---
title: "Shunt motor: resistance for 1200 rpm"
subject: "Electrical Machine"
chapter: 4
key: "semester_4/electrical_machine:shunt-motor-40a-1400-to-1200"
source: https://www.bctengineer.com/q/semester_4/electrical_machine/shunt-motor-40a-1400-to-1200
site: BCT Engineering
---

# Shunt motor: resistance for 1200 rpm

Electrical Machine · Chapter 4: DC Motor

- Asked in: 2070 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-40a-1400-to-1200
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machine/chapter-4

## Question

A 220 V dc shunt motor draws a current of 40 A at full load and runs with speed of 1400 rpm. Calculate the value of resistance required to be inserted in the armature circuit so that speed drops to 1200 rpm at constant load. Given that Ra = 0.02 ohm and Rf = 100 ohms.

## Short answer

Field unchanged, flux constant. At constant load torque, armature current remains 37.8A. E_b at 1400 rpm = 219.244V, at 1200 rpm = 187.923V. Required series resistance R = (220–187.923)/37.8 – 0.02 ≈ 0.83Ω.

## Answer

At constant load torque and constant flux (shunt field), the armature current remains unchanged, and the speed falls by the reduction of back emf.

#### Equivalent circuit
```text
 +--[Ra=0.02]--[R ?]--(Eb)--+
 |                           |
 +---- Rf = 100 ohm ---------+--- V = 220 V
```

#### Initial state
- I_sh = 220 / 100 = 2.2 A
- I_a = 40 − 2.2 = 37.8 A
- E_b1 = 220 − 37.8 × 0.02 = 219.244 V at 1400 rpm

#### New state (1200 rpm, same I_a)
E_b2 = 219.244 × (1200 / 1400) = 187.923 V

#### Resistance
220 = 187.923 + 37.8 (0.02 + R)

0.02 + R = 32.077 / 37.8 = 0.8486 Ω

**R = 0.8486 − 0.02 = 0.829 Ω**

**Answer: about 0.83 Ω must be inserted in series with the armature.**

## Check yourself

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

1. What is the armature current?
   A. 37.8 A
   B. 42.2 A
   C. 40 A
   D. 2.2 A
   Answer: A. I_sh = 2.2 A, so I_a = 40 - 2.2 = 37.8 A. (Key point: Armature current)

2. What does constant load imply?
   A. Speed stays the same
   B. Armature current stays the same
   C. Back emf stays the same
   D. Field current changes
   Answer: B. With constant load torque and flux, I_a is unchanged. (Key point: Constant load assumption)

3. What is E_b2 at 1200 rpm?
   A. 256 V
   B. 200 V
   C. 187.9 V
   D. 219.2 V
   Answer: C. E_b2 = 219.244 × 1200/1400 = 187.923 V. (Key point: New back emf)

4. What is the series resistance needed?
   A. 0.02 Ω
   B. 1.2 Ω
   C. 0.849 Ω
   D. 0.829 Ω
   Answer: D. 0.02 + R = 32.077/37.8 = 0.8486, so R = 0.829 Ω. (Key point: Resistance)

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