---
title: "Long shunt compound generator speed, 300 conductors"
subject: "Electrical Machines"
chapter: 3
key: "semester_4/electrical_machines:gen-long-shunt-10kw-300-conductors"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/gen-long-shunt-10kw-300-conductors
site: BCT Engineering
---

# Long shunt compound generator speed, 300 conductors

Electrical Machines · Chapter 3: DC Generator

- Asked in: 2070 Bhadra, 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-300-conductors
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-3

## Question

A 4 pole, 250 V long shunt dc compound generator supplies a load of 10 kW at the rated voltage. The armature, series and shunt field resistances are $0.1\ \Omega$, $0.15\ \Omega$ and $250\ \Omega$ respectively. The armature is lap wound with 300 conductors. If the flux per pole is 50 mWb, calculate the speed of the generator.

## Short answer

Armature current = 41 A, generated emf = 260.25 V. With lap winding (4 poles, 300 conductors, A=4), speed N = 60×4×260.25/(0.05×300×4) = 1041.0 rpm.

## Answer

In a long shunt machine the shunt field is across the load terminals, so the series field carries the full armature current.

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

#### Currents

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

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

#### Generated emf

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

#### Speed

For lap winding $A = P = 4$:

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

**Answer:** generator speed $= 1041.0\ \text{rpm}$ (brush drop not given, so neglected).

## Check yourself

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

1. For 10 kW at 250 V with Rsh = 250 ohm, IL and Ish are:
   A. 40 A and 1 A
   B. 10 A and 1 A
   C. 40 A and 2 A
   D. 41 A and 1 A
   Answer: A. IL = 10000/250 = 40 A; Ish = 250/250 = 1 A. (Key point: Load and shunt current)

2. In this long shunt generator the armature current is:
   A. 40 A
   B. 41 A
   C. 1 A
   D. 39 A
   Answer: B. Ia = IL + Ish = 40 + 1 = 41 A. (Key point: Armature current)

3. The generated emf is:
   A. 250 V
   B. 254.1 V
   C. 260.25 V
   D. 256.15 V
   Answer: C. E = 250 + 41(0.1 + 0.15) = 260.25 V. (Key point: Generated emf)

4. For lap winding with 4 poles, A equals 4, giving speed:
   A. 520.5 rpm
   B. 2082 rpm
   C. 1000 rpm
   D. 1041 rpm
   Answer: D. N = 60x4x260.25/(0.05x300x4) = 1041 rpm. (Key point: Lap speed)

## Close variants set in other exams

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

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## Related in other subjects

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