---
title: "Superposition, current IL through 2Ω"
subject: "Basic Electrical Engineering"
chapter: 3
key: "semester_1/basic_electrical_engineering:superposition-il-2-ohm-5a-10v-20v"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/superposition-il-2-ohm-5a-10v-20v
site: BCT Engineering
---

# Superposition, current IL through 2Ω

Basic Electrical Engineering · Chapter 3: Network Theorems

- Asked in: 2070 Asar, 8 marks
- Video: [Superposition Theorem Made Easy: Step-by-Step Circuit Analysis](https://www.youtube.com/watch?v=mT846tX8FFI) — Prof MAD
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/superposition-il-2-ohm-5a-10v-20v
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-3

## Question

Use Superposition theorem to find the current IL through 2 Ω resistors in figure below. [Figure: 5 A current source on the left with 5 Ω shunt; 10 Ω in series with a 10 V source on top; 3 Ω shunt; 2 Ω load carrying IL; 20 V source in the bottom branch.]

## Short answer

Apply superposition: with 5 A source alone, 0.926 A downward; with 10 V source, -0.370 A upward; with 20 V source, 4.444 A downward. Total IL = 5 A downward through the 2 Ω resistor.

## Answer

#### Reading the figure
Bottom rail = ground. Node 1: 5 A source (up) into node 1 and a 5 Ω shunt. From node 1, 10 Ω in series with the 10 V battery (+ on the 10 Ω side) to node 2. Node 2: 3 Ω shunt to ground and the 2 Ω load, whose bottom end is joined to ground through the 20 V battery (+ on the ground side). I_L flows downwards in the 2 Ω.

#### Case 1: 5 A alone (10 V and 20 V shorted)
Node 2 to ground: 3 Ω ∥ 2 Ω = 1.2 Ω
From node 1: 5 Ω ∥ (10 + 1.2) = 5 × 11.2/16.2 = 3.457 Ω
V₁ = 5 × 3.457 = 17.28 V
V₂ = 17.28 × 1.2/11.2 = 1.852 V
I′ = 1.852/2 = **0.926 A** (downwards)

#### Case 2: 10 V alone (5 A opened, 20 V shorted)
The 10 V source (with 10 Ω + 5 Ω = 15 Ω) feeds node 2. Its + is on the 10 Ω side, so it pulls node 2 negative:

V₂ (1/3 + 1/2 + 1/15) = −10/15 → V₂ = −0.7407 V
I″ = V₂/2 = **−0.370 A** (upwards)

#### Case 3: 20 V alone (5 A opened, 10 V shorted)
The 20 V battery has its + on the ground side, so the load's lower end is at −20 V:

V₂/3 + V₂/15 + (V₂ + 20)/2 = 0 → 0.9 V₂ = −10 → V₂ = −11.11 V
I‴ = (V₂ + 20)/2 = (−11.11 + 20)/2 = **4.444 A** (downwards)

#### Total
I_L = 0.926 − 0.370 + 4.444 = **5.0 A** (downwards through the 2 Ω)

(Check: 25/27 − 10/27 + 120/27 = 135/27 = 5 A.)

**Answer: I_L = 5 A**

## Check yourself

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

1. When the 5 A source acts alone in superposition, how are the 10 V and 20 V sources treated?
   A. Opened
   B. Replaced by 1 ohm resistors
   C. Replaced by short circuits
   D. Kept in the circuit
   Answer: C. Ideal voltage sources are shorted and current sources opened when they are switched off. (Key point: Handling other sources)

2. With only the 5 A source active, what is the resistance from node 2 to ground that is used first?
   A. 3 ohm in series with 2 ohm = 5 ohm
   B. 5 ohm in parallel with 10 ohm
   C. 3 ohm in parallel with 2 ohm = 1.2 ohm
   D. 10 ohm in parallel with 3 ohm
   Answer: C. The 3 ohm shunt and the 2 ohm load (to ground via the shorted 20 V) are in parallel, giving 1.2 ohm. (Key point: Case 1 method)

3. Why is the contribution due to the 10 V source negative (-0.370 A)?
   A. The 10 V source has a larger resistance
   B. Because current sources always give negative values
   C. Its positive terminal on the 10 ohm side pulls node 2 negative, so the load current flows upward
   D. Because it is applied last
   Answer: C. That source drives node 2 below ground, reversing the current direction in the 2 ohm relative to the assumed downward direction. (Key point: Sign of the contributions)

4. What is the final load current obtained by summing the three contributions algebraically?
   A. 0.926 - 0.370 - 4.444 = -3.89 A
   B. 0.926 + 0.370 + 4.444 = 5.74 A
   C. 0.926 - 0.370 + 4.444 = 5.0 A
   D. 4.444 - 0.926 - 0.370 = 3.15 A
   Answer: C. Contributions add algebraically with direction: 0.926 - 0.370 + 4.444 = 5.0 A downward. (Key point: Total current)

## Related in Basic Electrical Engineering

- [Superposition, current in branch AB](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Asuperposition-branch-ab-6v-12v) — Basic Electrical Engineering, chapter 3
- [Superposition theorem, current in RL](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Asuperposition-load-2-ohm-10v-5a) — Basic Electrical Engineering, chapter 3
- [Superposition theorem, 5 Ω current](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Asuperposition-5-ohm-15v-20v) — Basic Electrical Engineering, chapter 3
- [Nodal analysis current through 10Ω](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Anodal-current-through-10-ohm-5a) — Basic Electrical Engineering, chapter 2, asked 2×
- [Thevenin theorem, current in 10Ω](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Athevenin-current-10-ohm-20v-12v) — Basic Electrical Engineering, chapter 3
- [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×

## Related in other subjects

- [Superposition theorem and maximum power to RL](https://www.bctengineer.com/question.md?key=semester_1%2Ffundamental_of_electrical_and_electronics_engineering%3Asuperposition-max-power-transfer) — Fundamental of Electrical and Electronics Engineering, chapter 1
- [Superposition: branch currents, 6 V and 12 V](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Asuperposition-6v-12v) — Basic Electronics Engineering, chapter 1
- [Thevenin: current through 10 Ω](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Athevenin-10v-5a) — Basic Electronics Engineering, chapter 1
- [Laplace: current through 2 Ω, exponential source](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Alaplace-current-through-2ohm-exponential-source) — Electrical Circuit Theory, chapter 4