It was learned in the chapter "Inductor" that for the pure inductor of
[Fig. 1],
Fig. 1: Inductive ac circuit.
the voltage leads the current by $90^\circ$ and that the reactance of the coil $X_L$ is
determined by $wL$.
$$v = V_m \sin wt \Rightarrow V = V \angle{0^\circ}$$
By Ohm's law,
$$I = {V \angle{0^\circ} \over X_L \angle{\theta_L}} = {V \over X_L}\angle{0^\circ - \theta_L}$$
Since v leads i by $90^\circ$, i must have an angle of $90^\circ$ associated with it.
To satisfy this condition, $\theta_L$ must equal $+90^\circ$. Substituting $\theta_L = 90^\circ$, we
obtain
$$I = {V \angle{0^\circ} \over X_L \angle{90^\circ}} = {V \over X_L}\angle{0^\circ - 90^\circ} = {V \over X_L}\angle{- 90^\circ}$$
so that in the time domain,
$$i = \sqrt{2} {V \over X_L} \sin(wt - 90^\circ)$$
The fact that $\theta_L = 90^\circ$ will now be employed in the following polar
format for inductive reactance to ensure the proper phase relationship
between the voltage and current of an inductor.
$$\bbox[10px,border:1px solid grey]{Z_L = X_L \angle{- 90^\circ}} \tag{1}$$
The boldface roman quantity $Z_L$, having both magnitude and an
associated angle, is referred to as the impedance of an inductive element. It is measured in ohms and is a measure of how much the inductive element will "control or impede" the level of current through the
network (always keep in mind that inductive 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 $Z_L$ 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.
$$ \Rightarrow \text{Phasor Form } V = 16.968 \angle 0^\circ$$
$$ I = {V \over Z_L} = {V \angle 0^\circ \over X_L \angle 90^\circ}\\
= {16.968 \angle 0^\circ \over 3 \angle 90^\circ} = 5.656 \angle -90^\circ$$
and
$$i=\sqrt{2}(5.656) \sin (wt-90) = 8.0 \sin (wt-90) $$
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