Impedance of Inductive Reactance

Electrical Circuit Analysis > Series and Parallel ac Circuits > Impedance and the Phasor Diagram

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The inductive reactance, $X_L$, of a coil is an indication of the ability of a coil to produce self-induced voltage. Recall that inductive reactance is expressed mathematically as

$$\bbox[10px,border:1px solid grey]{X_L=2 \pi fL = wL}$$

It was learned in the chapter "Inductor" that for the pure inductor of [Fig. 1],

Fig. 1: Inductive ac circuit.

The voltage across inductor can be written as

$$v_L = V_m \sin wt \Rightarrow V_L = V \angle{0^\circ}$$

Since $v$ leads $i$ by $90^\circ$, $i$ must have an angle of $-90^\circ$ associated with it.

$$i_L = I_m \sin (wt - 90^\circ) \Rightarrow I_L = I \angle{-90^\circ}$$

By Ohm's law, impedance across inductor can be derived as

$$ \begin{split} Z_L &= {V_L \over I_L} \\ &= {V \angle {0^\circ} \over I \angle{-90^\circ}} \\ &= {V \over I} \angle{0^\circ + 90^\circ}\\ Z_L &= X_L \angle{+90^\circ} \end{split} $$

In the time domain,

$$i = \sqrt{2} {V \over X_L} \sin(wt - 90^\circ)$$

The fact that $-90^\circ$ 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.

$$v = 24 \sin wt $$

$$ \Rightarrow \text{Phasor Form } V_L = 16.968 \angle 0^\circ$$

$$ I_L = {V_L \over Z_L} = {V \angle 0^\circ \over X_L \angle 90^\circ}\\ = {16.968 \angle 0^\circ \over 3 \angle 90^\circ} = 5.656 \angle -90^\circ$$

and

$$i_L=\sqrt{2}(5.656) \sin (wt-90) = 8.0 \sin (wt-90) $$

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