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

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 solve this 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?

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

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?