---
title: "Long shunt compound generator emf, 80 A at 230 V"
subject: "Electrical Machines"
chapter: 3
key: "semester_4/electrical_machines:gen-long-shunt-80a-230v"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/gen-long-shunt-80a-230v
site: BCT Engineering
---

# Long shunt compound generator emf, 80 A at 230 V

Electrical Machines · Chapter 3: DC Generator

- Asked in: 2077 Chaitra, 8 marks
- Video: [DC Long - Shunt Compound Generator - Problem 1 (TAGALOG)](https://www.youtube.com/watch?v=Ew4Yj_7BV_A) — 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-long-shunt-80a-230v
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-3

## Question

A long shunt dc compound generator delivers a current of 80 A to the load at 230 V. The shunt field, series and armature winding resistances are $100\ \Omega$, $0.04\ \Omega$ and $0.2\ \Omega$ respectively. Calculate the emf generated by the armature.

## Short answer

I_sh=230/100=2.3 A, I_a=82.3 A. E=230+82.3×(0.2+0.04)=249.75 V.

## Answer

In a long shunt compound generator the shunt field is connected across the load terminals, so the series field carries the full armature current.

Given: $I_L = 80$ A, $V = 230$ V, $R_{sh} = 100\ \Omega$, $R_{se} = 0.04\ \Omega$, $R_a = 0.2\ \Omega$.

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

#### Step 1: Shunt field current

$$
I_{sh} = \frac{V}{R_{sh}} = \frac{230}{100} = 2.30\ \text{A}
$$

#### Step 2: Armature current

$$
I_a = I_{se} = I_L + I_{sh} = 80 + 2.30 = 82.30\ \text{A}
$$

#### Step 3: Generated emf

$$
E = V + I_a(R_a + R_{se}) = 230 + 82.30(0.2 + 0.04) = 230 + 19.752 = 249.75\ \text{V}
$$

**Answer:** emf generated by the armature $E = 249.75\ \text{V}$.

## Check yourself

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

1. The shunt field current is:
   A. 80 A
   B. V / Rsh = 2.3 A
   C. 0.43 A
   D. 2.3 A from E/Rsh
   Answer: B. The shunt is across terminal voltage 230 V: 230/100 = 2.3 A. (Key point: Shunt current)

2. The armature current in long shunt is:
   A. 80 A
   B. 77.7 A
   C. 82.3 A
   D. 2.3 A
   Answer: C. Ia = 80 + 2.3 = 82.3 A. (Key point: Armature current)

3. Generated emf is obtained from:
   A. E = V + IL(Ra + Rse)
   B. E = V - Ia(Ra + Rse)
   C. E = V + Ia Rsh
   D. E = V + Ia(Ra + Rse)
   Answer: D. Series field and armature are in series carrying Ia. (Key point: Emf equation)

4. The emf generated is:
   A. 249.75 V
   B. 230 V
   C. 245.4 V
   D. 251.2 V
   Answer: A. 230 + 82.3x0.24 = 249.75 V. (Key point: Result)

## Close variants set in other exams

- [Short shunt compound generator emf, 80 A at 220 V](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Agen-short-shunt-80a-220v) — Electrical Machines, chapter 3, asked 2×

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- [EMF of short shunt compound generator (0.5 Ω armature)](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Ashort-shunt-compound-50a-200v-0p5ohm) — Electrical Machines I, chapter 3, asked 2×

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