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Nodal analysis Va, Vb, Vc and 2.5Ω current

Short answer

Treat nodes a and c as a supernode with Vc = Va-12. Node b equation: (Vb-Va)/2 + Vb/5 = 8. Supernode equation: Va/5 + (Va-Vb)/2 + Vc/2.5 = 0. Solving yields Va=14.15 V, Vb=21.54 V, Vc=2.15 V. Current through 2.5 Ω = 0.862 A.

Use Nodal Analysis Method to determine the Va, Vb and Vc and calculate current through 2.5 Ω. [Figure: 12 V source on top between nodes Va and Vb? with 2 Ohm resistor between Va and Vb; 8 A current source between Vb and Vc; Va has 5 Ohm shunt in series with a 5 A current source; Vb has a 5 Ohm shunt; Vc has a 2.5 Ohm shunt in series with a 3 A current source.]

Reading the figure

Bottom rail = ground. Nodes a, b, c:

  • a: 5 Ω to ground, 5 A source into a, 2 Ω to b, and the 12 V battery to c (+ on the a side, so V_a = V_c + 12).
  • b: 5 Ω to ground, the 8 A source from c to b (into b).
  • c: 2.5 Ω to ground, 3 A source into c.

Because the battery joins a and c, treat a and c as one supernode with V_c = V_a − 12.

Equations

Node b (currents leaving b = 8 A coming in): (V_b − V_a)/2 + V_b/5 = 8 ⇒ −0.5 V_a + 0.7 V_b = 8 ... (1)

Supernode (a + c): currents leaving = sources entering (5 + 3 − 8 = 0) V_a/5 + (V_a − V_b)/2 + V_c/2.5 = 0 Put V_c = V_a − 12: 0.2 V_a + 0.5 V_a − 0.5 V_b + 0.4 V_a − 4.8 = 0 ⇒ 1.1 V_a − 0.5 V_b = 4.8 ... (2)

Solution

From (1) and (2): V_a = 14.15 V, V_b = 21.54 V V_c = V_a − 12 = 2.15 V

Current through 2.5 Ω

I = V_c/2.5 = 2.154/2.5 = 0.862 A

Answer: V_a = 14.15 V, V_b = 21.54 V, V_c = 2.15 V; I₂.₅Ω = 0.862 A (towards ground)

Assumption: 12 V battery has its + terminal at node a, as read from the plate lengths.

✓ Answer checked by Claude Opus 5.5 (high effort)

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