---
title: "Series motor speed at 25 A"
subject: "Electrical Machines I"
chapter: 4
key: "semester_4/electrical_machines_i:series-motor-230v-100a-25a-speed"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/series-motor-230v-100a-25a-speed
site: BCT Engineering
---

# Series motor speed at 25 A

Electrical Machines I · Chapter 4: DC Motor

- Asked in: 2078 Chaitra, 10 marks
- Video: [Understanding The DC Series Motor: GATE Electrical Machines](https://www.youtube.com/watch?v=_9Gus73253A) — Ekeeda GATE & ESE
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines_i/series-motor-230v-100a-25a-speed
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-4

## Question

A 230 V dc series motor draws a line current of 100 A from mains while running at 1000 rpm. Its armature resistance is 0.15 Ohms and field resistance is 0.1 Ohms. When the motor draws current of 25 A, calculate the speed.

## Short answer

Back EMF at 100 A: E_b1 = 230 - 100×(0.15+0.1) = 205 V. At 25 A: E_b2 = 230 - 25×0.25 = 223.75 V. Since flux ∝ I, new speed N_2 = 1000 × (223.75/205) × (100/25) ≈ 4366 rpm.

## Answer

For a DC series motor below saturation, flux is proportional to the current: $\phi \propto I$. Hence

$$
\frac{N_2}{N_1} = \frac{E_{b2}}{E_{b1}} \times \frac{\phi_1}{\phi_2} = \frac{E_{b2}}{E_{b1}} \times \frac{I_1}{I_2}
$$

**Given:** $V = 230$ V, $I_1 = 100$ A, $N_1 = 1000$ rpm, $R_a = 0.15\ \Omega$, $R_{se} = 0.1\ \Omega$, $I_2 = 25$ A. Brush drop neglected.

```text
  + o---Rse---Ra---(M)---o -
        I = Ia = Ise
```

#### Step 1: Back emf at 100 A
$$
\begin{aligned}
E_{b1} &= V - I_1 (R_a + R_{se}) \\
&= 230 - 100 \times 0.25 = 205\ \text{V}
\end{aligned}
$$

#### Step 2: Back emf at 25 A
$$
\begin{aligned}
E_{b2} &= 230 - 25 \times 0.25 \\
&= 230 - 6.25 = 223.75\ \text{V}
\end{aligned}
$$

#### Step 3: New speed
$$
\begin{aligned}
N_2 &= 1000 \times \frac{223.75}{205} \times \frac{100}{25} \\
&= 1000 \times 1.0915 \times 4 = 4365.9\ \text{rpm}
\end{aligned}
$$

| Quantity | 100 A | 25 A |
|---|---|---|
| $I(R_a + R_{se})$ drop | 25 V | 6.25 V |
| Back emf | 205 V | 223.75 V |
| Relative flux | 1 | 0.25 |
| Speed | 1000 rpm | 4366 rpm |

**Answer:** Speed at 25 A $\approx 4366$ rpm.

The speed rises more than four times when current falls to one quarter. This is the typical series characteristic: at light load the flux is weak and the speed becomes very high, so a series motor is always coupled to its load directly or through gears, never through a belt.

## Check yourself

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

1. What is assumed about flux in a series motor?
   A. Flux is proportional to line current (unsaturated)
   B. Flux is constant
   C. Flux is inversely proportional to current
   D. Flux is proportional to speed
   Answer: A. This gives N2/N1 = (Eb2/Eb1)(I1/I2). (Key point: Flux assumption)

2. What are Eb1 and Eb2?
   A. 230 V and 230 V
   B. 205 V and 223.75 V
   C. 205 V and 205 V
   D. 255 V and 236.25 V
   Answer: B. V - I(0.25) with I = 100 A and 25 A. (Key point: Back emf values)

3. What is the new speed?
   A. About 1092 rpm
   B. About 916 rpm
   C. About 4366 rpm
   D. About 250 rpm
   Answer: C. 1000×(223.75/205)×4. (Key point: Speed ratio)

4. Why does speed rise so much at light load?
   A. Back emf drops sharply
   B. Armature resistance increases
   C. Supply voltage increases
   D. Flux weakens with current, so speed rises
   Answer: D. Hence a series motor is never belt-coupled. (Key point: Light-load behaviour)

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