Question

Which expression gives the capacitor current in terms of peak current \[i_m\]?

विकल्प

• \[i=i_m\sin\left(\omega t-\frac{\pi}{2}\right)\]

• \[i=i_m\sin\omega t\]

• \[i=i_m\sin 2\omega t\]

• \[i=i_m\sin\left(\omega t+\frac{\pi}{2}\right)\]

Answer 1

The current is \[i=\omega Cv_m\cos\omega t\]. Rewriting cosine as a shifted sine gives \[i=i_m\sin\left(\omega t+\frac{\pi}{2}\right)\].