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A.C. through pure capacitance alone

The circuit containing a pure capacitor of capacitance C farad is shown in Fig. 26 (a). Let the alternating voltage applied across the circuit be given by the equation,

u = V max sin wt

Charge on the capacitor at any instant,

q = C u

i = d q / dt = d / dt ( CV max sin wt ) = w CV max cos wt

i = V max / 1 / w c sin ( wt + π / 2)

i = V max / XC sin (wt + π /2)

The denominator Xc = 1 / wC (opposition offered to the flow of alternating current by a pure capacitor) is known as capacitive reactance.

It is given in ohms if C is in farad and w in radian/second.

The value of current will be maximum when sin ( wt + π / 2) =1

I max = V max / XC

Substituting this value in eqn. (ii), we get

i = I max sin (wt + π /2)

Power. Refer Fig. 26 (c)

Instantaneous power,

P = ui = V max sin wt x I max sin (wt + π / 2)

= V max I max sin wt cos wt = V max / √2 . I max /√2 sin 2 wt

Power for the whole cycle =  V max / √2 . I max / √2 ʃ0 sin 2 wt = 0


This fact is graphically illustrated in Fig. 26 (c). It may be noted that, during the first quarter cycle, what so ever power or energy is supplied by the source is stored in the electric field set-up between the capacitor plates. During the next quarter cycle, the electric field collapses and the power or energy stored in the field is returned to the source. The process is repeated in each alternation and this circuit does not absorb any power.

Hence in a pure capacitive circuit, we have

1. I = V / XC = V x 2 πf C amps

2. Current always leads the applied voltage by 90°.

3. Power consumed is zero.


Variation of Xc and f:

Since Xc = 2πf C and if C is kept constant, than

Fig. 27. shows the variation. As the frequency increases Xc decreases, so the current increases.