First slide

Moving coil Galvanometer

Question

A galvanometer of resistance 50 $Ω$ is connected to a battery of 3 V 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

Moderate

A

5050 $Ω$
B

5550 $Ω$
C

6050 $Ω$
D

4450 $Ω$

Solution

initial current $i = \frac{3}{3000} = 10^{- 3} A$
$10^{- 3} A \to 30 divisions$
$20 div \to 10^{- 3} \times \frac{2}{3} A$
For current $\frac{2}{3} \times 10^{- 3}$ resistance supposed to be, $10^{- 3} \times \frac{2}{3} = \frac{3}{R + 50}$
$R + 50 = \frac{9}{2} \times 10^{3} = 4500$
$R = 4450 Ω$

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