\(I=10\sqrt{2}sin100\pi t\) trong đó \(\omega=100\pi\)(rad/s)
Ta có: \(Z_C=\dfrac{1}{\omega C}=\dfrac{1}{100\pi\cdot\dfrac{250}{\pi}}=4\cdot10^{-5}\Omega\)
\(U_0=I_0\cdot Z_C=10\sqrt{2}\cdot4\cdot10^{-5}=4\sqrt{2}\cdot10^{-4}V\)
Ta có pha=\(\dfrac{\pi}{2}\) nên:
\(U=U_0cos\left(\omega t+\varphi\right)=4\sqrt{2}\cdot10^{-4}\cdot cos\left(100\pi t-\dfrac{\pi}{2}\right)\)