Impedance of Capacitive Reactance

It was learned in the chapter "Capacitor" that for the pure capacitor of [Fig. 1],
Capacitive ac circuit.
Fig. 1: Capacitive ac circuit.
the current leads the voltage by $90^\circ$ and that the reactance of the capacitor $X_C$ is determined by $1/wC$.
$$v = V_m \sin wt \Rightarrow V = V \angle{0^\circ}$$
Applying Ohm's law and using phasor algebra, we find
$$I = {V \angle{0^\circ} \over X_C \angle{\theta_C}} = {V \over X_C}\angle{0^\circ - \theta_C}$$
Since i leads v by $90^\circ$, i must have an angle of $+90^\circ$ associated with it.
To satisfy this condition, $\theta_C$ must equal $-90^\circ$. Substituting $\theta_C=-90^\circ$, we obtain
$$I = {V \angle{0^\circ} \over X_C \angle{-90^\circ}} = {V \over X_C}\angle{0^\circ + 90^\circ} = {V \over X_C}\angle{90^\circ}$$
so that in the time domain,
$$i = \sqrt{2} {V \over X_C} \sin(wt + 90^\circ)$$
The fact that $\theta_C=-90^\circ$ will now be employed in the following polar format for capacitive reactance to ensure the proper phase relationship between the voltage and current of an capacitor.
$$\bbox[10px,border:1px solid grey]{Z_C = X_C \angle{- 90^\circ}} \tag{1}$$
The boldface roman quantity $Z_C$, having both magnitude and an associated angle, is referred to as the impedance of an capacitive element. It is measured in ohms and is a measure of how much the capacitive element will "control or impede" the level of current through the network (always keep in mind that capacitive elements are storage devices and do not dissipate like resistors). The above format, like that defined for the resistive element, will prove to be a useful "tool" in the analysis of ac networks. Again, be aware that $Z_C$ is not a phasor quantity, for the same reasons indicated for a resistive element.
Example 1: Using complex algebra, find the current i for the circuit of [Fig. 2]. Sketch the waveforms of v and i.
Fig. 2: For Example 1.
Solution: Note [Fig. 3]:
Fig. 3: Waveforms for Example 1.
$$v = 15 \sin wt $$
$$ \Rightarrow \text{Phasor Form } V = 10.605 \angle 0^\circ$$
$$ I = {V \over Z_C} = {V \angle 0^\circ \over X_C \angle -90^\circ}\\ = {10.605 \angle 0^\circ \over 2 \angle -90^\circ} = 5.303 \angle 90^\circ$$
$$i=\sqrt{2}(5.303) \sin (wt+90) = 7.5 \sin (wt+90) $$

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