---
title: "Long shunt compound generator emf, 40 A at 200 V"
subject: "Electrical Machines"
chapter: 3
key: "semester_4/electrical_machines:gen-long-shunt-40a-200v"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/gen-long-shunt-40a-200v
site: BCT Engineering
---

# Long shunt compound generator emf, 40 A at 200 V

Electrical Machines · Chapter 3: DC Generator

- Asked in: 2073 Bhadra (old course), 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-40a-200v
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-3

## Question

A dc long shunt compound generator has armature winding resistance of 0.4 ohm, series field winding resistance of 0.5 ohm and shunt-field winding resistance of 100 ohms. The generator delivers a current of 40 A to the load at 200 volt. Calculate the emf generated by the armature.

## Short answer

I_sh=200/100=2 A, I_a=42 A. E=200+42×(0.4+0.5)=237.8 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: $I_L = 40$ A, $V = 200$ V, $R_{sh} = 100\ \Omega$, $R_{se} = 0.5\ \Omega$, $R_a = 0.4\ \Omega$.

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

#### Step 1: Shunt field current

$$
I_{sh} = \frac{V}{R_{sh}} = \frac{200}{100} = 2.00\ \text{A}
$$

#### Step 2: Armature current

$$
I_a = I_{se} = I_L + I_{sh} = 40 + 2.00 = 42.00\ \text{A}
$$

#### Step 3: Generated emf

$$
E = V + I_a(R_a + R_{se}) = 200 + 42.00(0.4 + 0.5) = 200 + 37.800 = 237.80\ \text{V}
$$

**Answer:** emf generated by the armature $E = 237.80\ \text{V}$.

## Check yourself

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

1. With V = 200 V and Rsh = 100 ohm, Ish is:
   A. 0.5 A
   B. 2 A
   C. 200 A
   D. 1 A
   Answer: B. Ish = 200/100 = 2 A. (Key point: Shunt current)

2. In a long shunt generator the series field carries:
   A. IL = 40 A only
   B. Ish = 2 A
   C. Ia = IL + Ish = 42 A
   D. IL - Ish = 38 A
   Answer: C. The series field is in line with the armature. (Key point: Series field current)

3. The generated emf is:
   A. 218 V
   B. 217.6 V
   C. 236 V
   D. 237.8 V
   Answer: D. E = 200 + 42(0.4 + 0.5) = 237.8 V. (Key point: Emf)

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