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EMF per turn Et = K√kVA

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.

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

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×10−3=EtTI×10−3=Et AT×10−3Q = VI\times10^{-3} = E_tTI\times10^{-3} = E_t\,AT\times10^{-3}

where AT=ITAT = IT (ampere-turns per winding) and Et=4.44fϕmE_t = 4.44f\phi_m.

Let the ratio of magnetic to electric loading be

r=ϕmAT  ⇒  AT=ϕmrr = \frac{\phi_m}{AT} \;\Rightarrow\; AT = \frac{\phi_m}{r}

Then

Q=4.44fϕm⋅ϕmr×10−3=4.44fϕm2r×10−3ϕm=rQ×1034.44f\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

Et=4.44fϕm=4.44frQ×1034.44f=4.44f r×103 QEt=KQ,K=4.44f r×103\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 rr (chosen for minimum cost), KK is constant.

Typical values of K

TransformerK
1-ph shell type1.0–1.2
1-ph core type0.75–0.85
3-ph shell type1.3
3-ph core, distribution0.45
3-ph core, power0.6–0.7

✓ Answer checked by Claude Sonnet 5.5

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