A galvanometer of resistance 25 Ω is connected to a battery of 2 V along with a resistance of 3000 Ω . In this case, a full-scale deflection of 30 units is obtained in the galvanometer. In order to re

Question

A galvanometer of resistance 25 Ω is connected to a battery of 2 V along with a resistance of 3000 Ω . In this case, a full-scale deflection of 30 units is obtained in the galvanometer. In order to reduce this deflection to 10 units , how much more resistance (in Ω ) should be added to the circuit in series?

TestHub · Accepted Answer

The deflection in a galvanometer is directly proportional to the current.

First case:

$30 x = \frac{2}{25 + 3000}$

Second case:

$10 x = \frac{2}{25 + 3000 + R}$

$\Rightarrow 3 = \frac{25 + 3000 + R}{25 + 3000}$

On solving, we get

$R = 6050 Ω$

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