Confused with the formula for Power Factor Correction

Question

The formula provided in my lecture notes: $$C=\frac{Q_c}{\omega \:V_{RMS}^{\:2}}$$ Where $$Q_c=Q_{old}-Q_{new}=P\left(tan\left(\theta _{old}\right)-tan\left(\theta _{new}\:\right)\right)$$ Such that \$\theta\$ is the power factor angle, \$P\$ is the real power and \$Q\$ is the reactive power.

I've tried starting with $$Q_c=I^{\:2}_{\:RMS}\left(X_L-\left(\frac{X_LX_C}{X_L+X_C}\right)\right)$$ and then solving for \$C\$ but that didn't help.

Here is the circuit, \$C\$ is the capacitance of the parallel capacitor that should be added in order to get the desired power factor correction.

Answer 1

From comments - I'm not sure which formula you are confused with.

It's the first one that I'm trying to derive, but can't seem to reach anything. – Essam

Well real power (P) is \$\dfrac{V_{RMS}^2}{R}\$ and reactive power (Q) is \$\dfrac{V_{RMS}^2}{X_C}\$.

And, given that \$X_C= \dfrac{1}{\omega C}\$ we can say this: -

$$Q = \omega C \cdot V_{RMS}^2\hspace{2cm}\text{or}\hspace{2cm} C = \dfrac{Q}{\omega\cdot V_{RMS}^2}$$