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

# Shunt motor driving pump: torque proportional to speed squared

Electrical Machines · Chapter 4: DC Motor

- Asked in: 2068 Bhadra, 8 marks
- Video: [Numerical on speed control of DC shunt motors  #electricmotors #electricalengineering](https://www.youtube.com/watch?v=T7KmAY7FYA4) — EEE, SJBIT
- 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-torque-square
- 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 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 ($R_a$) = $0.1\ \Omega$ and field winding resistance ($R_f$) = $100\ \Omega$.

## Answer

For a fan or centrifugal pump, torque is proportional to the square of speed: $T\propto N^2$. The flux is constant, so $T\propto I_a$ and therefore $I_a\propto N^2$. The speed is lowered by an external armature resistance $R_x$, and $N\propto E_b$.

**Initial condition (1000 rpm):**

$$
I_f = \frac{200}{100} = 2\ \text{A}, \qquad I_{a1} = 50 - 2 = 48\ \text{A}
$$

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

**At 800 rpm:**

$$
I_{a2} = I_{a1}\left(\frac{800}{1000}\right)^2 = 48\times 0.64 = 30.72\ \text{A}
$$

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

$$
R_a + R_x = \frac{V - E_{b2}}{I_{a2}} = \frac{200 - 156.16}{30.72} = 1.4271\ \Omega
$$

$$
R_x = 1.4271 - 0.1 = 1.3271\ \Omega
$$

**Answer:** external armature resistance $R_x = 1.3271\ \Omega$ (compare with $0.8133\ \Omega$ for a constant-torque load).

## Check yourself

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

1. For a centrifugal pump, torque varies as:
   A. Constant
   B. N
   C. N^2
   D. 1/N
   Answer: C. Fan or pump load torque is proportional to the square of speed. (Key point: Load characteristic)

2. Ia2 is:
   A. 48 A
   B. 38.4 A
   C. 30.72 A
   D. 24 A
   Answer: C. 48 x 0.64 = 30.72 A, because Ia is proportional to N^2. (Key point: Armature current at 800 rpm)

3. Ra + Rx is found from:
   A. (V - Eb2)/Ia2 = 43.84/30.72
   B. (V - Eb2)/48
   C. Eb2/Ia2
   D. V/Ia2
   Answer: A. Voltage equation with the new Ia2 gives 1.4271 ohm. (Key point: Resistance calculation)

4. Compared with constant torque (0.8133 ohm), Rx here is:
   A. Smaller
   B. Equal
   C. Zero
   D. Larger, 1.3271 ohm, as current is lower
   Answer: D. Lower current needs a bigger resistance for the same voltage drop. (Key point: Comparison with constant torque)

## Close variants set in other exams

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

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## Related in other subjects

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