---
title: "Shunt motor speed with 50 Ω field resistance (1200 rpm)"
subject: "Electrical Machines I"
chapter: 4
key: "semester_4/electrical_machines_i:shunt-motor-250v-25a-half-load"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/shunt-motor-250v-25a-half-load
site: BCT Engineering
---

# Shunt motor speed with 50 Ω field resistance (1200 rpm)

Electrical Machines I · Chapter 4: DC Motor

- Asked in: 2072 Asoj, 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-25a-half-load
- 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 0.4 ohm and field winding resistance of 125 ohms. It draws a current of 25 amp at half load and the corresponding speed is 1200 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

Field current drops to 1.4286 A (flux ratio 0.7143). Armature current rises to 38.64 A, back emf 234.54 V. Speed = 1200 × (234.54/240.8) / 0.7143 ≈ 1636 rpm.

## Answer

**Given:** $V = 250$ V, $R_a = 0.4\ \Omega$, $R_{sh} = 125\ \Omega$, $I_{L1} = 25$ A at half load, $N_1 = 1200$ 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
$$
\begin{aligned}
I_{sh1} &= \frac{250}{125} = 2\ \text{A}, \quad I_{a1} = 25 - 2 = 23\ \text{A} \\
E_{b1} &= 250 - 23 \times 0.4 = 240.8\ \text{V}
\end{aligned}
$$

#### New field
$$
I_{sh2} = \frac{250}{125 + 50} = 1.4286\ \text{A}, \qquad \frac{\phi_2}{\phi_1} = \frac{125}{175} = 0.7143
$$

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

#### New speed
$$
\begin{aligned}
E_{b2} &= 250 - 38.64 \times 0.4 = 234.54\ \text{V} \\
N_2 &= 1200 \times \frac{234.54}{240.8} \times \frac{1}{0.7143} = 1636.4\ \text{rpm}
\end{aligned}
$$

**Answer:** New speed $\approx 1636$ rpm.

## Check yourself

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

1. What is the initial armature current of the motor drawing 25 A at half load (250 V, Rsh = 125 ohm)?
   A. 25 A
   B. 2 A
   C. 23 A
   D. 27 A
   Answer: C. The shunt field takes 250/125 = 2 A, and the armature current is line current minus field current: 25 - 2 = 23 A. (Key point: Armature current)

2. Adding 50 ohm in series with the field winding changes the flux by what factor (flux proportional to field current)?
   A. 1.4 times
   B. 0.7143 times
   C. 0.5 times
   D. 1.0, unchanged
   Answer: B. Field current falls from 2 A to 250/175 = 1.4286 A, so flux ratio = 125/175 = 0.7143. (Key point: Flux change from series resistance)

3. Why does the armature current rise to 38.64 A, more than the 20% torque increase alone would give (27.6 A)?
   A. Torque is proportional to flux x Ia, and flux has dropped, so Ia must rise by 1.2/0.7143
   B. Back emf has increased, which forces more current
   C. Supply voltage has increased
   D. The armature resistance has reduced
   Answer: A. T is proportional to phi x Ia. With phi at 0.7143 of its value and T at 1.2 times, Ia2 = 1.2 x 23 / 0.7143 = 38.64 A. (Key point: New armature current)

4. Which relation gives the new speed N2 = 1200 x (234.54/240.8) x (1/0.7143)?
   A. N is proportional to Eb x flux
   B. N is proportional to Eb / flux
   C. N is proportional to Ia / flux
   D. N is proportional to flux / Eb
   Answer: B. N is proportional to Eb/phi; weaker field and similar Eb give higher speed, about 1636 rpm. (Key point: Speed relation)

## Close variants set in other exams

- [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 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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