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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 :

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$\frac{nKS}{(G + S)}$

$\frac{nK (G + S)}{S}$

$\frac{(G + S)}{nKS}$

$\frac{KS}{(S + G)}$

answer is D.

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Detailed Solution

$Figure of merit \frac{I_{g}}{θ} = K \begin{aligned} I_{g} = Kn \\ I = \frac{I_{g}}{s} (G + S) \\ I = \frac{nK}{S} (G + S) \end{aligned}$

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