© Jonathan Bartlett 2020
J. BartlettElectronics for Beginnershttps://doi.org/10.1007/978-1-4842-5979-5_22

22. Reactance and Impedance

Jonathan Bartlett1 
(1)
Tulsa, OK, USA
 

22.1 Reactance

We have discussed resistance quite a bit in this book. Resistance is specifically about the ability of a component to be a good conductor of electricity. When a circuit encounters resistance, power is lost through the resistor.

However, another way of preventing current from flowing is known as reactance. With reactance, the power isn’t dissipated, but rather the current is prevented from flowing altogether.

Let’s think again about what happens when a voltage is connected to a capacitor in series. The capacitor starts to fill up. As the capacitor gets more and more full, there is less charge that can get onto the plate of the capacitor. This prevents the other side from filling up as well, and current cannot get through the capacitor. This acts in a similar way to resistance—it is preventing (impeding) the flow of current. However, it is not dissipating power because it is actually preventing the current from flowing. This is known as reactance . For capacitors, it is called capacitative reactance , and for inductors, it is called inductive reactance .

Reactance is usually frequency dependent. Again, going to the capacitor example, with high frequencies, the capacitor is continually charging and discharging, so it never really gets full, so it never impedes the current flow very much. Therefore, capacitors add very little reactance with high-frequency AC. The lower frequencies, however, give the capacitors time to get full, and when they are full, they impede the current flow. Therefore, for capacitors, lower frequencies create more reactance.

Reactance is measured in ohms, just like resistance. However, they cannot be simply added to resistances, so they are usually prefixed with the letter j. So 50 ohms of reactance is usually labeled as j50 Ω so that it is understood as a reactance. Reactances can be added to each other, and resistances can be added to each other. We will see how to combine them in Section 22.2, “Impedance.” Reactances are usually denoted using the letter X .

The reactance of a capacitor (XC) is given by the following formula:
$$ {X}_C=-frac{1}{2pi cdot fcdot C} $$
(22.1)
In this formula, f is the AC frequency of the signal (in hertz), and C is the capacitance. It might seem strange that this produces a negative value. The reason for this will make sense as we go forward. However, it is not producing negative impedance—we will see this when we combine reactance with resistance.
  • Example 22.21 What is the reactance of a 50 nF capacitor to a signal of 200 Hz? To find this, we merely use the formula:

$$ {displaystyle egin{array}{l}{X}_C=-frac{1}{2pi cdot fcdot C}\ {}kern0.96em =frac{1}{2pi cdot 200cdot 0.00000005}\ {}kern0.96em approx frac{1}{2cdot 3.14cdot 200cdot 0.00000005}\ {}kern0.96em =frac{1}{;0.00000624}\ {}kern1.08em approx -j160256;Omega kern0.84em end{array}} $$

You can see from this formula why it is said that a capacitor blocks DC. DC is, essentially, current that does not oscillate. In other words, the frequency is zero. Therefore, the formula will reduce to $$ frac{1}{0} $$, which is infinite. Therefore, it has infinite reactance against DC.

Also note what happens as the frequency increases. As the frequency increases, the denominator gets larger and larger. That means that the reactance is getting smaller and smaller—closer and closer to zero. As the frequency goes up, the reactance is essentially heading toward zero, but will never get there because the frequency can’t be infinite.

The formula for inductive reactance (XL) is very similar:
$$ {X}_L=2pi cdot fcdot L $$
(22.2)
In this equation, f is the signal frequency, and L is the inductance of the inductor in Henries.
  • Example 22.22 Let’s say that I have an 3 H inductor with a 50 Hz AC signal. How much reactance does the inductor have in this circuit?

$$ {displaystyle egin{array}{l}{X}_L=2pi cdot fcdot L\ {}kern0.84em =2cdot 3.14cdot 50cdot 3\ {}kern0.84em =j9420;Omega end{array}} $$
The reactance in this circuit is j9,420 Ω.
../images/488495_1_En_22_Chapter/488495_1_En_22_Fig1_HTML.png
Figure 22-1

Total Distance Traveled vs. Total Displacement

22.2 Impedance

In fact, resistance and reactance are usually combined together in a circuit to get a quantity known as impedance , which is simply the combination of resistive and reactive quantities. Impedance is often designated using the letter Z .

Resistance and reactance aren’t added together directly; instead, you can think of them acting at angles to each other. Let’s say that I start at my house and walk 10 feet out my front door. Then, I turn 90 degrees right and walk another 10 feet. While I have walked 20 feet, I am not 20 feet from my door. I am, instead, a little over 14 feet from my door. I can call this my displacement. Figure 22-1 shows what this looks like visually.

