---
title: "Shunt motor driving pump: constant torque resistance"
subject: "Electrical Machines"
chapter: 4
key: "semester_4/electrical_machines:dcm-shunt-200v-pump-const-torque"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/dcm-shunt-200v-pump-const-torque
site: BCT Engineering
---

# Shunt motor driving pump: constant torque resistance

Electrical Machines · Chapter 4: DC Motor

- Asked in: 2072 Asoj
- Video: [DC Motor Problems: Examples 1-4 (Motors #5)](https://www.youtube.com/watch?v=iQpDBn9rIeM) — Aaron Danner
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines/dcm-shunt-200v-pump-const-torque
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-4

## Question

A 200 V DC shunt motor drives a centrifugal pump where constant torque is required. The motor draws a current of 50 A when running at 1000 rpm. What value of resistance must be inserted in the armature circuit to reduce the speed to 800 rpm at constant torque? Given that armature winding resistance $R_a = 0.1\ \Omega$ and field winding resistance $R_f = 100\ \Omega$.

## Short answer

Constant torque requires the same armature current of 48 A. At the reduced speed of 800 rpm, the back emf is 156.16 V, so the total armature circuit resistance must be (200 – 156.16) / 48 = 0.9133 Ω, yielding an external resistance of 0.8133 Ω.

## Answer

The field is across the supply, so the flux is constant. Constant torque means constant armature current. The speed is lowered by an external armature resistance $R_x$, so $N\propto E_b$.

**Initial condition:**

$$
I_f = \frac{200}{100} = 2\ \text{A}, \qquad I_a = 50 - 2 = 48\ \text{A}
$$

$$
E_{b1} = 200 - 48(0.1) = 195.20\ \text{V}
$$

**At 800 rpm:**

$$
E_{b2} = E_{b1}\times\frac{800}{1000} = 195.20\times 0.8 = 156.16\ \text{V}
$$

The torque is constant, so $I_a = 48$ A:

$$
V = E_{b2} + I_a(R_a + R_x) \;\Rightarrow\; R_a + R_x = \frac{200 - 156.16}{48} = 0.9133\ \Omega
$$

$$
R_x = 0.9133 - 0.1 = 0.8133\ \Omega
$$

**Answer:** external armature resistance $R_x = 0.8133\ \Omega$.

## Check yourself

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

1. For a shunt motor with constant flux, constant torque means:
   A. Constant speed
   B. Constant back emf
   C. Constant armature current
   D. Constant line current only
   Answer: C. T is proportional to Phi Ia with Phi fixed. (Key point: Constant torque meaning)

2. Ia = 48 A gives Eb1 =
   A. 195.2 V
   B. 200 V
   C. 190 V
   D. 199.5 V
   Answer: A. 200 - 48 x 0.1 = 195.2 V. (Key point: Initial Eb)

3. Eb2 at 800 rpm is:
   A. 195.2 V
   B. 176 V
   C. 160 V
   D. 156.16 V
   Answer: D. 195.2 x 0.8 = 156.16 V. (Key point: Eb at 800 rpm)

4. The external armature resistance is:
   A. 0.9133 ohm
   B. 0.8133 ohm
   C. 1.3271 ohm
   D. 0.1 ohm
   Answer: B. (200 - 156.16)/48 = 0.9133; minus 0.1 gives 0.8133 ohm. (Key point: External resistance)

## Close variants set in other exams

- [Shunt motor driving pump: torque proportional to speed squared](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Adcm-shunt-200v-pump-torque-square) — Electrical Machines, chapter 4, asked 2×

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