---
title: "Nodal analysis current through 10Ω"
subject: "Basic Electrical Engineering"
chapter: 2
key: "semester_1/basic_electrical_engineering:nodal-current-through-10-ohm-5a"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/nodal-current-through-10-ohm-5a
site: BCT Engineering
---

# Nodal analysis current through 10Ω

Basic Electrical Engineering · Chapter 2: DC circuits

- Asked in: 2073 Chaitra, 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_engineering/nodal-current-through-10-ohm-5a
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-2

## Question

Use the nodal voltage method to find the current flowing through 10Ω resistor for the network shown below. [Figure: 5 A current source on the left; 5 Ω shunt; 10 Ω in series with a 10 V source on top; 3 Ω shunt; 2 Ω in series with a 20 V source on the right.]

## Short answer

Define node 1 (left) and node 2 (right) with bottom rail as ground. Nodal equations: V1/5 + (V1-10-V2)/10 = 5, (V2-V1+10)/10 + V2/3 + (V2+20)/2 = 0. Solving gives V1=16.67 V, V2=-10 V. Current through 10 Ω resistor = (V1-10-V2)/10 = 1.667 A.

## Answer

#### Reading the figure
Bottom rail = ground.
- Node 1 (left): 5 A source into the node, 5 Ω to ground, and the branch 10 Ω in series with the 10 V battery (+ on the node 1 side).
- Node 2: reached through that branch; 3 Ω to ground, and the 2 Ω branch in series with the 20 V battery (− on the 2 Ω side, + to ground).

#### Nodal equations (currents leaving the node)
Node 1: V₁/5 + (V₁ − 10 − V₂)/10 = 5
Node 2: (V₂ − V₁ + 10)/10 + V₂/3 + (V₂ + 20)/2 = 0

Simplify:
(1) 3 V₁ − V₂ = 60
(2) −3 V₁ + 28 V₂ = −330 ... (multiply node 2 by 30: 3V₂ − 3V₁ + 30 + 10V₂ + 15V₂ + 300 = 0)

Adding: 27 V₂ = −270, so V₂ = −10 V; then V₁ = 50/3 = 16.67 V.

#### Current in the 10 Ω
I = (V₁ − 10 − V₂)/10 = (16.67 − 10 + 10)/10 = 1.667 A, from node 1 towards node 2.

**Answer: I₁₀Ω = 1.667 A (5/3 A)**

Assumption: battery polarities read from the plates and + marks in the figure (10 V: + on the node 1 side; 20 V: + on the ground side).

## Check yourself

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

1. Node 1 equation is
   A. V1/5 + (V1 - V2)/10 = 10
   B. V1/5 + (V1 + 10 - V2)/10 = 0
   C. V1/5 + (V1 - 10 - V2)/10 = 5
   D. V1/5 + (V1 - 10 - V2)/10 = 0
   Answer: C. The 5 A source enters node 1, so it goes on the right side. (Key point: Node 1 equation)

2. Multiplying node 1 by 10 gives
   A. 3 V1 + V2 = 60
   B. 3 V1 - V2 = 60
   C. V1 - V2 = 60
   D. 3 V1 - V2 = 5
   Answer: B. 2V1 + V1 - 10 - V2 = 50, giving 3 V1 - V2 = 60. (Key point: Simplified form)

3. Adding the two equations gives 27 V2 = -270. So
   A. V2 = -10 V, V1 = 16.67 V
   B. V2 = 10 V, V1 = 23.3 V
   C. V2 = -10 V, V1 = 0 V
   D. V2 = 27 V, V1 = 29 V
   Answer: A. V2 = -10 V; from (1), V1 = (60 - 10)/3 = 16.67 V. (Key point: Solving)

4. The current in the 10 Ω is
   A. (V1 - V2)/10 = 2.67 A
   B. (V1 + 10 - V2)/10 = 3.67 A
   C. V1/10 = 1.667 A with no battery
   D. (V1 - 10 - V2)/10 = 1.667 A
   Answer: D. The 10 V battery in series is subtracted: (16.67 - 10 + 10)/10 = 1.667 A. (Key point: Current in 10 Ω)

## Close variants set in other exams

- [Nodal analysis current through 10Ω](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Anodal-current-10-ohm-25v-50v) — Basic Electrical Engineering, chapter 2, asked 2×

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