Current distribution in an ammeter

An ammeter is made by connecting a shunt resistance in parallel with a galvanometer.

Let the main current be $I$.

Given that $5\%$ of the main current passes through the galvanometer, $I_g = 0.05I = \frac{I}{20}$

Therefore, the current through the shunt is $I_s = I - I_g = I - \frac{I}{20} = \frac{19I}{20}$

Use parallel branch condition

Since galvanometer and shunt are in parallel, potential difference across both is same: $I_g G = I_s S$ where $S$ is the shunt resistance.

Substituting values: $\frac{I}{20}G = \frac{19I}{20}S$

Cancel $\frac{I}{20}$ from both sides: $G = 19S$ $S = \frac{G}{19}$

Find resistance of the ammeter

The ammeter resistance is the equivalent resistance of $G$ and $S$ in parallel: $R_A = \frac{GS}{G+S}$

Substitute $S = \frac{G}{19}$: $R_A = \frac{G\cdot \frac{G}{19}}{G + \frac{G}{19}}$

$R_A = \frac{\frac{G^2}{19}}{\frac{20G}{19}}$

$R_A = \frac{G}{20}$

Match with options

$R_A = \frac{G}{20}$

So the correct option is C.