---
title: "Mesh analysis, current in 5Ω"
subject: "Basic Electrical Engineering"
chapter: 2
key: "semester_1/basic_electrical_engineering:mesh-current-5-ohm-8v-10v-12v"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/mesh-current-5-ohm-8v-10v-12v
site: BCT Engineering
---

# Mesh analysis, current in 5Ω

Basic Electrical Engineering · Chapter 2: DC circuits

- Asked in: 2068 Chaitra, 6 marks
- Video: [Mesh Analysis for Circuits Explained](https://www.youtube.com/watch?v=eQpc2QRFv7Y) — 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/mesh-current-5-ohm-8v-10v-12v
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-2

## Question

Determine current in 5Ω resistor by mesh analysis in figure below. [Figure: 8 V source on the left; 20 Ω shunt; 10 V source on the top right; 3 Ω on the right; 2 Ω and 2 Ω resistors in the middle; 12 V source with 5 Ω in the bottom branch.]

## Short answer

Assign clockwise mesh currents I₁, I₂, I₃. Solve: 22I₁−20I₂−2I₃=8, −20I₁+25I₂−2I₃=10, −2I₁−2I₂+9I₃=12. The 5 Ω resistor carries I₃ = 560/167 ≈ 3.35 A.

## Answer

#### Reading the figure
- Left branch: 8 V battery (+ on top).
- Top middle: 20 Ω shunt between the top wire and the middle node M.
- Top right: 10 V battery (+ on the right) and 3 Ω in the right vertical.
- Middle row: 2 Ω (left) and 2 Ω (right) with the 20 Ω joining at M.
- Bottom branch: 12 V battery (+ on the left) in series with the 5 Ω.

Battery polarities are read from the plate lengths in the figure.

```text
 +--------+--[20]--+--(10V)--+
 |        |        |         |
(8V)     ...      ...       [3]
 |   I1   |   I2   |         |
 +--[2]---+---M----[2]-------+
 |            I3             |
 +--(12V)------[5]-----------+
```

#### Mesh currents (all clockwise)
- I₁: top left mesh (8 V, 20 Ω, 2 Ω)
- I₂: top right mesh (20 Ω, 10 V, 3 Ω, 2 Ω)
- I₃: bottom mesh (2 Ω, 2 Ω, 5 Ω, 12 V)

#### Mesh equations
Mesh 1: 8 = 20 (I₁ − I₂) + 2 (I₁ − I₃)  ⇒ 22 I₁ − 20 I₂ − 2 I₃ = 8
Mesh 2: 10 = 3 I₂ + 2 (I₂ − I₃) + 20 (I₂ − I₁)  ⇒ −20 I₁ + 25 I₂ − 2 I₃ = 10
Mesh 3: 12 = 2 (I₃ − I₁) + 2 (I₃ − I₂) + 5 I₃  ⇒ −2 I₁ − 2 I₂ + 9 I₃ = 12

#### Solution
I₁ = 4.679 A, I₂ = 4.411 A, I₃ = 3.353 A (Cramer's rule, determinant 1002)

The 5 Ω carries the mesh 3 current only.

**Answer: Current in the 5 Ω resistor = I₃ ≈ 3.35 A (560/167 A)**

## Check yourself

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

1. Which current is the current in the 5 ohm resistor?
   A. I1
   B. I2 - I3
   C. I1 + I2 + I3
   D. I3, since the 5 ohm lies only in mesh 3
   Answer: D. The 5 ohm is on the outer edge of mesh 3, so it carries I3 only. (Key point: Mesh current choice)

2. How is the voltage drop across a resistor shared by two meshes found?
   A. R times the difference of the two mesh currents
   B. R times the sum of the mesh currents
   C. R times the larger mesh current
   D. Zero
   Answer: A. Common-branch drop is R(Ia - Ib) for same-direction mesh currents, e.g. 20(I1 - I2). (Key point: Shared resistor)

3. What law is applied to each mesh to form the equations?
   A. KCL at each mesh
   B. KVL: sum of source voltages = sum of resistor drops
   C. Thevenin's theorem
   D. Superposition
   Answer: B. Mesh analysis writes KVL around each loop. (Key point: Method)

4. What is the 5 ohm current?
   A. 4.68 A
   B. 4.41 A
   C. About 3.35 A
   D. 1.00 A
   Answer: C. Cramer's rule gives I3 = 560/167, about 3.35 A. (Key point: Result)

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