---
title: "Series motor speed with 5 Ω series resistance"
subject: "Electrical Machines I"
chapter: 4
key: "semester_4/electrical_machines_i:series-motor-200v-800rpm-5ohm"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/series-motor-200v-800rpm-5ohm
site: BCT Engineering
---

# Series motor speed with 5 Ω series resistance

Electrical Machines I · Chapter 4: DC Motor

- Asked in: 2075 Baisakh, 8 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-200v-800rpm-5ohm
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-4

## Question

A 200 V d.c. series motor runs at 800 rpm when taking a line current of 15 A. The armature and field resistances are 0.6 Ω and 0.4 Ω respectively. Find the speed at which it will run when connected in series with a 5 Ω resistance and taking the same current at the same voltage.

## Short answer

Back emf without added resistance is 185 V, with 5 Ω it drops to 110 V. Constant current keeps flux unchanged, so speed is proportional to back emf: 800 × 110/185 ≈ 476 rpm.

## Answer

For a series motor, the same current means the same flux, so speed is directly proportional to back emf: $\dfrac{N_2}{N_1} = \dfrac{E_{b2}}{E_{b1}}$.

**Given:** $V = 200$ V, $I = 15$ A, $N_1 = 800$ rpm, $R_a = 0.6\ \Omega$, $R_{se} = 0.4\ \Omega$, extra series resistance 5 Ω.

#### Back emfs
$$
\begin{aligned}
E_{b1} &= V - I (R_a + R_{se}) = 200 - 15 \times 1.0 = 185\ \text{V} \\
E_{b2} &= 200 - 15 \times (1.0 + 5) = 200 - 90 = 110\ \text{V}
\end{aligned}
$$

#### New speed
$$
N_2 = 800 \times \frac{110}{185} = 475.7\ \text{rpm}
$$

**Answer:** Speed with 5 Ω in series $\approx 476$ rpm.

## Check yourself

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

1. Why is N2/N1 = Eb2/Eb1?
   A. Same current means same flux
   B. Torque is zero
   C. Speed is constant
   D. Voltage differs
   Answer: A. N ∝ Eb/φ. (Key point: Same current)

2. What is Eb1?
   A. 200 V
   B. 185 V
   C. 170 V
   D. 110 V
   Answer: B. 200 - 15×1. (Key point: Eb1)

3. What is Eb2?
   A. 185 V
   B. 125 V
   C. 110 V
   D. 90 V
   Answer: C. 200 - 15×6. (Key point: Eb2)

4. What is the new speed?
   A. About 800 rpm
   B. About 1345 rpm
   C. About 300 rpm
   D. About 476 rpm
   Answer: D. 800×110/185. (Key point: Speed)

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## Related in other subjects

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