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If only $2 \%$ of the total current passes through an ammeter having coil of resistance ${ }^{\prime} \mathrm{R}^{\prime}$ then the resistance of the shunt of an ammeter is
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$\frac{\mathrm{R}}{49}$
$\frac{\mathrm{I}_{\mathrm{g}}}{\mathrm{I}}=\frac{2}{100}=\frac{1}{50}$
$\frac{\mathrm{I}_{\mathrm{g}}}{\mathrm{I}}=\frac{\mathrm{S}}{\mathrm{S}+\mathrm{R}} \quad \therefore \quad \frac{1}{50}=\frac{\mathrm{S}}{\mathrm{S}+\mathrm{R}}$
$\mathrm{S}+\mathrm{R}=50 \mathrm{~S}$
$\mathrm{R}=50 \mathrm{~S}, \mathrm{R}=49 \mathrm{~S}$ or $\mathrm{S}=\frac{\mathrm{R}}{49}$
$\frac{\mathrm{I}_{\mathrm{g}}}{\mathrm{I}}=\frac{\mathrm{S}}{\mathrm{S}+\mathrm{R}} \quad \therefore \quad \frac{1}{50}=\frac{\mathrm{S}}{\mathrm{S}+\mathrm{R}}$
$\mathrm{S}+\mathrm{R}=50 \mathrm{~S}$
$\mathrm{R}=50 \mathrm{~S}, \mathrm{R}=49 \mathrm{~S}$ or $\mathrm{S}=\frac{\mathrm{R}}{49}$
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