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

# Short shunt compound generator, 100 A at 250 V

Electrical Machines · Chapter 3: DC Generator

- Asked in: 2071 Magh, 6 marks
- Video: [Short Shunt DC Compound Generator - Problem 1 (TAGALOG)](https://www.youtube.com/watch?v=zSnfNAfDAQs) — Engr. Aaron Pineda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines/gen-short-shunt-100a-250v
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/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 $130\ \Omega$, armature $0.1\ \Omega$ and the series field $0.1\ \Omega$. Find the emf generated and the armature current, if the brush drop is 1 V per brush.

## Short answer

V_sh=250+100×0.1=260 V; I_sh=260/130=2 A; I_a=102 A. With 2 V brush drop, E=260+102×0.1+2=272.2 V.

## Answer

In a short shunt compound generator the shunt field is connected across the armature only, so the series field carries the load current $I_L$ and the shunt field sees $V + I_LR_{se}$.

Given: $I_L = 100$ A, $V = 250$ V, $R_{sh} = 130\ \Omega$, $R_{se} = 0.1\ \Omega$, $R_a = 0.1\ \Omega$.

Brush drop $= 2$ V (1 V per brush, 2 brushes).

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

#### Step 1: Voltage across the shunt field

$$
V_{sh} = V + I_LR_{se} = 250 + 100(0.1) = 260.00\ \text{V}
$$

#### Step 2: Currents

$$
I_{sh} = \frac{V_{sh}}{R_{sh}} = \frac{260.00}{130} = 2.000\ \text{A}, \qquad I_a = I_L + I_{sh} = 100 + 2.000 = 102.000\ \text{A}
$$

#### Step 3: Generated emf

$$
E = V_{sh} + I_aR_a + 2 = 260.00 + (102.000)(0.1) + 2 = 272.20\ \text{V}
$$

**Answer:** generated emf $E = 272.20\ \text{V}$ and armature current $I_a = 102.000\ \text{A}$.

## Check yourself

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

1. In short shunt the voltage across the shunt field is:
   A. 250 V
   B. 240 V
   C. 272.2 V
   D. 260 V
   Answer: D. V + IL Rse = 250 + 100x0.1 = 260 V. (Key point: Short shunt Vsh)

2. The armature current is:
   A. 102 A
   B. 100 A
   C. 98 A
   D. 2 A
   Answer: A. Ish = 260/130 = 2 A so Ia = 102 A. (Key point: Armature current)

3. The generated emf, with 2 V total brush drop, is:
   A. 270.2 V
   B. 272.2 V
   C. 260 V
   D. 262.2 V
   Answer: B. E = 260 + 102x0.1 + 2 = 272.2 V. (Key point: Emf with brush drop)

## Related in Electrical Machines

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- [EMF of short shunt compound generator (0.5 Ω armature)](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Ashort-shunt-compound-50a-200v-0p5ohm) — Electrical Machines I, chapter 3, asked 2×
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## Related in other subjects

- [Short shunt compound generator, 100 A](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machine%3Ashort-shunt-compound-100a-250v) — Electrical Machine, chapter 3
- [EMF of long shunt compound generator](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Along-shunt-compound-40a-200v) — Electrical Machines I, chapter 3, asked 2×
- [Short shunt compound generator, 30 A](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machine%3Ashort-shunt-compound-30a-220v) — Electrical Machine, chapter 3
- [Short shunt compound generator, 80 A](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machine%3Ashort-shunt-compound-80a-220v) — Electrical Machine, chapter 3