Kirchhoff's Laws & Measuring Instruments question from JEE Main 2021 (31 Aug, Shift 1)

Answer: 6

1. Upper branch: $(4\|4) + 2 + (15\|10) = 2 + 2 + 6 = 10\ \Omega$.
2. Lower branch: $(8\|8) + (12\|12) = 4 + 6 = 10\ \Omega$.
3. The two branches in parallel give $5\ \Omega$; with the $1\ \Omega$: $R = 6\ \Omega$, $I = \dfrac{12}{6} = 2\text{ A}$.
4. Each branch carries $1\text{ A}$, so the drop across the $15\ \Omega\|10\ \Omega$ pair is $1 \times 6$.
5. Voltage across $15\ \Omega$ $= 6\text{ V}$