---
title: "Three phase output equation and circular windings"
subject: "Electric Machine Design"
chapter: 3
key: "semester_5/electric_machine_design:output-equation-circular-windings"
source: https://www.bctengineer.com/q/semester_5/electric_machine_design/output-equation-circular-windings
site: BCT Engineering
---

# Three phase output equation and circular windings

Electric Machine Design · Chapter 3: Transformer Design

- Asked in: 2076 Asoj, 6 marks
- Video: [Output equation of transformer part1 by amit sir](https://www.youtube.com/watch?v=UAfjDg8jhIE) — Xmer tech
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/electric_machine_design/output-equation-circular-windings
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-3

## Question

Derive the output equation of 3-phase transformer. Why are the windings of transformer made in circular form?

## Short answer

Output equation: Q = 3.33 f B_m δ K_w A_w A_i × 10⁻³ kVA, derived from EMF per phase and ampere‑turn relation from window area. Circular windings withstand short‑circuit radial forces as uniform hoop stress, need least copper, are easier to wind, and provide uniform insulation and cooling.

## Answer

#### Output equation of a 3-phase core type transformer
Let $A_i$ = net core area, $A_w$ = window area, $B_m$ = maximum flux density, $\delta$ = current density, $K_w$ = window space factor, $T$ = turns per phase, $I$ = phase current.

EMF per phase:

$$
V \approx E = 4.44fB_mA_iT
$$

There are three limbs and two windows. Each window holds one primary and one secondary of two adjacent limbs, so the copper area in one window is

$$
K_wA_w = 2\left(\frac{I_pT_p}{\delta} + \frac{I_sT_s}{\delta}\right) = \frac{4AT}{\delta} \;\Rightarrow\; AT = \frac{K_wA_w\delta}{4}
$$

Output:

$$
\begin{aligned}
Q &= 3VI\times10^{-3} = 3\times4.44fB_mA_i\,(IT)\times10^{-3}\\
&= 3\times4.44fB_mA_i\times\frac{K_wA_w\delta}{4}\times10^{-3}\\
Q &= 3.33\,fB_m\delta K_wA_wA_i\times10^{-3}\ \text{kVA}
\end{aligned}
$$

#### Why windings are made circular
1. **Short-circuit forces:** under a short circuit the radial forces act outward on the HV and inward on the LV winding. A circular coil takes them as uniform hoop stress and keeps its shape; a rectangular coil would bend at the flat sides.
2. **Least copper:** for a given enclosed area, a circle has the smallest perimeter, so the mean length of turn, copper weight, cost and $I^2R$ loss are least.
3. **Ease of winding:** circular coils are easily wound on formers on a lathe, with uniform tension.
4. **Uniform insulation and cooling:** clearances between core, LV and HV are uniform all round, and oil ducts are easy to provide.

Because the winding is circular, the core is made stepped (cruciform or multi-stepped) to fill the circle as fully as possible.

## Check yourself

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

1. Copper area in one window of a 3-phase core transformer is:
   A. Kw Aw = 2AT/delta
   B. Kw Aw = AT/delta
   C. Kw Aw = 4AT/delta
   D. Kw Aw = 3AT/delta
   Answer: C. Two primary and two secondary half-windings: 2(AT/delta + AT/delta) = 4AT/delta. (Key point: Copper area in window)

2. The output equation constant 3.33 comes from:
   A. 4.44 / 2
   B. 3 x 4.44 / 2
   C. 4.44 x 4
   D. 3 x 4.44 / 4
   Answer: D. Q = 3 x 4.44 f Bm Ai x Kw Aw delta/4 x 10^-3. (Key point: Output constant)

3. Why do circular coils withstand short-circuit forces better?
   A. They have less copper
   B. Radial forces give uniform hoop stress so the shape is kept
   C. They have lower resistance
   D. They need no insulation
   Answer: B. Rectangular coils would bend at flat sides under radial forces. (Key point: Short-circuit forces)

4. Why does a circular coil need the least copper?
   A. A circle has the smallest perimeter for a given enclosed area, so Lmt is least
   B. It uses thinner wire
   C. It has fewer turns
   D. It has larger area
   Answer: A. Shortest mean length of turn means lowest copper weight and I^2R loss. (Key point: Least copper)

## Related in Electric Machine Design

- [Output equation of three phase transformer](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Aoutput-equation-three-phase-transformer) — Electric Machine Design, chapter 3, asked 2×
- [Output equation of three phase induction motor](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Ainduction-motor-output-equation) — Electric Machine Design, chapter 4, asked 6×
- [Circular windings and two-stepped core](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Astepped-core-circular-windings) — Electric Machine Design, chapter 3
- [Output (kVA) equation of single phase transformer](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Aoutput-equation-single-phase-transformer) — Electric Machine Design, chapter 3, asked 3×
- [Design: 125 kVA 6600/400 V three phase transformer](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Adesign-125kva-three-phase-windings) — Electric Machine Design, chapter 3, asked 4×
- [Output equation and separation of D, L](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Aoutput-equation-separation-dl) — Electric Machine Design, chapter 4

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