Equivalent circuit, efficiency and regulation of 5 kVA transformer

OC test on LV gives the shunt branch directly on LV. SC test on HV gives series values on HV, which are referred to LV by $a^2 = (220/400)^2 = 0.3025$. Rated HV current $= 5000/400 = 12.5$ A, so the SC reading at 11.4 A must be scaled to full load.

i) Equivalent circuit parameters referred to LV

OC test:

$$\begin{aligned} \cos\phi_0 &= \frac{100}{220\times2} = 0.2273 \\ I_c &= 2\times0.2273 = 0.4545\ \text{A}, \quad I_m = 2\times0.9738 = 1.9477\ \text{A} \\ R_0 &= \frac{220}{0.4545} = 484.0\ \Omega, \quad X_0 = \frac{220}{1.9477} = 112.96\ \Omega \end{aligned}$$

SC test (HV side):

$$\begin{aligned} R_{eq,HV} &= \frac{200}{11.4^2} = 1.539\ \Omega, \quad Z_{eq,HV} = \frac{40}{11.4} = 3.509\ \Omega \\ X_{eq,HV} &= \sqrt{3.509^2 - 1.539^2} = 3.153\ \Omega \end{aligned}$$

Referred to LV:

$$R_{eq,LV} = 1.539\times0.3025 = 0.4655\ \Omega, \quad X_{eq,LV} = 3.153\times0.3025 = 0.9539\ \Omega$$

ii) Efficiency at full load, 0.9 pf lag

$$\begin{aligned} P_{cu,fl} &= 200\left(\frac{12.5}{11.4}\right)^2 = 240.46\ \text{W} \\ P_{out} &= 5000\times0.9 = 4500\ \text{W} \\ \eta &= \frac{4500}{4500 + 100 + 240.46} = 92.97\% \end{aligned}$$

Approximate regulation (on HV side, $\sin\phi = 0.4359$)

$$\begin{aligned} \%\text{Reg} &= \frac{I(R\cos\phi + X\sin\phi)}{V}\times100 \\ &= \frac{12.5(1.539\times0.9 + 3.153\times0.4359)}{400}\times100 \\ &= \frac{17.31 + 17.18}{400}\times100 = 8.62\% \end{aligned}$$

Answer: $R_0 = 484\ \Omega$, $X_0 = 112.96\ \Omega$, $R_{eq} = 0.466\ \Omega$, $X_{eq} = 0.954\ \Omega$ (all LV side); efficiency = 92.97%; regulation ≈ 8.62%.