---
title: "EMF of short shunt compound generator (0.5 Ω armature)"
subject: "Electrical Machines I"
chapter: 3
key: "semester_4/electrical_machines_i:short-shunt-compound-50a-200v-0p5ohm"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/short-shunt-compound-50a-200v-0p5ohm
site: BCT Engineering
---

# EMF of short shunt compound generator (0.5 Ω armature)

Electrical Machines I · Chapter 3: DC Generator

- Asked in: 2073 Magh, 8 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-compound-50a-200v-0p5ohm
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-3

## Question

A dc short shunt compound generator has armature winding resistance of 0.5 ohm, series field winding resistance of 0.3 ohm and shunt field winding resistance of 200 ohms. It supplies a current of 50 Amp to the load at 200 V. Calculate the emf generated by the armature.

## Short answer

For short shunt, shunt field voltage is V + I_L R_se = 200 + 50×0.3 = 215 V. Shunt current is 215/200 = 1.075 A, so armature current is 51.075 A. Generated emf = 215 + 51.075×0.5 = 240.54 V.

## Answer

**Given:** $I_L = 50$ A, $V = 200$ V, $R_a = 0.5\ \Omega$, $R_{se} = 0.3\ \Omega$, $R_{sh} = 200\ \Omega$; short-shunt connection.

```text
     +--Ra--+-------Rse-------+
     |      |                 |
    (Eg)   Rsh              Load 200 V
     |      |                 |
     +------+-----------------+
```

#### Step 1: Voltage across shunt field
Series field carries load current:

$$
V_{sh} = V + I_L R_{se} = 200 + 50 \times 0.3 = 215\ \text{V}
$$

#### Step 2: Shunt field and armature current
$$
\begin{aligned}
I_{sh} &= \frac{215}{200} = 1.075\ \text{A} \\
I_a &= I_L + I_{sh} = 50 + 1.075 = 51.075\ \text{A}
\end{aligned}
$$

#### Step 3: Generated emf
$$
\begin{aligned}
E_g &= V_{sh} + I_a R_a \\
&= 215 + 51.075 \times 0.5 \\
&= 215 + 25.54 = 240.54\ \text{V}
\end{aligned}
$$

Brush drop is neglected as it is not given.

**Answer:** EMF generated by the armature $E_g \approx 240.54$ V.

## Check yourself

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

1. What voltage is across the shunt field?
   A. 200 V
   B. 200 + 50 x 0.3 = 215 V
   C. 200 - 15 = 185 V
   D. 200 + 50 x 0.5 = 225 V
   Answer: B. In short shunt the shunt field is across the armature: V + I_L R_se, with the series field carrying I_L. (Key point: Shunt field voltage)

2. What is I_a?
   A. 50 A
   B. 50 + 1.075 = 51.075 A
   C. 50 - 1.075 = 48.925 A
   D. 50 + 1 = 51 A
   Answer: B. I_sh = 215/200 = 1.075 A and I_a = I_L + I_sh. (Key point: Armature current)

3. What is E_g?
   A. 200 + 51.075 x 0.5 = 225.5 V
   B. 215 + 51.075 x 0.5 = 240.54 V
   C. 215 + 50 x 0.5 = 240 V
   D. 200 + 51.075 x 0.8
   Answer: B. E_g = V_sh + I_a R_a = 215 + 25.54 = 240.54 V. (Key point: Generated emf)

## Close variants set in other exams

- [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×

## Related in Electrical Machines I

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## Related in other subjects

- [Long shunt compound generator emf, 40 A at 200 V](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Agen-long-shunt-40a-200v) — Electrical Machines, chapter 3
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