Question

A motor attached to a 120 V/60 Hz power line draws an 8.00 A current. Its average energy dissipation is 850 W.

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

Answer 1

If you add series capacitance, the voltage drop across the capacitor may decrease/increase the voltage across the motor to a level that the motor may not work. Power factor correction is typically done with the capacitor in parallel with the motor.

Apparent Power
S = V_S I = 120V * 8.00A = 960 V?A

Power Factor
pf = P / S = 850 W / 960 V?A = 0.885 lagging

Reactive Power
Q_L = ? ( S² - P² )= ? ( (960 V?A)² - (850W)² ) = 446 VAR

This is where it gets a little dicey for me (as an engineer). I'd do this with powers and parallel capacitor so that the voltage is known.

Inductive Reactance
Q_L = I² X_L
X_L = Q_L / I² = 446VAR / (8.0A)² = 6.97?

To bring the pf = 1, leading capacitive reactive power must equal lagging inductive reactive power:
X_C = X_L = 6.97?

Capacitance+
C = 1 / (2 ? f X_C) = 1 / (2 * ? * 60Hz * 6.97?) = 380.5 ?F