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A galvanometer of resistance $50 Ω$ is connected to a battery of 3V along with a resistance of $2950 Ω$ in series. A full scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 20 divisions, the resistance in series should be :

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$5050 Ω$

$4450 Ω$

$6050 Ω$

$5550 Ω$

answer is D.

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Detailed Solution

$i = \frac{3}{2950 + 50} = 10^{- 3} A$

If defection reduced to 20

$i = \frac{2}{3} \times 10^{- 3} A$

$10^{- 3} \times 3000 = \frac{2}{3} \times 10^{- 3} \times R$

$R = 4500 Ω$

series resistance= $4500 - 50 = 4450 Ω$

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