---
title: "Series motor current at 800 rpm"
subject: "Electrical Machine"
chapter: 4
key: "semester_4/electrical_machine:series-motor-current-at-800rpm"
source: https://www.bctengineer.com/q/semester_4/electrical_machine/series-motor-current-at-800rpm
site: BCT Engineering
---

# Series motor current at 800 rpm

Electrical Machine · Chapter 4: DC Motor

- Asked in: 2069 Bhadra
- Video: [Problem in DC series motor](https://www.youtube.com/watch?v=sYf7Uo5WxSo) — BeTECH SIVA
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machine/series-motor-current-at-800rpm
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machine/chapter-4

## Question

A dc series motor with armature resistance of 0.06 Ω, and field winding resistance of 0.04 Ω is supplied by a 220 V source. If the motor draws 25 A when running at 1200 rpm, calculate the current drawn by motor when running at 800 rpm.

## Short answer

Assuming linear magnetization, Eb ∝ I N. Solving 220 − 0.1 I = 217.5 × (I/25) × (800/1200) gives I ≈ 37.3 A.

## Answer

The load condition is not given, so it is assumed that the field is not saturated (Φ ∝ I). Then E_b ∝ I N.

#### Equivalent circuit
```text
 +--[Ra=0.06]--[Rse=0.04]--(Eb)--+
 |                I               |
 +-------------- 220 V -----------+
```
R = 0.06 + 0.04 = 0.1 Ω

#### First condition
E_b1 = 220 − 25 × 0.1 = 217.5 V at 1200 rpm, I₁ = 25 A

#### Second condition (800 rpm, I₂ unknown)
E_b2 = 220 − 0.1 I₂

E_b2 / E_b1 = (Φ₂ N₂) / (Φ₁ N₁) = (I₂/25) × (800/1200)

220 − 0.1 I₂ = 217.5 × (I₂/25) × (2/3) = 5.8 I₂

220 = 5.9 I₂

**I₂ = 37.29 A**

Check: E_b2 = 220 − 3.729 = 216.27 V; and 217.5 × (37.29/25) × (2/3) = 216.27 V ✓.

**Answer: the motor draws about 37.3 A at 800 rpm** (a lower speed means the load torque is higher, so more current is drawn).

## Check yourself

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

1. What assumption is made for the flux?
   A. Field is unsaturated so Φ ∝ I
   B. Flux is zero
   C. Flux is constant
   D. Flux ∝ 1/I
   Answer: A. Without saturation data, Φ ∝ I is assumed. (Key point: Assumption)

2. What is E_b1?
   A. 220 V
   B. 217.5 V
   C. 222.5 V
   D. 195 V
   Answer: B. 220 - 25 × 0.1 = 217.5 V. (Key point: Initial back emf)

3. Which equation is solved for I2?
   A. I2 = 25 × 800/1200
   B. 220 = 0.1 I2
   C. 220 - 0.1 I2 = 217.5 × (I2/25) × (800/1200)
   D. 220 - 0.1 I2 = 217.5 × (800/1200)
   Answer: C. E_b2/E_b1 = (I2/I1)(N2/N1). (Key point: Equation)

4. What is the current at 800 rpm?
   A. 25 A
   B. 16.7 A
   C. 50 A
   D. 37.3 A
   Answer: D. 220 = 5.9 I2 gives 37.29 A; lower speed means more load torque and current. (Key point: Result)

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