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?