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If only $1 \%$ of total current is passed through a galvanometer of resistance ' $\mathrm{G}$ ' then the resistance of the shunt is
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Verified Answer
The correct answer is:
$\frac{\mathrm{G}}{99} \Omega$
Given: $I_g=\frac{1}{100} I$
$\begin{aligned}
& \therefore \quad \frac{\mathrm{I}}{\mathrm{I}_{\mathrm{B}}}=100 \\
& \quad \mathrm{~S}=\frac{\mathrm{GI}_{\mathrm{B}}}{\mathrm{I}-\mathrm{I}_8}=\frac{\mathrm{G}}{\frac{\mathrm{I}}{\mathrm{I}_{\mathrm{B}}}-1} .
\end{aligned}$
Substituting, (i) intó (ii),
$\mathrm{S}=\frac{\mathrm{G}}{100-1}=\frac{\mathrm{G}}{99}$
$\begin{aligned}
& \therefore \quad \frac{\mathrm{I}}{\mathrm{I}_{\mathrm{B}}}=100 \\
& \quad \mathrm{~S}=\frac{\mathrm{GI}_{\mathrm{B}}}{\mathrm{I}-\mathrm{I}_8}=\frac{\mathrm{G}}{\frac{\mathrm{I}}{\mathrm{I}_{\mathrm{B}}}-1} .
\end{aligned}$
Substituting, (i) intó (ii),
$\mathrm{S}=\frac{\mathrm{G}}{100-1}=\frac{\mathrm{G}}{99}$
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