---
title: "Shunt motor speed with added resistance"
subject: "Electrical Machine"
chapter: 4
key: "semester_4/electrical_machine:shunt-motor-resistance-30-percent-torque"
source: https://www.bctengineer.com/q/semester_4/electrical_machine/shunt-motor-resistance-30-percent-torque
site: BCT Engineering
---

# Shunt motor speed with added resistance

Electrical Machine · Chapter 4: DC Motor

- Asked in: 2073 Magh, 8 marks
- Video: [Numerical on speed control of DC shunt motors  #electricmotors #electricalengineering](https://www.youtube.com/watch?v=T7KmAY7FYA4) — EEE, SJBIT
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machine/shunt-motor-resistance-30-percent-torque
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machine/chapter-4

## Question

A dc shunt motor is supplied by a source of 200 V. It draws a current of 20 A and runs at speed of 1500 rpm. The armature and field winding resistance are 0.08 Ω and 110 Ω respectively. A resistance of 0.02 Ω is added in series with armature and load torque is increased by 30%, calculate new speed.

## Short answer

Field current constant, flux constant, torque ∝ I_a. Initial I_a=18.18A, E_b=198.545V. After 30% torque rise, I_a=23.64A, total armature resistance 0.10Ω, E_b=197.636V. New speed = 1500 × (197.636/198.545) ≈ 1493 rpm.

## Answer

In a shunt motor the field is connected across the supply, so flux Φ is constant. Torque ∝ I_a, and speed N ∝ E_b.

#### Equivalent circuit
```text
 +--[Ra=0.08]--[Rext=0.02]--(Eb)--+
 |         Ia                      |
 +---- Rf = 110 ohm ---------------+--- V = 200 V
```

#### Initial condition
- I_sh = 200 / 110 = 1.818 A
- I_a1 = 20 − 1.818 = 18.18 A
- E_b1 = V − I_a1R_a = 200 − 18.18 × 0.08 = 198.545 V
- N₁ = 1500 rpm

#### New condition
- Torque rises by 30 % with Φ constant, so I_a2 = 1.3 × 18.18 = 23.64 A
- Total armature circuit resistance = 0.08 + 0.02 = 0.10 Ω
- E_b2 = 200 − 23.64 × 0.10 = 197.636 V

#### New speed
N₂ / N₁ = E_b2 / E_b1 (Φ unchanged)

N₂ = 1500 × 197.636 / 198.545 = **1493.1 rpm**

| Quantity | Before | After |
|---|---|---|
| I_a (A) | 18.18 | 23.64 |
| R in armature (Ω) | 0.08 | 0.10 |
| E_b (V) | 198.545 | 197.636 |
| N (rpm) | 1500 | 1493.1 |

**Answer: new speed ≈ 1493 rpm.**

## Check yourself

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

1. In a shunt motor with constant flux, how is torque related to armature current?
   A. T ∝ I_a
   B. T ∝ 1/I_a
   C. T ∝ I_a^2
   D. T ∝ N
   Answer: A. With Φ constant, T = KΦI_a is proportional to I_a. (Key point: Torque and current)

2. If torque rises by 30 %, what is the new I_a (initial 18.18 A)?
   A. 25.0 A
   B. 23.64 A
   C. 18.18 A
   D. 14.0 A
   Answer: B. I_a2 = 1.3 × 18.18 = 23.64 A. (Key point: New armature current)

3. Which relation gives the new speed?
   A. N2/N1 = E_b1/E_b2
   B. N2 = N1
   C. N2/N1 = E_b2/E_b1 since Φ is constant
   D. N2/N1 = I_a1/I_a2
   Answer: C. Speed is proportional to back emf when flux is constant. (Key point: Speed relation)

4. What is E_b2 with the added 0.02 Ω?
   A. 200 V
   B. 198.545 V
   C. 195.0 V
   D. 197.636 V
   Answer: D. E_b2 = 200 - 23.64 × 0.10 = 197.636 V, giving about 1493 rpm. (Key point: New back emf)

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