In an ammeter, $5 \%$ of the main current passes through the galvanometer. If resistance of the galvanometer is G, the resistance of ammeter will be :
JEE MAIN 2024, Diffcult
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$\mathrm{I}_{\mathrm{S}} \mathrm{S}=\mathrm{I}_{\mathrm{g}} \mathrm{G}$
$\frac{95}{100} \mathrm{IS}=\frac{5 \mathrm{I}}{100} \mathrm{G}$
$\mathrm{S}=\frac{\mathrm{G}}{19}$
$\mathrm{R}_{\mathrm{A}}=\frac{\mathrm{SG}}{\mathrm{S}+\mathrm{G}}=\frac{\frac{\mathrm{G}^2}{19}}{\frac{20 \mathrm{G}}{19}}$
$\mathrm{R}_{\mathrm{A}}=\frac{\mathrm{G}}{20}$
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