Physics - Current Electricity Question with Solution | TestHub

As potential drop across both branches would be the same, we get $I_{S}S=I_{g}G$

$\left(\frac{95}{100}I\right)S=\left(\frac{5I}{100}\right)G$

$\Rightarrow S=\frac{G}{19}$

Now, resistance of ammeter will be

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

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