---
title: "Short shunt cumulative compound generator, 7.5 kW"
subject: "Electrical Machines"
chapter: 3
key: "semester_4/electrical_machines:gen-short-shunt-cumulative-7-5kw"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/gen-short-shunt-cumulative-7-5kw
site: BCT Engineering
---

# Short shunt cumulative compound generator, 7.5 kW

Electrical Machines · Chapter 3: DC Generator

- Asked in: 2076 Baisakh, 8 marks
- Video: [#27 DC Generator -  Cumulative compound Generator #4 - Summary of compounding](https://www.youtube.com/watch?v=MRiiLlEFvao) — ELECTRICAL IS EASY
- 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-cumulative-7-5kw
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-3

## Question

A short shunt cumulative 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

I_L=7500/230≈32.609 A, R_L=230/32.609≈7.053 Ω. V_sh=230+32.609×0.3=239.783 V, I_sh=2.3978 A, I_a=35.007 A. Induced emf E=239.783+35.007×0.4=253.79 V.

## Answer

Short shunt: the shunt field is across the armature and the series field is in the load line. Cumulative: the series field aids the shunt field, which does not change the circuit calculation.

#### Load current and load resistance

$$
I_L = \frac{P}{V} = \frac{7500}{230} = 32.609\ \text{A}, \qquad R_L = \frac{V}{I_L} = \frac{230}{32.609} = 7.053\ \Omega
$$

#### Shunt field and armature currents

$$
V_{sh} = V + I_LR_{se} = 230 + (32.609)(0.3) = 239.783\ \text{V}
$$

$$
I_{sh} = \frac{239.783}{100} = 2.3978\ \text{A}, \qquad I_a = I_L + I_{sh} = 35.007\ \text{A}
$$

#### Induced emf

$$
E = V_{sh} + I_aR_a = 239.783 + (35.007)(0.4) = 253.79\ \text{V}
$$

**Answer:** induced emf $= 253.79\ \text{V}$; load resistance $= 7.053\ \Omega$.

## Check yourself

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

1. The load current for 7.5 kW at 230 V is:
   A. 30.67 A
   B. 35.0 A
   C. 32.609 A
   D. 7.5 A
   Answer: C. IL = 7500/230 = 32.609 A. (Key point: Load current)

2. The load resistance is:
   A. Vsh / IL
   B. V / Ia
   C. 230 / 100
   D. V / IL = 7.053 ohm
   Answer: D. RL = 230/32.609 = 7.053 ohm. (Key point: Load resistance)

3. Does the cumulative connection change the circuit calculation of the emf?
   A. No, it only affects flux, not the circuit equations used
   B. Yes, Rse is subtracted
   C. Yes, Ish doubles
   D. Yes, Ra is ignored
   Answer: A. Cumulative or differential only matters to flux; resistance drops are computed the same way. (Key point: Cumulative effect)

4. The induced emf is found by:
   A. E = V + Ia Ra
   B. E = Vsh + Ia Ra with Vsh = V + IL Rse
   C. E = V + IL Ra
   D. E = V - IL Rse + Ia Ra
   Answer: B. Vsh = 239.78 V and Ia = 35.007 A give E = 253.79 V. (Key point: Short shunt emf)

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