A 10 V battery of negligible internal resistance is connected with a 200 V battery and a set of parallel resistances as shown in the figure. The value of current in the circuit.

15 A

The correct answer is Option (2) → 15 A

Three resistors of $38\ \Omega$ are connected in parallel, so the equivalent resistance is

$R_{\text{eq}}=\frac{38}{3}\ \Omega$

The two batteries oppose each other, so the net emf across the resistor network is

$V_{\text{net}}=200\ \text{V}-10\ \text{V}=190\ \text{V}$

Current in the circuit:

$I=\frac{V_{\text{net}}}{R_{\text{eq}}}=\frac{190}{38/3}=190\times\frac{3}{38}=(190/38)\times3=5\times3=15\ \text{A}$

Final Answer: $I = 15\ \text{A}$