Kirchhoff's Laws & Measuring Instruments question from JEE Main 2023 (11 Apr, Shift 2)

Answer: (b)

1. Between C and A: $R_3 = 6\ \Omega$ in parallel with $R_2 + R_1 = 6\ \Omega$ gives $3\ \Omega$.
2. D to A: $(R_4 + 3) = 6\ \Omega$ in parallel with $R_7 = 3\ \Omega$ gives $2\ \Omega$.
3. Path E–D–A: $R_5 + 2 = 4\ \Omega$ across $8$ V, so $V_{DA} = 8 \times \frac{2}{4} = 4$ V.
4. $V_{CA} = 4 \times \frac{3}{6} = 2$ V, so the current in $R_2$ (via $R_2 + R_1 = 6\ \Omega$) is $\frac{2}{6}$.
5. So the answer is (b) $\frac{1}{3}$ A.