Active Power Drawn By Capacitor

Somebody might correct me, but I believe no active power is drawn by a pure capacitance. Not sure about 2nd question, but I suppose when the equipment is switched off or at no load the capacitor would affect the power factor of the complete system, maybe causing a leading PF, which isn't desirable.

find the capacitor's specified value for

loss factor = tan delta

and multiply it by the reactive power.

This will give you active power = power loss.

I guess you missed the lecture on power measurement. Use a precision wattmeter to measure the loss.

The lecture that you missed on basic AC circuit theory would have taught you that capacitors are used to raise the voltage on loaded lines, and that when the lines are unloaded and the capacitors are still there then you might raise the voltage beyond the equipment ratings.

harshipatel,

In addition to the answer by RAMConsult, to raise the voltage on the lines, capacitors are very often used to correct the power factor. Typical industrial loads have a significant inductive component, so the current and voltage are out of phase to each other. (Mathematically, the power factor is the cosine of the phase angle between voltage and current waveforms.) This results in significant inefficiencies in transformers and on the circuits and power lines. Capacitors, as you would learn in basic AC theory classes, have the opposite effect on the current and voltage relations (compared to inductors such as motors). Thus, a properly-sized capacitor bank will correct nearly all of the power factor problems caused by inductive loads.

Now, when the inductive load is turned off, a connected capacitor bank will tip the power factor the other way, which is equally bad (and can raise the circuit voltage). That is the primary reason why they are switched off at no-load conditions.

--JMM