Finda phase relation between current and voltage in an a.c. circuit containing a pure capacitance. A pure capacitor blocks directcurrent, why ?

Explanation

Let C is the capacitance of a capacitor which is connected to source of an alternating emf.
Let E = Instantaneous value of alternating emf and `E= E_0 sinomegat`
Let I = Current flowing through the circuit at any instant.
and q be the charge on the capacitor at that instant.

The instantaneous potential difference across the capacitor `=q/C`
Now, the instantaneous emf due to source
= Instantaneous potential difference across the capacitor `E =q/c`
`q = CE = CE_0sin omegat`
Differentiating both sides w.r.t., we have
`(dq)/(dt)=d/(dt)(CE_0sinomegat)`
or `I =CE_0(cosomegat)(omega)=E_0/(1//omega)cosomegat`
or `I=(E_0)/(1//omegaC)sin(omegat+pi/2)" " [:. sin(omegat+pi//2)= cosomegat]`
Now , `IomegaC` has got units of resistance and is called capacitive reactance . It is denoted by `X_C`
`implies I = E_0/X_C sin(omegat+pi/2)`
But `I=E_0/X_C=I_0`= The peak value of alternating current .
`I =I_0sin(omegat+pi/2)" "..(ii)`
From equations (i) and (ii), it follows that the alternating current leads the emf by phase angle `pi/2` when a.c. passes through a circuit containing capacitor.