Kirchhoff's Laws & Measuring Instruments question from JEE Main 2022 (29 Jun, Shift 1)

Answer: 10

1. Single loop: total resistance $= 1 + 2 = 3\ \Omega$, so $I = \dfrac{15}{3} = 5\text{ A}$.
2. The cell's positive plate faces B, so current leaves the cell towards B and returns through the $2\ \Omega$ resistor from B to A.
3. Current flows from higher to lower potential in a resistor: $V_B - V_A = I \times 2 = 5 \times 2$
4. $V_B - V_A = 10\text{ V}$