---
title: "Shunt motor pump: resistance for 800 rpm"
subject: "Electrical Machine"
chapter: 4
key: "semester_4/electrical_machine:shunt-motor-centrifugal-constant-torque"
source: https://www.bctengineer.com/q/semester_4/electrical_machine/shunt-motor-centrifugal-constant-torque
site: BCT Engineering
---

# Shunt motor pump: resistance for 800 rpm

Electrical Machine · Chapter 4: DC Motor

- Asked in: 2072 Asoj
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machine/shunt-motor-centrifugal-constant-torque
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machine/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, Ra = 0.1 Ω and field winding resistance, Rf = 100 Ω.

## Short answer

Shunt field constant, flux constant. At constant torque, armature current remains 48A. E_b at 1000 rpm = 195.2V, at 800 rpm = 156.16V. Required series resistance R = (200–156.16)/48 – 0.1 = 0.813Ω.

## Answer

At constant torque and constant flux (shunt field is unchanged), the armature current stays the same. So the speed is reduced by inserting resistance, which lowers the back emf.

#### Equivalent circuit
```text
 +--[Ra=0.1]--[R ?]--(Eb)--+
 |       Ia = 48 A          |
 +---- Rf = 100 ohm --------+--- V = 200 V
```

#### Step 1: initial state
- I_sh = 200 / 100 = 2 A
- I_a1 = 50 − 2 = 48 A
- E_b1 = 200 − 48 × 0.1 = 195.2 V at 1000 rpm

#### Step 2: new state (constant torque)
T ∝ Φ I_a, Φ unchanged, so I_a2 = I_a1 = 48 A.

E_b2 = E_b1 × (800/1000) = 195.2 × 0.8 = 156.16 V

#### Step 3: required resistance
V = E_b2 + I_a (R_a + R)

200 = 156.16 + 48 (0.1 + R)

0.1 + R = 43.84 / 48 = 0.9133 Ω

**R = 0.9133 − 0.1 = 0.813 Ω**

**Answer: a resistance of about 0.813 Ω must be inserted in series with the armature.**

## Check yourself

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

1. With constant torque and constant flux, what happens to I_a?
   A. It stays the same (48 A)
   B. It halves
   C. It rises in proportion to speed
   D. It becomes zero
   Answer: A. T ∝ ΦI_a, so constant T and Φ means constant I_a. (Key point: Armature current at constant torque)

2. What is E_b2 at 800 rpm?
   A. 124.9 V
   B. 156.16 V
   C. 195.2 V
   D. 200 V
   Answer: B. E_b2 = 195.2 × 0.8 = 156.16 V. (Key point: Back emf at 800 rpm)

3. Which equation gives the extra resistance?
   A. 200 = 195.2 + 48R
   B. 200 = 156.16 + 50(0.1 + R)
   C. 200 = 156.16 + 48(0.1 + R)
   D. 156.16 = 200 + 48R
   Answer: C. V = E_b2 + I_a(R_a + R) with I_a = 48 A. (Key point: Resistance equation)

4. What is the required series resistance?
   A. 1.327 Ω
   B. 0.1 Ω
   C. 0.913 Ω
   D. 0.813 Ω
   Answer: D. 0.1 + R = 43.84/48 = 0.9133, so R = 0.813 Ω. (Key point: Result)

## Close variants set in other exams

- [Shunt motor pump, torque ∝ speed²](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machine%3Ashunt-motor-centrifugal-torque-square) — Electrical Machine, chapter 4, asked 2×

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