Question

need correct answer please help ASAP!

A motor attached to a 120 V/60 Hz power line draws an 7.60 A current. Its average energy dissipation is 810 W.

How much series capacitance needs to be added to increase the power factor to 1.0?

Answer 1

power factor = true power/ Apparennt power

Apparent Power
S = Vs * I = 120V * 7.60A = 912 V⋅A

Power Factor
pf = P / S = 810 W / 912 V⋅A = 0.888 lagging

Reactive Power
QL = Sqrt( S² - P² )= sqrt ( (912 V⋅A)² - (810W)² ) = 419.09 VAR

Poor power factor can be corrected by adding another load to the circuit drawing an equal and opposite amount of reactive power, to cancel out the effects of the load’s inductive reactance. Inductive reactance can only be canceled by capacitive reactance, so we have to add a capacitor in series to the additional load , to "remove" the effect of leakage inductance that limits the output current.

If Capacitor is to be connected in series -

419 VAR = I^2 * x
x = 419/(7.6^2)
x = 7.25

Xc = 1/2* PI*F*C

7.25 = 1/2* PI*F*C
C = 1/( 2 * 7.25 * 3.14 * 60 )
C= 419.09 / (2 * 3.14 * 60 * 120^2)
C= 366.1uF

VALUE OF CAPACITANCE TO BE ADDED IN SERIES = 366.1uF