In a modified ammeter, precisely $0.5\%$ of the main circuit current passes through the galvanometer

Question

In a modified ammeter, precisely $0.5\%$ of the main circuit current passes through the galvanometer coil. If the resistance of the galvanometer is $G$, the equivalent resistance of the entire ammeter will be

Professor PC Thomas Classes · Answer

(b) : The current through the galvanometer is $I_g = 0.5\% 	ext{ of } I = \frac{0.5}{100}I = \frac{I}{200}$. The remaining current flows through the shunt: $I_s = I - I_g = I - \frac{I}{200} = \frac{199I}{200}$. Since they are in parallel, $I_g G = I_s S \implies \left(\frac{I}{200}ight)G = \left(\frac{199I}{200}ight)S$, which means $S = \frac{G}{199}$. The equivalent resistance of the ammeter $R_A$ is the parallel combination of $G$ and $S$: $R_A = \frac{GS}{G+S} = \frac{G(G/199)}{G + G/199} = \frac{G^2/199}{200G/199} = \frac{G}{200}$.