(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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.

3. 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.

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# Improving power factor using capacitors (what capacitance?)

Can you offer guidance or do you also need help?

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