---
title: "EMF of 25 kW short shunt generator with brush drop"
subject: "Electrical Machines I"
chapter: 3
key: "semester_4/electrical_machines_i:short-shunt-25kw-250v-brush-drop"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/short-shunt-25kw-250v-brush-drop
site: BCT Engineering
---

# EMF of 25 kW short shunt generator with brush drop

Electrical Machines I · Chapter 3: DC Generator

- Asked in: 2072 Asoj, 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-25kw-250v-brush-drop
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-3

## Question

A 25 kW short-shunt dc compound generator delivers rated power to the load at 250 V. The generator has shunt field, series field and armature resistance of 125 Ω, 0.5 Ω and 0.1 Ω respectively. Calculate the emf generated in armature winding, taking 2 V as total brush drop.

## Short answer

Load current is 25000/250 = 100 A. Shunt field voltage is 250 + 100×0.5 = 300 V, shunt current 2.4 A, armature current 102.4 A. Generated emf = 300 + 102.4×0.1 + 2 = 312.24 V.

## Answer

**Given:** $P = 25$ kW, $V = 250$ V, $R_{sh} = 125\ \Omega$, $R_{se} = 0.5\ \Omega$, $R_a = 0.1\ \Omega$, total brush drop $V_b = 2$ V; short-shunt connection.

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

#### Step 1: Load current
$$
I_L = \frac{25000}{250} = 100\ \text{A}
$$

#### Step 2: Voltage across shunt field
$$
V_{sh} = V + I_L R_{se} = 250 + 100 \times 0.5 = 300\ \text{V}
$$

#### Step 3: Shunt field and armature current
$$
\begin{aligned}
I_{sh} &= \frac{300}{125} = 2.4\ \text{A} \\
I_a &= I_L + I_{sh} = 100 + 2.4 = 102.4\ \text{A}
\end{aligned}
$$

#### Step 4: Generated emf
$$
\begin{aligned}
E_g &= V_{sh} + I_a R_a + V_b \\
&= 300 + 102.4 \times 0.1 + 2 \\
&= 300 + 10.24 + 2 = 312.24\ \text{V}
\end{aligned}
$$

**Answer:** EMF generated in the armature $E_g = 312.24$ V.

## Check yourself

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

1. What is I_L?
   A. 25000/250 = 100 A
   B. 25/250 = 0.1 A
   C. 250/25 = 10 A
   D. 25000/125 = 200 A
   Answer: A. I_L = P/V = 100 A. (Key point: Load current)

2. What is V_sh?
   A. 250 V
   B. 250 + 100 x 0.1 = 260 V
   C. 250 + 100 x 0.5 = 300 V
   D. 250 - 50 = 200 V
   Answer: C. V_sh = V + I_L R_se = 300 V, since the series field carries I_L. (Key point: Shunt field voltage)

3. What is I_a?
   A. 100 A
   B. 102.4 A
   C. 100 + 2 = 102 A
   D. 97.6 A
   Answer: B. I_sh = 300/125 = 2.4 A, so I_a = 100 + 2.4 = 102.4 A. (Key point: Armature current)

4. How is the total brush drop of 2 V used?
   A. Subtracted from E_g
   B. Ignored
   C. Added: E_g = V_sh + I_a R_a + 2 = 312.24 V
   D. Doubled to 4 V
   Answer: C. Brush drop is a loss, so the emf must exceed the terminal-side voltage by it. It is given as total 2 V. (Key point: Brush drop in emf)

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