---
title: "EMF, copper loss and efficiency of 30 kW shunt generator"
subject: "Electrical Machines I"
chapter: 3
key: "semester_4/electrical_machines_i:shunt-gen-30kw-200v-emf-losses"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/shunt-gen-30kw-200v-emf-losses
site: BCT Engineering
---

# EMF, copper loss and efficiency of 30 kW shunt generator

Electrical Machines I · Chapter 3: DC Generator

- Asked in: 2074 Bhadra, 8 marks
- 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_i/shunt-gen-30kw-200v-emf-losses
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/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 is 150 A, shunt current 4 A, armature current 154 A. Generated emf = 200 + 154×0.05 = 207.7 V. Copper losses are 1185.8 W armature plus 800 W shunt = 1985.8 W; efficiency = 30000/(30000+1985.8+1000) = 90.95%.

## Answer

**Given:** $P_{out} = 30$ kW, $V = 200$ V, $R_a = 0.05\ \Omega$, $R_{sh} = 50\ \Omega$, iron + friction loss = 1000 W.

#### Currents
$$
\begin{aligned}
I_L &= \frac{30000}{200} = 150\ \text{A} \\
I_{sh} &= \frac{200}{50} = 4\ \text{A} \\
I_a &= I_L + I_{sh} = 154\ \text{A}
\end{aligned}
$$

#### (i) Generated emf
$$
\begin{aligned}
E_g &= V + I_a R_a \\
&= 200 + 154 \times 0.05 = 207.7\ \text{V}
\end{aligned}
$$

#### (ii) Copper losses
$$
\begin{aligned}
P_{cu,a} &= I_a^2 R_a = 154^2 \times 0.05 = 1185.8\ \text{W} \\
P_{cu,sh} &= V I_{sh} = 200 \times 4 = 800\ \text{W} \\
P_{cu} &= 1185.8 + 800 = 1985.8\ \text{W}
\end{aligned}
$$

#### (iii) Efficiency
$$
\begin{aligned}
P_{loss} &= 1985.8 + 1000 = 2985.8\ \text{W} \\
P_{in} &= 30000 + 2985.8 = 32985.8\ \text{W} \\
\eta &= \frac{30000}{32985.8} \times 100 = 90.95\%
\end{aligned}
$$

| Quantity | Value |
|---|---|
| Generated emf | 207.7 V |
| Armature copper loss | 1185.8 W |
| Shunt copper loss | 800 W |
| Total copper loss | 1985.8 W |
| Efficiency | 90.95 % |

**Answer:** (i) $E_g = 207.7$ V, (ii) total copper loss $= 1985.8$ W, (iii) $\eta \approx 90.95\%$.

## Check yourself

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

1. What is I_a?
   A. 150 A
   B. 154 A
   C. 146 A
   D. 4 A
   Answer: B. I_L = 150 A, I_sh = 4 A, so I_a = I_L + I_sh = 154 A. (Key point: Armature current)

2. What is E_g?
   A. 200 + 150 x 0.05 = 207.5 V
   B. 200 + 154 x 0.05 = 207.7 V
   C. 200 - 154 x 0.05 = 192.3 V
   D. 200 + 4 x 0.05 = 200.2 V
   Answer: B. E_g = V + I_a R_a = 207.7 V for a generator. (Key point: Generated emf)

3. Which gives the total copper loss?
   A. 154^2 x 0.05 + 200 x 4 = 1185.8 + 800 = 1985.8 W
   B. 150^2 x 0.05 = 1125 W
   C. 1185.8 W only
   D. 1185.8 + 1000 W
   Answer: A. Copper loss is armature (I_a^2 R_a) plus shunt field (V I_sh); the 1000 W iron and friction is separate. (Key point: Copper losses)

4. What is the efficiency?
   A. 30000/31000 = 96.8%
   B. 30000/32985.8 = 90.95%
   C. 30000/(30000 - 2985.8)
   D. 30000/30000+1985.8 = 93.8%
   Answer: B. Input = 30000 + total losses 2985.8 W (copper plus iron and friction), so efficiency is 90.95%. (Key point: Efficiency)

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