---
title: "Nodal analysis current in branch a-b"
subject: "Basic Electrical and Electronics Engineering"
chapter: 1
key: "semester_1/basic_electrical_and_electronics_engineering:nodal-current-branch-ab"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_and_electronics_engineering/nodal-current-branch-ab
site: BCT Engineering
---

# Nodal analysis current in branch a-b

Basic Electrical and Electronics Engineering · Chapter 1: Fundamentals of Electrical and Electronics Circuits

- Asked in: 2081 Chaitra (new course), 6 marks
- Video: [KCL in just 10 min with best and easy way (Nodal Analysis)](https://www.youtube.com/watch?v=g4HwMC0qU4g) — Zarrar Khan
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/basic_electrical_and_electronics_engineering/nodal-current-branch-ab
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_and_electronics_engineering/chapter-1

## Question

Determine the current through branch a-b of the circuit given below using Nodal Analysis. [Figure: top branch from a to b has 10 Ω, then b to right has 2 Ω; left side has 2 V source then 5 Ω down from a; a 25 V source (+ at top) in the middle-left branch; a 4 Ω resistor from b down to c; a 50 V source (+ at top) on the right side; c is the bottom common node.]

## Short answer

With node c as reference, V_a = 25 V. KCL at node b: (Vb-25)/10 + Vb/4 + (Vb-50)/2 = 0 gives Vb = 32.35 V. Current in a-b = (25 - 32.35)/10 = -0.735 A, i.e., 0.735 A flows from b to a.

## Answer

**Assumptions (from the figure).** Point c is the reference node (0 V). The 25 V source (+ at top) is connected directly from node a to c, so $V_a=25$ V. Node b connects to a through 10 Ω, to c through 4 Ω, and to the 50 V source (+ at top) through the 2 Ω resistor. The left branch (2 V source with 5 Ω) is across the fixed node a, so it only changes the current of the 25 V source and does not affect $V_b$.

**KCL at node b** (currents leaving b equal zero):

$$
\frac{V_b-25}{10}+\frac{V_b}{4}+\frac{V_b-50}{2}=0
$$

Multiply by 20:

$$
2V_b-50+5V_b+10V_b-500=0
$$

$$
17V_b=550 \Rightarrow V_b=32.35\ \text{V}
$$

**Current in branch a–b (from a to b):**

$$
I_{ab}=\frac{V_a-V_b}{10}=\frac{25-32.35}{10}=-0.735\ \text{A}
$$

The negative sign means 0.735 A actually flows from b to a.

**Check:** current in the 4 Ω resistor is $32.35/4=8.09$ A (b to c). Current from the 50 V source through 2 Ω into b is $(50-32.35)/2=8.82$ A. Current from b to a is 0.735 A. KCL at b: $8.82 = 8.09+0.735$ ✓.

**Answer: $V_b=32.35$ V and $I_{ab}=0.735$ A, flowing from b to a.**

## Check yourself

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

1. Why is Va = 25 V directly?
   A. Node a is the reference
   B. The 25 V source connects node a straight to the reference node c
   C. The 10 Ω resistor drops 25 V
   D. The 2 V source adds to 23 V
   Answer: B. An ideal voltage source between a and the reference fixes Va = 25 V. (Key point: Fixed node voltage)

2. How does the left branch (2 V source with 5 Ω) affect Vb?
   A. It does not, since it is across the fixed node a
   B. It lowers Vb by 2 V
   C. It adds 5 Ω to the 10 Ω branch
   D. It makes Vb equal to 2 V
   Answer: A. It only changes the 25 V source's current; node a stays at 25 V. (Key point: Effect of left branch)

3. Which equation gives Vb?
   A. (Vb-25)/10 + Vb/4 + (Vb-50)/2 = 0, so Vb = 32.35 V
   B. (Vb-25)/10 + Vb/4 = (Vb-50)/2
   C. (25-Vb)/10 + Vb/4 + (50-Vb)/2 = 0 with Vb = 8 V
   D. (Vb-25)/10 + Vb/4 + (Vb+50)/2 = 0
   Answer: A. Currents leaving b through 10 Ω, 4 Ω and 2 Ω sum to zero, giving 17 Vb = 550. (Key point: KCL at b)

4. I_ab = (25 - 32.35)/10 = -0.735 A. What does the sign mean?
   A. The calculation is wrong
   B. 0.735 A flows from a to b
   C. 0.735 A flows from b to a
   D. No current flows in branch a-b
   Answer: C. A negative value means the actual current is opposite to the assumed a to b direction. (Key point: Direction of Iab)

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