---
title: "Compound generator emf, long and short shunt"
subject: "Electrical Machines"
chapter: 3
key: "semester_4/electrical_machines:gen-compound-50a-500v-long-short"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/gen-compound-50a-500v-long-short
site: BCT Engineering
---

# Compound generator emf, long and short shunt

Electrical Machines · Chapter 3: DC Generator

- Asked in: 2078 Chaitra, 8 marks
- 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_machines/gen-compound-50a-500v-long-short
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-3

## Question

A DC compound generator delivers 50 A to the load at 500 V. The armature, series field and shunt field windings resistance are $0.05\ \Omega$, $0.03\ \Omega$ and $250\ \Omega$ respectively. The voltage drop in carbon brush is 1 V per brush. Calculate the generated emf i) for long shunt compound ii) for short shunt compound.

## Short answer

Load 50 A at 500 V, total brush drop 2 V. Long shunt: I_sh=2 A, I_a=52 A, E = 500+52×(0.05+0.03)+2 = 506.16 V. Short shunt: V_sh=501.5 V, I_sh≈2.006 A, I_a≈52.006 A, E = 501.5+52.006×0.05+2 = 506.10 V.

## Answer

Brush drop $= 1\ \text{V}\times 2\ \text{brushes} = 2\ \text{V}$. Load current $I_L = 50$ A, terminal voltage $V = 500$ V.

#### (i) Long shunt

The shunt field is across the load terminals:

$$
I_{sh} = \frac{V}{R_{sh}} = \frac{500}{250} = 2\ \text{A}, \qquad I_a = I_{se} = 50 + 2 = 52\ \text{A}
$$

$$
\begin{aligned}
E &= V + I_a(R_a + R_{se}) + V_{brush} \\
&= 500 + 52(0.05 + 0.03) + 2 \\
&= 500 + 4.16 + 2 = 506.16\ \text{V}
\end{aligned}
$$

#### (ii) Short shunt

The series field carries only the load current. The shunt field is across the armature, so its voltage is

$$
V_{sh} = V + I_LR_{se} = 500 + 50(0.03) = 501.50\ \text{V}
$$

$$
I_{sh} = \frac{501.50}{250} = 2.006\ \text{A}, \qquad I_a = I_L + I_{sh} = 50 + 2.006 = 52.006\ \text{A}
$$

$$
\begin{aligned}
E &= V_{sh} + I_aR_a + V_{brush} \\
&= 501.50 + (52.006)(0.05) + 2 \\
&= 501.50 + 2.600 + 2 = 506.10\ \text{V}
\end{aligned}
$$

**Answer:** (i) long shunt $E = 506.16\ \text{V}$; (ii) short shunt $E = 506.10\ \text{V}$.

## Check yourself

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

1. With 1 V per brush, the total brush drop to include in E is:
   A. 1 V
   B. 0 V
   C. 2 V (two brushes)
   D. 4 V
   Answer: C. A machine has a positive and a negative brush set, so drop = 2 x 1 = 2 V. (Key point: Brush drop)

2. In the long shunt connection, Ish is 2 A and the armature (series) current is:
   A. 50 A
   B. 48 A
   C. 2 A
   D. 52 A
   Answer: D. Ia = IL + Ish = 50 + 2 = 52 A. (Key point: Long shunt current)

3. The long shunt emf is:
   A. 506.16 V
   B. 504.16 V
   C. 506.10 V
   D. 502 V
   Answer: A. E = 500 + 52(0.08) + 2 = 506.16 V. (Key point: Long shunt emf)

4. In short shunt, the shunt field voltage is:
   A. V = 500 V
   B. V + IL Rse = 501.5 V
   C. V + Ia Ra
   D. V - IL Rse = 498.5 V
   Answer: B. The shunt sits across the armature, so it sees terminal voltage plus the series drop at IL. (Key point: Short shunt voltage)

## Related in Electrical Machines

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