A galvanometer has resistance $80 \Omega$ and it is shunted with resistance $20 \Omega$. If $20 \%$ of the…
- 0.2 A
- 0.8 A
- 1 A
- 1.2 A
Solution

$\begin{aligned} & I=I_1+I_2 \\ & I_1=0.2 \mathrm{I} \quad \therefore \quad \mathrm{I}_2=0.8 \mathrm{I} \\ & \text { Also, } \mathrm{I}_1 \mathrm{G}=\mathrm{I}_2 \mathrm{~S} \\ \therefore \quad & 0.2 \mathrm{I} \times 80=0.8 \mathrm{I} \times 20 \\ \therefore \quad & \mathrm{I}=1 \mathrm{~A}\end{aligned}$
Asked in: MHT CET 2024 (15 May Shift 1)