Experiment: AC Circuit with a Capacitor

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 = E₀ 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 E₀ sin ωt) = ω CE₀. cos ωt

i = E₀/(1- ωC) sin (ωt + π/2)

i = I₀ sin (ωt + π/2) … … (2)

where I₀ = E₀ (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.