---
title: "Compound generator with 80 lamps"
subject: "Electrical Machine"
chapter: 3
key: "semester_4/electrical_machine:compound-generator-long-short-shunt-lamps"
source: https://www.bctengineer.com/q/semester_4/electrical_machine/compound-generator-long-short-shunt-lamps
site: BCT Engineering
---

# Compound generator with 80 lamps

Electrical Machine · Chapter 3: DC Generator

- Asked in: 2074 Bhadra, 6 marks
- Video: [long shunt compound generators](https://www.youtube.com/watch?v=CWJjNBdG8ow) — Stephen Mutale
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machine/compound-generator-long-short-shunt-lamps
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machine/chapter-3

## Question

In a 220 V compound generator, the armature, series and shunt windings have resistances of 0.3 Ω, 0.2 Ω and 60 Ω respectively. The load consists of 80 lamps, each rated at 60 W and 220 V. Find the total emf and armature current when the machine is connected for i) long shunt and ii) short shunt.

## Short answer

Load current = 4800/220 = 21.82 A. Long shunt: Ish = 3.667 A, Ia = 25.49 A, E = 232.74 V. Short shunt: Vsh = 224.36 V, Ish = 3.739 A, Ia = 25.56 A, E = 232.03 V.

## Answer

#### Load current
Load power = 80 × 60 = 4800 W at 220 V (terminal voltage V = 220 V).

I_L = 4800 / 220 = 21.82 A

Brush drop is not given, so it is neglected.

#### i) Long shunt
```text
 Ia --[Ra=0.3]--[Rse=0.2]--+------> IL=21.82 A
 (E)                       |
                    Rsh=60 (across armature+series)
```
The shunt field is connected across the load terminals (series field in the armature circuit).
- I_sh = V / R_sh = 220 / 60 = 3.667 A
- I_a = I_se = I_L + I_sh = 21.82 + 3.667 = **25.49 A**
- E = V + I_a(R_a + R_se) = 220 + 25.49 × (0.3 + 0.2) = 220 + 12.74 = **232.74 V**

#### ii) Short shunt
```text
 Ia --[Ra=0.3]--+--[Rse=0.2]--+--> IL
 (E)             |            |
                Rsh=60 (across armature only)
```
The shunt is connected across the armature, and the series field is in the line.
- Voltage drop in series field = I_L R_se = 21.82 × 0.2 = 4.364 V
- Shunt voltage = V_sh = 220 + 4.364 = 224.36 V
- I_sh = 224.36 / 60 = 3.739 A
- I_a = I_L + I_sh = 21.82 + 3.739 = **25.56 A**
- E = V_sh + I_aR_a = 224.36 + 25.56 × 0.3 = 224.36 + 7.67 = **232.03 V**

| Connection | I_sh (A) | I_a (A) | E (V) |
|---|---|---|---|
| Long shunt | 3.667 | 25.49 | 232.74 |
| Short shunt | 3.739 | 25.56 | 232.03 |

**Answer: Long shunt: I_a = 25.49 A, E = 232.74 V. Short shunt: I_a = 25.56 A, E = 232.03 V.**

## Check yourself

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

1. What is the load current for 80 lamps of 60 W at 220 V?
   A. 12.0 A
   B. 21.82 A
   C. 18.2 A
   D. 26.4 A
   Answer: B. I_L = 4800/220 = 21.82 A. (Key point: Load current)

2. In the long shunt connection, what is I_a?
   A. I_L = 21.82 A
   B. I_L - I_sh = 18.15 A
   C. I_L + I_sh = 25.49 A, with I_sh = 220/60
   D. I_sh only
   Answer: C. The shunt is across the terminals at 220 V, so I_sh = 3.667 A and I_a = 25.49 A. (Key point: Long shunt)

3. In the short shunt connection, what voltage is across the shunt field?
   A. 220 V
   B. 220 - 4.36 = 215.64 V
   C. 220 + I_a R_a
   D. 220 + I_L R_se = 224.36 V
   Answer: D. The shunt is across the armature, so it sees terminal voltage plus the series drop. (Key point: Short shunt)

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