---
title: "Short shunt compound generator emf, 80 A at 220 V"
subject: "Electrical Machines"
chapter: 3
key: "semester_4/electrical_machines:gen-short-shunt-80a-220v"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/gen-short-shunt-80a-220v
site: BCT Engineering
---

# Short shunt compound generator emf, 80 A at 220 V

Electrical Machines · Chapter 3: DC Generator

- Asked in: 2068 Magh, 8 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-80a-220v
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-3

## Question

A short shunt compound generator delivers a current of 80 A to the load at 220 V. The shunt field, series and armature winding resistances are $100\ \Omega$, $0.05\ \Omega$ and $0.1\ \Omega$ respectively. Calculate the emf generated by the armature.

## Short answer

Shunt field voltage is V + I_L R_se = 220 + 80×0.05 = 224 V. Shunt current 224/100 = 2.24 A, armature current 80 + 2.24 = 82.24 A. Generated emf E = 224 + 82.24×0.1 = 232.22 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: $I_L = 80$ A, $V = 220$ V, $R_{sh} = 100\ \Omega$, $R_{se} = 0.05\ \Omega$, $R_a = 0.1\ \Omega$.

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

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

$$
V_{sh} = V + I_LR_{se} = 220 + 80(0.05) = 224.00\ \text{V}
$$

#### Step 2: Currents

$$
I_{sh} = \frac{V_{sh}}{R_{sh}} = \frac{224.00}{100} = 2.240\ \text{A}, \qquad I_a = I_L + I_{sh} = 80 + 2.240 = 82.240\ \text{A}
$$

#### Step 3: Generated emf

$$
E = V_{sh} + I_aR_a = 224.00 + (82.240)(0.1) = 232.22\ \text{V}
$$

**Answer:** generated emf $E = 232.22\ \text{V}$.

## Check yourself

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

1. The shunt field voltage in short shunt is:
   A. 220 V
   B. 216 V
   C. 224 V
   D. 232.22 V
   Answer: C. 220 + 80x0.05 = 224 V. (Key point: Shunt voltage)

2. The armature current is:
   A. 80 A
   B. 77.76 A
   C. 2.24 A
   D. 82.24 A
   Answer: D. Ish = 224/100 = 2.24 A; Ia = IL + Ish. (Key point: Armature current)

3. The emf generated is:
   A. 232.22 V
   B. 228.2 V
   C. 224 V
   D. 230.0 V
   Answer: A. E = 224 + 82.24x0.1 = 232.22 V. (Key point: Emf)

## Close variants set in other exams

- [Long shunt compound generator emf, 80 A at 230 V](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Agen-long-shunt-80a-230v) — Electrical Machines, chapter 3, asked 2×

## Related in Electrical Machines

- [EMF of short shunt compound generator (50 A, 220 V)](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Ashort-shunt-50a-220v-emf) — Electrical Machines I, chapter 3
- [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

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