---
title: "Shunt motor: resistance to reduce speed to 1200 rpm"
subject: "Electrical Machines"
chapter: 4
key: "semester_4/electrical_machines:dcm-shunt-220v-40a-1400rpm-1200rpm"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/dcm-shunt-220v-40a-1400rpm-1200rpm
site: BCT Engineering
---

# Shunt motor: resistance to reduce speed to 1200 rpm

Electrical Machines · 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_machines/dcm-shunt-220v-40a-1400rpm-1200rpm
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/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 $R_a = 0.02\ \Omega$ and $R_f = 100\ \Omega$.

## Short answer

Field current is 2.2 A, armature current 37.8 A. Initial back emf is 219.244 V; at 1200 rpm back emf is 187.923 V. Required total armature resistance is 0.8486 ohm, so inserted resistance is 0.8286 ohm.

## Answer

The field is across the supply, so the flux is constant and $N\propto E_b$. Constant load torque means constant armature current.

**Initial condition:**

$$
I_f = \frac{220}{100} = 2.2\ \text{A}, \qquad I_a = 40 - 2.2 = 37.8\ \text{A}
$$

$$
E_{b1} = 220 - 37.8(0.02) = 219.244\ \text{V}
$$

**At 1200 rpm:**

$$
E_{b2} = E_{b1}\times\frac{1200}{1400} = 187.923\ \text{V}
$$

$$
R_a + R_x = \frac{V - E_{b2}}{I_a} = \frac{220 - 187.923}{37.8} = 0.8486\ \Omega
$$

$$
R_x = 0.8486 - 0.02 = 0.8286\ \Omega
$$

**Answer:** resistance to be inserted in the armature circuit $R_x = 0.8286\ \Omega$.

## Check yourself

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

1. Ia with 40 A line and If = 2.2 A is:
   A. 37.8 A
   B. 40 A
   C. 42.2 A
   D. 2.2 A
   Answer: A. 40 - 2.2 = 37.8 A. (Key point: Armature current)

2. Eb1 is:
   A. 220 V
   B. 219.244 V
   C. 218 V
   D. 187.9 V
   Answer: B. 220 - 37.8 x 0.02 = 219.244 V. (Key point: Eb1)

3. Eb2 at 1200 rpm is:
   A. 219.2 V
   B. 200 V
   C. 187.92 V
   D. 160 V
   Answer: C. 219.244 x 1200/1400 = 187.92 V. (Key point: Eb2)

4. The inserted resistance is:
   A. 0.8486 ohm
   B. 0.02 ohm
   C. 1.5 ohm
   D. 0.8286 ohm
   Answer: D. Total = (220 - 187.923)/37.8 = 0.8486; minus 0.02 gives 0.8286 ohm. (Key point: Resistance)

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