《电路》（英文版）9-3 Apparent power and power factor

89-3 Apparent power and power factor We assume v=Vm cos(ot +a)i=Im cos(at+B) The phase angle by which the voltage leads the current is 9=a-B P=-v cos 9 or P=VI cos 9 2 S=vI--(A)--apparent power The ratio of the real or average power to the apparent power is PF= called the power factor(PF). In the sinusoidal case: PF=cos 9 For a purely resistive load,9=0, PF=1. For a purely reactive load, 9=+90, PF=0

§9-3 Apparent power and power factor We assume  =Vm cos( t +) i = I m cos( t +  ) The phase angle by which the voltage leads the current is  = −  cos cos 2 1 P V I or P VI = m m = S =VI − −(VA) − −apparent power The ratio of the real or average power to the apparent power is called the power factor(PF). S P VI P PF = = In the sinusoidal case: PF = cos For a purely resistive load,  = 0 , PF=1. For a purely reactive load, 90 , PF=0.   = 

JPF=0.5→c0s9=0.5 9=609--inductive load 9=-600--capacitive load The term leading or lagging referring to the phase of the current with respect to the voltage Leading PF---capacitive load. Lagging PF--inductive load Examp 60 60 60 12∠-53.1°A j+1+j53+j45∠53.1 S.=60×12=720V4 十 1+j5 0×12c0s53.1°=432W Js=60∠00V load=122×2+122×1=432W PF=c0s9=cos53.,1°=0.6 a generator produce an output voltage of 200V at 60 Hz,PF=1,P=1000,I1=5A.IfPF=0.5,12=10A.When R=0.2,P1=5W;P2=20w(PF>0.85)

   = − − − = − − capacitive load inductive load     60 60 If PF = 0.5 → cos = 0.5 The term leading or lagging referring to the phase of the current with respect to the voltage. Leading PF---capacitive load. Lagging PF--inductive load. Example: I s • − =  + • V s V  60 0 2 − j1 1 + j5 A j j j Is   12 53.1 5 53.1 60 3 4 60 2 1 1 5 60 =  −  = + = − + + = • Ss = 6012 = 720VA Ps = 6012cos 53.1  = 432W Pload = 122 2 + 122 1 = 432W = cos = cos53.1 = 0.6  PF  A generator produce an output voltage of 200V at 60 Hz, PF=1,P=1000W, I1=5A. If PF=0.5,I2=10A. When R=0.2,P1=5W;P2=20W.(PF>0.85)