Kirchhoff's Laws & Measuring Instruments question from MHT CET 2021 (20 Sep, Shift 1)

Answer: (a)

1. Take $V_E = 0$. Reading the cells (long plate = +): $V_B = +6\text{ V}$, $V_A = V_F = -8\text{ V}$, $V_C = V_D = +12\text{ V}$.
2. Current A to B through $28\ \Omega$: $\frac{-8 - 6}{28} = -\frac{1}{2}\text{ A}$
3. Current C to B through $54\ \Omega$: $\frac{12 - 6}{54} = \frac{1}{9}\text{ A}$
4. Junction B: $I = -\frac{1}{2} + \frac{1}{9} = -\frac{7}{18}\text{ A}$
5. So the answer is (a) $-\frac{7}{18}$ A.