Question 343 of 380

A $72\Omega$ galvanometer is shunted by a resistance of $8\Omega$. The percentage of the total current which passes through the galvanometer is

A$0.1$B$10$C$25$D$0.25$

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Answer: (B) $10$

Let $G$ be resistance of galvanometer and $I'$ be the current passing through the galvanometer produces full scale deflection. If $I$ is the maximum current, and since, $G$ and $S$ are in parallel so, the potential difference between them will be same.
Then,
${I}^{'}G={I}_{2}S \Rightarrow 72{I}^{'}=8{I}_{2} \Rightarrow {I}_{2}=9{I}^{'} \Rightarrow I={I}^{'}+{I}_{2}=10{I}^{'}$
So the percentage of total current passes through galvanometer is
$\dfrac{{I}^{'}}{I}\times 100$$=\dfrac{{I}^{'}}{10{I}^{'}}\times 100=10$

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