Impedance of Resistive Elements

The impedance of an ideal resistor is purely real and is called resistive impedance: In this case, the voltage and current waveforms are proportional and in phase.
In the previous chapter, we found, for the purely resistive circuit of [Fig. 1], that v and i were in phase, and the magnitude
$$ I_m = {V_m \over R} \,\, \text{or} \,\, V_m = I_m R$$
Fig. 1: Resistive ac circuit.
In phasor form,
$$v = V_m \sin wt \Rightarrow V = V \angle{0^\circ}$$
where $V = 0.707 V_m$.
Applying Ohm's law and using phasor algebra, we have
$$I = {V \angle{0^\circ} \over R \angle{\theta^\circ}}={V \over R} \angle{0^\circ-\theta^\circ}$$
Since i and v are in phase, the angle associated with i also must be $0^\circ$.
To satisfy this condition, $\theta_R$ must equal $0^\circ$. Substituting $\theta_R = 0^\circ$, we find
$I = {V \angle{0^\circ} \over R \angle{0^\circ}}={V \over R} \angle{0^\circ-0^\circ} = {V \over R} \angle{0^\circ} $
so that in the time domain,
$$ i = \sqrt{2}{V \over R} \sin wt$$
The fact that $\theta_R = 0^\circ$ will now be employed in the following polar format to ensure the proper phase relationship between the voltage and current of a resistor:
$$ Z_R = R \angle 0^\circ$$
The boldface roman quantity $Z_R$, having both magnitude and an associated angle, is referred to as the impedance of a resistive element. It is measured in ohms and is a measure of how much the element will "impede" the flow of charge through the network. The above format will prove to be a useful "tool" when the networks become more complex and phase relationships become less obvious. It is important to realize, however, that $Z_R$ is not a phasor, even though the format $R \angle 0$ is very similar to the phasor notation for sinusoidal currents and voltages. The term phasor is reserved for quantities that vary with time, and R and its associated angle of $0^\circ$ are fixed, nonvarying quantities.
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.
$$v = 100 \sin wt $$
$$ \Rightarrow \text{Phasor Form } V = 70.71 \angle 0^\circ$$
$$ I = {V \over Z_R} = {V \angle 0^\circ \over R \angle 0^\circ}\\ = {70.71 \angle 0^\circ \over 5 \angle 0^\circ} = 14.14 \angle 0^\circ$$
$$i=\sqrt{2}(14.14) \sin wt = 20 \sin wt $$

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