---
title: "Loop current method, current in 10Ω"
subject: "Basic Electrical Engineering"
chapter: 2
key: "semester_1/basic_electrical_engineering:loop-current-10-ohm-10v-5v-20v"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/loop-current-10-ohm-10v-5v-20v
site: BCT Engineering
---

# Loop current method, current in 10Ω

Basic Electrical Engineering · Chapter 2: DC circuits

- Asked in: 2069 Chaitra, 8 marks
- Video: [Mesh Analysis or Maxwell’s Loop Current Method | Electrical Engineering](https://www.youtube.com/watch?v=E5M4WvJds_A) — Electrical and Electronics Engineering
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/loop-current-10-ohm-10v-5v-20v
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-2

## Question

Find the current through the 10 Ω resistor using loop-current method? [Figure: 10 V source on the left; 4 Ω, 2 Ω and 5 Ω along the top; 10 Ω shunt in series with a 5 V source; 6 Ω shunt; 20 V source on the right.]

## Short answer

Solving three mesh equations yields I₁ = 0.137 A, I₂ = −0.308 A. Current in the 10 Ω resistor is I₁ − I₂ = 0.445 A, flowing downward.

## Answer

#### Reading the figure
Three meshes side by side. All three batteries have the + terminal on top.
- Left mesh: 10 V, 4 Ω, then the shunt branch (10 Ω in series with the 5 V battery).
- Middle mesh: the 2 Ω on top, the 6 Ω shunt on the right.
- Right mesh: 5 Ω and the 20 V battery.

#### Loop currents (all clockwise)
I₁ (left), I₂ (middle), I₃ (right).

Loop 1: 10 = 4 I₁ + 10 (I₁ − I₂) + 5
⇒ 14 I₁ − 10 I₂ = 5

Loop 2 (going up through the 5 V from − to +, a rise of 5 V):
5 = 2 I₂ + 6 (I₂ − I₃) + 10 (I₂ − I₁)
⇒ −10 I₁ + 18 I₂ − 6 I₃ = 5

Loop 3 (going down through the 20 V from + to −, a drop of 20 V):
0 = 5 I₃ + 20 + 6 (I₃ − I₂)
⇒ −6 I₂ + 11 I₃ = −20

#### Solution
I₁ = 0.137 A, I₂ = −0.308 A, I₃ = −1.986 A

#### Current in the 10 Ω
I = I₁ − I₂ = 0.137 + 0.308 = 0.445 A, flowing downward (top to bottom)

**Answer: I₁₀Ω ≈ 0.445 A**

## Check yourself

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

1. Loop 1 gives
   A. 14 I1 - 10 I2 = 5
   B. 14 I1 + 10 I2 = 5
   C. 4 I1 - 10 I2 = 15
   D. 14 I1 - 10 I2 = 15
   Answer: A. 10 = 4 I1 + 10(I1 - I2) + 5, since 5 V is a drop. (Key point: Loop 1 equation)

2. In loop 2 the 5 V source appears on the left (5 = ...) because
   A. it is a drop
   B. it is a current source
   C. it is ignored
   D. going up through it from - to + is a rise
   Answer: D. Direction opposite to loop 1 gives opposite sign. (Key point: Loop 2 sign)

3. The 10 Ω current is
   A. I1 + I2
   B. I2 - I3
   C. I1 - I2
   D. I3
   Answer: C. It is shared by loops 1 and 2. (Key point: Shared resistor)

4. With I1 = 0.137 and I2 = -0.308 the current is
   A. 0.171 A
   B. 0.445 A downward
   C. 0.137 A
   D. 1.986 A
   Answer: B. 0.137 + 0.308 = 0.445. (Key point: Result)

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