Q79 / 90 · JEE Main 2024 (01 Feb Shift 2) · Physics

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:

Solution

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

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

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

Now, resistance of ammeter will be

$R_{A} = \frac{S G}{S + G} = \frac{\frac{G^{2}}{19}}{\frac{20 G}{19}}$

$\Rightarrow R_{A} = \frac{G}{20}$

Hence, the correct answer is option: 'A'.