---
title: "EMF of short shunt compound generator (6 kW, 250 V)"
subject: "Electrical Machines I"
chapter: 3
key: "semester_4/electrical_machines_i:short-shunt-compound-6kw-250v"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/short-shunt-compound-6kw-250v
site: BCT Engineering
---

# EMF of short shunt compound generator (6 kW, 250 V)

Electrical Machines I · Chapter 3: DC Generator

- Asked in: 2076 Bhadra, 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-6kw-250v
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-3

## Question

A dc short shunt compound generator delivers 6 kW to the load at 250 V. The armature winding, series field winding and shunt field winding have resistance of 0.1 Ω, 0.2 Ω and 250 Ω respectively. Calculate the emf generated by armature.

## Short answer

Load current 24 A, voltage across shunt field 254.8 V, shunt current 1.0192 A, armature current 25.02 A. Generated emf Eg = 254.8 + 25.02×0.1 = 257.3 V.

## Answer

In a short-shunt compound generator the shunt field is connected directly across the armature, and the series field carries the load current.

**Given:** $P = 6$ kW, $V = 250$ V, $R_a = 0.1\ \Omega$, $R_{se} = 0.2\ \Omega$, $R_{sh} = 250\ \Omega$.

```text
     +--Ra--+--------Rse--------+
     |      |                   |  +
    (Eg)   Rsh                 Load  V=250 V
     |      |                   |  -
     +------+-------------------+
```

#### Step 1: Load current
$$
I_L = \frac{P}{V} = \frac{6000}{250} = 24\ \text{A}
$$

#### Step 2: Voltage across the shunt field
The series field carries $I_L$, so

$$
\begin{aligned}
V_{sh} &= V + I_L R_{se} \\
&= 250 + 24 \times 0.2 = 254.8\ \text{V}
\end{aligned}
$$

#### Step 3: Shunt field current and armature current
$$
\begin{aligned}
I_{sh} &= \frac{V_{sh}}{R_{sh}} = \frac{254.8}{250} = 1.0192\ \text{A} \\
I_a &= I_L + I_{sh} = 24 + 1.0192 = 25.0192\ \text{A}
\end{aligned}
$$

#### Step 4: Generated emf
$$
\begin{aligned}
E_g &= V_{sh} + I_a R_a \\
&= 254.8 + 25.0192 \times 0.1 \\
&= 254.8 + 2.502 = 257.30\ \text{V}
\end{aligned}
$$

Brush drop is not given, so it is neglected.

**Answer:** EMF generated by the armature $E_g \approx 257.3$ V.

## Check yourself

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

1. What is the load current?
   A. I_L = 6000/250 = 24 A
   B. I_L = 250/6 = 41.7 A
   C. I_L = 6000 x 250 = 1.5 MA
   D. I_L = 6/250 = 0.024 A
   Answer: A. I_L = P/V = 6000/250 = 24 A. (Key point: Load current)

2. In a short-shunt connection, what is the voltage across the shunt field?
   A. Terminal voltage V only, 250 V
   B. V + I_L R_se = 254.8 V
   C. V - I_L R_se = 245.2 V
   D. V + I_a R_a
   Answer: B. The shunt field is across the armature, so it sees terminal voltage plus the drop in the series field, which carries I_L. (Key point: Shunt field voltage in short shunt)

3. How is armature current found?
   A. I_a = I_L = 24 A
   B. I_a = I_L - I_sh = 22.98 A
   C. I_a = V_sh/R_a
   D. I_a = I_L + I_sh = 24 + 1.0192 = 25.02 A
   Answer: D. The armature feeds the load and the shunt field: I_a = I_L + I_sh. (Key point: Armature current)

4. Which expression gives E_g here (no brush drop)?
   A. E_g = V + I_a R_a
   B. E_g = V_sh + I_a R_a = 254.8 + 2.5 = 257.3 V
   C. E_g = V + I_L R_se
   D. E_g = V_sh - I_a R_a
   Answer: B. E_g equals the voltage across the armature terminals (V_sh) plus the armature drop I_a R_a. (Key point: Generated emf)

## Related in Electrical Machines I

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