Question

What instantaneous current results from \[q=Cv_m\sin\omega t\]?

Options

• \[i=\frac{Cv_m}{\omega}\cos\omega t\]

• \[i=\omega Cv_m\cos\omega t\]

• \[i=\omega Cv_m\sin\omega t\]

• \[i=Cv_m\sin\omega t\]

Answer 1

Differentiate the charge with respect to time. Since the derivative of \[\sin\omega t\] is \[\omega\cos\omega t\], the current is \[i=\omega Cv_m\cos\omega t\].