A capacitor of capacity `C` is connected in `A.C` circuit. If the applied emf is `V = V_(0) sin omega t`, then the current is

Options

A. `I_(d) = (V_(0))/(omegaC) cos omega t `
B. ` I_(d) = (V_(0))/(omegaC) sin omega t`
C. `I_(d) = V_(0) omega C sin omega t`
D. ` I_(d) = V_(0) omega C cos omega t `

Answer

D

Explanation

Given `V = V_(0) sin omega t " "` . . . (i)
Now displacement current `I_(d) ` is given by
`I_(d) = C(dV)/(dt)`
`= C (d)/(dt) (V_(0) sin omega t) ` [by using equation (i)]
`= C (V_(0) omega) cos omega t`
`I_(d) = V_(0) omega C cos omega t`