---
title: "Shunt motor speed at armature current 30 A"
subject: "Electrical Machines"
chapter: 4
key: "semester_4/electrical_machines:dcm-shunt-60a-230v-speed-30a"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/dcm-shunt-60a-230v-speed-30a
site: BCT Engineering
---

# Shunt motor speed at armature current 30 A

Electrical Machines · Chapter 4: DC Motor

- Asked in: 2071 Magh, 6 marks
- Video: [⚙️⚡ DC Shunt Motor ⭐ Armature Current, Motor Speed & Developed Torque ✅ Example 1 #dcmotor](https://www.youtube.com/watch?v=kD9IhajL7Mc) — CAN Education
- 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-60a-230v-speed-30a
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-4

## Question

A dc shunt motor runs at 600 RPM taking 60 A from a 230 V supply. Armature resistance is $0.2\ \Omega$ and field resistance is $115\ \Omega$. Find the speed when the current through the armature is 30 A.

## Short answer

Field current is 2 A. At 600 rpm, Ia=58 A, Eb1=218.4 V. When armature current is 30 A, Eb2=224 V. Using N∝Eb, speed = 615.4 rpm.

## Answer

The field current is constant ($V$ and $R_f$ are fixed), so the flux is constant and $N\propto E_b$.

**Field current:** $I_f = \dfrac{230}{115} = 2$ A.

**Case 1 (600 rpm, line current 60 A):**

$$
I_{a1} = 60 - 2 = 58\ \text{A}, \qquad E_{b1} = 230 - 58(0.2) = 218.4\ \text{V}
$$

**Case 2 (armature current 30 A):**

$$
E_{b2} = 230 - 30(0.2) = 224.0\ \text{V}
$$

$$
N_2 = N_1\frac{E_{b2}}{E_{b1}} = 600\times\frac{224.0}{218.4} = 615.4\ \text{rpm}
$$

**Answer:** speed at armature current 30 A $= 615.4\ \text{rpm}$.

## Check yourself

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

1. Why is N proportional to Eb in this shunt motor problem?
   A. Armature current is constant
   B. The field current is fixed since V and Rf are fixed
   C. Torque is constant
   D. Ra is zero
   Answer: B. Shunt field across fixed V gives constant flux. (Key point: Constant flux)

2. If = 230/115 = 2 A. The armature current at 60 A line is:
   A. 60 A
   B. 62 A
   C. 30 A
   D. 58 A
   Answer: D. Ia = line current - field current = 60 - 2. (Key point: Armature current from line)

3. Eb1 at 600 rpm is:
   A. 218.4 V
   B. 230 V
   C. 224 V
   D. 211.6 V
   Answer: A. 230 - 58 x 0.2 = 218.4 V. (Key point: Back emf case 1)

4. The speed at Ia = 30 A is:
   A. 600 rpm
   B. 615.4 rpm
   C. 584.6 rpm
   D. 640 rpm
   Answer: B. Eb2 = 224 V; N2 = 600 x 224/218.4 = 615.4 rpm. (Key point: Final speed)

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