- #1

libelec

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## Homework Statement

The power factor of a circuit has to be improved from 0,6 to 0,9 by adding capacitors. Originally, it only has an inductor L, a resistence R and a generator V. The circuit requires an active power of 3000 VA, and that should remain the same. V

_{ef}= 220V.

## The Attempt at a Solution

The real power P = Pa*Power Factor = 1800 W.

The inductive reactive power Pr

_{l}= Pa *(Power Factor

^{2}- 1)

^{1/2}= 2400 VAR. This Pr

_{l}and Pa shall remain the same after including the capacitors.

Now, the capacitive reactive power reacts 180º to the inductive reactive power. Then, the total reactive power Pr

_{t}= Pr

_{l}- Pr

_{c}. Then Pr

_{c}= Pr

_{l}- Pr

_{t}.

Since the power factor shall be 0.9 and Pa remains the same, knowing that Pr

_{t}reacts 90º to the real power P, then Pr

_{t}= (Pa

^{2}+ (Pa*0,9)

^{2})

^{1/2}= 1307,67 VAR

Finally

Pr

_{c}= 2400 VAR - 1307,67 VAR = 1092,33 VAR.

Now, how can I find the capacitance C (total) of the capacitors added? Remember, I don't know L, R or w.

Thanks.