---
title: "EMF and load resistance of short shunt compound generator"
subject: "Electrical Machines I"
chapter: 3
key: "semester_4/electrical_machines_i:short-shunt-7p5kw-230v-emf-load"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/short-shunt-7p5kw-230v-emf-load
site: BCT Engineering
---

# EMF and load resistance of short shunt compound generator

Electrical Machines I · Chapter 3: DC Generator

- Asked in: 2070 Magh, 8 marks
- Video: [How to Calculate Generated EMF in a Short Shunt Compound DC Generator (Step-by-Step Solved Problem) ](https://www.youtube.com/watch?v=VGVbSigl4LE) — 5 Minute Electrical 
- 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-7p5kw-230v-emf-load
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-3

## Question

A short shunt compound dc generator supplies 7.5 kW at 230 V. The shunt field, series field and armature resistance are 100, 0.3 and 0.4 ohms respectively. Calculate the induced emf and the load resistance.

## Short answer

Load current 32.61 A, shunt voltage 239.78 V, armature current 35.01 A. Induced emf is 253.79 V and load resistance 7.05 Ω.

## Answer

**Given:** $P = 7.5$ kW, $V = 230$ V, $R_{sh} = 100\ \Omega$, $R_{se} = 0.3\ \Omega$, $R_a = 0.4\ \Omega$; short-shunt connection.

```text
     +--Ra--+-------Rse-------+
     |      |                 |
    (Eg)   Rsh              Load 230 V
     |      |                 |
     +------+-----------------+
```

#### Step 1: Load current
$$
I_L = \frac{7500}{230} = 32.609\ \text{A}
$$

#### Step 2: Shunt field voltage and current
$$
\begin{aligned}
V_{sh} &= V + I_L R_{se} = 230 + 32.609 \times 0.3 = 239.78\ \text{V} \\
I_{sh} &= \frac{239.78}{100} = 2.398\ \text{A}
\end{aligned}
$$

#### Step 3: Armature current and induced emf
$$
\begin{aligned}
I_a &= I_L + I_{sh} = 32.609 + 2.398 = 35.007\ \text{A} \\
E_g &= V_{sh} + I_a R_a = 239.78 + 35.007 \times 0.4 \\
&= 239.78 + 14.00 = 253.79\ \text{V}
\end{aligned}
$$

#### Step 4: Load resistance
$$
R_L = \frac{V}{I_L} = \frac{230}{32.609} = 7.053\ \Omega
$$

(Or $R_L = V^2/P = 230^2/7500 = 7.053\ \Omega$.)

**Answer:** Induced emf $E_g \approx 253.79$ V; load resistance $R_L \approx 7.05\ \Omega$.

## Check yourself

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

1. What is I_L?
   A. 7500 x 230
   B. 7500/230 = 32.61 A
   C. 230/7.5 = 30.7 A
   D. 7.5/230 A
   Answer: B. I_L = P/V = 32.61 A. (Key point: Load current)

2. What is E_g?
   A. 230 + 35 x 0.4 = 244 V
   B. 239.78 + 35.007 x 0.4 = 253.79 V
   C. 230 + 32.609 x 0.7 = 252.8 V
   D. 239.78 - 14 V
   Answer: B. V_sh = 239.78 V, I_sh = 2.398 A, I_a = 35.007 A, E_g = V_sh + I_a R_a. (Key point: Induced emf)

3. What is the load resistance?
   A. V/I_a = 6.57 ohm
   B. V/I_L = V^2/P = 7.05 ohm
   C. V_sh/I_L = 7.35 ohm
   D. R_sh in parallel
   Answer: B. Load resistance is terminal voltage over load current, 230/32.609 = 7.053 ohm. (Key point: Load resistance)

## Related in Electrical Machines I

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## Related in other subjects

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