A SKVA, 200/400 volt, 50Hz, single phase transformer gave the following test results.

Draw the equivalent circuit of the transformer referred to LV side insert all parameters values.

Effectively at 0.9 power factor lending if operating at rated load.

Given:

$P_{APPARENT}= 5 KVA, 200/400 volt, 50 Hz$

V1= 220V, I0= 0.7A, WI=60 W

VSC= 22V, I2= 16, WSC=120W

From OC test:-

W1=V1 I0 cos Ø0

$\cos \emptyset_0=\dfrac{W_i}{V_1I_0}=\dfrac{60}{200\times0.7}=0.4285 A \\ \emptyset_0=64.623^0 \\ \sin \emptyset=0.904 \\ I_w=I_0 \cos \emptyset_0 \\=(0.7)(0.4286) \\ =0.2999 A \\ I_u=I_0 \sin \emptyset_0 \\ =(0.7)(0.935) \\ =0.6324 A \\ R_0=\dfrac{V_1}{I_w}=\dfrac{200}{0.2999}=666..89 \Omega \\ X_0=\dfrac{V_i}{I_u}=\dfrac{200}{0.6324}=316.255 \Omega$

Form SC test :-

$W_{SC}=I^2_2R_{02} \\ R_{02}=\dfrac{W_{SC}}{I_2^2}=\dfrac{120}{16^2}=0.46875 \Omega \\ Z_{02}=\dfrac{V_{SC}}{I_2}=\dfrac{22}{16}=1.375 \Omega \\ X_{02}=\sqrt{Z_{02}^2-R_{02}^2}=1.2926 \Omega \\ Now \ \ \ k= \dfrac{E_2}{E_1}=\dfrac{400}{200}=2 \\ R_{01}=\dfrac{R_{02}}{k^2}=0.1172 \Omega \\ X_{01}=\dfrac{X_{02}}{K^2}=0.32315 \Omega$

Now pf 0.9 X=1...........rated load

$\% η=\dfrac{x\times\text{full load KVL}\times pf}{(x\times\text{full load KVL}\times pf)++W_p\text{(in KW)}+x^2w_{cu}\text{(in KW)}} \\ η=\dfrac{1\times5\times0.9}{(1\times5\times0.9)+(0.06)+1^2(0.07324)\times100} \\ η=0..9712\times100 \\ η=97.12 \% \\ \therefore η=0.9712$