---
title: "Short shunt compound generator, 100 A"
subject: "Electrical Machine"
chapter: 3
key: "semester_4/electrical_machine:short-shunt-compound-100a-250v"
source: https://www.bctengineer.com/q/semester_4/electrical_machine/short-shunt-compound-100a-250v
site: BCT Engineering
---

# Short shunt compound generator, 100 A

Electrical Machine · Chapter 3: DC Generator

- Asked in: 2071 Magh, 6 marks
- Video: [Example of short shunt compound generator](https://www.youtube.com/watch?v=QOGmMJNpfmE) — Dr Warid Alzihaymee
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machine/short-shunt-compound-100a-250v
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machine/chapter-3

## Question

A short shunt compound generator supplies a load current of 100 A at 250 V. The generator has the following winding resistances: shunt field 130 Ω, armature 0.1 Ω and the series field 0.1 Ω. Find the emf generated and the armature current, if the brush drop is 1 V per brush.

## Short answer

Series drop = 10 V, Vsh = 260 V, Ish = 2 A, Ia = 102 A. Adding armature drop and 2 V brush drop yields E = 272.2 V, Ia = 102 A.

## Answer

In a short shunt connection the shunt field is across the armature and the series field is in the load line.

#### Equivalent circuit
```text
 Ia --[Ra=0.1]--+--[Rse=0.1]--+--> IL=100 A  V=250 V
 (E)             |            |
               Rsh=130 ohm (across armature)
```

#### Calculation
1. Series field drop = 100 × 0.1 = 10 V
2. Voltage across the shunt branch (armature terminals):
   V_sh = 250 + 10 = 260 V
3. Shunt current: I_sh = 260 / 130 = 2 A
4. **Armature current**: I_a = 100 + 2 = **102 A**
5. Armature drop = 102 × 0.1 = 10.2 V
6. Brush drop = 2 × 1 = 2 V
7. **EMF generated**: E = 260 + 10.2 + 2 = **272.2 V**

| Step | Value |
|---|---|
| Series drop | 10 V |
| V across armature terminals | 260 V |
| I_sh | 2 A |
| I_a | 102 A |
| Armature drop | 10.2 V |
| Brush drop | 2 V |
| E | 272.2 V |

**Answer: E = 272.2 V and I_a = 102 A.**

## Check yourself

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

1. What is the voltage drop in the series field?
   A. 0.1 V
   B. 25 V
   C. 10 V (100 A x 0.1 ohm)
   D. 1 V
   Answer: C. The series field carries the load current in short shunt. (Key point: Series drop)

2. What is I_a?
   A. 100 A
   B. 98 A
   C. 2 A
   D. 102 A
   Answer: D. I_sh = 260/130 = 2 A; I_a = 100 + 2. (Key point: Armature current)

3. What is the generated emf?
   A. 272.2 V
   B. 262.2 V
   C. 260 V
   D. 250 V
   Answer: A. E = 260 + 102 x 0.1 + 2 = 272.2 V. (Key point: Emf)

## Related in Electrical Machine

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