---
title: "Shunt motor pump, torque ∝ speed²"
subject: "Electrical Machine"
chapter: 4
key: "semester_4/electrical_machine:shunt-motor-centrifugal-torque-square"
source: https://www.bctengineer.com/q/semester_4/electrical_machine/shunt-motor-centrifugal-torque-square
site: BCT Engineering
---

# Shunt motor pump, torque ∝ speed²

Electrical Machine · Chapter 4: DC Motor

- Asked in: 2068 Bhadra, 8 marks
- 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-torque-square
- 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 torque is proportional to the square of speed. 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. Given: Armature resistance (Ra) = 0.1 Ω and field winding resistance (Rf) = 100 Ω.

## Short answer

For T ∝ N², armature current ∝ N². Initial I_a=48A. At 800 rpm, I_a=30.72A, E_b=156.16V. Required series resistance R = (200–156.16)/30.72 – 0.1 ≈ 1.327Ω.

## Answer

Here the load torque is not constant: T ∝ N². Flux is constant (shunt motor), so T ∝ I_a. Therefore I_a ∝ N².

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

#### Initial condition (1000 rpm)
- I_sh = 200/100 = 2 A; I_a1 = 50 − 2 = 48 A
- E_b1 = 200 − 48 × 0.1 = 195.2 V

#### At 800 rpm
- I_a2 = I_a1 (N₂/N₁)² = 48 × (0.8)² = 48 × 0.64 = 30.72 A
- E_b2 = E_b1 × (800/1000) = 156.16 V

#### Resistance
V = E_b2 + I_a2 (R_a + R)

200 = 156.16 + 30.72 (0.1 + R)

0.1 + R = 43.84 / 30.72 = 1.4271 Ω

**R = 1.4271 − 0.1 = 1.327 Ω**

| Case | I_a (A) | E_b (V) | Series R (Ω) |
|---|---|---|---|
| Constant torque (for comparison) | 48 | 156.16 | 0.813 |
| T ∝ N² | 30.72 | 156.16 | 1.327 |

**Answer: R ≈ 1.327 Ω must be inserted in the armature circuit.**

## Check yourself

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

1. For T ∝ N^2 and constant flux, how does I_a vary with speed?
   A. I_a ∝ N^2
   B. I_a ∝ N
   C. I_a ∝ 1/N
   D. I_a is constant
   Answer: A. T ∝ I_a with constant flux, so I_a ∝ N^2. (Key point: Torque-current link)

2. What is I_a at 800 rpm?
   A. 24 A
   B. 30.72 A
   C. 38.4 A
   D. 48 A
   Answer: B. I_a2 = 48 × 0.8^2 = 30.72 A. (Key point: New armature current)

3. Why is the required resistance larger than for constant torque?
   A. Supply voltage is larger
   B. Back emf is larger
   C. Armature current is smaller, so a bigger R is needed to drop the same voltage
   D. Flux is larger
   Answer: C. The same voltage drop of 43.84 V at smaller current needs more resistance. (Key point: Comparison)

4. What is the resistance to insert?
   A. 1.427 Ω
   B. 0.1 Ω
   C. 0.813 Ω
   D. 1.327 Ω
   Answer: D. 0.1 + R = 43.84/30.72 = 1.4271, so R = 1.327 Ω. (Key point: Result)

## Close variants set in other exams

- [Shunt motor pump: resistance for 800 rpm](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machine%3Ashunt-motor-centrifugal-constant-torque) — Electrical Machine, chapter 4, asked 2×

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