---
title: "DC shunt motor armature torque"
subject: "Basic Electrical and Electronics Engineering"
chapter: 2
key: "semester_1/basic_electrical_and_electronics_engineering:dc-shunt-motor-torque"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_and_electronics_engineering/dc-shunt-motor-torque
site: BCT Engineering
---

# DC shunt motor armature torque

Basic Electrical and Electronics Engineering · Chapter 2: Electrical Machines

- Asked in: 2081 Kartik (new course), 4 marks
- Video: [How to Calculate Armature Torque and Shaft Torque in a DC Shunt Motor](https://www.youtube.com/watch?v=afLFA6htsOU) — 5 Minute Electrical 
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/basic_electrical_and_electronics_engineering/dc-shunt-motor-torque
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_and_electronics_engineering/chapter-2

## Question

A 4 pole 230 V dc shunt motor, lap wound has 980 conductors. The flux per pole is 20 mWb. Determine the torque developed by the armature in Nm when the current drawn by the motor is 26 A. The armature resistance is 0.12 Ω and the field resistance is 130 Ω.

## Short answer

Armature current is24.23 A. Torque is computed using Ta = (P φ Z)/(2π A) Ia, giving75.59 N·m.

## Answer

**Given:** $P=4$, lap winding so $A=P=4$, $Z=980$, $\phi=20$ mWb, $V=230$ V, line current $I_L=26$ A, $R_a=0.12\ \Omega$, $R_{sh}=130\ \Omega$.

**Shunt field current:**

$$
I_{sh}=\frac{V}{R_{sh}}=\frac{230}{130}=1.769\ \text{A}
$$

**Armature current:**

$$
I_a=I_L-I_{sh}=26-1.769=24.23\ \text{A}
$$

**Armature torque** (using $T_a=\frac{P\phi Z}{2\pi A}I_a$):

$$
T_a=\frac{4\times0.02\times980}{2\pi\times4}\times24.23=\frac{78.4}{25.13}\times24.23=3.119\times24.23=75.59\ \text{N·m}
$$

**Check through back emf and speed:**

$$
E_b=V-I_aR_a=230-24.23\times0.12=227.09\ \text{V}
$$

$$
N=\frac{60AE_b}{P\phi Z}=\frac{60\times4\times227.09}{4\times0.02\times980}=695.2\ \text{rpm}
$$

$$
T_a=\frac{E_bI_a}{2\pi N/60}=\frac{227.09\times24.23}{72.80}=75.59\ \text{N·m}
$$

**Answer: Armature torque $T_a=75.59$ N·m (at about 695 rpm).**

## Check yourself

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

1. How is the armature current of the shunt motor found?
   A. Ia = IL = 26 A
   B. Ia = IL + Ish = 27.8 A
   C. Ia = V/Ra = 1917 A
   D. Ia = IL - Ish = 26 - 230/130 = 24.23 A
   Answer: D. In a shunt motor the line current splits into field and armature currents. (Key point: Armature current)

2. For a 4-pole lap-wound armature, how many parallel paths A are there?
   A. A = 4
   B. A = 2
   C. A = 1
   D. A = 8
   Answer: A. In lap winding, A = P. (Key point: Lap winding paths)

3. Which expression gives the armature torque?
   A. Ta = P phi Z IL/(2 pi A)
   B. Ta = P phi Z Ia/(2 pi A) = 75.59 N·m
   C. Ta = P phi Z Ia/A = 474.9 N·m
   D. Ta = Eb/Ia
   Answer: B. Torque uses the armature current only: 3.119 x 24.23 = 75.59 N·m. (Key point: Torque formula)

4. How is the back emf used to verify the torque?
   A. Eb = V + Ia Ra
   B. Eb = V = 230 V
   C. Eb = V - Ia Ra = 227.09 V, and Ta = Eb Ia/omega gives the same value
   D. Eb = Ia Ra
   Answer: C. Eb = 230 - 24.23 x 0.12 = 227.09 V; Eb Ia divided by omega at 695 rpm also equals 75.59 N·m. (Key point: Back emf check)

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