AC Through Pure Capacitance Alone
When an alternating voltage is applied to the plates of a capacitor (AC Through Pure Capacitance Alone), the capacitor is charged first in one direction and then in the opposite direction. When reference to Figure (A), let
v = p.d. developed between plates at any instant
q = Charge on plates at that instant.
Then q = Cv ………………………..where C is the capacitance
= C Vm sin ωt ……………………..putting the value of v.
Now, current i is given by the rate of flow of charge
The denominator XC = 1/ωC is known as capacitive reactance and is in ohms if C is in farad and ω in radian/second. It is seen that if the applied voltage is given by v = Vmsin ωt, then the current is given by i = Im sin (ωt + π/2). Hence, we find that the current in a pure capacitor leads its voltage by a quarter cycle as shown in Figure (B) or phase difference between its voltage and current is π/2 with the current leading. Vector representation is given in Figure (B). Note that VC is taken along the reference axis.
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Power : Instantaneous power
p = vi = Vm sin ωt. Im sin (ωt + 90°)
This fact is graphically illustrated in Figure (C). We find that in a purely capacitive circuit, the average demand of power from supply is zero (as in a purely inductive circuit). Again, it is seen that power wave is a sine wave of frequency double that of the voltage and current waves. The maximum value of the instantaneous power is VmIm/2.
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