AC Circuit Containing Capacitance Only

Let an alternating voltage, E = E0jωt be applied across the plates of a capacitor C. Let q be the charge at any instant across the plates of the capacitor. The p.d. across the plates at any instant must be equal to the applies emf at that instant. The q/c = E0ejwt


The current through the circuit is given by,
        I = dq/dt = d(CE0  ejwt)/dt = jωC E0 ejωt
Or        I = E0/1/ωC ej(ωt + π/2) = I0 ej(ωt + π/2)
Here,         I0 = E0 /1/ωC is the peak value of alternating current.
The actual current in the circuit is
        I = I0 sin (ωt + π/2)

In a pure capacitor, the current leads the emf by π/2