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Current Electricity MCQs for NEET — Physics Questions with Answers

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If in the experiment of Wheatstone's bridge, the positions of cells and galvanometer are interchanged, then balance points will

Change

Remain unchanged

Depend on the internal resistance of cell and resistance of galvanometer

None of these

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Explanation

The actual circuit is same.

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Two cells when connected in series are balanced on 8m on a potentiometer. If the cells are connected with polarities of one of the cells reversed, they balance on 2m. The ratio of e.m.f.'s of the two cells is

3 : 5

5 : 3

3 : 4

4 : 3

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Explanation

$\frac{E_{1}}{E_{2}} = \frac{l_{1} + l_{2}}{l_{1} - l_{2}} = \frac{(8 + 2)}{(8 - 2)} = \frac{5}{3}$

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Which is a wrong statement :

The Wheatstone bridge is most sensitive when all the four resistances are of the same order

In a balanced Wheatstone bridge, interchanging the positions of galvanometer and cell affects the balance of the bridge

Kirchhoff's first law (for currents meeting at a junction in an electric circuit) expresses the conservation of charge

The rheostat can be used as a potential divider

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Explanation

In balanced Wheatstone bridge, the arms of galvanometer and cell can be interchanged without affecting the balance of the bridge.

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Constantan wire is used in making standard resistances because of its :

Specific resistance is low

Density is high

Temperature coefficient of resistance is negligible

Melting point is high

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Explanation

Due to the negligible temperature co-efficient of resistance of constantan wire, there is no change in it's resistance value with change in temperature.

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A potentiometer consists of a wire of length 4 m and resistance 10 Ω. It is connected to a cell of e.m.f. 2 V. The potential difference per unit length of the wire will be :

0.5 V/m

2 V/m

5 V/m

10 V/m

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Explanation

Since potential difference for full length of wire = 2 V

∴ P.D. per unit length of wire $= \frac{2}{4} = 0 .5 \frac{V}{m}$

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In a potentiometer experiment, the galvanometer shows no deflection when a cell is connected across 60 cm of the potentiometer wire. If the cell is shunted by a resistance of 6 Ω, the balance is obtained across 50 cm of the wire. The internal resistance of the cell is

0.5 Ω

0.6 Ω

1.2 Ω

1.5 Ω

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Explanation

$r = \frac{(l_{1} - l_{2})}{l_{2}} \times R' = (\frac{60 - 50}{50}) \times 6 = 1 .2 Ω$

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We have a galvanometer of resistance 25 Ω. It is shunted by a 2.5 Ω wire. The part of total current that flows through the galvanometer is given as

$\frac{I}{I_{0}} = \frac{1}{11}$

$\frac{I}{I_{0}} = \frac{1}{10}$

$\frac{I}{I_{0}} = \frac{3}{11}$

$\frac{I}{I_{0}} = \frac{4}{11}$

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Explanation

$\frac{i}{i_{g}} = \frac{G + S}{S}$ ⇒ $\frac{i_{g}}{i} = \frac{S}{G + S} = \frac{2 .5}{27 .5} = \frac{1}{11}$

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A voltmeter has resistance of 2000 ohms and it can measure upto 2V. If we want to increase its range to 10 V, then the required resistance in series will be

2000 Ω

4000 Ω

6000 Ω

8000 Ω

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Explanation

Here $n = \frac{10}{2} = 5$

∴ $R = (n - 1) G = (5 - 1) 2000 = 8000 Ω$

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The resistance of a galvanometer coil is R. What is the shunt resistance required to convert it into an ammeter of range 4 times

$\frac{R}{5}$

$\frac{R}{4}$

$\frac{R}{3}$

4 R

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Explanation

$\frac{i}{i_{g}} = 1 + \frac{G}{S} \Rightarrow \frac{4}{1} = 1 + \frac{R}{S} \Rightarrow S = \frac{R}{3}$ .

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A galvanometer has a resistance of 25 ohm and a maximum of 0.01 A current can be passed through it. In order to change it into an ammeter of range 10 A, the shunt resistance required is

5/999 ohm

10/999 ohm

20/999 ohm

25/999 ohm

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Explanation

$i_{g} = i \frac{S}{G + S}$ ⇒ $0 .01 = 10 \frac{S}{25 + S}$

$\Rightarrow 1000 S = 25 + S$ ⇒ $S = \frac{25}{999} Ω$ .

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