A 10 V battery having 3 Ω internal resistance is connected with a 200 V battery and a 35 Ω resistance as shown in the figure. The terminal potential drop at 10 V battery will be:

25 V

The correct answer is Option (2) → 25 V

Given:

Battery 1: $E_1 = 10 \, \text{V}$, internal resistance $r = 3 \, \Omega$

Battery 2: $E_2 = 200 \, \text{V}$

External resistor: $R = 35 \, \Omega$

Total series resistance: $R_\text{total} = R + r = 35 + 3 = 38 \, \Omega$

Assuming batteries are aiding: $E_\text{net} = E_1 + E_2 = 210 \, \text{V}$

Current in circuit: $I = \frac{E_\text{net}}{R_\text{total}} = \frac{210}{38} \approx 5.53 \, \text{A}$

Terminal potential difference across 10 V battery: $V = E_1 + I r = 10 + (5.53 \cdot 3) \approx 26.6 \, \text{V}$

Answer: $V \approx 25 \, \text{V}$