---
title: "Shunt motor speed with series armature resistance"
subject: "Electrical Machines"
chapter: 4
key: "semester_4/electrical_machines:dcm-shunt-half-full-load-torque"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/dcm-shunt-half-full-load-torque
site: BCT Engineering
---

# Shunt motor speed with series armature resistance

Electrical Machines · Chapter 4: DC Motor

- Asked in: 2071 Magh (old course), 4+4 marks
- Video: [Speed Control of DC Motor | Voltage Control | Flux Control | Armature Resistance Control](https://www.youtube.com/watch?v=A4JAueEf_PA) — Electrical lectures
- 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-half-full-load-torque
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-4

## Question

A 240 V dc shunt motor has armature winding resistance of $0.4\ \Omega$ and field winding resistance of $120\ \Omega$. It draws a current of 27 A at half load and the corresponding speed is 600 rpm.
i) If a resistance of $1\ \Omega$ connected in series with the armature winding keeping the load torque constant to half load torque, calculate the new speed.
ii) If a resistance of $1\ \Omega$ connected in series with the armature winding and the load torque is increased to full load torque, calculate the new speed.

## Short answer

With flux constant and a1 Ω series armature resistance: (i) at half-load torque, armature current remains 25 A, back emf drops to 205 V, giving 534.8 rpm; (ii) at full-load torque, armature current doubles to 50 A, back emf falls to 170 V, giving 443.5 rpm.

## Answer

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

**Field current:** $I_f = 240/120 = 2$ A.

**Half-load condition:**

$$
I_{a1} = 27 - 2 = 25\ \text{A}, \qquad E_{b1} = 240 - 25(0.4) = 230\ \text{V}, \qquad N_1 = 600\ \text{rpm}
$$

With the extra 1 Ω, the armature circuit resistance is $0.4 + 1 = 1.4\ \Omega$.

#### (i) Load torque constant at half-load value

The torque is unchanged, so $I_{a2} = I_{a1} = 25$ A.

$$
E_{b2} = 240 - 25(1.4) = 205\ \text{V}
$$

$$
N_2 = 600\times\frac{205}{230} = 534.8\ \text{rpm}
$$

#### (ii) Load torque increased to full-load torque

Full-load torque is twice the half-load torque, so

$$
I_{a2} = 2\times 25 = 50\ \text{A}
$$

$$
E_{b2} = 240 - 50(1.4) = 170\ \text{V}
$$

$$
N_2 = 600\times\frac{170}{230} = 443.5\ \text{rpm}
$$

**Answer:** (i) $534.8\ \text{rpm}$; (ii) $443.5\ \text{rpm}$.

## Check yourself

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

1. At half load, Ia and Eb1 are:
   A. 27 A, 229 V
   B. 25 A, 230 V
   C. 25 A, 240 V
   D. 2 A, 239 V
   Answer: B. Ia = 27 - 2 = 25 A; Eb1 = 240 - 25 x 0.4 = 230 V. (Key point: Half-load Ia and Eb)

2. With torque constant, the armature current in (i) is:
   A. 12.5 A
   B. 50 A
   C. 27 A
   D. 25 A, unchanged
   Answer: D. Flux constant, so same torque means same Ia. (Key point: Case (i) current)

3. Speed in (i) is:
   A. 534.8 rpm
   B. 443.5 rpm
   C. 600 rpm
   D. 470 rpm
   Answer: A. Eb2 = 240 - 25 x 1.4 = 205 V; 600 x 205/230 = 534.8 rpm. (Key point: Case (i) speed)

4. At full-load torque (twice half-load), the speed is:
   A. 534.8 rpm
   B. 600 rpm
   C. 443.5 rpm
   D. 380 rpm
   Answer: C. Ia = 50 A; Eb2 = 240 - 70 = 170 V; 600 x 170/230 = 443.5 rpm. (Key point: Case (ii))

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