---
title: "DC chopper: input power, speed, torque, speed range"
subject: "Utilization of Electrical Energy"
chapter: 3
key: "semester_7/utilization_of_electrical_energy:chopper-400v-input-power-speed-torque"
source: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/chopper-400v-input-power-speed-torque
site: BCT Engineering
---

# DC chopper: input power, speed, torque, speed range

Utilization of Electrical Energy · Chapter 3: Control of Electric Drive

- Asked in: 2079 Baisakh, 8 marks; 2070 Chaitra (asked 2 times)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/chopper-400v-input-power-speed-torque
- Whole chapter: https://www.bctengineer.com/md/semester_7/utilization_of_electrical_energy/chapter-3

## Question

A DC chopper fed from 400V supply runs a separately excited motor. The armature resistance Ra = 0.1 Ω. The motor voltage constant is 4 Vsec/radian, the average armature current Ia = 150 A. The armature current is continuous and has negligible ripple. Determine
a) The input power
b) The motor speed and
c) The developed torque for a duty cycle of 60%
Neglect the losses in chopper.
If the duty cycle of chopper varies between 10% and 90%, find maximum and minimum speed.

## Short answer

With Va = kVs = 240 V, input power is Vs(kIa) = 36 kW. Motor speed is Eg/Kv = (240-15)/4 = 56.25 rad/s = 537 rpm, and torque is KvIa = 600 N-m. For duty cycle 10-90%, speed ranges from 59.7 to 823.6 rpm.

## Answer

**Data:** $V_s = 400$ V, $R_a = 0.1\ \Omega$, $K_v = 4$ V-s/rad, $I_a = 150$ A (continuous, ripple-free), duty cycle $k = 0.6$. For a step-down chopper, $V_a = kV_s$ and the average source current is $kI_a$.

#### a) Input power

$$
\begin{aligned}
V_a &= kV_s = 0.6\times400 = 240\ \text{V} \\
P_i &= V_s(kI_a) = 400\times0.6\times150 = 36{,}000\ \text{W} = 36\ \text{kW}
\end{aligned}
$$

#### b) Motor speed

$$
\begin{aligned}
E_g &= V_a - I_aR_a = 240 - 150\times0.1 = 225\ \text{V} \\
\omega &= \frac{E_g}{K_v} = \frac{225}{4} = 56.25\ \text{rad/s} \\
N &= \frac{56.25\times60}{2\pi} = 537.1\ \text{rpm}
\end{aligned}
$$

#### c) Developed torque

$$
T = K_vI_a = 4\times150 = 600\ \text{N-m}
$$

#### Speed range for $k$ = 0.1 to 0.9 (same $I_a = 150$ A)

| Duty cycle | $V_a$ (V) | $E_g = V_a - 15$ (V) | $\omega$ (rad/s) | $N$ (rpm) |
|---|---|---|---|---|
| 0.1 | 40 | 25 | 6.25 | 59.7 |
| 0.9 | 360 | 345 | 86.25 | 823.6 |

**Answer:** (a) 36 kW; (b) 56.25 rad/s = 537.1 rpm; (c) 600 N-m. Speed range: minimum 59.7 rpm (6.25 rad/s) at 10 %, maximum 823.6 rpm (86.25 rad/s) at 90 % duty cycle.

## Check yourself

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

1. Input power at k = 0.6 is:
   A. 36 kW
   B. 54 kW
   C. 60 kW
   D. 21.6 kW
   Answer: A. Pi = Vs (k Ia) = 400 x 0.6 x 150 = 36 kW. (Key point: Input power)

2. Speed at k = 0.6 uses:
   A. Eg = Va - Ia Ra = 225 V, w = 225/4 = 56.25 rad/s
   B. Eg = Va = 240 V
   C. w = Va/Ra
   D. Eg = Vs
   Answer: A. Eg = 240 - 15 = 225 V; N = 537.1 rpm. (Key point: Speed)

3. Developed torque is:
   A. 600 N-m
   B. 150 N-m
   C. 960 N-m
   D. 225 N-m
   Answer: A. T = Kv Ia = 4 x 150. (Key point: Torque)

4. Minimum speed at k = 0.1 is:
   A. 59.7 rpm
   B. 6.25 rpm
   C. 823.6 rpm
   D. 40 rpm
   Answer: A. Eg = 40 - 15 = 25 V, w = 6.25 rad/s = 59.7 rpm. (Key point: Speed range)

## Related in Utilization of Electrical Energy

- [Chopper ON time for separately excited DC motor](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Achopper-on-time-separately-excited-motor) — Utilization of Electrical Energy, chapter 3
- [Speed of separately excited DC motor (given J, B)](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Aseparately-excited-motor-speed-load-100nm) — Utilization of Electrical Energy, chapter 3
- [Six-pulse converter: supply voltage and firing range](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Asix-pulse-converter-firing-range) — Utilization of Electrical Energy, chapter 3
- [DC shunt motor speed at reduced voltage and starting resistance](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Adc-shunt-reduced-voltage-starting-resistance) — Utilization of Electrical Energy, chapter 3
- [Plugging of 200 V, 1000 rpm separately excited motor](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Aseparately-excited-plugging-200v-1000rpm) — Utilization of Electrical Energy, chapter 3
- [Plugging of 220 V separately excited motor (100 A)](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Aseparately-excited-plugging-220v-100a) — Utilization of Electrical Energy, chapter 3

## Related in other subjects

- [25 kW separately excited DC machine as motor or generator](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Asep-excited-25kw-125v-terminal-voltage) — Electrical Machines I, chapter 3
- [Separately excited elevator motor with field flux reduced](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Aelevator-motor-300nm-flux-reduced) — Electrical Machines I, chapter 4
- [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×