---
title: "Nodal analysis branch currents"
subject: "Basic Electrical Engineering"
chapter: 2
key: "semester_1/basic_electrical_engineering:nodal-branch-currents-6v-5v-10v"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/nodal-branch-currents-6v-5v-10v
site: BCT Engineering
---

# Nodal analysis branch currents

Basic Electrical Engineering · Chapter 2: DC circuits

- Asked in: 2073 Shrawan, 8 marks
- Video: [Nodal Analysis for Circuits Explained](https://www.youtube.com/watch?v=f-sbANgw4fo) — Engineer4Free
- 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-branch-currents-6v-5v-10v
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-2

## Question

Find the branch currents in the circuit of figure below by using nodal analysis. [Figure: 6 V source with 6 Ω on the left; 5 V source with 2 Ω in the middle top; 4 Ω on the right top; 3 Ω shunt; 4 Ω and 10 V source on the right side.]

## Short answer

Using nodal analysis, node voltages are V1 = 1.333 V, V2 = 5.667 V. Branch currents computed via Ohm’s law: 0.778 A (6 V branch), 0.444 A (3 Ω shunt), 0.333 A (5 V branch), 1.417 A (4 Ω shunt), 1.083 A (10 V branch).

## Answer

#### Reading the figure
Bottom rail = ground (0 V). Two unknown nodes:
- Node 1: top of the 3 Ω shunt. Left branch: 6 Ω in series with the 6 V battery (+ at top).
- Node 2: top of the 4 Ω shunt. Right branch: 4 Ω in series with the 10 V battery (+ at top).
- Between 1 and 2: 5 V battery (+ on the 2 Ω side) in series with 2 Ω.

Polarities are read from the plate lengths in the figure.

#### Nodal equations (currents leaving the node)
Node 1: (V₁ − 6)/6 + V₁/3 + (V₁ + 5 − V₂)/2 = 0
⇒ 1.0 V₁ − 0.5 V₂ = 1 − 2.5 → V₁ − 0.5 V₂ = −1.5 ... (i)

Node 2: (V₂ − V₁ − 5)/2 + V₂/4 + (V₂ − 10)/4 = 0
⇒ −0.5 V₁ + 1.0 V₂ = 2.5 + 2.5 = 5 ... (ii)

Check of (i): multiply node 1 by 6: (V₁ − 6) + 2V₁ + 3 (V₁ + 5 − V₂) = 0 → 6 V₁ − 3 V₂ = −9, same as V₁ − 0.5 V₂ = −1.5.

Solving (i) and (ii): V₁ = 1.333 V, V₂ = 5.667 V

#### Branch currents
| Branch | Current |
|---|---|
| 6 V battery and 6 Ω (towards node 1) | (6 − 1.333)/6 = 0.778 A |
| 3 Ω shunt (node 1 down) | 1.333/3 = 0.444 A |
| 5 V battery and 2 Ω (node 1 to node 2) | (1.333 + 5 − 5.667)/2 = 0.333 A |
| 4 Ω shunt (node 2 down) | 5.667/4 = 1.417 A |
| 10 V battery and 4 Ω (towards node 2) | (10 − 5.667)/4 = 1.083 A |

KCL check at node 1: 0.778 = 0.444 + 0.333. At node 2: 0.333 + 1.083 = 1.417.

**Answer: 0.778 A, 0.444 A, 0.333 A, 1.417 A and 1.083 A as tabulated.**

## Check yourself

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

1. Which is the nodal equation at node 1?
   A. (V1 + 6)/6 + V1/3 + (V1 - V2)/2 = 0
   B. (V1 - 6)/6 + V1/3 = 0
   C. (V1 - 6)/6 + V1/3 + (V1 + 5 - V2)/2 = 0
   D. (V1 - 6)/6 + V1/3 + (V2 - V1)/2 = 5
   Answer: C. Sum the currents leaving node 1 through the 6 V branch, the 3 Ω shunt and the 5 V + 2 Ω branch. (Key point: Node 1 equation)

2. After simplification, the two nodal equations are
   A. V1 + 0.5 V2 = 1.5 and 0.5 V1 + V2 = 5
   B. V1 - 0.5 V2 = -1.5 and -0.5 V1 + V2 = 5
   C. V1 - V2 = -1.5 and V1 + V2 = 5
   D. 2 V1 - V2 = 1.5 and V1 - 2 V2 = 5
   Answer: B. Multiplying node 1 by 6 gives 6 V1 - 3 V2 = -9, i.e. V1 - 0.5 V2 = -1.5. (Key point: Simplified equations)

3. With V1 = 1.333 V, the 3 Ω shunt current is
   A. 0.444 A
   B. 0.778 A
   C. 4.0 A
   D. 1.333 A
   Answer: A. I = V1/3 = 1.333/3 = 0.444 A. (Key point: Current in the 3 Ω shunt)

4. The KCL check at node 1 is
   A. 0.778 A = 0.444 A - 0.333 A
   B. 1.417 A = 0.444 A + 0.333 A
   C. 0.778 A + 0.444 A = 0.333 A
   D. 0.778 A = 0.444 A + 0.333 A
   Answer: D. The current entering from the 6 V branch splits into the shunt and the 5 V branch. (Key point: KCL check)

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