---
title: "Shunt motor speed after load reduced by 20%"
subject: "Electrical Machines"
chapter: 4
key: "semester_4/electrical_machines:dcm-shunt-load-reduced-20"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/dcm-shunt-load-reduced-20
site: BCT Engineering
---

# Shunt motor speed after load reduced by 20%

Electrical Machines · Chapter 4: DC Motor

- Asked in: 2073 Bhadra (old course), 8 marks
- Video: [speed control of dc motor | speed control of dc shunt motor | dc motor speed control | motor control](https://www.youtube.com/watch?v=BPGt3zxUn_8) — Mind of Engineer 
- 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-load-reduced-20
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-4

## Question

A 240 V dc shunt motor has armature resistance of $0.4\ \Omega$ and field resistance of $120\ \Omega$. It runs at 1500 rpm and draws a current of 5 A with certain load on its shaft. A resistance of $0.1\ \Omega$ is connected in series with armature winding and the load on the shaft is reduced by 20%, calculate the new speed of the motor.

## Short answer

With constant field flux, a 20 % reduction in load torque reduces armature current from 3 A to 2.4 A. Adding 0.1 Ω armature resistance gives a back emf of 238.8 V, identical to the initial value, so speed stays 1500 rpm.

## Answer

The field is across the supply, so the flux is constant. Then torque $\propto I_a$ and speed $\propto E_b$.

**Initial condition:**

$$
I_f = \frac{240}{120} = 2\ \text{A}, \qquad I_{a1} = 5 - 2 = 3\ \text{A}
$$

$$
E_{b1} = 240 - 3(0.4) = 238.8\ \text{V}
$$

**New condition:** load torque falls by 20%, so $T_2 = 0.8T_1$ and

$$
I_{a2} = 0.8\times 3 = 2.4\ \text{A}
$$

With the extra 0.1 Ω in the armature circuit, the armature circuit resistance is $0.4 + 0.1 = 0.5\ \Omega$:

$$
E_{b2} = 240 - 2.4(0.5) = 238.8\ \text{V}
$$

$$
N_2 = N_1\times\frac{E_{b2}}{E_{b1}} = 1500\times\frac{238.8}{238.8} = 1500\ \text{rpm}
$$

**Answer:** the new speed is $1500\ \text{rpm}$ (the series resistance drop exactly offsets the reduced armature drop, so the speed is unchanged).

## 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, the load reduced by 20% means:
   A. Ia2 = 0.8 Ia1
   B. Ia2 = 1.2 Ia1
   C. Ia2 = Ia1
   D. Ia2 = 0.2 Ia1
   Answer: A. T is proportional to Ia. (Key point: Torque vs Ia)

2. Ia1 with 5 A line and 2 A field current is:
   A. 5 A
   B. 7 A
   C. 2 A
   D. 3 A
   Answer: D. 5 - 2 = 3 A. (Key point: Initial Ia)

3. Eb2 with Ia = 2.4 A and R = 0.5 ohm is:
   A. 240 V
   B. 238.8 V
   C. 239.4 V
   D. 237 V
   Answer: B. 240 - 2.4 x 0.5 = 238.8 V. (Key point: Effect of extra resistance)

4. Hence the new speed is:
   A. 1200 rpm
   B. 1800 rpm
   C. 1500 rpm, unchanged
   D. 1250 rpm
   Answer: C. Eb2 = Eb1 so speed stays 1500 rpm. (Key point: Result)

## Related in Electrical Machines

- [Shunt motor speed with 0.1 Ω armature resistance, less load](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Ashunt-motor-240v-5a-armature-0p1) — Electrical Machines I, chapter 4, asked 3×
- [Armature resistance to reduce shunt motor speed to 1250 rpm](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Ashunt-motor-250v-30a-1250rpm-armature) — Electrical Machines I, chapter 4
- [Shunt motor speed with series armature resistance](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Adcm-shunt-half-full-load-torque) — Electrical Machines, chapter 4
- [Shunt motor speed with 50 Ω field resistance (1200 rpm)](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Ashunt-motor-250v-25a-half-load) — Electrical Machines I, chapter 4, asked 5×
- [Shunt motor speed with field resistance added (250 V, 1400 rpm)](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Ashunt-motor-250v-0p2ohm-1400rpm-50ohm) — Electrical Machines I, chapter 4, asked 5×
- [Shunt motor speed with 50 Ω field resistance and lower torque](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Ashunt-motor-220v-25a-50ohm-torque) — Electrical Machines I, chapter 4, asked 4×

## Related in other subjects

- [DC shunt motor speed at reduced voltage and starting resistance](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Adc-shunt-reduced-voltage-starting-resistance) — Utilization of Electrical Energy, chapter 3
- [Speed of 220 V series motor at reduced load current](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Aseries-motor-220v-38a-speed) — Electrical Machines I, chapter 4
- [Shunt motor speed with 30 Ω field resistance added](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Ashunt-motor-220v-30ohm-field-torque) — Electrical Machines I, chapter 4, asked 5×
- [Shunt motor speed with added resistance](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machine%3Ashunt-motor-resistance-30-percent-torque) — Electrical Machine, chapter 4