---
title: "Nodal analysis current I"
subject: "Basic Electrical Engineering"
chapter: 2
key: "semester_1/basic_electrical_engineering:nodal-current-i-1a-18v-12v"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/nodal-current-i-1a-18v-12v
site: BCT Engineering
---

# Nodal analysis current I

Basic Electrical Engineering · Chapter 2: DC circuits

- Asked in: 2069 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-i-1a-18v-12v
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-2

## Question

Find the current I in the circuit of figure below by applying nodal voltage method. [Figure: 1 A current source across the top 3 Ω resistor carrying I; 6 Ω on the left in series with an 18 V source; 6 Ω shunt; 6 Ω shunt; 6 Ω on top right; 12 V source on the right.]

## Short answer

Nodal analysis gives node voltages V₁ = 9 V, V₂ = 6 V. Current I through the 3 Ω resistor is (V₁ − V₂)/3 = 1 A.

## Answer

#### Reading the figure
Bottom rail = ground. Two unknown nodes:
- V₁: junction of the left 6 Ω (to the 18 V battery, + on top), the 6 Ω to ground and the 3 Ω.
- V₂: junction of the 3 Ω, the 6 Ω to ground and the right 6 Ω (to the 12 V battery, + on top).
- The 1 A source is in parallel with the 3 Ω and pushes current from node 2 to node 1 (arrow to the left).
- I is the current in the 3 Ω from node 1 to node 2.

#### Nodal equations (currents leaving the node)
Node 1: (V₁ − 18)/6 + V₁/6 + (V₁ − V₂)/3 − 1 = 0
Multiply by 6: V₁ − 18 + V₁ + 2 V₁ − 2 V₂ − 6 = 0
⇒ 4 V₁ − 2 V₂ = 24 ... (1)

Node 2: (V₂ − 12)/6 + V₂/6 + (V₂ − V₁)/3 + 1 = 0
Multiply by 6: V₂ − 12 + V₂ + 2 V₂ − 2 V₁ + 6 = 0
⇒ −2 V₁ + 4 V₂ = 6 ... (2)

#### Solution
From (1): V₂ = 2 V₁ − 12. In (2): −2 V₁ + 8 V₁ − 48 = 6, so V₁ = 9 V and V₂ = 6 V.

#### Current I
I = (V₁ − V₂)/3 = (9 − 6)/3 = 1 A

**Answer: I = 1 A**

## Check yourself

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

1. The 1 A source pushes current into node 1, so in the node 1 equation
   A. - 1 appears on the left
   B. +1 on the left
   C. it is ignored
   D. it is divided by 3
   Answer: A. Current entering reduces the net leaving. (Key point: Source in node 1)

2. Node 1 simplified is
   A. 4 V1 + 2 V2 = 24
   B. 2 V1 - V2 = 24
   C. 4 V1 - 2 V2 = 0
   D. 4 V1 - 2 V2 = 24
   Answer: D. Multiplying by 6. (Key point: Simplified)

3. The nodal solution is
   A. V1 = 6 V, V2 = 9 V
   B. V1 = 18, V2 = 12
   C. V1 = 9 V, V2 = 6 V
   D. V1 = 3, V2 = 0
   Answer: C. V2 = 2V1 - 12, then V1 = 9. (Key point: Solution)

4. I is
   A. the 1 A source current
   B. the current in the 3 Ω only: (9 - 6)/3 = 1 A
   C. V1/6
   D. (V1 + V2)/3
   Answer: B. It happens to equal 1 A. (Key point: Meaning of I)

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