OC/SC test of 2200/110 V, 40 kVA transformer

The OC test (LV side) gives the shunt branch; the SC test (HV side) gives the series impedance referred to HV. Rated HV current $= 40000/2200 = 18.18$ A, so the SC test (18.2 A) is practically at full load.

(i) $R_0$ and $X_0$ (LV side)

$$\begin{aligned} \cos\phi_0 &= \frac{320}{110\times8.4} = 0.3463 \\ I_c &= 8.4\times0.3463 = 2.909\ \text{A}, \quad I_m = 8.4\times\sin(69.74^\circ) = 7.880\ \text{A} \\ R_0 &= \frac{110}{2.909} = 37.81\ \Omega, \quad X_0 = \frac{110}{7.880} = 13.96\ \Omega \end{aligned}$$

SC test values (HV side)

$$\begin{aligned} R_{eq} &= \frac{650}{18.2^2} = 1.962\ \Omega, \quad Z_{eq} = \frac{85}{18.2} = 4.670\ \Omega \\ X_{eq} &= \sqrt{4.670^2 - 1.962^2} = 4.238\ \Omega \end{aligned}$$

(ii) Full-load efficiency at 0.85 pf lag

Full-load copper loss (corrected to rated current 18.18 A):

$$\begin{aligned} P_{cu,fl} &= 650\left(\frac{18.18}{18.2}\right)^2 = 648.7\ \text{W} \\ P_{out} &= 40000\times0.85 = 34000\ \text{W} \\ \eta &= \frac{34000}{34000 + 320 + 648.7} = \frac{34000}{34968.7} = 97.23\% \end{aligned}$$

(iii) Approximate voltage regulation (0.85 pf lag, $\sin\phi = 0.527$)

$$\begin{aligned} \%R &= \frac{I(R_{eq}\cos\phi + X_{eq}\sin\phi)}{V}\times100 \\ &= \frac{18.18(1.962\times0.85 + 4.238\times0.527)}{2200}\times100 \\ &= \frac{30.33 + 40.59}{2200}\times100 = 3.22\% \end{aligned}$$

Answer: $R_0 = 37.81\ \Omega$, $X_0 = 13.96\ \Omega$ (LV side); full-load efficiency = 97.23%; regulation ≈ 3.22%.