In an ammeter, $0.25 \%$ of main current passes through the galvanometer. If the resistance of the galvanometer is ' $G$ ', the resistance of ammeter will be

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Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{I}=\mathrm{I}_{\mathrm{g}}+\mathrm{I}_{\mathrm{s}} \\ & \mathrm{I}=\frac{0.25 \mathrm{I}}{100}+\mathrm{I}_{\mathrm{s}} \\ & \mathrm{I}_{\mathrm{s}}=\frac{399}{400} \mathrm{I} ...(i)\\ & \mathrm{I}_{\mathrm{g}} \mathrm{G}=\mathrm{I}_{\mathrm{s}} \mathrm{~S} \\ & \mathrm{~S}=\frac{0.25}{100} \times \frac{400}{399} \times \mathrm{G} \\ & \mathrm{~S}=\frac{1}{399} \mathrm{G}...(ii) \end{aligned}$ ...[From (i)] Total resistance of ammeter $\begin{aligned} R & =\frac{G S}{G+S} \\ & =\frac{G\left(\frac{1}{399}\right)G}{\left(G+\frac{1}{399}\right) G} \\ & =\frac{\frac{1}{\left(\frac{399}{400}\right)}}{399} G \\ R & =\frac{1}{400} G \end{aligned}$ ...[From (ii)]

In an ammeter, $0.25 \%$ of main current passes through the galvanometer. If the resistance of the…

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