---
title: "EMF and efficiency of short shunt compound generator (50 A, 200 V)"
subject: "Electrical Machines I"
chapter: 3
key: "semester_4/electrical_machines_i:short-shunt-compound-50a-200v"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/short-shunt-compound-50a-200v
site: BCT Engineering
---

# EMF and efficiency of short shunt compound generator (50 A, 200 V)

Electrical Machines I · Chapter 3: DC Generator

- Asked in: 2079 Chaitra, 8 marks; 2077 Chaitra, 8 marks (asked 2 times)
- Video: [Short shunt dc generator Induced emf and terminal voltage equation | Mruduraj](https://www.youtube.com/watch?v=NKNd4KZtXxI) — Learning Vibes
- 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-compound-50a-200v
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-3

## Question

A DC short shunt compound generator supplies a current of 50 A to the load at 200 V. The armature, series and shunt field resistances are 0.08 Ω, 0.05 Ω and 100 Ω respectively. Calculate the emf generated by the armature and efficiency of the generator. Given that the no-load loss of the generator is 300 W.

## Short answer

Generated EMF = 206.66 V; efficiency = 90.48%. Vsh = 200 + 50×0.05 = 202.5 V, Ish = 2.025 A, Ia = 52.025 A, EMF = Vsh + Ia×0.08. Output 10 kW, total copper + stray losses 1051.6 W.

## Answer

In a **short-shunt** compound generator the shunt field is connected directly across the armature, and the series field is in series with the load.

```text
       Ia            Rse = 0.05     IL = 50 A
   +---->----+-------/\/\/\-----+--->---o +
   |         |                  |
 ( E )     [Rsh]                       V = 200 V
 Ra=0.08   100 ohm                load
   |         |                  |
   +---------+------------------+-------o -
```

#### Step 1: Voltage across the shunt field
$$
V_{sh} = V + I_LR_{se} = 200 + 50 \times 0.05 = 202.5\ \text{V}
$$

#### Step 2: Shunt field and armature currents
$$
\begin{aligned}
I_{sh} &= \frac{V_{sh}}{R_{sh}} = \frac{202.5}{100} = 2.025\ \text{A} \\
I_a &= I_L + I_{sh} = 50 + 2.025 = 52.025\ \text{A}
\end{aligned}
$$

#### Step 3: Generated EMF
$$
\begin{aligned}
E_g &= V_{sh} + I_aR_a = 202.5 + 52.025 \times 0.08 \\
&= 202.5 + 4.162 = 206.66\ \text{V}
\end{aligned}
$$

#### Step 4: Losses
The no-load (rotational: iron + friction) loss of 300 W is taken as the stray loss; the field and armature copper losses are added separately.
$$
\begin{aligned}
\text{Armature Cu loss} &= I_a^2R_a = 52.025^2 \times 0.08 = 216.53\ \text{W} \\
\text{Series field loss} &= I_L^2R_{se} = 50^2 \times 0.05 = 125\ \text{W} \\
\text{Shunt field loss} &= V_{sh}I_{sh} = 202.5 \times 2.025 = 410.06\ \text{W} \\
\text{Stray (no-load) loss} &= 300\ \text{W} \\
\text{Total loss} &= 216.53 + 125 + 410.06 + 300 = 1051.59\ \text{W}
\end{aligned}
$$
Check: $E_gI_a = 206.66 \times 52.025 = 10751.6$ W = output + copper losses ($10000 + 751.6$).

#### Step 5: Efficiency
$$
\begin{aligned}
\text{Output} &= VI_L = 200 \times 50 = 10000\ \text{W} \\
\eta &= \frac{10000}{10000 + 1051.59}\times 100 = 90.48\%
\end{aligned}
$$

**Answer:** Generated EMF $E_g =$ **206.66 V**; efficiency $\approx$ **90.48%**.

## Check yourself

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

1. In a short-shunt compound generator, voltage across the shunt field is:
   A. V + IL Rse = 202.5 V
   B. V = 200 V
   C. V + Ia Rse
   D. V - IL Rse
   Answer: A. Shunt field is across the armature, so it sees terminal voltage plus series field drop: 200 + 50 x 0.05. (Key point: Short shunt voltage)

2. The armature current is:
   A. 50 A
   B. 52.025 A
   C. 52.5 A
   D. 48 A
   Answer: B. Ish = 202.5/100 = 2.025 A, Ia = IL + Ish = 52.025 A. (Key point: Armature current)

3. Generated EMF Eg is:
   A. 200 V
   B. 204.2 V
   C. 206.66 V
   D. 210 V
   Answer: C. Eg = Vsh + Ia Ra = 202.5 + 52.025 x 0.08 = 206.66 V. (Key point: Generated EMF)

4. Which losses are added for total loss?
   A. Armature, series field, shunt field copper loss plus 300 W stray
   B. Only armature copper loss
   C. Only 300 W
   D. Armature and series loss only
   Answer: A. Total = 216.53 + 125 + 410.06 + 300 = 1051.59 W, giving 90.48% efficiency. (Key point: Efficiency losses)

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