---
title: "Shunt motor speed with 250 ohm in field circuit"
subject: "Electrical Machines"
chapter: 4
key: "semester_4/electrical_machines:dcm-shunt-field-resistance-250"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/dcm-shunt-field-resistance-250
site: BCT Engineering
---

# Shunt motor speed with 250 ohm in field circuit

Electrical Machines · Chapter 4: DC Motor

- Asked in: 2076 Baisakh, 8 marks
- Video: [Speed Control of DC Shunt Motor by Field Control Method](https://www.youtube.com/watch?v=3nHgNE1Rlaw) — ZERO TO MASTER
- 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-field-resistance-250
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-4

## Question

A 250 V dc shunt motor draws an armature current of 20 A and runs with a speed of 1500 rpm. If a resistance of $250\ \Omega$ is inserted in series with field winding keeping the load torque constant, find the new speed. Given that armature winding resistance is $0.25\ \Omega$ and field winding resistance is $250\ \Omega$.

## Short answer

Inserting 250 Ω in the field circuit halves the field current to 0.5 A, halving flux. To maintain constant torque, armature current doubles to 40 A, back emf becomes 240 V, and the new speed is 1500 × (240/245) × 2 ≈ 2939 rpm.

## Answer

Assume the flux is proportional to field current. The load torque is constant, so $\Phi I_a$ is constant. Field weakening therefore increases the armature current.

**Initial:**

$$
I_{f1} = \frac{250}{250} = 1\ \text{A}, \qquad E_{b1} = 250 - 20(0.25) = 245\ \text{V}
$$

**After inserting 250 Ω in the field:**

$$
I_{f2} = \frac{250}{250 + 250} = 0.5\ \text{A}, \qquad \frac{\Phi_2}{\Phi_1} = \frac{0.5}{1} = 0.5
$$

**Constant torque** ($T\propto\Phi I_a$):

$$
I_{a2} = I_{a1}\times\frac{\Phi_1}{\Phi_2} = 20\times 2 = 40\ \text{A}
$$

$$
E_{b2} = 250 - 40(0.25) = 240\ \text{V}
$$

**Speed** (from $E_b\propto\Phi N$):

$$
N_2 = N_1\times\frac{E_{b2}}{E_{b1}}\times\frac{\Phi_1}{\Phi_2} = 1500\times\frac{240}{245}\times 2 = 2939\ \text{rpm}
$$

**Answer:** new speed $\approx 2939\ \text{rpm}$.

## Check yourself

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

1. Adding 250 ohm to a 250 ohm field gives flux ratio Phi2/Phi1 of:
   A. 2
   B. 1
   C. 0.5
   D. 0.25
   Answer: C. If halves from 1 A to 0.5 A, so flux halves. (Key point: Flux ratio)

2. At constant torque, Ia2 is:
   A. 10 A
   B. 20 A
   C. 30 A
   D. 40 A
   Answer: D. Ia2 = Ia1 x Phi1/Phi2 = 20 x 2. (Key point: New armature current)

3. Eb2 is:
   A. 240 V
   B. 245 V
   C. 250 V
   D. 230 V
   Answer: A. 250 - 40 x 0.25 = 240 V. (Key point: New Eb)

4. The new speed is about:
   A. 1469 rpm
   B. 2939 rpm
   C. 1500 rpm
   D. 750 rpm
   Answer: B. N2 = 1500 x (240/245) x 2 = 2939 rpm. (Key point: Speed)

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