---
title: "4-pole shunt generator, 3 kW at 120 V"
subject: "Electrical Machine"
chapter: 3
key: "semester_4/electrical_machine:shunt-generator-4pole-wave-3kw"
source: https://www.bctengineer.com/q/semester_4/electrical_machine/shunt-generator-4pole-wave-3kw
site: BCT Engineering
---

# 4-pole shunt generator, 3 kW at 120 V

Electrical Machine · Chapter 3: DC Generator

- Asked in: 2070 Magh, 8 marks; 2069 Bhadra (asked 2 times)
- Video: [DC. Shunt generator : problem & solution](https://www.youtube.com/watch?v=lU9cq1u_hLE) — Sumit Kumar Maitra
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machine/shunt-generator-4pole-wave-3kw
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machine/chapter-3

## Question

A 4-pole dc shunt generator has wave wound armature. The armature and field winding resistance are 0.2 Ω and 60 Ω respectively. The brush contact drop is 1 V per brush. The generator is delivering a power of 3 kW at 120 V. Calculate:
i) Total armature current coming out from the brush
ii) Current in each armature conductor
iii) Generated EMF (E)

## Short answer

Load current 25 A, field 2 A, armature current Ia=27 A. Wave winding (A=2): conductor current=13.5 A. Generated EMF=120+27×0.2+2=127.4 V.

## Answer

Load current is found from P/V, then field current is added to get the armature current. A wave winding has A = 2 parallel paths whatever the number of poles.

#### Equivalent circuit
```text
        Ia=27A      Ra=0.2
   +----->---[====]---(E)---+---->--- IL=25A ---+
   |                        |                   |
   |   brush drop 2 x 1 V   |                [LOAD]  V=120 V
   |                        |                   |
   +---> If=2A ---[Rsh=60]--+-------------------+
```

#### i) Armature current
- I_L = 3000 / 120 = 25 A
- I_sh = 120 / 60 = 2 A
- **I_a = I_L + I_sh = 25 + 2 = 27 A**

#### ii) Current in each armature conductor
Wave winding: A = 2 parallel paths.

I_conductor = I_a / A = 27 / 2 = **13.5 A**

#### iii) Generated EMF
E = V + I_a R_a + brush drop
= 120 + (27 × 0.2) + (2 × 1)
= 120 + 5.4 + 2 = **127.4 V**

| Quantity | Value |
|---|---|
| Load current | 25 A |
| Field current | 2 A |
| Armature current | 27 A |
| Armature drop | 5.4 V |
| Brush drop (2 brushes) | 2 V |
| Generated emf | 127.4 V |

**Answer: I_a = 27 A, conductor current = 13.5 A, E = 127.4 V.**

## Check yourself

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

1. How is the armature current of a shunt generator found?
   A. Ia = IL − Ish
   B. Ia = IL + Ish
   C. Ia = IL
   D. Ia = Ish
   Answer: B. Ia = 25 + 2 = 27 A (load plus field current). (Key point: Armature current)

2. How many parallel paths does a 4-pole wave winding have?
   A. 4
   B. 1
   C. 8
   D. 2
   Answer: D. Wave winding always has A = 2. (Key point: Wave parallel paths)

3. What is the current in each armature conductor?
   A. 13.5 A
   B. 27 A
   C. 6.75 A
   D. 25 A
   Answer: A. Ia/A = 27/2 = 13.5 A. (Key point: Conductor current)

4. How is the generated emf computed?
   A. E = V − Ia Ra
   B. E = V + Ia Ra + brush drop = 127.4 V
   C. E = V + Ia Ra = 125.4 V
   D. E = V + IL Ra + 2 = 127 V
   Answer: B. 120 + 5.4 + 2 = 127.4 V. (Key point: Generated EMF)

## Related in Electrical Machine

- [4-pole shunt generator, wave wound, 3 kW](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Agen-wave-wound-3kw-120v) — Electrical Machines, chapter 3, asked 3×
- [EMF of 25 kW short shunt generator with brush drop](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Ashort-shunt-25kw-250v-brush-drop) — Electrical Machines I, chapter 3
- [Wave-wound shunt generator, power absorbed by load](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Agen-shunt-378-conductors-load-power) — Electrical Machines, chapter 3
- [Long shunt compound generator speed, lap and wave](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Agen-long-shunt-10kw-lap-wave) — Electrical Machines, chapter 3, asked 2×
- [Shunt generator feeding six lamps](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Ashunt-gen-lap-six-lamps) — Electrical Machines I, chapter 3
- [Starting current; speed and % drop of shunt motor](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Adcm-high-starting-current-speed-drop) — Electrical Machines, chapter 4

## Related in other subjects

- [Armature dimensions of 50 kW shunt generator](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Adc-generator-50kw-armature) — Electric Machine Design, chapter 5
- [30 kW 240 V generator, 2.5% drop](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Adc-generator-30kw-240v-2-5pct) — Electric Machine Design, chapter 5, asked 3×
- [30 kW 240 V generator, 3% drop](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Adc-generator-30kw-240v-3pct) — Electric Machine Design, chapter 5, asked 3×
- [DC shunt motor armature torque](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_and_electronics_engineering%3Adc-shunt-motor-torque) — Basic Electrical and Electronics Engineering, chapter 2