---
title: "Shunt motor speed with field resistance added (250 V, 1400 rpm)"
subject: "Electrical Machines I"
chapter: 4
key: "semester_4/electrical_machines_i:shunt-motor-250v-0p2ohm-1400rpm-50ohm"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/shunt-motor-250v-0p2ohm-1400rpm-50ohm
site: BCT Engineering
---

# Shunt motor speed with field resistance added (250 V, 1400 rpm)

Electrical Machines I · Chapter 4: DC Motor

- Asked in: 2071 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_i/shunt-motor-250v-0p2ohm-1400rpm-50ohm
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-4

## Question

A 250 V dc shunt motor has armature-winding resistance of 0.2 ohm and field winding resistance of 125 ohms. It draws a current of 25 amp at half load and the corresponding speed is 1400 rpm. If a resistance of 50 ohm is connected in series with the field winding and load torque is increased by 20%, calculate the new speed.

## Short answer

With field resistance increased, flux ratio is 0.7143. Armature current becomes 38.64 A, back emf 242.27 V. Speed = 1400 × (242.27/245.4) / 0.7143 ≈ 1935 rpm.

## Answer

**Given:** $V = 250$ V, $R_a = 0.2\ \Omega$, $R_{sh} = 125\ \Omega$, $I_{L1} = 25$ A at half load, $N_1 = 1400$ rpm; 50 Ω added in field; load torque +20%.

For a shunt motor, take flux proportional to field current (no saturation). Then

$$
T \propto \phi I_a, \qquad N \propto \frac{E_b}{\phi}, \qquad E_b = V - I_a R_a
$$

#### Initial condition
$$
\begin{aligned}
I_{sh1} &= \frac{250}{125} = 2\ \text{A}, \quad I_{a1} = 25 - 2 = 23\ \text{A} \\
E_{b1} &= 250 - 23 \times 0.2 = 245.4\ \text{V}
\end{aligned}
$$

#### New field
$$
I_{sh2} = \frac{250}{175} = 1.4286\ \text{A}, \qquad \frac{\phi_2}{\phi_1} = \frac{125}{175} = 0.7143
$$

#### New armature current ($T_2 = 1.2\,T_1$)
$$
I_{a2} = \frac{1.2 \times 23}{0.7143} = 38.64\ \text{A}
$$

#### New speed
$$
\begin{aligned}
E_{b2} &= 250 - 38.64 \times 0.2 = 242.27\ \text{V} \\
N_2 &= 1400 \times \frac{242.27}{245.4} \times \frac{1}{0.7143} = 1935.0\ \text{rpm}
\end{aligned}
$$

**Answer:** New speed $\approx 1935$ rpm.

## Check yourself

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

1. For Ra = 0.2 ohm, Ia1 = 23 A at 250 V, what is the initial back emf?
   A. 245.4 V
   B. 250 V
   C. 254.6 V
   D. 240.8 V
   Answer: A. Eb1 = V - Ia Ra = 250 - 23 x 0.2 = 245.4 V. (Key point: Initial back emf)

2. After adding 50 ohm to the field circuit (Rsh = 125 ohm), what is phi2/phi1?
   A. 0.4
   B. 1.2
   C. 0.7143
   D. 1.4
   Answer: C. Flux is proportional to field current: 125/175 = 0.7143. (Key point: Flux ratio)

3. Which equation gives the new armature current Ia2 when load torque rises 20% and flux falls?
   A. Ia2 = 1.2 x Ia1 x phi2/phi1
   B. Ia2 = 1.2 x Ia1 / (phi2/phi1)
   C. Ia2 = Ia1 / 1.2
   D. Ia2 = Ia1 x (phi2/phi1)
   Answer: B. T is proportional to phi Ia, so Ia2 = 1.2 Ia1 / 0.7143 = 38.64 A. (Key point: Torque condition)

4. What happens to speed after the field is weakened and torque raised, and what is the result?
   A. It falls to about 1000 rpm
   B. It stays at 1400 rpm
   C. It rises to about 1935 rpm
   D. It rises to about 1500 rpm
   Answer: C. N2 = 1400 x (242.27/245.4) / 0.7143 = 1935 rpm; the flux reduction dominates and raises the speed. (Key point: Final speed)

## Close variants set in other exams

- [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 1 Ω armature resistance added](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Ashunt-motor-240v-27a-1ohm-torque) — Electrical Machines I, chapter 4, asked 5×
- [Shunt motor speed with 1.2 Ω armature resistance added](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Ashunt-motor-240v-30a-1p2ohm-torque) — Electrical Machines I, chapter 4, asked 5×
- [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×

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