---
title: "30 kW shunt generator: emf, copper loss, efficiency"
subject: "Electrical Machines"
chapter: 3
key: "semester_4/electrical_machines:gen-shunt-30kw-loss-1000w"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/gen-shunt-30kw-loss-1000w
site: BCT Engineering
---

# 30 kW shunt generator: emf, copper loss, efficiency

Electrical Machines · Chapter 3: DC Generator

- Asked in: 2074 Bhadra (old course)
- Video: [17 Solved Problems #1 - Losses and Efficiency of DC Generator](https://www.youtube.com/watch?v=zmLePFQwS28) — SP Engg. ACADEMY
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines/gen-shunt-30kw-loss-1000w
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-3

## Question

A dc shunt generator gives full load output of 30 kW at a terminal voltage of 200 V. The armature and shunt field resistances are 0.05 ohm and 50 ohm respectively. The iron and friction losses are 1000 W. Calculate: (i) generated emf; (ii) copper losses; (iii) efficiency.

## Short answer

Load current 150 A, shunt current 4 A, armature current 154 A. Generated emf E = 200 + 154×0.05 = 207.70 V. Copper losses: armature 1185.8 W, shunt 800 W, total 1985.8 W. Efficiency = 30000/(30000+1985.8+1000) × 100 = 90.95%.

## Answer

For a shunt generator $I_a = I_L + I_{sh}$ and $E = V + I_aR_a$.

#### (i) Generated emf

$$
I_L = \frac{30000}{200} = 150\ \text{A}, \qquad I_{sh} = \frac{200}{50} = 4\ \text{A}
$$

$$
I_a = I_L + I_{sh} = 154\ \text{A}
$$

$$
E = V + I_aR_a = 200 + (154)(0.05) = 207.70\ \text{V}
$$

#### (ii) Copper losses

$$
\begin{aligned}
\text{Armature copper loss} &= I_a^2R_a = (154)^2(0.05) = 1185.8\ \text{W} \\
\text{Shunt field copper loss} &= I_{sh}^2R_{sh} = (4)^2(50) = 800.0\ \text{W} \\
\text{Total copper loss} &= 1185.8 + 800.0 = 1985.8\ \text{W}
\end{aligned}
$$

#### (iii) Efficiency

Total losses $=$ copper loss $+$ iron and friction loss $= 1985.8 + 1000 = 2985.8\ \text{W}$.

$$
\eta = \frac{\text{output}}{\text{output} + \text{losses}} = \frac{30000}{30000 + 2985.8}\times 100 = 90.95\%
$$

**Answer:** (i) $E = 207.70\ \text{V}$; (ii) copper loss $= 1985.8\ \text{W}$; (iii) $\eta = 90.95\%$.

## Check yourself

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

1. For the shunt generator, Ia equals:
   A. IL = 150 A
   B. IL - Ish = 146 A
   C. Ish = 4 A
   D. IL + Ish = 154 A
   Answer: D. IL = 150 A and Ish = 4 A add at the armature. (Key point: Armature current)

2. The generated emf is:
   A. 207.7 V
   B. 200 V
   C. 207.0 V
   D. 208.2 V
   Answer: A. E = 200 + 154x0.05 = 207.7 V. (Key point: Generated emf)

3. Total copper loss is:
   A. 1185.8 W only
   B. 1985.8 W (armature 1185.8 W + shunt 800 W)
   C. 800 W only
   D. 154^2x50 W
   Answer: B. Ia^2Ra + Ish^2Rsh = 1185.8 + 800. (Key point: Copper losses)

4. Efficiency uses total losses of:
   A. copper loss only
   B. iron and friction loss only
   C. copper loss plus iron and friction = 2985.8 W, giving 90.95%
   D. output minus copper loss
   Answer: C. eta = output/(output + total losses) with total losses including iron and friction. (Key point: Efficiency)

## Close variants set in other exams

- [30 kW shunt generator: emf, copper loss, efficiency (100 W)](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Agen-shunt-30kw-loss-100w) — Electrical Machines, chapter 3, asked 2×

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