Efficiency and regulation of 5 kVA transformer

Only the short-circuit data is given; the open-circuit result is needed for iron loss. Assumption: the usual OC data for this textbook problem, 220 V, 2 A, 100 W on the LV side, i.e. iron loss $P_i = 100$ W. (If a different OC wattmeter reading is given, replace 100 W.)

Rated HV current: $I_2 = \frac{5000}{400} = 12.5$ A. The SC test was done at 11.4 A, so its copper loss must be scaled to full load.

Parameters from SC test (HV side)

$$\begin{aligned} Z_{02} &= \frac{40}{11.4} = 3.509\ \Omega \\ R_{02} &= \frac{200}{11.4^2} = 1.539\ \Omega \\ X_{02} &= \sqrt{3.509^2 - 1.539^2} = 3.153\ \Omega \end{aligned}$$

Full-load copper loss

$$P_{cu,FL} = 200 \times \left(\frac{12.5}{11.4}\right)^2 = I_2^2R_{02} = 12.5^2 \times 1.539 = 240.5\ \text{W}$$

Efficiency at full load, 0.9 pf lagging

$$\begin{aligned} \text{Output} &= 5000 \times 0.9 = 4500\ \text{W} \\ \text{Losses} &= P_i + P_{cu} = 100 + 240.5 = 340.5\ \text{W} \\ \eta &= \frac{4500}{4500 + 340.5} \times 100 = 92.97\% \end{aligned}$$

Voltage regulation at full load, 0.9 pf lagging

$\cos\phi = 0.9$, $\sin\phi = 0.4359$.

$$\begin{aligned} \%\text{Reg} &= \frac{I_2(R_{02}\cos\phi + X_{02}\sin\phi)}{V_2}\times 100 \\ &= \frac{12.5\,(1.539 \times 0.9 + 3.153 \times 0.4359)}{400}\times 100 \\ &= \frac{12.5 \times (1.385 + 1.375)}{400}\times 100 = 8.62\% \end{aligned}$$

The regulation does not depend on the OC test, so 8.62% holds whatever the iron loss is.

Answer: Full-load efficiency at 0.9 pf lagging $\approx$ 92.97% (with $P_i$ = 100 W); voltage regulation = 8.62%.