---
title: "Plugging resistance for 30 kW DC shunt motor"
subject: "Utilization of Electrical Energy"
chapter: 3
key: "semester_7/utilization_of_electrical_energy:dc-shunt-30kw-plugging-resistance"
source: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/dc-shunt-30kw-plugging-resistance
site: BCT Engineering
---

# Plugging resistance for 30 kW DC shunt motor

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

- Asked in: 2073 Shrawan
- Video: [Mani V | Electrical Machines and Drives | Plugging of DC Shunt Motor | SNS Institutions](https://www.youtube.com/watch?v=xv-bANdVLeM) — Mani V
- 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-30kw-plugging-resistance
- Whole chapter: https://www.bctengineer.com/md/semester_7/utilization_of_electrical_energy/chapter-3

## Question

A 30 kW, 220V, dc shunt motor with a full load speed of 535 rev/min is to be braked by plugging. Estimate the value of resistance which should be placed in series with it to limit the initial braking current to 200A. What would be the initial value of the electric braking torque and the value when the speed has fallen to half its full load value? (Given: armature resistance of motor = 0.086 Ω, Full load armature current = 150A)

## Short answer

For plugging, total armature circuit resistance is (220+207.1)/200 = 2.1355 ohm, so series resistance is 2.05 ohm. Initial braking torque is 739.3 N·m; at half speed current is 151.5 A and torque is 560.1 N·m.

## Answer

Data: $V = 220$ V, $N = 535$ rpm, $R_a = 0.086\ \Omega$, $I_a = 150$ A (full load), limit $I_b = 200$ A.

#### Back emf at full-load speed
$$
\begin{aligned} E_b &= V - I_a R_a = 220 - 150\times0.086 = 207.1\ \text{V} \\ \omega &= \frac{2\pi\times535}{60} = 56.03\ \text{rad/s} \\ K\phi &= \frac{207.1}{56.03} = 3.6966\ \text{V s/rad} \end{aligned}
$$

#### Series resistance for plugging
At the instant of plugging, the reversed supply voltage and the back emf act in the same direction round the armature circuit:
$$
\begin{aligned} R_a + R &= \frac{V + E_b}{I_b} = \frac{220 + 207.1}{200} = 2.1355\ \Omega \\ R &= 2.1355 - 0.086 = 2.05\ \Omega \end{aligned}
$$

#### Initial braking torque
$$
T_{b0} = K\phi\, I_b = 3.6966 \times 200 = 739.3\ \text{N·m}
$$

#### Torque at half speed (267.5 rpm)
$$
\begin{aligned} E &= \frac{207.1}{2} = 103.55\ \text{V} \\ I &= \frac{220 + 103.55}{2.1355} = 151.51\ \text{A} \\ T &= 3.6966 \times 151.51 = 560.1\ \text{N·m} \end{aligned}
$$

**Answer: series resistance = 2.05 Ω; initial braking torque = 739.3 N·m; torque at half speed = 560.1 N·m.**

(Full-load torque for comparison $= 3.6966 \times 150 = 554.5$ N·m, so braking starts at 1.33 × full-load torque.)

## Check yourself

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

1. Back emf at full load is:
   A. 220 V
   B. 207.1 V
   C. 232.9 V
   D. 150 V
   Answer: B. Eb = 220 - 150 x 0.086 = 207.1 V. (Key point: Back emf)

2. How is the total resistance for plugging found?
   A. (V - Eb)/Ib
   B. V/Ib
   C. (V + Eb)/Ib = 427.1/200 = 2.1355 Ω, so R = 2.05 Ω
   D. Eb/Ib
   Answer: C. Supply and back emf add after reversal; subtract Ra = 0.086 Ω. (Key point: Series resistance)

3. At half speed the braking torque is:
   A. Larger than initial
   B. Zero
   C. 560.1 N·m, smaller than the initial 739.3 N·m
   D. 739.3 N·m unchanged
   Answer: C. E = 103.55 V, I = 323.55/2.1355 = 151.5 A, T = 3.6966 x 151.5 = 560.1 N·m. (Key point: Half-speed torque)

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