---
title: "Nodal analysis, current through 15Ω"
subject: "Basic Electrical Engineering"
chapter: 2
key: "semester_1/basic_electrical_engineering:nodal-current-15-ohm-8a-48v"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/nodal-current-15-ohm-8a-48v
site: BCT Engineering
---

# Nodal analysis, current through 15Ω

Basic Electrical Engineering · Chapter 2: DC circuits

- Asked in: 2068 Chaitra, 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-15-ohm-8a-48v
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-2

## Question

Use Nodal analysis method to calculate the current through the 15Ω resistor in the figure shown below. [Figure: 15 Ω on the left; 8 Ω, 12 Ω and 10 Ω along the top; 6 Ω in series with an 8 A current source in the second branch; 5 Ω shunt; 48 V source with 2 Ω on the right.]

## Short answer

Set up nodal equations at three nodes: 23V₁−15V₂=0, −3V₁+5V₂−2V₃=−192, −5V₂+22V₃=240. Solutions: V₁=−42.87 V. Current through 15 Ω = V₁/15 = −2.86 A (magnitude 2.86 A, flowing upward).

## Answer

#### Reading the figure
The bottom rail is ground. Nodes:
- V₁: top of the 15 Ω (the 15 Ω hangs between V₁ and ground), joined by 8 Ω to V₂.
- V₂: junction of the 8 Ω, the branch 6 Ω + 8 A source (current leaving V₂, downward) and the 12 Ω.
- V₃: junction of the 12 Ω, the 5 Ω shunt and the 10 Ω going to the right.
- Right branch: 10 Ω, then 2 Ω in series with the 48 V battery (+ on top), so it behaves as a 48 V source with 12 Ω in series.

The 6 Ω is in series with the current source, so it does not affect the equations.

#### Nodal equations (currents leaving each node)
Node 1: V₁/15 + (V₁ − V₂)/8 = 0
Node 2: (V₂ − V₁)/8 + (V₂ − V₃)/12 + 8 = 0
Node 3: (V₃ − V₂)/12 + V₃/5 + (V₃ − 48)/12 = 0

Simplified (multiply by 120, 24 and 60):
- 23 V₁ − 15 V₂ = 0
- −3 V₁ + 5 V₂ − 2 V₃ = −192
- −5 V₂ + 22 V₃ = 240

From the first equation, V₁ = 15 V₂/23; from the third, V₃ = (240 + 5 V₂)/22. Substituting both into the second equation gives 655 V₂ = −43056, so

**V₂ = −65.73 V, V₁ = −42.87 V, V₃ = −4.03 V**

#### Current in the 15 Ω
I = V₁/15 = −42.87/15 = −2.858 A

The negative sign shows that the current flows from ground up through the 15 Ω towards node 1 (and on through the 8 Ω to the 8 A source).

**Answer: I₁₅Ω ≈ 2.86 A**

## Check yourself

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

1. Why does the 6 ohm not appear in the nodal equations?
   A. It is in series with a current source, which fixes the branch current
   B. It is shorted by the 8 A source
   C. It carries no current
   D. It is in parallel with the 48 V source
   Answer: A. A current source sets the branch current regardless of any series resistor. (Key point: Series element with current source)

2. How is the 10 ohm + 2 ohm + 48 V branch treated at node 3?
   A. As a 48 V source with 10 ohm only
   B. As a 48 V source in series with 12 ohm, giving the term (V3 - 48)/12
   C. As a 2 ohm shunt to ground
   D. As a 12 A current source
   Answer: B. The 10 ohm and 2 ohm add in series, so the branch is 48 V with 12 ohm. (Key point: Source transformation of right branch)

3. What are the node voltages found?
   A. V1 = 42.87 V, V2 = 65.73 V, V3 = 4.03 V
   B. V1 = -2.86 V, V2 = -65.73 V, V3 = 48 V
   C. V1 = -42.87 V, V2 = -65.73 V, V3 = -4.03 V
   D. V1 = 0, V2 = -8 V, V3 = 48 V
   Answer: C. Solving the three simplified equations gives these negative values. (Key point: Result)

4. How is the 15 ohm current found and what does its sign mean?
   A. I = V1/8 = -5.4 A, flowing down
   B. I = V1/15 = -2.86 A, flowing from node 1 down to ground
   C. I = (V1 - V2)/15 = 1.5 A
   D. I = V1/15 = -2.86 A, so current flows from ground up toward node 1
   Answer: D. The 15 ohm is between V1 and ground; a negative V1 means current flows up through it. Magnitude 2.86 A. (Key point: Current in 15 ohm)

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