# Confused with the formula for Power Factor Correction

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.

• FYI, EE.SE uses $ instead of just  for inline math. Commented Oct 22, 2020 at 19:38 • I'm not sure which formula you are confused with. The first one is very basic... Commented Oct 22, 2020 at 19:59 • It's the first one that I'm trying to derive, but can't seem to reach anything. Commented Oct 22, 2020 at 20:04 • If there is any circuits provided in your lecture ,post that also otherwise all these variables are confusing Commented Oct 22, 2020 at 20:20 • Yes, I've just edited the question with that. Commented Oct 22, 2020 at 20:29 ## 1 Answer 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}$$ • But for$Q_c$the reactance isn't just$X_c$I always thought it should be$\left(X_L-\left(\frac{X_LX_C}{X_L+X_C}\right)\right)$i.e.$X_{old}-X_{new}$, perhaps if the connection was series then we would instead have$X_L-(X_L+X_C) = -X_C\\$ Commented Oct 22, 2020 at 20:48
• No, the capacitor is used in parallel with the reactive load in order to create power factor creation. Basically it forms a tuned resonant circuit (there is no difference in the math) and therefore, the capacitive reactance directly produces a reactive power and that is what the 1st formula describes. Commented Oct 22, 2020 at 21:24
• I'm not sure why I followed an entirely different train of thought, but thinking of it in that way makes sense, thank you. Commented Oct 23, 2020 at 7:02