Impedance of Inductive Reactance

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.
$$v = 24 \sin wt$$ $$\Rightarrow \text{Phasor Form } V = 16.968 \angle 0^\circ$$ $$I = {V \over Z_L} = {V \angle 0^\circ \over X_L \angle 90^\circ}\\ = {16.968 \angle 0^\circ \over 3 \angle 90^\circ} = 5.656 \angle -90^\circ$$ and $$i=\sqrt{2}(5.656) \sin (wt-90) = 8.0 \sin (wt-90)$$