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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, then the deflection reduces to 1 division. The galvanometer resistance is
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Verified Answer
The correct answer is:
$99 \Omega$
Given, shunt resistance, $R_{S}=1 \Omega$
Total current, $I=100 \mathrm{~A}$
Galvanometer current, $I_{g}=1 \mathrm{~A}$
We know that,
$$
\begin{array}{ll}
& I_{g}=\frac{I R_{\mathrm{S}}}{R_{\mathrm{S}}+G} \\
\Rightarrow & 1=\frac{100 \times 1}{1+G} \\
\Rightarrow & 1+G=100 \\
\Rightarrow & G=99 \Omega
\end{array}
$$
Total current, $I=100 \mathrm{~A}$
Galvanometer current, $I_{g}=1 \mathrm{~A}$
We know that,
$$
\begin{array}{ll}
& I_{g}=\frac{I R_{\mathrm{S}}}{R_{\mathrm{S}}+G} \\
\Rightarrow & 1=\frac{100 \times 1}{1+G} \\
\Rightarrow & 1+G=100 \\
\Rightarrow & G=99 \Omega
\end{array}
$$
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