---
title: "Thevenin theorem, current in 13Ω"
subject: "Basic Electrical Engineering"
chapter: 3
key: "semester_1/basic_electrical_engineering:thevenin-13-ohm-200v-50v-180v"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/thevenin-13-ohm-200v-50v-180v
site: BCT Engineering
---

# Thevenin theorem, current in 13Ω

Basic Electrical Engineering · Chapter 3: Network Theorems

- Asked in: 2069 Asar, 8 marks
- Video: [Thevenin's theorem Solved Example | Electric Circuits | Network Analysis | Network Theory](https://www.youtube.com/watch?v=tH9boIFwHW8) — 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/thevenin-13-ohm-200v-50v-180v
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-3

## Question

Applying Thevenin's theorem, calculate the magnitude and direction of current in the 13Ω resistor in the circuit shown in the following figure. [Figure: 200 V source and 50 V source on the left; 20 Ω and 10 Ω resistors; 15 Ω in series with a 180 V source on top; 13 Ω load between terminals a and b.]

## Short answer

Find Thevenin voltage: Vth = 70 V (b positive to a) with Rth = 0 Ω. The current in the 13 Ω resistor is 70/13 = 5.38 A from b to a.

## Answer

#### Reading the figure
- Left branch: 200 V battery (+ on top) joins node N₁ (bottom) to the top node T (V_T = V_N1 + 200).
- 50 V battery (+ on top) below N₁: its + end is at N₁ and its − end is terminal a.
- 15 Ω from T to the middle node N₂; 20 Ω from N₁ to N₂; 10 Ω from N₂ to the right node N₃.
- 180 V battery (+ on the T side) between T and N₃ (V_N3 = V_T − 180).
- The load 13 Ω is between a and b, and b is wired to N₃.

Take N₁ as reference (0 V).

#### Step 1: Thevenin voltage (13 Ω removed)
- V_T = 200 V
- V_N3 = V_b = 200 − 180 = 20 V
- No current flows in the a-b path, so V_a = −50 V (50 V battery, − terminal)

V_b − V_a = 20 − (−50) = 70 V

**V_th = 70 V** (b is positive with respect to a).

(The resistors do not change V_b or V_a: V_N2 settles at 70.8 V by KCL at N₂, but it does not enter V_ab.)

#### Step 2: Thevenin resistance (sources off)
With the three batteries replaced by short circuits, a is joined to N₁, N₁ to T, T to N₃ and N₃ to b by wires. So a and b become the same point:

**R_th = 0 Ω**

#### Step 3: Current in the 13 Ω
I = V_th/(R_th + 13) = 70/13 = **5.385 A**

Direction: V_b > V_a, so the current flows through the 13 Ω from b to a (right to left).

**Answer: I = 5.38 A, from b to a**

## Check yourself

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

1. How is V_th determined with the 13 ohm removed?
   A. V_th = 180 - 50 = 130 V
   B. V_b = 200 - 180 = 20 V and V_a = -50 V, so V_th = 70 V
   C. V_th = 200 - 180 - 50 = -30 V
   D. V_th = 200 + 180 + 50 = 430 V
   Answer: B. No current flows through the a-b path, so V_a = -50 V and V_b = 20 V, giving V_b - V_a = 70 V. (Key point: V_th calculation)

2. Why is R_th = 0 ohm?
   A. Because the 15 ohm cancels the 20 ohm
   B. Because the 13 ohm is included
   C. Because the resistors are all in parallel
   D. With all batteries shorted, terminals a and b are connected by wires only, so they become one point
   Answer: D. Ideal batteries become shorts, joining a to b directly; the resistors are not between a and b. (Key point: R_th with sources off)

3. What is the current in the 13 ohm?
   A. 70/(13+45) = 1.21 A
   B. 30/13 = 2.31 A
   C. 70/13 = 5.385 A
   D. 70/(13+15) = 2.5 A
   Answer: C. I = V_th/(R_th + R_L) = 70/(0 + 13). (Key point: Load current)

4. Why does current flow from b to a through the 13 ohm?
   A. a is positive with respect to b
   B. The 180 V battery is largest
   C. b is at a higher potential than a (V_b > V_a)
   D. Current always flows left to right
   Answer: C. Current flows from higher to lower potential through the load; V_b = 20 V is above V_a = -50 V. (Key point: Direction)

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