A 23 kVA 2300230 V 60 Hz step down transformer has the follo

A 23 kVA, 2300/230 V, 60 Hz, step down transformer has the following resistance and leakage reactance values: R_1 = 4 Ohm, R_2 = 0.04 Ohm X_1 = 12 Ohm, and X_2 = 0.12 Ohm. The transformer is operating at 0.75 of its rated load. If the power factor of load is 0.87 leading and no-load current is negligible, determine: The current of secondary winding The induced emf in the secondary winding The induced emf and the current on the primary side considering transformation ratio The source voltage The power supplied to the load The power input The efficiency

Solution

k= 230/2300=1/10 R1=4ohms, R2= 0.04ohms

X1=12OHMS, X2=0.012OHMS, POWER FACTOR- 0.87

high voltage side is obviously primary side

R01= R1+R2\'=R1+R2/K²=4+0.04/(1/10)²=8ohms

X01=X1+X2\'=X1+X2/K²=12+0.012/(1/10)²=13.2

Z01= sqrt of{R01²+X01²}= sqrt of {8²+13.2²}= 15.43ohms

R02=R2+R1\'= R2+K²R1=0.04+[1/10]²x4= 0.08 ohms

X02=X2+X1\'=0.012+{1/10}²x12=0.00144

Z02=sqrt of {R02²+X02²}= sqrt of {0.08²+0.00144²}= 0.0800ohms

emf= 4.44fp.f

4.44x50x0.87=193.14v