---
title: "Nodal analysis current in 6 Ω"
subject: "Basic Electrical and Electronics Engineering"
chapter: 1
key: "semester_1/basic_electrical_and_electronics_engineering:nodal-analysis-6-ohm-current"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_and_electronics_engineering/nodal-analysis-6-ohm-current
site: BCT Engineering
---

# Nodal analysis current in 6 Ω

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

- Asked in: 2081 Kartik (new course), 6 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_and_electronics_engineering/nodal-analysis-6-ohm-current
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_and_electronics_engineering/chapter-1

## Question

Calculate the current flowing through 6 Ω resistor using Nodal analysis. [Figure: 2 A current source in parallel with a 12 Ω resistor R1 on the left, feeding node D; between D and A is 8 Ω (R2), between A and B is 4 Ω (R3), between D and B is 6 Ω (R4) as the bridge element, between D and C is 20 Ω (R5), between C and B is 10 Ω (R6); C is connected back to the source bottom rail.]

## Short answer

Set node C as reference. KCL equations yield VD = 9.767 V, VB = 6.977 V. Current through 6 Ω resistor = (VD - VB)/6 = 0.465 A, direction from D to B.

## Answer

**Circuit reading (from the figure).** Node C is joined to the bottom rail, so take C as the reference (0 V). The 2 A source and the 12 Ω resistor $R_1$ are both between node D and the rail. Between D and B there are two paths in parallel: the 6 Ω bridge $R_4$ and the series path D–A–B ($8+4=12\ \Omega$; node A has no other connection, so no separate equation is needed). The 20 Ω resistor joins D to the reference and the 10 Ω resistor joins B to the reference.

Unknowns: $V_D$ and $V_B$.

**KCL at node D** (2 A flows in, the rest leaves through the branches):

$$
2=\frac{V_D}{12}+\frac{V_D}{20}+\frac{V_D-V_B}{6}+\frac{V_D-V_B}{12}
$$

**KCL at node B:**

$$
\frac{V_B-V_D}{6}+\frac{V_B-V_D}{12}+\frac{V_B}{10}=0
$$

**Collect terms** (conductances in siemens):

$$
\begin{aligned} 0.3833\,V_D-0.25\,V_B&=2\\ -0.25\,V_D+0.35\,V_B&=0 \end{aligned}
$$

From the second equation, $V_B=0.7143\,V_D$. Substituting in the first:

$$
0.3833V_D-0.1786V_D=2 \Rightarrow V_D=9.767\ \text{V},\quad V_B=6.977\ \text{V}
$$

**Current through the 6 Ω resistor:**

$$
I_{6\Omega}=\frac{V_D-V_B}{6}=\frac{9.767-6.977}{6}=0.465\ \text{A}\ (D\to B)
$$

**Answer: $I_{6\Omega}=0.465$ A, from D to B.**

## Check yourself

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

1. Which node is chosen as the reference and what are the unknowns?
   A. Node D is the reference; unknowns VA, VB, VC
   B. Node C is the reference (0 V); unknowns VD and VB
   C. Node B is the reference; unknowns VA and VD
   D. Node A is the reference; unknowns VD and VC
   Answer: B. C is joined to the bottom rail, so it is 0 V; only VD and VB remain unknown. (Key point: Reference node)

2. Why does node A need no separate equation?
   A. It is at 0 V
   B. It is connected to the current source
   C. Its voltage always equals VD
   D. It has no other connection, so 8 Ω and 4 Ω act as one 12 Ω path from D to B
   Answer: D. A is only joined to the 8 Ω and 4 Ω resistors, so they carry the same current and combine as 12 Ω. (Key point: Series path D-A-B)

3. Which KCL equation applies at node B?
   A. (VB-VD)/6 + (VB-VD)/12 + VB/10 = 0
   B. (VB-VD)/6 + VB/10 = 2
   C. (VD-VB)/6 + (VB-VD)/12 + VB/10 = 0
   D. VB/6 + VB/12 + VB/10 = 0
   Answer: A. Currents leaving B go through 6 Ω and 12 Ω towards D and through 10 Ω to ground, summing to zero. (Key point: KCL at B)

4. With VD = 9.767 V and VB = 6.977 V, what is the current in the 6 Ω resistor?
   A. 1.628 A from B to D
   B. 0.465 A from D to B
   C. 0.465 A from B to D
   D. 2.79 A from D to B
   Answer: B. I = (VD - VB)/6 = 2.79/6 = 0.465 A, flowing from the higher-potential D to B. (Key point: Current in 6 Ω)

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