# Capacitor Correcting Power Factor

#### jegues

Joined Sep 13, 2010
733
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!

#### Peytonator

Joined Jun 30, 2008
105

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:

#### mrmount

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

#### nyasha

Joined Mar 23, 2009
90
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.

#### t_n_k

Joined Mar 6, 2009
5,455
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.
If you post the problem - perhaps as a new thread then you will most likely get some good advice.