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Regulation and zero-regulation power factor of 100 kVA transformer

Short answer

Use percent regulation = epsilon_r cos phi + epsilon_x sin phi with epsilon_x = sqrt(10^2 - 5^2) = 8.66%. At 0.8 lagging regulation is 9.20%; zero regulation occurs at 0.866 leading pf.

A single phase, 100 kVA, 2000/200 V, 50 Hz transformer has an impedance drop of 10% and resistance drop of 5%. Calculate: a) The regulation at full load 0.8 power factor lagging. b) The value of power factor at which regulation is zero.

Per-unit resistance and impedance drops are given: εr=5%\varepsilon_r = 5\%, εz=10%\varepsilon_z = 10\%, so the reactance drop is

εx=102−52=8.66%\varepsilon_x = \sqrt{10^2 - 5^2} = 8.66\%

a) Regulation at full load, 0.8 pf lagging

%Reg=εrcos⁡ϕ+εxsin⁡ϕ=5×0.8+8.66×0.6=4.0+5.196=9.20%\begin{aligned} \%\text{Reg} &= \varepsilon_r\cos\phi + \varepsilon_x\sin\phi \\ &= 5\times0.8 + 8.66\times0.6 \\ &= 4.0 + 5.196 = 9.20\% \end{aligned}

b) Power factor for zero regulation

Regulation is zero for a leading pf, when

εrcos⁡ϕ−εxsin⁡ϕ=0tan⁡ϕ=εrεx=58.66=0.577⇒ϕ=30∘cos⁡ϕ=εxεz=8.6610=0.866\begin{aligned} \varepsilon_r\cos\phi - \varepsilon_x\sin\phi &= 0 \\ \tan\phi &= \frac{\varepsilon_r}{\varepsilon_x} = \frac{5}{8.66} = 0.577 \Rightarrow \phi = 30^\circ \\ \cos\phi &= \frac{\varepsilon_x}{\varepsilon_z} = \frac{8.66}{10} = 0.866 \end{aligned}

Answer: (a) regulation = 9.20%; (b) zero regulation at pf = 0.866 leading.

✓ Answer checked by Claude Opus 5.5 (high effort)

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