Since this is a right triangle, the distance from the start to the end is found on the hypotenuse of this triangle. We can calculate the total displacement using the Pythagorean theorem (A2 + B2 = C2). If we solve for C (total displacement), we get
$$ C=sqrt{A^2+{B}^2} $$
(22.3)
So, in our distance example, if I went forward 10 feet, turned left, and went another 10 feet, the total distance traveled would be
$$ {displaystyle egin{array}{l}C=sqrt{A^2+{B}^2}\ {}kern0.48em =sqrt{10^2+{10}^2}\ {}kern0.48em =sqrt{100+100}\ {}kern0.48em =sqrt{200}\ {}kern0.48em approx 14.14;end{array}} $$

The total impedance is like the total distance from your door. Resistance and reactance are like different walking directions (at right angles to each other), and impedance is the total displacement. That’s why we use the letter j to signify reactance—it is just like resistance but in a different direction.

So how do we calculate impedance? In fact, it is calculated precisely like the displacement calculation in Equation 22.3:
$$ mathrm{impedance}=sqrt{{mathrm{resistance}}^2+{mathrm{reactance}}^2} $$
(22.4)
Or, using their common abbreviations, we can say:
$$ Z=sqrt{R^2+{X}^2} $$
(22.5)
Let’s see how we can use Equation 22.4 to calculate total impedance. If I have a circuit that has 30 Ω of resistance and j20 Ω of reactance, then the formula for total impedance is
$$ {displaystyle egin{array}{l}mathrm{impedance}=sqrt{{mathrm{resistance}}^2+{mathrm{reactance}}^2}\ {}kern2.759999em =sqrt{30^2+{20}^2}\ {}kern2.759999em =sqrt{900+400}\ {}kern2.759999em =sqrt{1300}\ {}kern2.759999em approx 36.1;Omega end{array}} $$
As you can see, when using this formula, we are calculating a total impedance, so the j drops away.
  • Example 22.23 If I have a 1 kΩ resistor in series with a 100 nF capacitor with a 800 Hz signal, what is the total impedance to the signal that my circuit is giving?

  • To find out impedance, we need both resistance and reactance. We already have resistance—1 kΩ. The reactance is found by using Equation 22.1:

$$ {displaystyle egin{array}{l}{X}_C=-frac{1}{2pi cdot fcdot C}\ {}kern0.96em =frac{1}{2pi cdot 800cdot 0.0000001}\ {}kern0.96em approx frac{1}{;0.0005024}\ {}kern0.96em approx -j1990;Omega end{array}} $$

So the reactance is about − j1990 Ω.

So, if the resistance is 1000 Ω and the reactance is − j1990 Ω, what is the impedance? The impedance is found by using Equation 22.4:
$$ {displaystyle egin{array}{l}mathrm{impedance}=sqrt{{mathrm{resistance}}^2+{mathrm{reactance}}^2}\ {}kern2.759999em =sqrt{1000^2+{left(-1990
ight)}^2}\ {}kern2.759999em =sqrt{1000000+3960100}\ {}kern2.759999em =sqrt{4960100}\ {}kern2.759999em approx 2227;Omega end{array}} $$

22.3 RLC Circuits

So far we have discussed RC (resistor-capacitor) and RL (resistor-inductor) circuits. When you combine all of these components together, you get an RLC (resistor-inductor-capacitor) circuit.

When you calculate the impedance of such circuits, you have to be sure you include the reactance of both the capacitative and the inductive components. While inductors and capacitors both offer reactance to certain frequencies, their reactances actually oppose each other. That is, the reactance of one cancels out the reactance of the other. This is why the capacitative reactance is negative and the inductive reactance is positive.

Therefore, when calculating reactances that include both inductance and capacitance, you can add the reactances just like you would add resistances. However, since the capacitive reactances are negative and the inductive reactances are positive, they wind up canceling each other out to some degree.
  • Example 22.24 If I have an inductor of 5 mH and a capacitor of 5 μF in series with a 200 Ω resistor, what is the impedance of the circuit for a frequency of 320 Hz?

  • To solve this problem, we need to first find the capacitative reactance (XC) and the inductive reactance(XL). To get the capacitative reactance, we use Equation 22.1:

