4.3 Pure capacitor circuit

  Alternating Current

The circuit containing only a pure capacitor is known as pure capacitor circuit.

Consider a pure capacitor of capacitance $C$ is connected to an alternating current as shown in the figure.

Let the instantaneous charge on the capacitor be $q$ at any time $t$.

So, from Kirchoff's loop law we can write,

$$V - \frac{q}{C} = 0$$$$q = C{V_0}\sin \omega t = 0$$Differentiating with respect to time $t$ we get, $$\frac{{dq}}{{dt}} = C{V_0}\omega \cos \omega t$$$$i = \frac{{{V_0}}}{{\left( {\frac{1}{{\omega C}}} \right)}}\cos \omega t$$$$i = \frac{{{V_0}}}{{\left( {\frac{1}{{\omega C}}} \right)}}\sin \left( {\omega t + \frac{\pi }{2}} \right)$$$$i = \frac{{{V_0}}}{{{X_C}}}\sin \left( {\omega t + \frac{\pi }{2}} \right)\quad ...(iii)$$ or $$i = {i_0}\sin \left( {\omega t + \frac{\pi }{2}} \right)$$

Thus, the alternating current leads the voltage by a phase angle of $\left( {\frac{\pi }{2}} \right)$, when alternating current flows through a capacitor.

where,$${i_0} = \frac{{{V_0}}}{{\left( {\frac{1}{{\omega C}}} \right)}} = \frac{{{V_0}}}{{{X_C}}}$$

Capacitive reactance: It is the opposition offered by the capacitor to the flow of alternating current through it.

$${X_C} = \frac{1}{{\omega C}} = \frac{1}{{2\pi fC}}$$

The capacitance reactance is infinite for D.C. $\left( {f = 0} \right)$ and has a finite value for A.C.

So, we can draw the phasor diagram as,

If we are intrested only in phase relationship, the phasor diagram may be represented as,

Equation $(iii)$ can also be written as,

$$i = \frac{{{V_0}}}{Z}\sin \left( {\omega t + \frac{\pi }{2}} \right)$$

For pure capacitor circuit, impedance $Z$ is equal to the capacitive reactance $\left( {{X_C} = \frac{1}{{\omega C}}} \right)$ of the circuit.

Mathematically, $$Z = {X_C} = \frac{1}{{\omega C}}$$