---
title: "EMF per turn Et = K√kVA"
subject: "Electric Machine Design"
chapter: 3
key: "semester_5/electric_machine_design:emf-per-turn-kva"
source: https://www.bctengineer.com/q/semester_5/electric_machine_design/emf-per-turn-kva
site: BCT Engineering
---

# EMF per turn Et = K√kVA

Electric Machine Design · Chapter 3: Transformer Design

- Asked in: 2072 Chaitra, 4 marks
- Video: [Relation between Emf per turn Et and Transformer rating Q #designofelectricalmachines #sdyelgatte](https://www.youtube.com/watch?v=JNvypbNbwEk) — S D Yelgatte 📖 ⚡
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/electric_machine_design/emf-per-turn-kva
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-3

## Question

For a transformer show that the emf per turn Et is given by Et = K√kVA where kVA = rating of transformer.

## Short answer

With rating Q = E_t·AT ×10⁻³ and a constant ratio r = φ_m/AT, we get φ_m = √(r Q×10³/(4.44f)). Then E_t = 4.44f φ_m = √(4.44 f r×10³) √Q, so E_t = K √(kVA) where K depends on frequency and the magnetic‑to‑electric loading ratio.

## Answer

The emf per turn is fixed by the ratio of magnetic loading to electric loading, which designers keep nearly constant for a given type of transformer.

#### Proof (single-phase)
Rating in kVA:

$$
Q = VI\times10^{-3} = E_tTI\times10^{-3} = E_t\,AT\times10^{-3}
$$

where $AT = IT$ (ampere-turns per winding) and $E_t = 4.44f\phi_m$.

Let the ratio of magnetic to electric loading be

$$
r = \frac{\phi_m}{AT} \;\Rightarrow\; AT = \frac{\phi_m}{r}
$$

Then

$$
\begin{aligned}
Q &= 4.44f\phi_m\cdot\frac{\phi_m}{r}\times10^{-3} = \frac{4.44f\phi_m^2}{r}\times10^{-3}\\
\phi_m &= \sqrt{\frac{rQ\times10^3}{4.44f}}
\end{aligned}
$$

Hence

$$
\begin{aligned}
E_t &= 4.44f\phi_m = 4.44f\sqrt{\frac{rQ\times10^3}{4.44f}} = \sqrt{4.44f\,r\times10^3}\,\sqrt{Q}\\
E_t &= K\sqrt{Q},\qquad K = \sqrt{4.44f\,r\times10^3}
\end{aligned}
$$

For a given frequency and a fixed ratio $r$ (chosen for minimum cost), $K$ is constant.

#### Typical values of K
| Transformer | K |
|---|---|
| 1-ph shell type | 1.0–1.2 |
| 1-ph core type | 0.75–0.85 |
| 3-ph shell type | 1.3 |
| 3-ph core, distribution | 0.45 |
| 3-ph core, power | 0.6–0.7 |

## Check yourself

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

1. The kVA rating can be written as:
   A. Q = Et / AT
   B. Q = Et^2 AT
   C. Q = AT/Et
   D. Q = Et x AT x 10^-3
   Answer: D. Q = V I = Et T I = Et AT (x10^-3). (Key point: Rating relation)

2. The ratio r in the proof is:
   A. phi_m / AT
   B. AT / phi_m
   C. Bm / delta
   D. Bm x delta
   Answer: A. r = magnetic loading / electric loading = phi_m / AT. (Key point: Loading ratio)

3. K = sqrt(4.44 f r x 10^3) is constant because:
   A. Q is fixed
   B. f and r are fixed for a given type
   C. Bm is zero
   D. Et is fixed
   Answer: B. Frequency and the chosen ratio r are constant. (Key point: Constant K)

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