---
title: "Current in 5Ω using Thevenin theorem"
subject: "Basic Electrical Engineering"
chapter: 3
key: "semester_1/basic_electrical_engineering:thevenin-5-ohm-cd-1a-8v"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/thevenin-5-ohm-cd-1a-8v
site: BCT Engineering
---

# Current in 5Ω using Thevenin theorem

Basic Electrical Engineering · Chapter 3: Network Theorems

- Asked in: 2072 Kartik, 8 marks
- Video: [Thevenin's Theorem | Engineering Circuit Analysis | Solved Examples](https://www.youtube.com/watch?v=jpO37lVBe-o) — Electrical Electronics Applications
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/thevenin-5-ohm-cd-1a-8v
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-3

## Question

Find out the current through 5 ohm resistor connected across the terminal c and d in the network shown below using the Thevenin's theorem. [Figure: 1 A current source on the left with 2 Ω shunt; 5 Ω between terminals c and d; 2 Ω shunt after d; 2 Ω in series with an 8 V source on the right.]

## Short answer

Thevenin voltage is −2 V and Thevenin resistance is 3 Ω. The current through the 5 Ω resistor is 2/(3+5) = 0.25 A, flowing from terminal d to c.

## Answer

#### Reading the figure
Bottom rail = ground. The 1 A source (arrow up) feeds node c. c has a 2 Ω shunt. The 5 Ω is connected between c and d. d has a 2 Ω shunt, and a branch with 2 Ω in series with the 8 V battery (+ on top, towards the 2 Ω).

#### Step 1: Remove the 5 Ω; find V_th across c-d
- V_c = 1 A × 2 Ω = 2 V
- Node d: the 8 V source with 2 Ω and the 2 Ω shunt form a divider:
 V_d = 8 × 2/(2 + 2) = 4 V

V_th = V_c − V_d = 2 − 4 = −2 V (d is 2 V higher than c)

#### Step 2: Thevenin resistance (sources off)
1 A opened, 8 V shorted:
- At c: 2 Ω
- At d: 2 Ω ∥ 2 Ω = 1 Ω

R_th = 2 + 1 = **3 Ω**

```text
  c o--[3 Ω]--(2 V)--[5 Ω]--o d      (polarity: d is positive)
```

#### Step 3: Current in the 5 Ω
I = |V_th|/(R_th + 5) = 2/(3 + 5) = **0.25 A**, flowing from d to c through the 5 Ω.

**Answer: I₅Ω = 0.25 A (from d to c)**

## Check yourself

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

1. What is V_c with the 5 ohm removed?
   A. 2 V = 1 A x 2 ohm
   B. 1 V
   C. 4 V
   D. 8 V
   Answer: A. The 1 A flows through the 2 ohm shunt. (Key point: V_c)

2. What is V_d?
   A. 8 V
   B. 4 V by divider 8 x 2/(2+2)
   C. 2 V
   D. 6 V
   Answer: B. 2 ohm series with source and 2 ohm shunt split the 8 V. (Key point: V_d)

3. What is R_th?
   A. 2 + 2 + 2 = 6 ohm
   B. 1 ohm
   C. 2 + (2 || 2) = 3 ohm
   D. 5 ohm
   Answer: C. At c: 2 ohm; at d: 2||2 = 1 ohm. (Key point: R_th)

4. What is the 5 ohm current and direction?
   A. 0.4 A from c to d
   B. 0.25 A from c to d
   C. 2 A from d to c
   D. 2/(3+5) = 0.25 A from d to c
   Answer: D. d is 2 V higher than c so current flows d to c. (Key point: Current)

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