A 10 V battery of negligible internal resistance is being charged by a 200 V battery with a resistance of \( 38\,\Omega \) in series. The value of the current in the circuit is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 5 A
While charging, the two batteries are connected positive to positive, so their emfs oppose each other in the loop. The net emf drives the current through the series resistance (Kirchhoff's loop rule).
\( \displaystyle I = \frac{E_1 - E_2}{R} = \frac{200 - 10}{38} = \frac{190}{38} = 5\,\mathrm{A} \)