### Impedance

When alone in an AC circuit, inductors, capacitors, and resistors all impede current. How do they behave when all three occur together? Interestingly, their individual resistances in ohms do not simply add. Because inductors and capacitors behave in opposite ways, they partially to totally cancel each other’s effect. Figure 6.48 shows an *RLC *series circuit with an AC voltage source, the behavior of which is the subject of this section. The crux of the analysis of an *RLC* circuit is the frequency dependence of ${X}_{L}$ and ${X}_{C}\text{,}$ and the effect they have on the phase of voltage versus current (established in the preceding section). These give rise to the frequency dependence of the circuit, with important *resonance* features that are the basis of many applications, such as radio tuners.

The combined effect of resistance $R\text{,}$ inductive reactance ${X}_{L}\text{,}$ and capacitive reactance ${X}_{C}$ is defined to be impedance—an AC analogue to resistance in a DC circuit. Current, voltage, and impedance in an *RLC* circuit are related by the following AC version of Ohm’s law:

6.63 $${I}_{0}=\frac{{V}_{0}}{Z}\phantom{\rule{0.25em}{0ex}}\text{or}\phantom{\rule{0.25em}{0ex}}{I}_{\text{rms}}=\frac{{V}_{\text{rms}}}{Z}\text{.}$$

Here, ${I}_{0}$ is the peak current, ${V}_{0}$ the peak source voltage, and $Z$ is the impedance of the circuit. The units of impedance are ohms, and its effect on the circuit is as you might expect: the greater the impedance, the smaller the current. To get an expression for $Z$ in terms of $R\text{,}$${X}_{L}\text{,}$ and ${X}_{C}\text{,}$ we will now examine how the voltages across the various components are related to the source voltage. Those voltages are labeled ${V}_{R}\text{,}$${V}_{L}\text{,}$ and ${V}_{C}$ in Figure 6.48.

Conservation of charge requires current to be the same in each part of the circuit at all times, so that we can say the currents in $R\text{,}$$L\text{,}$ and $C$ are equal and in phase. But we know from the preceding section that the voltage across the inductor ${V}_{L}$ leads the current by one-fourth of a cycle, the voltage across the capacitor ${V}_{C}$ follows the current by one-fourth of a cycle, and the voltage across the resistor ${V}_{R}$ is exactly in phase with the current. Figure 6.49 shows these relationships in one graph, as well as showing the total voltage around the circuit $V={V}_{R}+{V}_{L}+{V}_{C}\text{,}$ where all four voltages are the instantaneous values. According to Kirchhoff’s loop rule, the total voltage around the circuit $V$ is also the voltage of the source.

You can see from Figure 6.49 that while ${V}_{R}$ is in phase with the current, ${V}_{L}$ leads by $\text{90\xba}\text{,}$ and ${V}_{C}$ follows by $\text{90\xba}\text{.}$ Thus, ${V}_{L}$ and ${V}_{C}$ are $\text{180\xba}$ out of phase (crest to trough) and tend to cancel, although not completely unless they have the same magnitude. Since the peak voltages are not aligned (not in phase), the peak voltage ${V}_{0}$ of the source does *not* equal the sum of the peak voltages across $R\text{,}$$L\text{,}$ and $C\text{.}$ The actual relationship is

6.64 $${V}_{0}=\sqrt{{V}_{0R}^{\phantom{\rule{1.25em}{0ex}}2}+({V}_{0L}-{V}_{0C}{)}^{2}},$$

where ${V}_{0R\text{,}}$${V}_{0L}\text{,}$ and ${V}_{0C}$ are the peak voltages across $R\text{,}$$L\text{,}$ and $C\text{,}$ respectively. Now, using Ohm’s law and definitions from Reactance, Inductive and Capacitive, we substitute ${V}_{0}={I}_{0}Z$ into the above, as well as ${V}_{0R}={I}_{0}R\text{,}$${V}_{0L}={I}_{0}{X}_{L}\text{,}$ and ${V}_{0C}={I}_{0}{X}_{C}\text{,}$ yielding

6.65 $${I}_{0}Z=\sqrt{{I}_{0}^{\phantom{\rule{0.50em}{0ex}}2}{R}^{2}+({I}_{0}{X}_{L}-{I}_{0}{X}_{C}{)}^{2}}={I}_{0}\sqrt{{R}^{2}+({X}_{L}-{X}_{C}{)}^{2}}\text{.}$$

${I}_{0}$ cancels to yield an expression for $Z\text{:}$

6.66 $$Z=\sqrt{{R}^{2}+({X}_{L}-{X}_{C}{)}^{2}}\text{,}$$

which is the impedance of an *RLC* series AC circuit. For circuits without a resistor, take $R=\text{0;}$ for those without an inductor, take ${X}_{L}=0\text{;}$ and for those without a capacitor, take ${X}_{C}=0\text{.}$

