---
title: "Shunt motor speed with 1.2 Ω armature resistance added"
subject: "Electrical Machines I"
chapter: 4
key: "semester_4/electrical_machines_i:shunt-motor-240v-30a-1p2ohm-torque"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/shunt-motor-240v-30a-1p2ohm-torque
site: BCT Engineering
---

# Shunt motor speed with 1.2 Ω armature resistance added

Electrical Machines I · Chapter 4: DC Motor

- Asked in: 2070 Bhadra, 8 marks
- Video: [Speed Control Of DC Motor | Armature resistance control | DC machines | Lec-60](https://www.youtube.com/watch?v=7lQaKwoz91Q) — Education 4u
- 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-30a-1p2ohm-torque
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-4

## Question

A 240 V dc shunt motor has armature-winding resistance of 0.4 ohm and field winding resistance of 120 ohms. It draws a current of 30 amp at half load and the corresponding speed is 1400 rpm. If a resistance of 1.2 ohm is connected in series with the armature winding and load torque is decreased by 20%, calculate the new speed.

## Short answer

Flux constant; torque proportional to armature current, speed to back emf. Initial Ia=28 A, Eb1=228.8 V. With torque reduced 20%, Ia2=22.4 A. Total armature resistance 1.6 Ω gives Eb2=204.16 V. New speed =1400×204.16/228.8 ≈ 1249 rpm.

## Answer

For a shunt motor the field is across the constant supply, so flux is constant. Hence $T \propto I_a$ and $N \propto E_b$.

**Given:** $V = 240$ V, $R_a = 0.4\ \Omega$, $R_{sh} = 120\ \Omega$, $I_{L1} = 30$ A, $N_1 = 1400$ rpm, $R_{ext} = 1.2\ \Omega$, $T_2 = 0.8\,T_1$.

#### Step 1: Initial condition

$$
\begin{aligned}
I_{sh} &= \frac{240}{120} = 2\ \text{A} \\
I_{a1} &= 30 - 2 = 28\ \text{A} \\
E_{b1} &= 240 - 28 \times 0.4 = 240 - 11.2 = 228.8\ \text{V}
\end{aligned}
$$

#### Step 2: New armature current

$$
I_{a2} = 0.8 \times I_{a1} = 0.8 \times 28 = 22.4\ \text{A}
$$

#### Step 3: New back emf

Total armature-circuit resistance $= 0.4 + 1.2 = 1.6\ \Omega$:

$$
\begin{aligned}
E_{b2} &= 240 - 22.4 \times 1.6 \\
&= 240 - 35.84 = 204.16\ \text{V}
\end{aligned}
$$

#### Step 4: New speed

$$
\begin{aligned}
N_2 &= N_1 \times \frac{E_{b2}}{E_{b1}} \\
&= 1400 \times \frac{204.16}{228.8} = 1249.23\ \text{rpm}
\end{aligned}
$$

| Quantity | Before | After |
|---|---|---|
| Armature current | 28 A | 22.4 A |
| Armature circuit R | 0.4 Ω | 1.6 Ω |
| Back emf | 228.8 V | 204.16 V |
| Speed | 1400 rpm | 1249.2 rpm |

Even though the load torque falls (which alone would raise the speed slightly), the large drop in the added 1.2 Ω resistance makes the speed fall.

Assumptions: constant field flux, no armature reaction, brush drop neglected.

**Answer: New speed $\approx 1249$ rpm.**

## Check yourself

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

1. With torque decreased by 20% and constant flux, what is Ia2 (Ia1 = 28 A)?
   A. 33.6 A
   B. 28 A
   C. 24.4 A
   D. 22.4 A
   Answer: D. T is proportional to Ia, so Ia2 = 0.8 x 28 = 22.4 A. (Key point: New armature current)

2. What is Eb2?
   A. 240 - 22.4 x 0.4 = 231 V
   B. 240 - 28 x 1.6 = 195.2 V
   C. 240 - 22.4 x 1.6 = 204.16 V
   D. 240 - 22.4 x 1.2 = 213.1 V
   Answer: C. Total armature resistance is 0.4 + 1.2 = 1.6 ohm, so Eb2 = 204.16 V. (Key point: Back emf)

3. Why does the speed fall even though load torque has decreased?
   A. Flux has been reduced
   B. The added resistor drop lowers Eb more than the lighter load raises it
   C. Torque decrease always lowers speed
   D. Supply voltage drops
   Answer: B. Eb falls from 228.8 V to 204.16 V because of the drop in the added 1.2 ohm, and N is proportional to Eb. (Key point: Speed direction)

4. What is the new speed?
   A. About 1249 rpm
   B. About 1400 rpm
   C. About 1569 rpm
   D. About 1119 rpm
   Answer: A. N2 = 1400 x 204.16/228.8 = 1249 rpm. (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 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 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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- [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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