In an ammeter, 5 % of the main current passes through the...

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

\left(\dfrac{95}{100}I\right)S = \left(\dfrac{5I}{100}\right)G$</p><p>$⇒ S = \dfrac{G}{19}$</p><p>Now, resistance of ammeter will be</p><p>${R}_{A} = \dfrac{SG}{S + G} = \dfrac{\dfrac{{G}^{2}}{19}}{\dfrac{20G}{19}}$</p><p>$⇒ {R}_{A} = \dfrac{G}{20}