---
title: "Short shunt compound generator with heater load"
subject: "Electrical Machines"
chapter: 3
key: "semester_4/electrical_machines:gen-short-shunt-175a-heater"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/gen-short-shunt-175a-heater
site: BCT Engineering
---

# Short shunt compound generator with heater load

Electrical Machines · Chapter 3: DC Generator

- Asked in: 2079 Jestha, 8 marks
- Video: [Short Shunt DC Compound Generator - Problem 1 (TAGALOG)](https://www.youtube.com/watch?v=zSnfNAfDAQs) — Engr. Aaron Pineda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines/gen-short-shunt-175a-heater
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-3

## Question

A short shunt compound dc generator supplies a load current of 175 A to a series of parallel heater load whose effective resistance is $1.4\ \Omega$. The generator has armature, series and shunt field resistances are $15\ \Omega$, $0.1\ \Omega$ and $100\ \Omega$ respectively. Calculate emf generated, copper losses and electrical efficiency of generator if carbon brush drop is 2 V per brush.

## Short answer

Terminal voltage V=175×1.4=245 V; V_sh=262.5 V; I_sh=2.625 A; I_a=177.625 A. Generated emf E=262.5+177.625×15+4=2930.9 V. Copper loss total = 477011 W; electrical efficiency = 8.24%.

## Answer

The heater load takes $I_L = 175$ A, so the terminal voltage follows from Ohm's law. In a short shunt machine the series field carries the load current and the shunt field is across the armature.

```text
       Ia                 Rse   IL
  +--->---+----------+--[~~~]--->--+--o +
  |       |          |            |
 (E,Ra)   |        [Rsh]       Heaters
  |       |          |            |
  +-------+----------+------------+--o -
```

#### Terminal voltage

$$
V = I_L R_L = 175 \times 1.4 = 245.0\ \text{V}
$$

#### Shunt and armature currents

$$
V_{sh} = V + I_LR_{se} = 245.0 + 175(0.1) = 262.5\ \text{V}
$$

$$
I_{sh} = \frac{V_{sh}}{R_{sh}} = \frac{262.5}{100} = 2.625\ \text{A}, \qquad I_a = I_L + I_{sh} = 177.625\ \text{A}
$$

#### Generated emf

Brush drop $= 2\ \text{V}\times 2 = 4\ \text{V}$.

$$
E = V_{sh} + I_aR_a + V_{brush} = 262.5 + (177.625)(15) + 4 = 2930.9\ \text{V}
$$

#### Copper losses

$$
\begin{aligned}
P_a &= I_a^2R_a = (177.625)^2(15) = 473259.6\ \text{W} \\
P_{se} &= I_L^2R_{se} = 175^2(0.1) = 3062.5\ \text{W} \\
P_{sh} &= I_{sh}^2R_{sh} = (2.625)^2(100) = 689.1\ \text{W} \\
P_{cu} &= 473259.6 + 3062.5 + 689.1 = 477011.2\ \text{W}
\end{aligned}
$$

Brush contact loss $= V_{brush}I_a = 4 \times 177.625 = 710.5\ \text{W}$, so the total electrical loss is $477011.2 + 710.5 = 477721.7\ \text{W}$.

#### Electrical efficiency

$$
\text{Output} = VI_L = 245.0 \times 175 = 42875\ \text{W}
$$

$$
\text{Electrical power generated} = EI_a = 2930.9 \times 177.625 = 520596.7\ \text{W} \;(= 42875 + 477721.7)
$$

$$
\eta_{elec} = \frac{VI_L}{EI_a} = \frac{42875}{520596.7}\times 100 = 8.24\%
$$

**Answer:** $E = 2930.9\ \text{V}$; copper loss $= 477011.2\ \text{W}$ (plus $710.5\ \text{W}$ brush contact loss); electrical efficiency $= 8.24\%$. (The armature resistance of $15\ \Omega$ is used as given. It is almost certainly a misprint, since a practical armature resistance is a few hundredths of an ohm, which is why the emf is so high and the efficiency so low.)

## Check yourself

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

1. The terminal voltage of the heater load is found from:
   A. V = 175 x 15
   B. V = IL x RL = 175 x 1.4 = 245 V
   C. V = 175 x 100
   D. V = Ia x Ra
   Answer: B. The heater is the load, so Ohm's law with load current and effective resistance gives V. (Key point: Terminal voltage)

2. In short shunt the shunt field voltage is:
   A. 245 V
   B. 227.5 V
   C. V + IL Rse = 262.5 V
   D. E
   Answer: C. 262.5 = 245 + 175 x 0.1. (Key point: Shunt voltage)

3. Using Ia = 177.625 A and Ra = 15 ohm with 4 V brush drop, the emf is about:
   A. 267 V
   B. 2927 V
   C. 262.5 V
   D. 2931 V
   Answer: D. E = 262.5 + 177.625x15 + 4 = 2930.9 V. (Key point: Generated emf)

4. Electrical efficiency is calculated as:
   A. output / (output + copper losses)
   B. output / copper loss
   C. input / output
   D. (output - copper loss)/output + 1
   Answer: A. With output 42875 W and Pcu = 477011 W, efficiency is only 8.25%. (Key point: Electrical efficiency)

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