# Capacitor Correcting Power Factor

Discussion in 'Homework Help' started by jegues, Sep 24, 2011.

1. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
735
43
I feel fairly comfortable with this question, however I have some concerns about the signs for certain values throughout the problem.

First concern, the negative sign on the angle θ from the PF, does it matter that I put it to -36.87°?

The voltage across the load sits at 0°, and since the PF is lagging, the current should be behind the voltage, hence my reason for the negative sign.

Note that I did this for the angle for the new PF in part 2), which brings my next question.

Since I made θ and α negative angles, my reactive portions of the power come out as negative, which is okay but how do I get rid of the j in the impeadance for the capacitor in the expression,

$Q_{cap} = V^{2}(j\omega C)$

Are we only considering the magnitude?

If that's the case then the j disapeers, but how do I get rid of the negative on the LHS?

Is this the right way of thinking about it?

EDIT: I realized I've made mistakes in the first portion of the question.

File size:
16.4 KB
Views:
10
• ###### QZ1'.JPG
File size:
50.4 KB
Views:
10
Last edited: Sep 24, 2011
2. ### mik3 Senior Member

Feb 4, 2008
4,846
63
Note that the angle of I is negative but not the angle used in cos() and sin() since S=VI*.

To get rid of j you can set it as a 90 degrees angle.