---
title: "Nodal analysis, currents in 5Ω and 3Ω"
subject: "Basic Electrical Engineering"
chapter: 2
key: "semester_1/basic_electrical_engineering:nodal-current-5-ohm-3-ohm-6a-4a"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/nodal-current-5-ohm-3-ohm-6a-4a
site: BCT Engineering
---

# Nodal analysis, currents in 5Ω and 3Ω

Basic Electrical Engineering · Chapter 2: DC circuits

- Asked in: 2067 Asoj, 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-current-5-ohm-3-ohm-6a-4a
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-2

## Question

Using nodal analysis find the current through the 5 ohm and 3 ohm resistors of the following circuit. [Figure: 6 A current source on top; 5 Ω and 4 Ω and 2 Ω in the upper part; 3 Ω with a 5 V source on the left; 4 A current source in the middle; 1 Ω with a 2 V source on the right.]

## Short answer

Nodal equations at L, R, M yield V_L ≈ 15.99 V, V_R ≈ 2.34 V. Then I₅Ω = (15.99−2.34)/5 = 2.73 A; I₃Ω = (15.99−5)/3 = 3.66 A.

## Answer

#### Reading the figure
Bottom rail = ground. Three nodes:
- L (left): 3 Ω in series with the 5 V battery (+ on top) to ground; 5 Ω to R; 2 Ω to M; the 6 A source (arrow left, from R to L) delivers current into L.
- R (right): 1 Ω in series with the 2 V battery (+ on top) to ground; 4 Ω to M; the 6 A source takes current out of R.
- M (middle): 2 Ω to L, 4 Ω to R, and the 4 A source (arrow up) injecting current into M from ground.

#### Nodal equations (currents leaving each node)
Node L: (V_L − 5)/3 + (V_L − V_R)/5 + (V_L − V_M)/2 = 6
Node R: (V_R − 2)/1 + (V_R − V_L)/5 + (V_R − V_M)/4 = −6
Node M: (V_M − V_L)/2 + (V_M − V_R)/4 = 4

Multiply by 30, 20 and 4:
- 31 V_L − 6 V_R − 15 V_M = 230
- −4 V_L + 29 V_R − 5 V_M = −80
- −2 V_L − V_R + 3 V_M = 16

Solving: **V_L = 15.99 V, V_R = 2.34 V, V_M = 16.77 V**

#### Required currents
- 5 Ω: I = (V_L − V_R)/5 = (15.99 − 2.34)/5 = **2.73 A** (from L to R)
- 3 Ω: I = (V_L − 5)/3 = (15.99 − 5)/3 = **3.66 A** (from L down into the 5 V battery)

Other branch currents: 2 Ω (L to M) = −0.39 A, 4 Ω (M to R) = 3.61 A, 1 Ω = 0.34 A.

**Answer: I₅Ω ≈ 2.73 A; I₃Ω ≈ 3.66 A**

## Check yourself

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

1. How is the 3 ohm + 5 V branch current written at node L?
   A. V_L/3
   B. (V_L + 5)/3
   C. (V_L - 5)/3
   D. 5/3
   Answer: C. The branch has V_L at one end and the 5 V battery (+ on top) in series, so the drop across 3 ohm is V_L - 5. (Key point: Branch with battery)

2. How does the 6 A source appear in the equation of node R?
   A. +6
   B. 0
   C. As a 6 ohm resistor
   D. As -6 on the right side, because it takes current out of R
   Answer: D. It is the opposite of its effect at L, where it injects current. (Key point: Current source direction)

3. What is the 5 ohm current?
   A. (V_L - V_R)/5 = 2.73 A, from L to R
   B. V_L/5 = 3.2 A
   C. (V_L - 5)/5 = 2.2 A
   D. V_R/5 = 0.47 A
   Answer: A. Voltage difference across the 5 ohm is 13.65 V. (Key point: Required current in 5 ohm)

4. What is the 3 ohm current?
   A. 5.33 A
   B. 3.66 A, from (V_L - 5)/3
   C. 15.99/3 = 5.33 A
   D. 1.28 A
   Answer: B. The drop across the 3 ohm is V_L minus the 5 V battery, 10.99 V. (Key point: Current in 3 ohm)

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