---
title: "Shunt motor speed with 0.1 Ω armature resistance added"
subject: "Electrical Machines I"
chapter: 4
key: "semester_4/electrical_machines_i:shunt-motor-220v-20a-0p1ohm-armature"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/shunt-motor-220v-20a-0p1ohm-armature
site: BCT Engineering
---

# Shunt motor speed with 0.1 Ω armature resistance added

Electrical Machines I · Chapter 4: DC Motor

- Asked in: 2075 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-220v-20a-0p1ohm-armature
- 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 20 A and runs at 1400 rpm with certain load on the shaft. The armature winding resistance and field winding resistance are 0.2 Ω and 220 Ω respectively. If a resistance of 0.1 Ω is added in series with armature and the load torque is increased by 10%, calculate the new speed.

## Answer

Adding resistance in the armature circuit does not change the field current, so flux is constant. Then $T \propto I_a$ and $N \propto E_b$.

**Given:** $V = 220$ V, $I_{L1} = 20$ A, $N_1 = 1400$ rpm, $R_a = 0.2\ \Omega$, $R_{sh} = 220\ \Omega$, extra 0.1 Ω in armature, torque +10%.

#### Initial condition
$$
\begin{aligned}
I_{sh} &= \frac{220}{220} = 1\ \text{A}, \quad I_{a1} = 20 - 1 = 19\ \text{A} \\
E_{b1} &= 220 - 19 \times 0.2 = 216.2\ \text{V}
\end{aligned}
$$

#### New condition
Flux constant, torque up 10%:

$$
\begin{aligned}
I_{a2} &= 1.1 \times 19 = 20.9\ \text{A} \\
E_{b2} &= 220 - 20.9 \times (0.2 + 0.1) = 220 - 6.27 = 213.73\ \text{V}
\end{aligned}
$$

#### New speed
$$
N_2 = N_1 \times \frac{E_{b2}}{E_{b1}} = 1400 \times \frac{213.73}{216.2} = 1384.0\ \text{rpm}
$$

**Answer:** New speed $\approx 1384$ rpm.

## Check yourself

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

1. Adding resistance in the armature circuit does what to flux?
   A. Leaves it unchanged
   B. Reduces it
   C. Increases it
   D. Reverses it
   Answer: A. Field current is unchanged. (Key point: Flux)

2. With flux constant and torque +10%, what is Ia2?
   A. 19 A
   B. 20.9 A
   C. 22 A
   D. 17.3 A
   Answer: B. 1.1×19. (Key point: Ia2)

3. What is Eb2?
   A. 215.9 V
   B. 216.2 V
   C. 213.73 V
   D. 220 V
   Answer: C. 220 - 20.9×0.3. (Key point: Eb2)

4. What is the new speed?
   A. About 1400 rpm
   B. About 1540 rpm
   C. About 1100 rpm
   D. About 1384 rpm
   Answer: D. 1400×213.73/216.2. (Key point: Speed)

## Close variants set in other exams

- [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×
- [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, 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×

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