A 30 KVA, 2400/120V, 50Hz transformer has high-voltage winding resistance of 0.1Ω and leakage reactance of 0.22Ω. The low voltage winding reactance is 0.035Ω & leakage reactance is 0.012Ω. Calculate equivalent resistance as referred to primary and secondary, equivalent reactance as referred to primary and secondary, equivalent impedance as referred to primary and secondary. Copper loss at full load and at 75% of full load. -

R1=0.1Ω; X1=0.22Ω; R2=0.035Ω; X2=0.012Ω

Equivalent Resistance ref. to primary,

$R1e=R_1+R^2_1 \\ = 0.1+0.035\bigg(\dfrac{2400}{120}\bigg)^2 \\ =14.1Ω$

Equivalent Reaction ref. to primary,

$X1=X1+X^2_1 \\ =0.22+0.012\bigg(\dfrac{2400}{120}\bigg)^2 \\ =5.02Ω$

$Z_1=\sqrt{R_1^2+X_1^2}=0.2417 \Omega ; \hspace{2cm} Z_2=\sqrt{R_2^2+X_2^2}=0.037 \Omega$

Equation Impedance ref. to primary,

$Z_{1s}=z_1+Z_2^1 \\ =0.2417+0.037\times\bigg(\dfrac{2400}{120}\bigg)^2 \\ = 15.0415 \Omega \\ or Z_{ie}=\sqrt{R_{ie}^2+X_1^2}=14.967\Omega$

Equation Reaction ref. to Secondary,

$X_{2e}=X_2+X_1^1 \\ =0.012+0.22 \times\bigg(\dfrac{120}{2400}\bigg)^2 \\ =0.01255 \Omega$

Equation Resistance ref. to secondary,

$R_{2e}=R_2+R_1^1 \\ =0.035+0.1\times\bigg(\dfrac{120}{2400}\bigg)^2 \\ =0.03525 \Omega$

Equation Impendance ref. to secondary, $Z2e=0.03742Ω$

$I_{1ft}=\dfrac{30\times10^3}{2400}=12.5 A \\ I_{2ft}=\dfrac{30\times10^3}{120}=250 A$ $\hspace{2cm}$ $P_{cuFi}=I_{1ft}^2R_1+I_{2ft}^2R_2 \\ \; \\ \; P_{cuFi}=2.203kw$

At 75% of full load x=0.75

$P_{cu}=X2X P_{cuFi}=1.2393kw$