# Improving power factor using capacitors (what capacitance?)

1. Dec 10, 2009

### libelec

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. Vef = 220V.

3. The attempt at a solution

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

The inductive reactive power Prl = Pa *(Power Factor2 - 1)1/2 = 2400 VAR. This Prl 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 Prt = Prl - Prc. Then Prc = Prl - Prt.

Since the power factor shall be 0.9 and Pa remains the same, knowing that Prt reacts 90º to the real power P, then Prt = (Pa2 + (Pa*0,9)2)1/2 = 1307,67 VAR

Finally

Prc = 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.

Can you offer guidance or do you also need help?
Draft saved Draft deleted

Similar Threads - Improving power factor Date
Find maximum dissipated power Thursday at 4:33 PM
Voltage drop and transformer improvement Jan 26, 2016
Improving the experiment Mar 28, 2013
Limitations and improvements Mar 24, 2012
How did the Bohr Model improve the Quantum Mechanical Model? Jan 26, 2012