---
title: "Nodal analysis Va, Vb, Vc and 2.5Ω current"
subject: "Basic Electrical Engineering"
chapter: 2
key: "semester_1/basic_electrical_engineering:nodal-va-vb-vc-2-5-ohm"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/nodal-va-vb-vc-2-5-ohm
site: BCT Engineering
---

# Nodal analysis Va, Vb, Vc and 2.5Ω current

Basic Electrical Engineering · Chapter 2: DC circuits

- Asked in: 2072 Kartik, 8 marks
- Video: [Apply nodal analysis to determine Va, Vb, and Vc in the circuit - Innk](https://www.youtube.com/watch?v=gum3ZDbItPg) — Writton Innk
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/nodal-va-vb-vc-2-5-ohm
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-2

## Question

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.]

## 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.

## Answer

#### 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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Why are nodes a and c treated as a supernode?
   A. They are connected by a resistor
   B. They are joined by a current source
   C. They are joined by the 12 V battery whose current is unknown
   D. They have the same voltage
   Answer: C. An ideal voltage source between two non-reference nodes forms a supernode. (Key point: Supernode)

2. The battery gives the relation
   A. V_c = V_a + 12
   B. V_c = V_a - 12
   C. V_c = 12
   D. V_c = V_a/12
   Answer: B. The + terminal is at a, so c is 12 V lower than a. (Key point: Constraint)

3. Node b equation is
   A. (V_b - V_a)/2 + V_b/5 = 8
   B. (V_b - V_a)/2 + V_b/5 = 0
   C. (V_a - V_b)/2 - V_b/5 = 8 with no 5 Ω
   D. V_b/5 = 8
   Answer: A. The 8 A source enters b, so it equals the current leaving through the 2 Ω and 5 Ω. (Key point: Node b equation)

4. With V_c = 2.15 V, the current through 2.5 Ω is
   A. 5.4 A
   B. 2.15 A
   C. 14.15 A
   D. 0.862 A
   Answer: D. I = V_c/2.5 = 2.154/2.5 = 0.862 A. (Key point: Final current)

## Related in Basic Electrical Engineering

- [Nodal analysis, current through 15Ω](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Anodal-current-15-ohm-8a-48v) — Basic Electrical Engineering, chapter 2
- [Nodal analysis current through 10Ω](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Anodal-current-through-10-ohm-5a) — Basic Electrical Engineering, chapter 2, asked 2×
- [Nodal analysis current through 10Ω](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Anodal-current-10-ohm-25v-50v) — Basic Electrical Engineering, chapter 2, asked 2×
- [Nodal analysis current I](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Anodal-current-i-1a-18v-12v) — Basic Electrical Engineering, chapter 2
- [Nodal analysis currents in all resistors](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Anodal-currents-23a-2a-10-ohm) — Basic Electrical Engineering, chapter 2
- [Nodal analysis, power in 5Ω](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Anodal-power-5-ohm-5a-3a-40v-20v) — Basic Electrical Engineering, chapter 2

## Related in other subjects

- [Nodal analysis current in branch a-b](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_and_electronics_engineering%3Anodal-current-branch-ab) — Basic Electrical and Electronics Engineering, chapter 1
- [Nodal analysis current in 6 Ω](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_and_electronics_engineering%3Anodal-analysis-6-ohm-current) — Basic Electrical and Electronics Engineering, chapter 1
- [Nodal analysis: current in each branch](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Anodal-analysis-branch-currents-dependent) — Electrical Circuit Theory, chapter 1
- [Classical: Vc(t) with 12 V, 0.5 H, 2 F](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Aclassical-vc-12v-1ohm-05h-2f) — Electrical Circuit Theory, chapter 3