
In the circuit, the cells are having negligible resistances. If the galvanometer shows null defection then…

- $12 \mathrm{~V}$
- $6 \mathrm{~V}$
- $4 \mathrm{~V}$
- $2 \mathrm{~V}$
Solution

Apply the Kirchoff law $\begin{aligned} & -12+500 \mathrm{i}+100 \mathrm{i}=0 \\ & 600 \mathrm{i}=12 \\ & \mathrm{i}=\frac{12}{600}=0.02 \mathrm{~A} \\ & \mathrm{~V}=\mathrm{iR}=100 \times 0.02=2 \mathrm{~V} \end{aligned}$
Asked in: AP EAMCET 2023 (17 May Shift 1)