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The ampermeter of the following circuit shows 10A (AC) and the voltmeter 220V (AC).

enter image description here

Furthermore its known that:

$$ cos\varphi=\frac{2}{3} \\ f=50Hz $$

I calculated the following: $$ P=UIcos\varphi =1466.67W \\ Q=UIsin\varphi = UIsin(arccos(\frac{2}{3}))=1639.78var \\ R=\frac{P}{I^2}=14.67\Omega $$

Now I also want to know the capacitance C. I found the solution to be: $$ C=\frac{I^2}{2\pi fQ}=\frac{I}{2\pi fUsin\varphi}=94,1\mu F $$ But I do not really understand the formula. Can someone explain me how it is derived?

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  • \$\begingroup\$ Which formula don't you understand and what is cos(squiggly thing) meant to represent? \$\endgroup\$
    – Andy aka
    Apr 24, 2016 at 19:53
  • \$\begingroup\$ Well, the last one, for the capacitance. The squiggly thing is a lowercase phi, the cos of that represents the power factor. \$\endgroup\$
    – Daiz
    Apr 24, 2016 at 19:59

1 Answer 1

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From impedance of capacitor $$Z_{c}=\frac{1}{j\omega C}$$, then the reactance is $$X{c}=\frac{1}{\omega C}$$ and reactive power is $$Q=I^2 X=\frac{U^2}{X}=\frac{I^2}{\omega C}$$ $$C=\frac{I^2}{\omega Q}$$

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