---
title: "200 V shunt motor power, torque and efficiency"
subject: "Electrical Machines I"
chapter: 4
key: "semester_4/electrical_machines_i:shunt-motor-200v-rotational-loss"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/shunt-motor-200v-rotational-loss
site: BCT Engineering
---

# 200 V shunt motor power, torque and efficiency

Electrical Machines I · Chapter 4: DC Motor

- Asked in: 2082 Kartik (new course), 6 marks
- Video: [Calculating Back EMF, Power and Torque for a DC Shunt Motor](https://www.youtube.com/watch?v=ndseqKIlR9c) — Engineers Academy
- 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-200v-rotational-loss
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-4

## Question

The armature and field resistances of a 200 V DC shunt motor are 0.12 Ω and 250 Ω respectively. The rotational loss of the motor is 250 W. The full load line current is 10 A and motor rpm is 1500. Determine: (i) The mechanical power developed (ii) The power output (iii) The load torque (iv) The full load efficiency

## Short answer

With Ish=0.8 A and Ia=9.2 A, Eb=198.9 V. Mechanical power developed is 1829.8 W; output after 250 W rotational loss is 1579.8 W. Load torque is 10.06 N·m and full-load efficiency is 79.0%.

## Answer

**Given:** $V = 200$ V, $R_a = 0.12\ \Omega$, $R_{sh} = 250\ \Omega$, rotational loss = 250 W, $I_L = 10$ A, $N = 1500$ rpm.

#### Currents and back emf
$$
\begin{aligned}
I_{sh} &= \frac{200}{250} = 0.8\ \text{A} \\
I_a &= I_L - I_{sh} = 10 - 0.8 = 9.2\ \text{A} \\
E_b &= V - I_a R_a = 200 - 9.2 \times 0.12 = 198.896\ \text{V}
\end{aligned}
$$

#### (i) Mechanical power developed
$$
P_m = E_b I_a = 198.896 \times 9.2 = 1829.84\ \text{W}
$$

#### (ii) Power output
$$
P_{out} = P_m - P_{rot} = 1829.84 - 250 = 1579.84\ \text{W}
$$

#### (iii) Load (shaft) torque
$$
\begin{aligned}
\omega &= \frac{2\pi N}{60} = \frac{2\pi \times 1500}{60} = 157.08\ \text{rad/s} \\
T_L &= \frac{P_{out}}{\omega} = \frac{1579.84}{157.08} = 10.06\ \text{N·m}
\end{aligned}
$$

#### (iv) Full-load efficiency
$$
\begin{aligned}
P_{in} &= V I_L = 200 \times 10 = 2000\ \text{W} \\
\eta &= \frac{1579.84}{2000} \times 100 = 78.99\%
\end{aligned}
$$

Check of losses: armature copper $= 9.2^2 \times 0.12 = 10.16$ W, field copper $= 200 \times 0.8 = 160$ W, rotational 250 W; total 420.16 W $= 2000 - 1579.84$ W.

**Answer:** (i) $P_m \approx 1829.8$ W, (ii) $P_{out} \approx 1579.8$ W, (iii) $T_L \approx 10.06$ N·m, (iv) $\eta \approx 79.0\%$.

## Check yourself

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

1. How is armature current found for a shunt motor given line current?
   A. Ia = IL - Ish, where Ish = V/Rsh
   B. Ia = IL + Ish
   C. Ia = V/Ra
   D. Ia = IL
   Answer: A. Ish = 200/250 = 0.8 A, so Ia = 9.2 A. (Key point: Armature current)

2. What is the mechanical power developed?
   A. V·IL = 2000 W
   B. Eb·Ia = about 1829.8 W
   C. V·Ia = 1840 W
   D. Eb·IL = 1830 W
   Answer: B. Eb = 200 - 9.2×0.12 = 198.896 V, so Pm = 198.896×9.2. (Key point: Mechanical power)

3. How is the output power obtained from Pm?
   A. Add the rotational loss to Pm
   B. Subtract the field copper loss from Pm
   C. Subtract the rotational loss: Pout = Pm - 250 W
   D. Pout equals Pm
   Answer: C. Rotational (iron + friction) loss is taken from the developed power, giving 1579.8 W. (Key point: Output power)

4. What is the full-load efficiency?
   A. About 91.4%
   B. About 68.2%
   C. About 82.6%
   D. About 79.0%
   Answer: D. η = 1579.84/2000 = 78.99%. (Key point: Efficiency)

## Related in Electrical Machines I

- [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×
- [Full-load torque, rotor current and copper loss of 440 V motor](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Aim-440v-1425rpm-rotor-impedance) — Electrical Machines I, chapter 5
- [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 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 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 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×

## Related in other subjects

- [DC shunt motor armature torque](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_and_electronics_engineering%3Adc-shunt-motor-torque) — Basic Electrical and Electronics Engineering, chapter 2
- [Shunt motor speed with 250 ohm in field circuit](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Adcm-shunt-field-resistance-250) — Electrical Machines, chapter 4
- [Shunt motor: field resistance to raise speed to 1100 rpm](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Adcm-shunt-field-resistance-1100rpm) — Electrical Machines, chapter 4
- [Armature size, 10 kW 900 rpm motor](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Adc-motor-10kw-900rpm) — Electric Machine Design, chapter 5, asked 5×