A battery of $e m f 8 \mathrm{~V}$ and internal resistance $0.5 \Omega$ is being charged by a 120 V dc…
- 11.5 V
- 1.15 V
- 115 V
- 0.5 V
Solution

$I=\frac{120-8}{15.5+0.5}=7 \mathrm{~A}$ $\therefore \quad$ The potential difference across 8 V battery is $\mathrm{V}=\mathrm{E}+\mathrm{I}_{\mathrm{r}}=8+7 \times 0.5=11.5 \mathrm{~V}$
Asked in: AP EAMCET 2024 (22 May Shift 2)