An alternating source of emf is connected across a capacitor of capacitance C (Figure: a). It is charged first in one direction and then in the other direction.

**Fig: Capacitive circuit**

The instantaneous value of the applied emf is given by

**e = E _{0} sin ωt …(1)**

At any instant, the potential difference across the capacitor will be equal to the applied emf

∴ e = q/C, where q is the charge in the capacitor

But i = dq/dt = d/dt (Ce)

i = d/dt (C E_{0} sin ωt) = ω CE_{0}. cos ωt

i = E_{0}/(1- ωC) sin (ωt + π/2)

**i = I _{0} sin (ωt + π/2) … … (2)**

where I_{0} = E_{0} (1/ωC)

1/ωC = X_{C} is the resistance offered by the capacitor. It is called capacitive reactance. Its unit is ohm.

From equations (1) and (2), it follows that in an a.c. circuit with a capacitor, the current leads the voltage by a phase angle of π/2. In other words, the emf lags behind the current by a phase angle of π/2.

This is represented graphically in Fig: b. Fig: c represents the phasor diagram of a.c. the circuit containing only C.

**∴ X _{C }= 1/ωC = 1/ (2π νC)**

where ν is the frequency of the a.c. supply. In a d.c. circuit ν = 0

**∴ X _{C} = ∞**

Thus a capacitor offers infinite resistance to d.c. For an a.c. the capacitive reactance varies inversely as the frequency of a.c. and also inversely as the capacitance of the capacitor.