Explain AC circuit with only capacitor.

Explanation

A pure capacitor connected in AC circuit is shown in figure. Capacitor is connected with ac voltage `V =V_(m) sin omega t`.

When a capacitor is connected to a voltage source in a dc circuit, current will flow for the short time required to charge the capacitor.
As charge accumulates on the capacitor plates, the voltage across them increases opposing the current.
When the capacitor is fully charged, the current in the circuit falls to zero.
When the capacitor is connected to an across source, it limits or regulates the current but does not completely prevent the flow of charge.
The capacitor is alternately charged and discharged as the current reverses each half cycle.
Let q be the charge on the capacitor at any time t. The instantaneous voltage V across the capacitor is `V = ( q)/( C )` where C is the capacitance of capacitor.
From the Kirchhoff.s loop rule,
`V _(m)sin omega t- ( q )/( c ) = 0`
`:. V_(m) sin omegat = ( q)/( C )`
`:. CV_(m) sin omega t q `
NOw `I = ( dq)/( dt )`
`:. I = ( d )/( dt) [ V_(m) C sin omega t ]`
`:. I = V_(m) C omega cos omega t `
`:. I = ( V _(m))/((1)/( omega C )) . cos omega t `
Now putting `cos omega t = sin ( omega t + ( pi )/( 2))`
`:. I = I_(m) sin ( omega t + (pi )/( 2))` .......(2)
where, `I_(m) = ( V_(m))/((1)/( omega C ))` is the amplitude of the oxcillating current.
Comparing this equation to `I_(m) = ( V_(m))/( R )`
`(1)/( omega C )` plays the role of resistance. It is called capacitance reactance and is denoted by `X_(C )`.
`:. X_(C ) = ( 1)/( omega C )`
So that the amplitude of the current is `I_(m) = ( V_(m))/( X_(C ))`
SI unit of capacitance reactance is Ohm `( Omega )` .
The capacitive reactance limits the amlitude of the current in the purely capacitive circuit in the same way as the resistance limits the current in a purely resistive circuit.
In a pure capacitive circuit amplitude of current is inversely proportional to the frequency and the capacitance.
A comparison of `V = V(m) sin omega t ` and `I = I_(m) sin ( omega t + ( pi )/(2))` shows the current is `( pi )/(2)` ahead of voltage.
Below figure shows the phasor diagram at an instant `t_(1)`.

Here the current phasor I is `( pi )/(2)` ahead of the voltage phasor V as they rotate counter clockwise. ( Lead)
Figure (b) shows that variation of voltage and current with time. The current reaches its maximum value earlier than the voltage by one -fourth of a period.
One-fourth of a period.
`(T)/(4) = ( T )/( 2pi ) xx ( 2pi )/( 4) = ( (pi )/( 2)) /( omega ) = ( pi )/( 2 omega )`