Find the current through 4 Ω resistance using Nodal analysis[Circuit: a common bottom wire (reference). Left branch: a 6 V battery (longer plate, +, at the top) in series with a 3 Ω resistor, going up from the bottom wire to the top-left corner; the…

Assumption: the direction of the 5 A source is not marked in the figure; it is taken as delivering 5 A upward into node B (the usual form of this problem).

Reference = bottom wire. Unknowns V_A, V_B. The 6 V source with 3 Ω feeds node A: current from A into that branch = (V_A − 6)/3.

KCL at A (sum of currents leaving = 0): \( \dfrac{V_A-6}{3} + \dfrac{V_A}{6} + \dfrac{V_A-V_B}{4} = 0 \)

×12: \( 4V_A - 24 + 2V_A + 3V_A - 3V_B = 0 \Rightarrow 9V_A - 3V_B = 24 \Rightarrow 3V_A - V_B = 8 \) … (1)

KCL at B (5 A enters): \( \dfrac{V_B-V_A}{4} + \dfrac{V_B}{2} = 5 \)

×4: \( V_B - V_A + 2V_B = 20 \Rightarrow -V_A + 3V_B = 20 \) … (2)

From (1): V_B = 3V_A − 8. Substituting in (2): −V_A + 9V_A − 24 = 20 ⇒ 8V_A = 44 ⇒ V_A = 5.5 V, V_B = 8.5 V.

Current through 4 Ω: \( I = \dfrac{V_B - V_A}{4} = \dfrac{8.5-5.5}{4} = \) 0.75 A, flowing from B to A.

Check at A: (5.5 − 6)/3 + 5.5/6 + (5.5 − 8.5)/4 = −0.1667 + 0.9167 − 0.75 = 0 ✓. At B: 0.75 + 4.25 = 5 A ✓.

(If the 5 A source were directed downward, the same method gives V_A = 0.5 V, V_B = −6.5 V and I = (0.5 − (−6.5))/4 = 1.75 A from A to B.)