---
title: "Long shunt compound generator speed, lap and wave"
subject: "Electrical Machines"
chapter: 3
key: "semester_4/electrical_machines:gen-long-shunt-10kw-lap-wave"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/gen-long-shunt-10kw-lap-wave
site: BCT Engineering
---

# Long shunt compound generator speed, lap and wave

Electrical Machines · Chapter 3: DC Generator

- Asked in: 2078 Poush, 8 marks
- Video: [Problems on DC Generators, shunt and compound generators](https://www.youtube.com/watch?v=69kUeAA1dYk) — Electrical and Electronics engg classes made easy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines/gen-long-shunt-10kw-lap-wave
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-3

## Question

A 4 pole, 250 V dc long shunt compound generator supplies a load of 10 kW at rated voltage. The armature, series field and shunt field resistances are $0.1\ \Omega$, $0.15\ \Omega$ and $250\ \Omega$ respectively. The armature is lap wound with 50 slots, each slot containing 6 conductors. If the flux per pole is 50 mWb, calculate the speed of generator. What would be the speed of same generator if armature is wave wound?

## Short answer

With 40 A load at 250 V, shunt field current 1 A, armature current 41 A, generated emf = 260.25 V. For lap winding (300 conductors, A=4): speed = 1041.0 rpm. For wave winding (A=2): speed = 520.5 rpm.

## Answer

In a long shunt compound generator the shunt field is across the **armature plus series field combination** (i.e. across the load terminals), so the series field carries the armature current.

```text
       Ia        Rse
  +--->---[~~~~]---+---------+----o +
  |                |         |
 (E,Ra)          [Rsh]     Load
  |                |         |
  +----------------+---------+----o -
```

#### Currents

$$
I_L = \frac{10000}{250} = 40\ \text{A}, \qquad I_{sh} = \frac{250}{250} = 1\ \text{A}
$$

$$
I_a = I_{se} = I_L + I_{sh} = 40 + 1 = 41\ \text{A}
$$

#### Generated emf

$$
E_g = V + I_a(R_a + R_{se}) = 250 + 41(0.1 + 0.15) = 260.25\ \text{V}
$$

#### Speed with lap winding

Slots $\times$ conductors per slot: $Z = 50\times 6 = 300$. For lap winding $A = P = 4$.

$$
E_g = \frac{\Phi Z N P}{60A} \;\Rightarrow\; N = \frac{60AE_g}{\Phi Z P} = \frac{60\times 4\times 260.25}{0.05\times 300\times 4} = 1041.0\ \text{rpm}
$$

#### Speed with wave winding

For wave winding $A = 2$; $E_g$, $\Phi$, $Z$ and $P$ are unchanged.

$$
N = \frac{60\times 2\times 260.25}{0.05\times 300\times 4} = 520.5\ \text{rpm}
$$

The wave-wound machine needs only half the speed because it has half the parallel paths (twice the conductors in series per path).

**Answer:** lap wound: $N = 1041.0\ \text{rpm}$; wave wound: $N = 520.5\ \text{rpm}$. (Brush drop is not given, so it is neglected.)

## Check yourself

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

1. In a long shunt compound generator, the armature current Ia equals:
   A. I_L only
   B. I_L - I_sh
   C. I_L + I_sh, and the series field carries the same current
   D. I_sh only
   Answer: C. The shunt is across armature plus series, so the series carries Ia = IL + Ish. (Key point: Long shunt currents)

2. Which expression gives the generated emf of a long shunt generator (brush drop neglected)?
   A. E = V + Ia Ra
   B. E = V - Ia(Ra + Rse)
   C. E = V + IL(Ra + Rse)
   D. E = V + Ia(Ra + Rse)
   Answer: D. Both armature and series field resistances carry Ia and drop voltage. (Key point: Generated emf)

3. With E = 260.25 V, Phi = 0.05 Wb, Z = 300, P = 4 and lap winding (A = 4), the speed is:
   A. 1041 rpm
   B. 520.5 rpm
   C. 2082 rpm
   D. 260 rpm
   Answer: A. N = 60x4x260.25/(0.05x300x4) = 1041 rpm. (Key point: Lap speed)

4. Why does the same generator need only about half the speed when wave wound?
   A. Flux per pole doubles
   B. A = 2 instead of 4, so conductors in series per path double
   C. Generated emf halves
   D. Armature resistance halves
   Answer: B. N is proportional to A, so changing A from 4 to 2 halves the speed to 520.5 rpm. (Key point: Wave winding speed)

## Close variants set in other exams

- [Long shunt compound generator speed, 300 conductors](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Agen-long-shunt-10kw-300-conductors) — Electrical Machines, chapter 3, asked 2×

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## Related in other subjects

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