JEE Main 2016 — Current Electricity Question with Solution

As shown in figure a shunt resistance $\left(S\right)$ is connected parallel to the galvanometer to convert it into an ammeter. Let $G$ is the resistance of the galvanometer.

Where $G = 100 \Omega$ and full-scale deflection current is $I_g = 10 mA = 10\times {10}^{-3} A$

For galvanometer $V_g = I_{g }G = 10\times {10}^{-3} \times 10 = 0.1 volt$

From the figure, the shunt is parallel to galvanometer

$\therefore$$V_S = V_g \Rightarrow I_S S = V_g ...\left(1\right)$

For ammeter full-scale deflection current is 

$I_{S }\left(=I-I_g\right) \approx 10 A$

$\therefore$ from equation $\left(1\right)$

$10 S = 0·1 \Rightarrow S = 0·01 \Omega$