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A galvanometer, whose resistance is 50 ohm, has 25 divisions in it. When a current of $4 \times 10^{- 4}$ A passes through it, its needle (pointer) deflects by one division. To use this galvanometer as a voltmeter of range 2.5V, it should be connected to a resistance of

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200 ohm

6250 ohm

6200 ohm

250 ohm

answer is A.

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Detailed Solution

The full scale deflection current

$I_{g} = 25 \times 4 \times 10^{- 4} A$

Let R be the resistance connected

$I_{g} = \frac{V}{R + R_{g}} \Rightarrow 25 \times 4 \times 10^{- 4} = \frac{2.5}{R + 50}$

$\Rightarrow R = 200 Ω$

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