Question 344 of 380

If $n$ represents the actual number of deflections in a converted galvanometer of resistance $G$ and shunt resistance $S$. Then the total current $I$ when its figure of merit is $K$ will be

A$\dfrac{KS}{\left(S+G\right)}$B$\dfrac{\left(G+S\right)}{nKS}$C$\dfrac{nKS}{\left(G+S\right)}$D$\dfrac{nK\left(G+S\right)}{S}$

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Answer: (D) $\dfrac{nK\left(G+S\right)}{S}$

Let current through galvanometer be ${I}_{G}$ and current through shunt resistance be $I-{I}_{G}$.

Given figure of merit $K=\dfrac{{I}_{G}}{n}$
Then, total current, $I=\dfrac{{I}_{G}\left(G+S\right)}{S}$
$I=\dfrac{nK\left(G+S\right)}{S}$