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A certain current on passing through a galvanometer produces a deflection of 100 divisions. When a shunt of one ohm is connected, the deflection reduces to 1 division. The galvanometer resistance is
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Verified Answer
The correct answer is:
$99 \Omega$
Shunt is connected to the galvanometer
$$
\begin{aligned}
i_{g} &=\frac{i S}{S+G} \\
1 &=\frac{100 \times 1}{(1+G)} \\
G &=99 \Omega
\end{aligned}
$$
$$
\Rightarrow \quad G=99 \Omega
$$
$$
\begin{aligned}
i_{g} &=\frac{i S}{S+G} \\
1 &=\frac{100 \times 1}{(1+G)} \\
G &=99 \Omega
\end{aligned}
$$
$$
\Rightarrow \quad G=99 \Omega
$$
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