reads 0.6V. The error in the voltmeter is

0.1V
0.025V
0.235V
0.18V

Explanation:-

Answer: Option A. -> 0.1V
:
A

i ρ = \frac{E}{l} = \frac{1.5}{3.75} \frac{V}{m} V^{1} = (i ρ) l = \frac{1.5}{3.75} \times 1.25 V^{1} = 0.5 V E r r o r = V - V^{1} = 0.6 - 0.5 = 0.1 V

Question 10. 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