---
title: "DC generator losses and efficiency"
subject: "Electrical Machine"
chapter: 3
key: "semester_4/electrical_machine:dc-generator-losses-efficiency"
source: https://www.bctengineer.com/q/semester_4/electrical_machine/dc-generator-losses-efficiency
site: BCT Engineering
---

# DC generator losses and efficiency

Electrical Machine · Chapter 3: DC Generator

- Asked in: 2069 Poush
- Video: [4. Losses , efficiency and voltage regulation in DC generator](https://www.youtube.com/watch?v=TztyY9CpP9k) — IOE Syllabus
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machine/dc-generator-losses-efficiency
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machine/chapter-3

## Question

What are different types of losses in DC generator? Derive the expression for the efficiency of DC generator.

## Short answer

Losses are copper (armature and field), iron (hysteresis and eddy current), mechanical, and stray. Efficiency η = V IL / (V IL + IL² Ra + Wc). Maximum efficiency occurs when variable loss IL²Ra equals constant loss Wc.

## Answer

The power lost inside the machine while converting mechanical power into electrical power is called loss. The ratio of electrical output to mechanical input is the efficiency.

#### Types of losses
```text
Losses
 |-- Copper losses (variable)
 |     armature I_a^2 R_a | field I_f^2 R_f | brush contact loss
 |-- Iron (core) losses (constant)
 |     hysteresis | eddy current
 |-- Mechanical losses (constant)
 |     bearing friction | brush friction | windage
 +-- Stray load loss (small)
```
1. **Copper losses**: armature winding (I_a²R_a), shunt field (I_sh²R_sh or V·I_sh) and series field (I_se²R_se); brush contact loss is brush drop × I_a. Armature copper loss is about 30 to 40 % of full-load losses.
2. **Iron losses**: hysteresis loss (K_h B^1.6 f V) and eddy-current loss (K_e B² f² t²V) in the armature core. Constant for constant flux and speed.
3. **Mechanical losses**: friction at bearings and brushes, windage loss (air friction).
4. **Stray load losses**: due to flux distortion and other effects.

Constant losses W_c = iron + mechanical + shunt field loss. Variable losses = I_a²R_a.

#### Efficiency
- Mechanical efficiency: η_m = E_g I_a / (mechanical input)
- Electrical efficiency: η_e = V I_L / (E_g I_a)
- **Overall efficiency: η = output / input = output / (output + losses)**

#### Derivation (shunt generator)
Let output = V I_L, armature current I_a = I_L + I_sh ≈ I_L (I_sh is small).
```text
Armature copper loss  = I_a^2 R_a ≈ I_L^2 R_a
Constant loss         = W_c
η = V I_L / (V I_L + I_L^2 R_a + W_c)
```
Dividing numerator and denominator by I_L:

η = V / (V + I_L R_a + W_c / I_L)

#### Condition for maximum efficiency
η is maximum when the denominator is minimum: d/dI_L (V + I_L R_a + W_c/I_L) = 0

R_a − W_c / I_L² = 0, so **I_L² R_a = W_c**

That is, the efficiency is maximum when the **variable loss equals the constant loss**. The load current at maximum efficiency is I_L = √(W_c / R_a).

## Check yourself

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

1. Which losses are classed as constant?
   A. Armature copper loss and brush loss
   B. Stray load loss only
   C. Iron, mechanical and shunt field loss
   D. Series field copper loss
   Answer: C. W_c = iron + mechanical + shunt field loss; armature copper loss varies with I_a². (Key point: Constant losses)

2. How is overall efficiency defined?
   A. Input / output
   B. Losses / input
   C. Output x losses
   D. Output / (output + losses)
   Answer: D. η = output/input = output/(output + losses). (Key point: Overall efficiency)

3. For a shunt generator, which expression is η?
   A. V / (V + I_L R_a + W_c/I_L)
   B. V / (V + I_L² R_a)
   C. I_L R_a / V
   D. V / (V - W_c)
   Answer: A. Divide V I_L / (V I_L + I_L²R_a + W_c) by I_L. (Key point: Shunt generator efficiency)

4. What is the condition for maximum efficiency?
   A. Constant loss is zero
   B. Variable loss equals constant loss (I_L² R_a = W_c)
   C. Output is maximum
   D. Armature copper loss is twice iron loss
   Answer: B. Setting the derivative of the denominator to zero gives R_a = W_c/I_L². (Key point: Max efficiency)

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