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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.

Image

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.

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    \$\begingroup\$ FYI, EE.SE uses \$ instead of just $ for inline math. \$\endgroup\$
    – The Photon
    Oct 22 '20 at 19:38
  • \$\begingroup\$ I'm not sure which formula you are confused with. The first one is very basic... \$\endgroup\$
    – Andy aka
    Oct 22 '20 at 19:59
  • \$\begingroup\$ It's the first one that I'm trying to derive, but can't seem to reach anything. \$\endgroup\$
    – Essam
    Oct 22 '20 at 20:04
  • \$\begingroup\$ If there is any circuits provided in your lecture ,post that also otherwise all these variables are confusing \$\endgroup\$
    – user215805
    Oct 22 '20 at 20:20
  • \$\begingroup\$ Yes, I've just edited the question with that. \$\endgroup\$
    – Essam
    Oct 22 '20 at 20:29
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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: -

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

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  • \$\begingroup\$ 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\$ \$\endgroup\$
    – Essam
    Oct 22 '20 at 20:48
  • \$\begingroup\$ 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. \$\endgroup\$
    – Andy aka
    Oct 22 '20 at 21:24
  • \$\begingroup\$ I'm not sure why I followed an entirely different train of thought, but thinking of it in that way makes sense, thank you. \$\endgroup\$
    – Essam
    Oct 23 '20 at 7:02

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