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Only $4 \%$ of the total current in the circuit passed through a galvanometer. If the resistance of the galvanometer is $\mathrm{G}$, then the shunt resistance connected to the galvanometer is
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The correct answer is:
$\frac{\mathrm{G}}{24}$
Relationship between the shunt resistance and galvanometer resistance is
$\mathrm{S}=\frac{\mathrm{I}_0 \times \mathrm{G}}{\mathrm{I}-\mathrm{I}_{\mathrm{G}}}$ ....(i)
$\mathrm{I}_{\mathrm{g}}=\frac{4}{100} \mathrm{I}=0.04 \mathrm{I}$
$\therefore \quad \frac{\mathrm{I}}{\mathrm{I}_{\mathrm{g}}}=25$ ....(ii)
Putting (ii) into (i)
$\mathrm{S}=\frac{\mathrm{G}}{\frac{\mathrm{I}}{\mathrm{I}_{\mathrm{g}}}-1}=\frac{\mathrm{G}}{25-1}=\frac{\mathrm{G}}{24}$
$\mathrm{S}=\frac{\mathrm{I}_0 \times \mathrm{G}}{\mathrm{I}-\mathrm{I}_{\mathrm{G}}}$ ....(i)
$\mathrm{I}_{\mathrm{g}}=\frac{4}{100} \mathrm{I}=0.04 \mathrm{I}$
$\therefore \quad \frac{\mathrm{I}}{\mathrm{I}_{\mathrm{g}}}=25$ ....(ii)
Putting (ii) into (i)
$\mathrm{S}=\frac{\mathrm{G}}{\frac{\mathrm{I}}{\mathrm{I}_{\mathrm{g}}}-1}=\frac{\mathrm{G}}{25-1}=\frac{\mathrm{G}}{24}$
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