---
title: "Shunt motor speed with 50 Ω field resistance and lower torque"
subject: "Electrical Machines I"
chapter: 4
key: "semester_4/electrical_machines_i:shunt-motor-220v-25a-50ohm-torque"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/shunt-motor-220v-25a-50ohm-torque
site: BCT Engineering
---

# Shunt motor speed with 50 Ω field resistance and lower torque

Electrical Machines I · Chapter 4: DC Motor

- Asked in: 2076 Baisakh, 8 marks
- Video: [EE213 - 26a - Speed control of shunt DC motors - Changing field resistance](https://www.youtube.com/watch?v=qmsgMsE1xzo) — MAFarooqi
- 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-220v-25a-50ohm-torque
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-4

## Question

A 220 V dc shunt motor draws a current of 25 A and runs at 1200 rpm with certain load on its shaft. The armature winding resistance and field winding resistances are 0.1 Ω and 110 Ω respectively. If a resistance of 50 Ω is connected in series with the field winding and load torque on the shaft is reduced by 10%, calculate the new speed.

## Short answer

Field current drops from 2 A to 1.375 A, flux ratio 0.6875. Torque reduced 10% gives I_a = (0.9×23)/0.6875 ≈ 30.11 A. E_b = 220 – 30.11×0.1 = 216.99 V. New speed = 1200 × (216.99/217.7) / 0.6875 ≈ 1740 rpm.

## Answer

**Given:** $V = 220$ V, $I_{L1} = 25$ A, $N_1 = 1200$ rpm, $R_a = 0.1\ \Omega$, $R_{sh} = 110\ \Omega$; 50 Ω added in field; load torque reduced by 10%.

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
$$
\begin{aligned}
I_{sh1} &= \frac{220}{110} = 2\ \text{A}, \quad I_{a1} = 25 - 2 = 23\ \text{A} \\
E_{b1} &= 220 - 23 \times 0.1 = 217.7\ \text{V}
\end{aligned}
$$

#### New field
$$
I_{sh2} = \frac{220}{110 + 50} = 1.375\ \text{A}, \qquad \frac{\phi_2}{\phi_1} = \frac{1.375}{2} = 0.6875
$$

#### New armature current ($T_2 = 0.9\,T_1$)
$$
I_{a2} = \frac{0.9 \times 23}{0.6875} = 30.11\ \text{A}
$$

#### New speed
$$
\begin{aligned}
E_{b2} &= 220 - 30.11 \times 0.1 = 216.99\ \text{V} \\
N_2 &= 1200 \times \frac{216.99}{217.7} \times \frac{1}{0.6875} = 1739.8\ \text{rpm}
\end{aligned}
$$

**Answer:** New speed $\approx 1740$ 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?
   A. 23 A
   B. 25 A
   C. 2 A
   D. 27 A
   Answer: A. Ia = 25 - 2. (Key point: Initial Ia)

2. What is φ2/φ1?
   A. 1.455
   B. 0.6875
   C. 0.5
   D. 0.9
   Answer: B. 1.375/2. (Key point: Flux ratio)

3. How is Ia2 found for 10% less torque?
   A. 0.9×23 = 20.7 A
   B. 23/0.9 A
   C. Ia2 = 0.9×23/0.6875 = 30.1 A
   D. 23×0.6875 A
   Answer: C. T ∝ φIa. (Key point: New Ia)

4. What is the new speed?
   A. About 1200 rpm
   B. About 826 rpm
   C. About 1810 rpm
   D. About 1740 rpm
   Answer: D. 1200×(216.99/217.7)/0.6875. (Key point: New speed)

## Close variants set in other exams

- [Shunt motor speed with 0.1 Ω armature resistance, less load](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Ashunt-motor-240v-5a-armature-0p1) — Electrical Machines I, chapter 4, asked 3×
- [Shunt motor speed with 0.1 Ω armature resistance added](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Ashunt-motor-220v-20a-0p1ohm-armature) — Electrical Machines I, chapter 4, asked 4×
- [Shunt motor speed with 0.05 Ω armature resistance added](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Ashunt-motor-220v-25a-0p05ohm-torque) — Electrical Machines I, chapter 4, asked 3×

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