### Example 6.12 Calculating Impedance and Current

An *RLC *series circuit has a $\text{40.0 \Omega}$ resistor, a 3.00 mH inductor, and a
$\text{5.00 \mu F}$ capacitor. (a) Find the circuit’s impedance at 60.0 Hz and 10.0 kHz, noting that these frequencies and the values for
$L$ and
$C$
are the same as in Example 6.10 and Example 6.11. (b) If the voltage source has ${V}_{\text{rms}}=\text{120}\phantom{\rule{0.25em}{0ex}}\text{V}\text{,}$ what is ${I}_{\text{rms}}$ at each frequency?

**Strategy**

For each frequency, we use $Z=\sqrt{{R}^{2}+({X}_{L}-{X}_{C}{)}^{2}}$ to find the impedance and then Ohm’s law to find current. We can take advantage of the results of the previous two examples rather than calculate the reactances again.

**Solution for (a)**

At 60.0 Hz, the values of the reactances were found in Example 6.10 to be ${X}_{L}=1\text{.}\text{13}\phantom{\rule{0.25em}{0ex}}\Omega $ and in Example 6.11 to be ${X}_{C}=\text{531}\phantom{\rule{0.25em}{0ex}}\Omega \text{.}$ Entering these and the given $\text{40.0 \Omega}$ for resistance into $Z=\sqrt{{R}^{2}+({X}_{L}-{X}_{C}{)}^{2}}$ yields

6.67 $$\begin{array}{lll}Z& =& \sqrt{{R}^{2}+({X}_{L}-{X}_{C}{)}^{2}}\\ & =& \sqrt{(\text{40}\text{.}0\phantom{\rule{0.25em}{0ex}}\Omega {)}^{2}+(1\text{.}\text{13}\phantom{\rule{0.25em}{0ex}}\Omega -\text{531}\phantom{\rule{0.25em}{0ex}}\Omega {)}^{2}}\\ & =& \text{531}\phantom{\rule{0.25em}{0ex}}\Omega \text{at60}\text{.}\text{0 Hz}\text{.}\end{array}$$

Similarly, at 10.0 kHz, ${X}_{L}=\text{188}\phantom{\rule{0.25em}{0ex}}\Omega $ and ${X}_{C}=3\text{.}\text{18}\phantom{\rule{0.25em}{0ex}}\Omega \text{,}$ so that

6.68 $$\begin{array}{lll}Z& =& \sqrt{(\text{40}\text{.}0\phantom{\rule{0.25em}{0ex}}\Omega {)}^{2}+(\text{188}\phantom{\rule{0.25em}{0ex}}\Omega -3\text{.}\text{18}\phantom{\rule{0.25em}{0ex}}\Omega {)}^{2}}\\ & =& \text{190}\phantom{\rule{0.25em}{0ex}}\Omega \text{at10}\text{.}\text{0 kHz.}\end{array}$$

**Discussion for (a)**

In both cases, the result is nearly the same as the largest value, and the impedance is definitely not the sum of the individual values. It is clear that ${X}_{L}$ dominates at high frequency and ${X}_{C}$ dominates at low frequency.

**Solution for (b)**

The current ${I}_{\text{rms}}$ can be found using the AC version of Ohm’s law in Equation $${I}_{rms}={V}_{rms}\text{/}\mathrm{Z:}$$

6.69 $${I}_{\text{rms}}=\frac{{V}_{\text{rms}}}{Z}=\frac{\text{120}\phantom{\rule{0.25em}{0ex}}\text{V}}{\text{531}\phantom{\rule{0.25em}{0ex}}\Omega}=0\text{.}\text{226}\phantom{\rule{0.25em}{0ex}}\text{A}$$

at 60.0 Hz.

Finally, at 10.0 kHz, we find

6.70 $${I}_{\text{rms}}=\frac{{V}_{\text{rms}}}{Z}=\frac{\text{120}\phantom{\rule{0.25em}{0ex}}\text{V}}{\text{190}\phantom{\rule{0.25em}{0ex}}\Omega}=0\text{.}\text{633}\phantom{\rule{0.25em}{0ex}}\text{A}$$

at 10.0 kHz.

**Discussion for (a)**

The current at 60.0 Hz is the same (to three digits) as that found for the capacitor alone in Example 6.11. The capacitor dominates at low frequency. The current at 10.0 kHz is only slightly different from that found for the inductor alone in Example 6.10. The inductor dominates at high frequency.