Question

Now consider a capacitor with capacitance c connected to an ac source (diagram 3). If the ac source provides a voltage vc(t)=v0cos(Ωt), what is the current ic(t) through the capacitor as a function of time?

Answer 1

The current through a capacitor connected to an AC source with voltage V_0cos(Ωt) is ΩCV_0cos(Ωt), leading the voltage by a quarter cycle or π/2 radians.

Step-by-step explanation:

Considering a capacitor connected to an AC voltage source with the voltage vc(t) = V0cos(Ωt), we can derive the current ic(t) through the capacitor as a function of time. From Kirchhoff's loop rule and the relationship Q = CV, the instantaneous charge on the capacitor is q(t) = CV0sin(Ωt). Calculating the rate of change of charge with respect to time gives us the current, ic(t) = d[q(t)]/dt = ΩCV0cos(Ωt). We identify that the current leads the voltage by π/2 radians (phase difference), which means that when plotted as functions of time, the current ic peaks a quarter of a cycle before the voltage across the capacitor.