---
title: "4-pole shunt generator, wave wound, 3 kW"
subject: "Electrical Machines"
chapter: 3
key: "semester_4/electrical_machines:gen-wave-wound-3kw-120v"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/gen-wave-wound-3kw-120v
site: BCT Engineering
---

# 4-pole shunt generator, wave wound, 3 kW

Electrical Machines · Chapter 3: DC Generator

- Asked in: 2071 Magh (old course), 2+3+3 marks; 2070 Magh, 8 marks; 2069 Bhadra (asked 3 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_machines/gen-wave-wound-3kw-120v
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-3

## Question

A 4-pole dc shunt generator has wave wound armature. The armature and field winding resistances are $0.2\ \Omega$ and $60\ \Omega$ respectively. The brush contact drop is 1 volt per brush. The generator is delivering a power of 3 kW at 120 V. Calculate:
(a) Total armature current coming out from the brush.
(b) Current in each armature conductor.
(c) Generator EMF (E).

## Short answer

Load current I_L = 3000/120 = 25 A, shunt field current I_sh = 120/60 = 2 A, so armature current I_a = 27 A. Wave winding has 2 parallel paths, so conductor current = 27/2 = 13.5 A. Generated emf E = V + I_a R_a + brush drop = 120 + 27×0.2 + 2 = 127.4 V.

## Answer

In a shunt generator the armature current supplies both the load and the field: $I_a = I_L + I_{sh}$. The generated emf must also cover the armature drop and the brush drop.

#### (a) Armature current

Load current:

$$
I_L = \frac{P}{V} = \frac{3000}{120} = 25\ \text{A}
$$

Shunt field current:

$$
I_{sh} = \frac{V}{R_{sh}} = \frac{120}{60} = 2\ \text{A}
$$

$$
I_a = I_L + I_{sh} = 25 + 2 = 27\ \text{A}
$$

#### (b) Current in each armature conductor

A wave winding has $A = 2$ parallel paths regardless of the number of poles.

$$
I_{\text{conductor}} = \frac{I_a}{A} = \frac{27}{2} = 13.5\ \text{A}
$$

#### (c) Generated emf

Brush drop $= 1\ \text{V} \times 2\ \text{brushes (one positive, one negative)} = 2\ \text{V}$.

$$
\begin{aligned}
E &= V + I_aR_a + V_{brush} \\
&= 120 + (27)(0.2) + 2 \\
&= 120 + 5.40 + 2 = 127.40\ \text{V}
\end{aligned}
$$

**Answer:** (a) $I_a = 27\ \text{A}$; (b) current per conductor $= 13.5\ \text{A}$; (c) $E = 127.40\ \text{V}$.

## Check yourself

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

1. For the 3 kW, 120 V shunt generator (Rsh = 60 Ω), the armature current is
   A. 25 A
   B. 27 A
   C. 23 A
   D. 2 A
   Answer: B. Ia = IL + Ish = 3000/120 + 120/60 = 25 + 2 = 27 A. (Key point: Armature current in shunt generator)

2. For the 4-pole wave-wound armature, the current in each conductor is
   A. 27/4 = 6.75 A
   B. 27 A
   C. 27/2 = 13.5 A
   D. 27 x 2 = 54 A
   Answer: C. A wave winding always has A = 2 parallel paths, whatever the number of poles, so each conductor carries Ia/2. (Key point: Current per conductor in wave winding)

3. Total brush contact drop (1 V per brush) is
   A. 1 V
   B. 2 V, one positive and one negative brush
   C. 4 V, one per pole
   D. 0.5 V
   Answer: B. Current passes through one positive and one negative brush, so brush drop = 2 V. (Key point: Brush drop)

4. The generated emf is
   A. 120 + 5.4 = 125.4 V
   B. 120 + 5.4 + 2 = 127.4 V
   C. 120 - 5.4 - 2 = 112.6 V
   D. 120 + 0.2 x 25 + 2 = 127 V
   Answer: B. E = V + Ia Ra + Vbrush = 120 + 27 x 0.2 + 2 = 127.4 V. Note Ia is 27 A, not the load current. (Key point: Generated emf)

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## Related in other subjects

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