Answer to Question #210538 in Electrical Engineering for Zwi

Question

The no-load voltage of a single phase transformer is 3300 / 400 V. It takes 7A at a power factor of 0.3 lagging on no-load. Determine the following: 

(a) The magnetizing current of the transformer.

(b) The magnetizing reactance of the transformer.

(c) The core loss current of the transformer.

(d) The core loss resistance of the transformer

(e) The iron losses of the transformer.

(f) The primary current if the load on the secondary is 80 A at a power factor of 0.85 leading.

Answer 1

(a)

The magnetizing current of the transformer is

$I_m=I_o\sin\phi\\ I_o=7A\\ \cos\phi=0.3\\ \phi=\cos^{-1}0.3\\ \phi=72.5\degree\\ \sin\phi=0.95\\ I_m=7\times 0.95=6.68A \\$

(b)

The magnetizing reactance of the transformer is

$X_o=\frac{E_1}{I_m}\\ E_1=3300V\\ I_m=6.68A\\ X_o=\frac{3300}{6.68}\\ X_o=494\Omega$

(c)The core loss current of the transformer is

$I_w=I_o\cos\phi\\ I_w=7\times 0.3\\ I_w=2.1A\\$

(d) The core loss resistance of the transformer is

$R_o=\frac{E_1}{I_w}\\ R_o=\frac{3300}{2.1}\\ R_o=1571.4\Omega\\$

(e) The iron losses of the transformewr is

$P_i=I_w^2R_o\\ P_i=2.1^2\times1571.4\\ P_i=6390W\\$

(f) The primary current I1 if the load on the secondary is 80 A at a power factor of 0.85 leading is calculated as follows

$\phi_1=72.5\degree\\ \phi_2=\cos^{-1}0.85=31.8\degree\\ \phi=72.5+31.8=104.3\degree\\ I_2'=(K)I_2\\ K=\frac{400}{3300}=\frac{4}{33}\\ I_2=80A\\ I_2'=9.7A$

$I_1=\sqrt{I_o^2+I_2'^2+2I_oI_2'+\cos\phi}\\ I_1=\sqrt{7^2+9.7^2+2(7)(9.7)+\cos104.3}\\ I_1=16.7A$