could anyone help a single phase motor takes 50 A at a power factor of 0.6 lagging from a 240V, 50 Hz supply. Determin a.) the current taken by the capacitor in parrallel with the motor to correct the power factor to unity, and b.) the value of the supply current after power factor correction. i worked out the Ic by pf=cosO 0=cos -1 (0.6) 0=53.13 true power=Vcoc0 =240 cos 53.13 144W Q=true power * tan 0 Q=144 * tan 53.13 Q=191.99 VA Ic=Q/V Ic=191.99/240 Ic=0.799A but am struggling with the value of the supply current question b i have calculated Xc as 300.38 ohms and C as 0.106uF but dont know where to go from here can anyone help cheers
is the answer to b just simply supply corrected to power factor of 0.6 = 144W/(240V*0.6) =1A ?????????
redshaw, I'll try. The apparent power is 50*240 = 12000 volt-amps The average power is 12000*.6 = 7200 watts The reactive capacitor power is sqrt(12000^2-7200^2) = 9600 vars The reactance of the capacitor is (240^2)/9600 = 6 ohms The reactive current is sqrt(9600/6) = 40 amps The current after correction is 7200/240 = 30 amps So correcting the power factor reduced the current load of the power supply by 20 amps. Ask if you have any questions. Ratch