A SINGLE - PHASE ELECTRIC MOTOR IS CONNECTED TO A POWER LINE WITH THE MEASURING

Question

A SINGLE - PHASE ELECTRIC MOTOR IS CONNECTED TO A POWER LINE WITH THE MEASURING INSTRUMENTS AS DEPICTED IN THE DIAGRAM SHOWN BELOW THE INSTRUMENT READINGS ARE AS FOLLOWS: WATTMETER (W) = 2500 W VOLTMETER (V) = 200 V AMETER (A) = 25 A GIVEN THAT THE LINE FREQUENCY (f) is, f = 60 Hz determine, THE MOTOR POWER FACTOR (p.f.), THE MOTOR REACTIVE POWER, Q, AND THE EQUIVALENT ELECTRICAL IMPEDANCE, WHAT IS THE VALUE OF THE CAPACITOR (C) REQUIRED IN PARALLEL WITH THE MOTOR TO CORRECT THE TOTAL POWER FACTOR TO UNITY? WHAT WILL BE THE LINE CURRENT [I (line)] AFTER THE POWER FACTOR CORRECTION IN "d"? ALL ELECTRICAL QUANTITIES IN RMS

Explanation / Answer

aparent power=S=VI=200*25=5000=5.0KVA

Power=5000w

power factor=P/S=2500/5000=0.5

we khnow that

Q=(E*E)/X

X=(E*E)/Q

Q=Sqr root{(S*S)-(P*P)}=sqr root{(5000*5000)-(2500*2500)=4330.12VAR

Xc=(200*200)/4330.12=9.237ohm

c=1/2*3.1416*f*Xc=1/2*3.1416*60*9.237=287microF

Z=sqr root{(R*R)+(Xc*Xc)}

new line current at unity power factor=P/(1.732*V*P.F)=7.217A

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