The deflection in a galvanometer falls to lef | vedclass

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(D) Let $G$ be the resistance of the galvanometer and $I$ be the total current. When shunted by $S_1 = 3 \Omega$,the current through the galvanometer

Vedclass · Accepted Answer

$\left(\frac{1}{8.5}\right)^{th}$

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(D) Let $G$ be the resistance of the galvanometer and $I$ be the total current. When shunted by $S_1 = 3 \Omega$,the current through the galvanometer is $I_g = \frac{I}{4}$.Using the current divider rule: $I_g = I \left( \frac{S_1}{G + S_1} ight) \Rightarrow \frac{I}{4} = I \left( \frac{3}{G + 3} ight)$.Solving for $G$: $G + 3 = 12 \Rightarrow G = 9 \Omega$.Now,an additional shunt $S_2 = 2 \Omega$ is connected in parallel to $S_1 = 3 \Omega$. The equivalent shunt resistance $S_{eq}$ is:$S_{eq} = \frac{S_1 	imes S_2}{S_1 + S_2} = \frac{3 	imes 2}{3 + 2} = \frac{6}{5} = 1.2 \Omega$.The new current through the galvanometer $I_g'$ is:$I_g' = I \left( \frac{S_{eq}}{G + S_{eq}} ight) = I \left( \frac{1.2}{9 + 1.2} ight) = I \left( \frac{1.2}{10.2} ight) = I \left( \frac{12}{102} ight) = I \left( \frac{1}{8.5} ight)$.Thus,the deflection falls to $\left(\frac{1}{8.5}ight)^{th}$ of the initial value.

The deflection in a galvanometer falls to $\left(\frac{1}{4}\right)^{th}$ of its initial value when it is shunted by a resistor of $3 \Omega$. If an additional shunt of $2 \Omega$ is connected in parallel to the earlier shunt,the deflection in the galvanometer falls to:

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