Question

Show mathematically that in an a.c. circuit only containing capacitance, the current leads e.m.f. by a phase angle π/2.

Answer 1

Let a capacitor of capacitance C be connected to an alternating source as shown in Fig. (a).

Let instantaneous alternating e.m.f. be given by

E = E₀ sin ωt ...........(1)

If I and q are current and charge in the circuit at any time t, then

Instantaneous e.m.f. across the capacitor

E = \(\frac{q}{C}\) 

or q = CE

Using Eq. (1). we get

q = CE₀ sin ωt

The instantaneous current is the rate of growth of charge on the capacitor

So I = \(\frac{dq}{dt} = \frac{d}{dt} \) (CE₀ sin ωt)

= CE₀ \(\frac{d}{dt} \)  sin ωt

= ω CE₀ cos ωt

= \(\frac{E_0}{1/\omega C}\)  cos ωt

or I = I₀ cos ωt ............(2)

where I₀ = \(\frac{E_0}{1/\omega C}\)  is the peak value of a.c. and \(\frac{1}{\omega C}\)  is the resistance offered by the capacitor and is called capacitive reactance denoted by X_C(= \(\frac{1}{\omega C}\)) 

Eq.(2) can be written as

I = I₀ sin (ωt + \(\frac{\pi}{2}\)) ............(3)

From Eq.(2) and (3), we find that the current leads the e.m.f. by a phase \(\frac{\pi}{2}\). Fig. (b) is the graphical representation of time variation of E and I.

Fig. (c) is the phasor diagram representing the phase relationship between alternating e.m.f. and current.