A single phase transformer has 1000 turns on the primary and 200 turns on the secondary. The no load current is 3A at a power factor of 0.2 lag and the secondary current is 280A at a power factor of 0.8 lag. Neglect R_{2 }and X_{2} calculate (1) Magnetizing component and loss component of no load current. (2) primary current (3) input power factor .Draw phasor diagram showing all the currents.

#### Solution

`I_P/I_S=N_S/N_P`

Therefore, `I_P=N_S/N_PxxI_S=200/1000xx280=56A`

cosπ_{2}=0.8 cosπ_{0}=0.2 sinπ_{2}=0.6 sinπ_{0}=0.98

solve for horizontal and vertical components

πΌ_{1}πππ π_{1}=πΌ_{2}πππ π_{2}+πΌ_{0}πππ π_{0} =(56×0.8)+(3×0.2)=45.4π΄

πΌ_{1}π πππ_{1}=πΌ_{2}π πππ_{2}+πΌ_{0}π πππ_{0} =(56×0.6)+(3×0.98) = 36.54A

πΌ_{1}=`sqrt(45.4^2+36.54^2)`=58.3π΄

π°_{π}=π°_{π}ππππ_{π}=(π×π.π)=π.πA

tanπ_{1} `(36.54)/(45.4)`= 0.805

π_{π}=ππ°

Power factor cosπ_{π}=πππππ°=π.ππ lagging.