JEE Main 2015 · 4 Apr · Shift 1, Physics question 18

Take Q at 0 and P at V. The 6 V cell has its negative plate towards P; the 9 V cell has its positive plate at the top. Currents from P to Q:

• Left path (6 V and 3 Ω): $\dfrac{V+6}3$.
• Right path (2 Ω, 9 V, 3 Ω): $\dfrac{V-9}5$.
• Through 1 Ω: $V$.

They sum to zero: $5(V+6)+3(V-9)+15V=0$, so $V=-\frac3{23}\approx-0.13$ V. The current in 1 Ω is 0.13 A from Q to P.