---
title: "Shunt motor speed with 1 Ω armature resistance added"
subject: "Electrical Machines I"
chapter: 4
key: "semester_4/electrical_machines_i:shunt-motor-240v-27a-1ohm-torque"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/shunt-motor-240v-27a-1ohm-torque
site: BCT Engineering
---

# Shunt motor speed with 1 Ω armature resistance added

Electrical Machines I · Chapter 4: DC Motor

- Asked in: 2070 Magh, 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-27a-1ohm-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 27 amp at half load and the corresponding speed is 1200 rpm. If a resistance of 1 ohm is connected in series with the armature winding and load torque is increased by 20%, calculate the new speed.

## Short answer

Field flux constant, so torque is proportional to armature current and speed to back emf. Initial armature current 25 A, back emf 230 V. With 20% higher load torque, armature current becomes 30 A. Added 1 ohm gives total armature resistance 1.4 ohm, back emf 198 V. New speed = 1200 × 198/230 ≈ 1033 rpm.

## Answer

In a shunt motor the field current, and so the flux, stays constant. Then $T \propto I_a$ and $N \propto E_b$.

**Given:** $V = 240$ V, $R_a = 0.4\ \Omega$, $R_{sh} = 120\ \Omega$, $I_{L1} = 27$ A, $N_1 = 1200$ rpm. Extra series resistance $R_{ext} = 1\ \Omega$. New torque $T_2 = 1.2\,T_1$.

#### Step 1: Initial condition

$$
\begin{aligned}
I_{sh} &= \frac{V}{R_{sh}} = \frac{240}{120} = 2\ \text{A} \\
I_{a1} &= I_{L1} - I_{sh} = 27 - 2 = 25\ \text{A} \\
E_{b1} &= V - I_{a1} R_a = 240 - 25 \times 0.4 = 230\ \text{V}
\end{aligned}
$$

#### Step 2: New armature current

With constant flux, $T \propto I_a$:

$$
I_{a2} = 1.2 \times I_{a1} = 1.2 \times 25 = 30\ \text{A}
$$

#### Step 3: New back emf

The armature circuit resistance is now $R_a + R_{ext} = 0.4 + 1 = 1.4\ \Omega$:

$$
\begin{aligned}
E_{b2} &= V - I_{a2}(R_a + R_{ext}) \\
&= 240 - 30 \times 1.4 = 198\ \text{V}
\end{aligned}
$$

#### Step 4: New speed

With constant flux, $N \propto E_b$:

$$
\begin{aligned}
N_2 &= N_1 \times \frac{E_{b2}}{E_{b1}} \\
&= 1200 \times \frac{198}{230} = 1033.04\ \text{rpm}
\end{aligned}
$$

Assumptions: the field current stays at 2 A (the extra resistance is in the armature only), armature reaction and brush drop are neglected. The 20% rise is taken on the half-load torque.

**Answer: New speed $\approx 1033$ rpm (armature current 30 A, back emf 198 V).**

## Check yourself

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

1. What are Ia1 and Eb1 for this motor (240 V, 27 A, Ra = 0.4, Rsh = 120)?
   A. 27 A and 229.2 V
   B. 25 A and 230 V
   C. 25 A and 240 V
   D. 29 A and 228.4 V
   Answer: B. Ish = 2 A, Ia1 = 25 A, Eb1 = 240 - 25 x 0.4 = 230 V. (Key point: Initial Ia and Eb)

2. Why is flux constant when 1 ohm is placed in series with the armature?
   A. Field current is set by V/Rsh, unaffected by armature resistance
   B. Armature reaction is large
   C. Torque is constant
   D. Field resistance rises
   Answer: A. The shunt field is across the supply, so flux is unchanged and T is proportional to Ia. (Key point: Constant flux)

3. What is Eb2 with Ia2 = 30 A and total armature resistance 1.4 ohm?
   A. 240 - 30 x 0.4 = 228 V
   B. 240 - 30 x 1.4 = 198 V
   C. 240 - 25 x 1.4 = 205 V
   D. 240 - 30 x 1 = 210 V
   Answer: B. The extra resistor is in the armature circuit, so use 1.4 ohm: 240 - 42 = 198 V. (Key point: New back emf)

4. What is the new speed?
   A. About 1033 rpm
   B. About 1200 rpm
   C. About 1394 rpm
   D. About 860 rpm
   Answer: A. N2 = 1200 x 198/230 = 1033 rpm because N is proportional to Eb at constant flux. (Key point: New 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 1.2 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×

## Related in Electrical Machines I

- [Armature resistance to reduce shunt motor speed to 1250 rpm](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Ashunt-motor-250v-30a-1250rpm-armature) — Electrical Machines I, chapter 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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## Related in other subjects

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- [Shunt motor: resistance for 1200 rpm](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machine%3Ashunt-motor-40a-1400-to-1200) — Electrical Machine, chapter 4