---
title: "Series motor current at 800 rpm"
subject: "Electrical Machines"
chapter: 4
key: "semester_4/electrical_machines:dcm-series-220v-current-800rpm"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/dcm-series-220v-current-800rpm
site: BCT Engineering
---

# Series motor current at 800 rpm

Electrical Machines · Chapter 4: DC Motor

- Asked in: 2069 Bhadra; 2066 Magh (old course), 8 marks (asked 2 times)
- 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_machines/dcm-series-220v-current-800rpm
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-4

## Question

A dc series motor with armature resistance of $0.06\ \Omega$ and field winding resistance of $0.04\ \Omega$ 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

With total resistance 0.1 Ω, Eb1 = 220−25×0.1 = 217.5 V. For series motor Eb ∝ IN, so (220−0.1I2)/217.5 = (I2/25)×(800/1200). Solving gives I2 ≈ 37.3 A.

## Answer

In a series motor the field current is the armature current, so (unsaturated) the flux is proportional to the current: $\Phi\propto I$. The back emf then follows $E_b \propto \Phi N \propto I N$.

Total resistance $R = R_a + R_{se} = 0.06 + 0.04 = 0.1\ \Omega$.

#### Condition 1: 1200 rpm, 25 A

$$
E_{b1} = V - I_1R = 220 - 25(0.1) = 217.5\ \text{V}
$$

#### Condition 2: 800 rpm, current $I_2$

$$
E_{b2} = 220 - 0.1I_2
$$

Using $\dfrac{E_{b2}}{E_{b1}} = \dfrac{\Phi_2N_2}{\Phi_1N_1} = \dfrac{I_2N_2}{I_1N_1}$:

$$
\frac{220 - 0.1I_2}{217.5} = \frac{I_2}{25}\times\frac{800}{1200}
$$

$$
220 - 0.1I_2 = 5.800\,I_2 \;\Rightarrow\; I_2 = \frac{220}{0.1 + 5.800} = 37.29\ \text{A}
$$

Check: $E_{b2} = 220 - 0.1(37.29) = 216.27\ \text{V}$.

**Answer:** the motor draws $I_2 = 37.29\ \text{A}$ at 800 rpm (assuming an unsaturated magnetic circuit).

## Check yourself

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

1. In an unsaturated series motor, flux is proportional to:
   A. Armature current
   B. Supply voltage
   C. Speed
   D. Constant
   Answer: A. The field carries Ia. (Key point: Series flux)

2. The total resistance in the circuit is:
   A. 0.06 ohm
   B. Ra + Rse = 0.1 ohm
   C. 0.04 ohm
   D. 0.0024 ohm
   Answer: B. Series field is in series with the armature. (Key point: Total resistance)

3. The back emf in condition 1 is:
   A. 220 V
   B. 222.5 V
   C. 220 - 25x0.1 = 217.5 V
   D. 217.5 x 1200
   Answer: C. Eb = V - IR. (Key point: Back emf at 1200 rpm)

4. The current at 800 rpm is:
   A. 25 A
   B. 16.7 A
   C. 50 A
   D. 37.29 A
   Answer: D. Eb2/Eb1 = I2N2/(I1N1) gives 220 - 0.1I2 = 5.8 I2, so I2 = 37.29 A. (Key point: Result)

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