---
title: "EMF of short shunt compound generator (50 A, 220 V)"
subject: "Electrical Machines I"
chapter: 3
key: "semester_4/electrical_machines_i:short-shunt-50a-220v-emf"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/short-shunt-50a-220v-emf
site: BCT Engineering
---

# EMF of short shunt compound generator (50 A, 220 V)

Electrical Machines I · Chapter 3: DC Generator

- Asked in: 2071 Bhadra, 5 marks
- Video: [How to Calculate Generated EMF in a Short Shunt Compound DC Generator (Step-by-Step Solved Problem) ](https://www.youtube.com/watch?v=VGVbSigl4LE) — 5 Minute Electrical 
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines_i/short-shunt-50a-220v-emf
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-3

## Question

A short shunt dc compound generator supplies a current of 50 A at 220 V. The shunt field, series field and armature winding resistances are 100 ohm, 0.05 ohm and 0.1 ohm respectively. Calculate the emf generated by the armature.

## Short answer

Shunt field voltage is 222.5 V, shunt current 2.225 A, armature current 52.225 A. Generated emf is 227.72 V.

## Answer

**Given:** $I_L = 50$ A, $V = 220$ V, $R_{sh} = 100\ \Omega$, $R_{se} = 0.05\ \Omega$, $R_a = 0.1\ \Omega$; short-shunt connection.

```text
     +--Ra--+-------Rse-------+
     |      |                 |
    (Eg)   Rsh              Load 220 V
     |      |                 |
     +------+-----------------+
```

$$
\begin{aligned}
V_{sh} &= V + I_L R_{se} = 220 + 50 \times 0.05 = 222.5\ \text{V} \\
I_{sh} &= \frac{222.5}{100} = 2.225\ \text{A} \\
I_a &= I_L + I_{sh} = 50 + 2.225 = 52.225\ \text{A} \\
E_g &= V_{sh} + I_a R_a = 222.5 + 52.225 \times 0.1 \\
&= 222.5 + 5.22 = 227.72\ \text{V}
\end{aligned}
$$

Brush drop is neglected as it is not given.

**Answer:** EMF generated $E_g \approx 227.72$ V.

## Check yourself

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

1. What is V_sh?
   A. 220 V
   B. 220 + 50 x 0.05 = 222.5 V
   C. 220 + 50 x 0.1 = 225 V
   D. 220 - 2.5 = 217.5 V
   Answer: B. The shunt field sees V plus the series field drop I_L R_se. (Key point: Shunt field voltage)

2. What is I_a?
   A. 50 A
   B. 52.225 A
   C. 47.775 A
   D. 52.5 A
   Answer: B. I_sh = 222.5/100 = 2.225 A; I_a = 50 + 2.225. (Key point: Armature current)

3. What is E_g?
   A. 222.5 + 5.22 = 227.72 V
   B. 220 + 5.22 = 225.22 V
   C. 222.5 - 5.22 = 217.28 V
   D. 220 + 52.225 x 0.15
   Answer: A. E_g = V_sh + I_a R_a = 222.5 + 52.225 x 0.1 = 227.72 V. (Key point: Generated emf)

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