When a galvanometer is shunted with a 4,Omega | vedclass

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(B) Let $G$ be the resistance of the galvanometer and $i$ be the main current.When shunted with $S_1 = 4\,\Omega$,the current through the galvanometer

Vedclass · Accepted Answer

$\frac{5}{13}$ of the deflection when shunted with $4\,\Omega$ only

Vedclass · Answer

(B) Let $G$ be the resistance of the galvanometer and $i$ be the main current.When shunted with $S_1 = 4\,\Omega$,the current through the galvanometer is $i_{g1} = i/5$.Using the shunt formula: $i_{g1} = i \cdot \frac{S_1}{G + S_1} \implies \frac{i}{5} = i \cdot \frac{4}{G + 4}$.Solving for $G$: $G + 4 = 20 \implies G = 16\,\Omega$.Now,the galvanometer is further shunted with $S_2 = 2\,\Omega$ in parallel with $S_1 = 4\,\Omega$.The equivalent shunt resistance $S_{eq}$ is: $\frac{1}{S_{eq}} = \frac{1}{4} + \frac{1}{2} = \frac{1+2}{4} = \frac{3}{4} \implies S_{eq} = \frac{4}{3}\,\Omega$.The new current through the galvanometer is $i_{g2} = i \cdot \frac{S_{eq}}{G + S_{eq}} = i \cdot \frac{4/3}{16 + 4/3} = i \cdot \frac{4/3}{52/3} = \frac{4}{52}i = \frac{1}{13}i$.The ratio of the new deflection to the previous deflection is $\frac{i_{g2}}{i_{g1}} = \frac{i/13}{i/5} = \frac{5}{13}$.Thus,the new deflection is $\frac{5}{13}$ of the deflection when shunted with $4\,\Omega$ only.

When a galvanometer is shunted with a $4\,\Omega$ resistance,the deflection is reduced to one-fifth. If the galvanometer is further shunted with a $2\,\Omega$ wire,the further reduction in the deflection will be (the main current remains same).

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