# Capacitor Correcting Power Factor

#### jegues

Joined Sep 13, 2010
733
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.

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#### mik3

Joined Feb 4, 2008
4,843
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.