---
title: "Shunt motor speed with 0.1 Ω armature resistance, less load"
subject: "Electrical Machines I"
chapter: 4
key: "semester_4/electrical_machines_i:shunt-motor-240v-5a-armature-0p1"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/shunt-motor-240v-5a-armature-0p1
site: BCT Engineering
---

# Shunt motor speed with 0.1 Ω armature resistance, less load

Electrical Machines I · Chapter 4: DC Motor

- Asked in: 2073 Bhadra, 8 marks
- Video: [speed control of dc motor | speed control of dc shunt motor | dc motor speed control | motor control](https://www.youtube.com/watch?v=BPGt3zxUn_8) — Mind of Engineer 
- 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-240v-5a-armature-0p1
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-4

## Question

A 240 V dc shunt motor has armature resistance of 0.4 Ω and field winding resistance of 120 Ω. It runs at 1500 rpm and draws a current of 5 A with certain load on its shaft. A resistance of 0.1 Ω is connected in series with armature winding and the load on the shaft is reduced by 20%, calculate the new speed of the motor.

## Short answer

Fixed field current gives constant flux. Initial back emf 238.8 V; with extra 0.1 Ω and reduced armature current (2.4 A), back emf stays 238.8 V, so speed remains 1500 rpm.

## Answer

Adding resistance in series with the armature leaves the field current unchanged, so flux is constant; then $T \propto I_a$ and $N \propto E_b$.

**Given:** $V = 240$ V, $R_a = 0.4\ \Omega$, $R_{sh} = 120\ \Omega$, $N_1 = 1500$ rpm, $I_{L1} = 5$ A; extra 0.1 Ω in armature; load torque reduced by 20%.

#### Initial condition
$$
\begin{aligned}
I_{sh} &= \frac{240}{120} = 2\ \text{A}, \quad I_{a1} = 5 - 2 = 3\ \text{A} \\
E_{b1} &= 240 - 3 \times 0.4 = 238.8\ \text{V}
\end{aligned}
$$

#### New condition
$$
\begin{aligned}
I_{a2} &= 0.8 \times 3 = 2.4\ \text{A} \\
E_{b2} &= 240 - 2.4 \times (0.4 + 0.1) = 240 - 1.2 = 238.8\ \text{V}
\end{aligned}
$$

#### New speed
$$
N_2 = 1500 \times \frac{238.8}{238.8} = 1500\ \text{rpm}
$$

The extra drop in the added resistance exactly balances the reduced drop from the smaller current ($3 \times 0.4 = 2.4 \times 0.5 = 1.2$ V), so the speed is unchanged.

**Answer:** New speed $= 1500$ rpm (no change).

## Check yourself

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

1. Does adding armature resistance change flux?
   A. No, field current is unchanged
   B. Yes, it weakens it
   C. Yes, it strengthens it
   D. It reverses it
   Answer: A. So T ∝ Ia. (Key point: Flux)

2. What is Ia2 for 20% less torque?
   A. 3.6 A
   B. 2.4 A
   C. 2 A
   D. 3 A
   Answer: B. 0.8×3. (Key point: Ia2)

3. What is Eb2?
   A. 239.04 V
   B. 240 V
   C. 238.8 V
   D. 237.6 V
   Answer: C. 240 - 2.4×0.5. (Key point: Eb2)

4. What is the new speed?
   A. 1560 rpm
   B. 1440 rpm
   C. 1200 rpm
   D. 1500 rpm, unchanged
   Answer: D. Eb1 = Eb2. (Key point: Result)

## Close variants set in other exams

- [Shunt motor speed with 50 Ω field resistance and lower torque](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Ashunt-motor-220v-25a-50ohm-torque) — Electrical Machines I, chapter 4, asked 4×
- [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×

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