---
title: "Nodal analysis: current in each branch"
subject: "Electrical Circuit Theory"
chapter: 1
key: "semester_3/electrical_circuit_theory:nodal-analysis-branch-currents-dependent"
source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/nodal-analysis-branch-currents-dependent
site: BCT Engineering
---

# Nodal analysis: current in each branch

Electrical Circuit Theory · Chapter 1: Network Analysis of AC circuit & Dependent Sources

- Asked in: 2070 Asar (old course), 8 marks
- Video: [Using node voltage analysis, find current in each branch of the circuit.](https://www.youtube.com/watch?v=LKDbFawhrd0) — Engg-Course-Made-Easy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/nodal-analysis-branch-currents-dependent
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-1

## Question

Find the current in each branch of the following figure using nodal analysis.
[Figure: 10 V source (+ at top) in series with a 3 Ω resistor feeding node N1. From N1: a 2 Ω resistor in series with a dependent voltage source of value I2 (+ at top) going down to the bottom wire; and a 5 Ω resistor carrying current I2 to node N2. From N2: a 4 Ω resistor to the bottom wire and a 2 A current source (arrow pointing up into N2) to the bottom wire.]

## Short answer

With V1, V2, I2 = (V1−V2)/5, nodal analysis gives V1 = 4.25 V, V2 = 6.333 V, I2 = −0.4167 A. Branch currents: 3 Ω source branch 1.917 A, 2 Ω + dependent source 2.333 A, 5 Ω I2 = −0.417 A, 4 Ω 1.583 A.

## Answer

**Assumptions from the figure:** 10 V source (+ at top) with 3 Ω feeds node N1. From N1, the branch 2 Ω in series with a dependent voltage source of value I₂ volts (+ at top) goes to ground. A 5 Ω resistor carries I₂ from N1 to N2. At N2, a 4 Ω resistor goes to ground and a 2 A source pushes current into N2.

#### Nodal equations (V₁ at N1, V₂ at N2)

Control variable: I₂ = (V₁ − V₂)/5.

Node N2 (current in = current out):

```text
I₂ + 2 = V₂/4
(V₁ − V₂)/5 + 2 = V₂/4        ... (1)
```

Node N1: current from source branch = current in 2 Ω branch + I₂. The 2 Ω branch current is I_b = (V₁ − I₂)/2:

```text
(10 − V₁)/3 = (V₁ − I₂)/2 + I₂        ... (2)
```

#### Solving

From (1): I₂ = V₂/4 − 2, and V₁ = V₂ + 5I₂ = 2.25V₂ − 10.

Put in (2): 2(10 − V₁) = 3(V₁ + I₂) → 20 = 5V₁ + 3I₂

```text
20 = 5(2.25V₂ − 10) + 3(V₂/4 − 2) = 12V₂ − 56
V₂ = 6.333 V
V₁ = 4.25 V
I₂ = 6.333/4 − 2 = −0.4167 A
```

#### Branch currents

| Branch | Current |
|---|---|
| 3 Ω (source branch) | (10 − 4.25)/3 = 1.917 A |
| 2 Ω + dependent source | (4.25 + 0.4167)/2 = 2.333 A |
| 5 Ω (I₂) | −0.4167 A (flows N2 to N1) |
| 4 Ω | 6.333/4 = 1.583 A |
| 2 A source | 2 A |

Check: 1.917 = 2.333 − 0.4167 at N1 ✓; 1.583 = 2 − 0.4167 at N2 ✓.

**Answer: I(3 Ω) = 1.917 A, I(2 Ω) = 2.333 A, I₂ = −0.417 A, I(4 Ω) = 1.583 A.**

## Check yourself

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

1. How is I2 expressed in nodal voltages?
   A. I2 = V2/4
   B. I2 = (10 − V1)/3
   C. I2 = V1/5
   D. I2 = (V1 − V2)/5
   Answer: D. I2 flows through the 5 Ω resistor between N1 and N2. (Key point: Controlling variable)

2. Which equation holds at N2?
   A. I2 + 2 = V2/4
   B. I2 = V2/4 + 2
   C. I2 + V2/4 = 2
   D. I2 − 2 = V2/4
   Answer: A. Currents in (I2 and the 2 A source) equal the current out through 4 Ω. (Key point: KCL at N2)

3. I2 = −0.417 A means...
   A. the actual current flows from N1 to N2
   B. the actual current flows from N2 to N1
   C. the 5 Ω resistor is open
   D. the calculation failed
   Answer: B. A negative value means opposite to the assumed direction. (Key point: Negative I2)

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