---
title: "Short shunt compound generator, 30 A"
subject: "Electrical Machine"
chapter: 3
key: "semester_4/electrical_machine:short-shunt-compound-30a-220v"
source: https://www.bctengineer.com/q/semester_4/electrical_machine/short-shunt-compound-30a-220v
site: BCT Engineering
---

# Short shunt compound generator, 30 A

Electrical Machine · Chapter 3: DC Generator

- Asked in: 2072 Magh
- 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-30a-220v
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machine/chapter-3

## Question

A short shunt compound generator delivers a load current of 30 A at 220 V and has armature, series and shunt field resistances of 0.05 Ω, 0.03 Ω and 200 Ω respectively. Calculate the induced emf and armature current. Allow 1 V per brush contact drop.

## Short answer

Series drop = 0.9 V, Vsh = 220.9 V, Ish = 1.1045 A, Ia = 31.10 A. Total drop including 2 V brush drop gives E = 224.46 V, Ia = 31.10 A.

## Answer

In a short shunt compound generator the shunt field is across the armature only, and the series field carries the load current.

#### Equivalent circuit
```text
 Ia --[Ra=0.05]--+--[Rse=0.03]--+--> IL=30 A  V=220 V
 (E)              |             |
                 Rsh=200 ohm (across armature)
```

#### Calculation
1. Series field drop = I_L R_se = 30 × 0.03 = 0.9 V
2. Voltage across shunt (and armature terminals):
   V_sh = 220 + 0.9 = 220.9 V
3. Shunt current: I_sh = 220.9 / 200 = 1.1045 A
4. **Armature current**: I_a = I_L + I_sh = 30 + 1.1045 = **31.10 A**
5. Armature drop: I_aR_a = 31.1045 × 0.05 = 1.555 V
6. Brush drop = 2 × 1 V = 2 V (two brushes)
7. **Induced emf**: E = V_sh + I_aR_a + brush drop = 220.9 + 1.555 + 2 = **224.46 V**

| Quantity | Value |
|---|---|
| Series drop | 0.9 V |
| Shunt voltage | 220.9 V |
| I_sh | 1.1045 A |
| I_a | 31.10 A |
| E | 224.46 V |

**Answer: armature current = 31.10 A and induced emf = 224.46 V.**

## Check yourself

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

1. In a short shunt connection, what is the voltage across the shunt field?
   A. 220 V
   B. 219.1 V
   C. 222 V
   D. 220.9 V (220 V + 30 x 0.03)
   Answer: D. The series drop 0.9 V adds to the terminal voltage. (Key point: Shunt voltage)

2. What is the armature current?
   A. 31.10 A
   B. 30 A
   C. 28.9 A
   D. 31.10 mA
   Answer: A. I_sh = 220.9/200 = 1.1045 A; I_a = 30 + 1.1045. (Key point: Armature current)

3. How is E found, and what is it?
   A. E = 220 + 31.1 = 251.1 V
   B. E = V_sh + I_aR_a + brush drop = 224.46 V
   C. E = 220 - 1.555 = 218.4 V
   D. E = 220.9 V
   Answer: B. 220.9 + 1.555 + 2 V for two brushes = 224.46 V. (Key point: Emf)

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## Related in other subjects

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