(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.

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Improving power factor using capacitors (what capacitance?)

Can you offer guidance or do you also need help?

Draft saved
Draft deleted

**Physics Forums | Science Articles, Homework Help, Discussion**