Impedance of Capacitive Reactance
It was learned in the chapter "Capacitor" that for the pure capacitor of [Fig. 1],
the current leads the voltage by $90^\circ$ and that the reactance of the capacitor $X_C$ is
determined by $1/wC$.
Applying Ohm's law and using phasor algebra, we find
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
so that in the time domain,
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.
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.
Fig. 1: Capacitive ac circuit.
$$v = V_m \sin wt \Rightarrow V = V \angle{0^\circ}$$
$$I = {V \angle{0^\circ} \over X_C \angle{\theta_C}} = {V \over X_C}\angle{0^\circ - \theta_C}$$
To satisfy this condition, $\theta_C$ must equal $-90^\circ$. Substituting $\theta_C=-90^\circ$, we obtain
$$I = {V \angle{0^\circ} \over X_C \angle{-90^\circ}} = {V \over X_C}\angle{0^\circ + 90^\circ} = {V \over X_C}\angle{90^\circ}$$
$$i = \sqrt{2} {V \over X_C} \sin(wt + 90^\circ)$$
$$\bbox[10px,border:1px solid grey]{Z_C = X_C \angle{- 90^\circ}} \tag{1}$$
Example 1:
Using complex algebra, find the current i for the circuit of [Fig. 2]. Sketch the waveforms of v and i.
Solution: Note [Fig. 3]:
and
Fig. 2: For Example 1.
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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