A 10 V battery of negligible internal resistance is being charged by a 200 V battery with a resistance of 38 Ω 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
Effective voltage driving the current:
$V = 200 - 10 = 190 \; \text{V}$
Series resistance: $R = 38 \; \Omega$
Current: $I = \frac{V}{R} = \frac{190}{38} = 5 \; \text{A}$