A motor attached to a 120V60Hz power line draws an 850A curr

A motor attached to a 120V/60Hz power line draws an 8.50A current. Its average energy dissipation is 770W. What is the rms resistor voltage? What is the motor\'s resistance? How much series capacitance needs to be added to increase the power factor to 1.0?

Solution

Part A:

Voltage V = 120 V

Current I = 8.5 A

Power dissipated P = 770 W

hence the power factor,

power factor = 770/(120*8.5) = 0.754

Part B:

resistor voltage = V cos(theta) = 120*0.754 = 90.48 V

Part C:

resistance = V/I = 90.48/8.5 = 10.64 Ohm

Part D:

Old reactance = R tan(phi) = 10.64*0.8711 = 9.268 ohm

New reactance = R tan(phi\') = 0

hence capacitance required = ?

1/omega C = 9.268

C = 1/(9.268*2pi*60) = 0.0002862 F oR 286.2 u F