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.
Best Answer
From comments - I'm not sure which formula you are confused with.
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: -
$$Q = \omega C \cdot V_{RMS}^2\hspace{2cm}\text{or}\hspace{2cm} C = \dfrac{Q}{\omega\cdot V_{RMS}^2}$$