JEE Main 2026 · 5 Apr · Shift 1, Physics question 34

Node potentials. Call the corners L (left), T (top) and B (bottom), and the midpoint of the vertical wire M.

• The 5 V cell joins T and B through wires with no resistance, so take $V_B=0$, $V_T=5$ V.
• The 10 V cell has its positive terminal towards L, so in the L-M branch $V_L=V_M+10-I_1(1)$, with $I_1$ flowing from M to L.

KCL at M.
$$\frac{5-V_M}2+\frac{0-V_M}2=I_1.$$

KCL at L.
$$\frac{V_L-5}4+\frac{V_L}4=I_1.$$

Solving. These give $V_M=0$, $V_L=7.5$ V and $I_1=2.5$ A.

Branch currents.

• $I_3=\dfrac{7.5-0}4=1.875$ A.
• Into T from L: $\dfrac{2.5}4=0.625$ A. Out of T through the upper 2 Ω: $\dfrac52=2.5$ A.
• So the 5 V branch supplies $I_2=2.5-0.625=1.875$ A.

(A)