A 50KVA, 4400/220 volt transformer has R1=3.45Ω, R2=0.009Ω. The reactance are X1=5.2Ω and X2=0.015Ω. Calculate for the transformer. - Full load current on primary and secondary side. - Equivalent resistance, reactance, impedances, referred to primary side and secondary side and Total Copper Loss using individual resistance and equivalent resistances. -

Given:

50 KVA

4400/220 volt transformer i.e.

E1=4400V

E2=220V

R1=3.45Ω

R2=0.009Ω

X1=52Ω

X2=0.015Ω

$K=\dfrac{E_2}{E_1} =0.05$

$I_1$ $=\dfrac{\text{KVA rating}\times1000}{E_2} \\ =\dfrac{50\times1000}{4400}\\=11.3636A$

$I_2$ $=\dfrac{\text{KVA rating}\times1000}{E_2} \\ =227.2727$

In reference to primary side

Resistance, R01=R1+R21

$R_{01}=R_1+\dfrac{R_2}{K^2}=3.45+\dfrac{0.009}{(0.05)^2}=7.05\Omega$

Reactance, X01=X1+X21

$X_{01}=X_1+\dfrac{X_2}{K^2}=5.2+\dfrac{0.015}{(0.05)^2}=11.2\Omega$

Impedance, $Z_{01}=\sqrt{R_{01}^2+X _{01}^2 } =13.23Ω$

In reference to secondary side,

Resistance,

$R_{02}$ $=R_2+R_1^1 \\ =R_2+K^2R_1 \\ =0.009+(0.05)^2 \times 3.45 \\ =0.017625 \Omega$

Reactance,

$X_{02}$ $=X_2+X_1^1 \\ =X_2+K^2X_1 \\ =0.015+(0.05)^2\times5.2 \\=0.028 \Omega \\ Z_{02}=\sqrt{R_{02}^2+x_{02}^2}=0.033^2$

Total Copper Loss, Using individual resistances,

$W_{cu}$ $=I_1^2R_1+I_2^2R_2 \\=(11.3636)^23.45+(227.2727)^20.009 \\=910.379 watt$

Using equivalent resistance,$R_{01}$

$W_{cu}$ $=(I_1)^2R_{01} \\ =(11.3636)^2\times7.05 \\ 910.367 watt$

Using equivalent resistance,$R_{02}$

$W_{cu}$ $=(I_2)^2R_{02} \\ =(227.2727)^2\times0.017625 \\ = 910.382 watt$