---
title: "EMF and efficiency of 20 kW compound generator"
subject: "Electrical Machines I"
chapter: 3
key: "semester_4/electrical_machines_i:compound-gen-20kw-220v-emf-efficiency"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/compound-gen-20kw-220v-emf-efficiency
site: BCT Engineering
---

# EMF and efficiency of 20 kW compound generator

Electrical Machines I · Chapter 3: DC Generator

- Asked in: 2069 Bhadra, 8 marks
- Video: [20 Solved Problems #4 - Losses and Efficiency of DC Generator](https://www.youtube.com/watch?v=o1QmJI4q-OA) — 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/compound-gen-20kw-220v-emf-efficiency
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-3

## Question

A shunt compound dc generator has the following parameters: Armature winding resistance (Ra) = 0.02 Ohms, Series field winding resistance (Rse) = 0.04 Ohms, Shunt field winding resistance (Rsh) = 100 Ohms, Iron and friction loss = 250 Watts. The generator is delivering 20 kW to the load at 220 V dc. Calculate emf generated by the armature and efficiency of the generator.

## Short answer

Assuming long-shunt, armature current is 93.11 A, generated emf 225.59 V. Total losses are 1254.2 W, giving efficiency 94.1%.

## Answer

**Given:** $P_{out} = 20$ kW, $V = 220$ V, $R_a = 0.02\ \Omega$, $R_{se} = 0.04\ \Omega$, $R_{sh} = 100\ \Omega$, iron + friction loss = 250 W.

**Assumption:** the type of compound connection is not stated, so the **long-shunt** connection is taken (the usual assumption). The short-shunt result is given at the end for comparison.

```text
     +--Ra--Rse--+-----------+
     |           |           |
    (Eg)        Rsh        Load 220 V
     |           |           |
     +-----------+-----------+
```

#### Step 1: Currents
$$
\begin{aligned}
I_L &= \frac{20000}{220} = 90.909\ \text{A} \\
I_{sh} &= \frac{220}{100} = 2.2\ \text{A} \\
I_a &= I_{se} = I_L + I_{sh} = 93.109\ \text{A}
\end{aligned}
$$

#### Step 2: Generated emf
$$
\begin{aligned}
E_g &= V + I_a (R_a + R_{se}) \\
&= 220 + 93.109 \times 0.06 \\
&= 220 + 5.587 = 225.59\ \text{V}
\end{aligned}
$$

#### Step 3: Losses
$$
\begin{aligned}
P_{cu,a} &= I_a^2 R_a = 93.109^2 \times 0.02 = 173.39\ \text{W} \\
P_{cu,se} &= I_a^2 R_{se} = 93.109^2 \times 0.04 = 346.77\ \text{W} \\
P_{cu,sh} &= V I_{sh} = 220 \times 2.2 = 484\ \text{W} \\
P_{loss} &= 173.39 + 346.77 + 484 + 250 = 1254.16\ \text{W}
\end{aligned}
$$

#### Step 4: Efficiency
$$
\begin{aligned}
P_{in} &= 20000 + 1254.16 = 21254.16\ \text{W} \\
\eta &= \frac{20000}{21254.16} \times 100 = 94.10\%
\end{aligned}
$$

**Answer:** $E_g \approx 225.59$ V, efficiency $\eta \approx 94.1\%$.

(If short shunt is taken: $V_{sh} = 223.64$ V, $I_{sh} = 2.236$ A, $I_a = 93.15$ A, $E_g \approx 225.50$ V, total loss $\approx 1254.2$ W, $\eta \approx 94.1\%$.)

## Check yourself

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

1. What is I_a?
   A. 90.909 A
   B. 93.109 A
   C. 88.709 A
   D. 2.2 A
   Answer: B. I_L = 90.909 A plus I_sh = 2.2 A. (Key point: Armature current)

2. What is E_g?
   A. 220 + 93.109 x 0.06 = 225.59 V
   B. 220 + 93.109 x 0.02 = 221.9 V
   C. 220 + 90.909 x 0.06
   D. 220 - 5.587
   Answer: A. Long shunt: E_g = V + I_a(R_a + R_se). (Key point: Generated emf)

3. What is the efficiency?
   A. 20000/21254 = 94.1%
   B. 20000/20250 = 98.8%
   C. 20000/(20000 - 1254)
   D. 93.109/220
   Answer: A. Losses: copper 173.39 + 346.77 + 484 plus 250 W = 1254 W, so input is 21254 W. (Key point: Efficiency)

## Related in Electrical Machines I

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## Related in other subjects

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