# Capacitor Correcting Power Factor

Discussion in 'Homework Help' started by jegues, Sep 13, 2011.

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

Sep 13, 2010
735
43
Everything in this problem is working out fine in my brain except for the last part where he determines the capcitance.

$Q_{cap} = 2.551 kVAR$

but how does

$Q_{cap} = V^{2} \omega_{c} \quad \text{?}$

I'm thinking,

$V_{c} = \frac{Q}{C}$

What is he doing here?

I know that $\omega_{c} = \omega$ since we assume steady state but where is that coming from?

Thanks again!

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2. ### Peytonator Active Member

Jun 30, 2008
105
3
At steady state:

Qc = (Vc^2)/Zc = 2.551 kVAR.

Now,

Zc = 1/(2*pi*f*C)

Remember that

w = 2*pi*f

Thus,

2.551*10^3 = (2*pi*f*C) * V^2 = V^2 * w * C

Solve for C and you get the answer.

Last edited: Sep 13, 2011
3. ### mrmount Active Member

Dec 5, 2007
59
7
The Qcap he has referred to in the formula is reactive power (not charge).

4. ### nyasha Active Member

Mar 23, 2009
90
1
I am doing a similar question from the textbook. However, l am having a hard time coming up with a relationshiop between the reactive power of the capacitor and the reactive power of the inductor. When the power factor is unity it simply means the reactive powers cancel each other. But when the power factor to lagging improves by 0.95 l don't know how to calculate that.

5. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
If you post the problem - perhaps as a new thread then you will most likely get some good advice.