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A galvanometer of resistance $50\,\Omega$ is connected to a battery of 3 V along with a resistance of $2950\,\Omega$ 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 -
A
$6050\,\Omega$
B
$4450\,\Omega$
C
$5050\,\Omega$
D
$5550\,\Omega$
Detailed Solution
Current through the galvanometer for full scale deflection: $I = \dfrac{3}{50 + 2950} = \dfrac{3}{3000} = 10^{-3}$ A
So the current for 30 divisions $= 10^{-3}$ A.
Deflection is proportional to current, so the current for 20 divisions $= \dfrac{20}{30} \times 10^{-3} = \dfrac{2}{3} \times 10^{-3}$ A
Let the new series resistance be $R$: $\dfrac{2}{3} \times 10^{-3} = \dfrac{3}{50 + R}$
$50 + R = \dfrac{3 \times 3}{2 \times 10^{-3}} = 4500\,\Omega$
$R = 4500 - 50 = 4450\,\Omega$
So the current for 30 divisions $= 10^{-3}$ A.
Deflection is proportional to current, so the current for 20 divisions $= \dfrac{20}{30} \times 10^{-3} = \dfrac{2}{3} \times 10^{-3}$ A
Let the new series resistance be $R$: $\dfrac{2}{3} \times 10^{-3} = \dfrac{3}{50 + R}$
$50 + R = \dfrac{3 \times 3}{2 \times 10^{-3}} = 4500\,\Omega$
$R = 4500 - 50 = 4450\,\Omega$