---
title: "Reverse current (plugging) braking of 20 kW DC shunt motor"
subject: "Utilization of Electrical Energy"
chapter: 3
key: "semester_7/utilization_of_electrical_energy:dc-shunt-reverse-current-braking-20kw"
source: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/dc-shunt-reverse-current-braking-20kw
site: BCT Engineering
---

# Reverse current (plugging) braking of 20 kW DC shunt motor

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

- Asked in: 2080 Baisakh, 8 marks
- Video: [plugging braking of dc motor |  braking in induction motor | in hindi | rheostatic braking | L-13](https://www.youtube.com/watch?v=L5xKZteSjJE) — Mind of Engineer 
- 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/dc-shunt-reverse-current-braking-20kw
- Whole chapter: https://www.bctengineer.com/md/semester_7/utilization_of_electrical_energy/chapter-3

## Question

A 220 V, 20 kW dc shunt motor running at its rated speed of 1200 rpm is to be braked by reverse current braking. The armature resistance is 0.1 Ω and the rated efficiency of the motor is 88%. Calculate:
i) The resistance to be connected in series with the armature to limit the initial braking current to twice the rated current.
ii) The initial braking torque.
iii) The torque when motor falls to 400 rpm.

## Short answer

Rated current 103.3 A, back emf 209.7 V. (i) Series resistance to limit initial braking current to twice rated is 1.98 Ω. (ii) Initial braking torque 344.7 N·m. (iii) Torque when speed falls to 400 rpm is 232.6 N·m.

## Answer

Data: $V = 220$ V, $P = 20$ kW, $N = 1200$ rpm, $R_a = 0.1\ \Omega$, $\eta = 88\ \%$. Shunt field current is not given, so it is neglected ($I_a \approx I_L$).

#### Rated current and back emf
$$
\begin{aligned} I_a &= \frac{P}{\eta V} = \frac{20000}{0.88\times220} = 103.31\ \text{A} \\ E_b &= V - I_a R_a = 220 - 10.33 = 209.67\ \text{V} \\ \omega &= \frac{2\pi\times1200}{60} = 125.66\ \text{rad/s} \\ K\phi &= \frac{E_b}{\omega} = \frac{209.67}{125.66} = 1.6685\ \text{V s/rad} \end{aligned}
$$

#### i) Series resistance for $I_b = 2I_a$
In reverse current braking (plugging) the armature is reversed, so the supply voltage and back emf add:
$$
\begin{aligned} R_a + R &= \frac{V + E_b}{2I_a} = \frac{220 + 209.67}{206.61} = 2.0796\ \Omega \\ R &= 2.0796 - 0.1 = 1.98\ \Omega \end{aligned}
$$

#### ii) Initial braking torque
$$
T_b = K\phi\, I_b = 1.6685 \times 206.61 = 344.7\ \text{N·m}
$$

#### iii) Torque at 400 rpm
$$
\begin{aligned} E &= 209.67\times\frac{400}{1200} = 69.89\ \text{V} \\ I &= \frac{220 + 69.89}{2.0796} = 139.40\ \text{A} \\ T &= 1.6685 \times 139.40 = 232.6\ \text{N·m} \end{aligned}
$$

**Answer: (i) 1.98 Ω, (ii) 344.7 N·m, (iii) 232.6 N·m.**

The braking torque falls as the motor slows, but does not become zero at standstill ($I = 220/2.0796 = 105.8$ A), so the supply must be switched off at zero speed or the motor will start in reverse.

## Check yourself

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

1. How is rated armature current found?
   A. 20000/220
   B. 20000/(0.88 x 220) = 103.3 A
   C. 20000 x 0.88/220
   D. 220/0.1
   Answer: B. Input power = P/eta, so Ia = P/(eta V) = 103.3 A. (Key point: Rated current)

2. Why does total resistance equal (V + Eb)/Ib?
   A. Supply and back emf oppose each other
   B. Armature reversal makes supply voltage and back emf add
   C. Back emf is zero
   D. Only V acts
   Answer: B. After reversing the armature the two act in the same direction; (220 + 209.67)/206.61 = 2.08 Ω, so R = 1.98 Ω. (Key point: Plugging equation)

3. How is initial braking torque found?
   A. T = V Ib
   B. T = K phi x Ib = 1.6685 x 206.61 = 344.7 N·m
   C. T = Eb Ib
   D. T = Ib^2 R
   Answer: B. Torque equals the flux constant times the braking current. (Key point: Torque)

4. What happens if supply is not disconnected at zero speed?
   A. Current falls to zero
   B. Torque remains (I = 105.8 A) and motor starts in reverse
   C. Motor stays at rest
   D. Back emf rises
   Answer: B. At zero speed Eb = 0, so I = 220/2.08 = 105.8 A and torque persists. (Key point: At standstill)

## Related in Utilization of Electrical Energy

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## Related in other subjects

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