Nodal analysis current in branch a-b

Assumptions (from the figure). Point c is the reference node (0 V). The 25 V source (+ at top) is connected directly from node a to c, so $V_a=25$ V. Node b connects to a through 10 Ω, to c through 4 Ω, and to the 50 V source (+ at top) through the 2 Ω resistor. The left branch (2 V source with 5 Ω) is across the fixed node a, so it only changes the current of the 25 V source and does not affect $V_b$.

KCL at node b (currents leaving b equal zero):

$$\frac{V_b-25}{10}+\frac{V_b}{4}+\frac{V_b-50}{2}=0$$

Multiply by 20:

$$2V_b-50+5V_b+10V_b-500=0$$ $$17V_b=550 \Rightarrow V_b=32.35\ \text{V}$$

Current in branch a–b (from a to b):

$$I_{ab}=\frac{V_a-V_b}{10}=\frac{25-32.35}{10}=-0.735\ \text{A}$$

The negative sign means 0.735 A actually flows from b to a.

Check: current in the 4 Ω resistor is $32.35/4=8.09$ A (b to c). Current from the 50 V source through 2 Ω into b is $(50-32.35)/2=8.82$ A. Current from b to a is 0.735 A. KCL at b: $8.82 = 8.09+0.735$ ✓.

Answer: $V_b=32.35$ V and $I_{ab}=0.735$ A, flowing from b to a.