$$ {displaystyle egin{array}{l}{X}_C=-frac{1}{2pi cdot fcdot C}\ {}kern0.96em =frac{1}{2pi cdot 320cdot 0.000005}\ {}kern0.96em approx frac{1}{;0.01}\ {}kern0.84em approx -j1990;Omega end{array}} $$
The inductive reactance is found using Equation 22.2:
$$ {displaystyle egin{array}{l}{X}_L=2pi cdot fcdot L\ {}kern0.84em =2pi cdot 320cdot 0.005\ {}kern0.84em approx j10Omega end{array}} $$
Now, we can just add these reactances together:
$$ {displaystyle egin{array}{l}{X}_{total}={X}_L-{X}_C\ {}kern1.32em =j;10;Omega +-j;100;Omega \ {}kern1.32em =-j90;Omega end{array}} $$
The fact that this is negative is not a problem because it will be squared (which will get rid of the negative) in the next step. Now that we know the resistance (200 Ω) and the reactance (− j90 Ω), we just need to use Equation 22.4 to calculate the total impedance:
$$ {displaystyle egin{array}{l}mathrm{impedance}=sqrt{{mathrm{resistance}}^2+{mathrm{reactance}}^2}\ {}kern2.759999em =sqrt{200^2+{left(-90
ight)}^2}\ {}kern2.759999em =sqrt{40000+8100}\ {}kern2.759999em =sqrt{48100}\ {}kern2.759999em approx 219;Omega end{array}} $$

So the total impedance (opposition to current) in this circuit is 219 Ω.

22.4 Ohm’s Law for AC Circuits

In an AC circuit, the current and voltage are continually varying. Therefore, using the traditional Ohm’s law, you would have to calculate Ohm’s law over and over again in order to find out the relationships between voltage, current, and resistance.

However, there is a form of Ohm’s law that works directly on AC circuits of a given frequency. That is, it is essentially a summary of the voltages and currents that happen on each cycle. Ohm’s law for AC circuits is basically identical to the previous Ohm’s law, but the terms are slightly different:
$$ {V}_{RMS}={I}_{RMS}cdot Z $$
(22.6)

Here, the voltage we are referring to (VRMS) is an average voltage through one cycle of AC. This average is known as the RMS average. It is a little different than the typical average you might think of. If an AC voltage is swinging back and forth from positive to negative, the actual average voltage is probably around zero. However, RMS voltage is about calculating the average amount of push in any direction—positive or negative. Therefore, the RMS voltage will always yield a positive answer.1

Likewise, the current refers to the RMS current (IRMS). Just like the RMS voltage, the RMS current will always be positive, because it is the measure of the average amount of flow in any direction.

Finally, the impedance Z is calculated as we have noted in this chapter—by combining resistances and reactances together into an impedance.

Thus, Ohm’s law for AC circuits can be used to express summary relationships about average voltage, average current, and impedance in an AC signal.
  • Example 22.25 I have an AC circuit whose RMS voltage is 10 V. I have calculated the impedance of this circuit to be 20 Ω. What is the RMS current of this circuit?

  • To find this out, we just rearrange Ohm’s law a little bit:

$$ {displaystyle egin{array}{l}{V}_{RMS}={I}_{RMS}cdot Z\ {}{I}_{RMS}=frac{V_{RMS}}{Z}end{array}} $$
Now I just use the values given to fill in the blanks:
$$ {I}_{RMS}=frac{V_{RMS}}{Z}kern2.04em =frac{10}{20}=0.5;mathrm{A} $$
In this circuit, we would have an average of half an amp (500 milliamps) flowing through the circuit.
  • Example 22.26 I have an AC voltage source with an RMS voltage of 5 V running at 200 Hz. It is connected in series with a 50 kΩ resistor and a 50 nF capacitor. What is the current in this circuit?

  • To find this out, we have to first find the total impedance of the circuit. That means we have to use Equation 22.1 to find the reactance of the capacitor:

$$ {displaystyle egin{array}{l}{X}_C=-frac{1}{2pi cdot fcdot C}\ {}kern0.96em =frac{1}{2cdot 3.14cdot 200cdot 0.000005}\ {}kern0.96em =frac{1}{;0.00000628}\ {}kern0.84em approx -j159236;Omega end{array}} $$
Now that we have the resistance (50 kΩ) and the reactance (− j159236 Ω), we can use them in Equation 22.5:
$$ {displaystyle egin{array}{l}mathrm{Z}=sqrt{R^2+{X}^2}\ {}kern0.48em =sqrt{50000^2+{159236}^2}\ {}kern0.48em =sqrt{2500000000+25356103696}\ {}kern0.48em =sqrt{27856103696}\ {}kern0.6em approx 166901;Omega end{array}} $$
Now we can use Ohm’s law for AC circuits (Equation 22.6 to find the current):
$$ {displaystyle egin{array}{l}{I}_{RMS}=frac{V_{RMS}}{Z};\ {}kern1.32em =frac{5}{166901}\ {}kern1.44em approx 0.00003;mathrm{A}end{array}} $$

This means that the average current (IRMS) will be approximately 0.00003 A, or 30 μA.

22.5 Resonant Frequencies of RLC Circuits

As we have seen, the capacitative reactance goes closer to zero when the frequency goes up. Likewise, the inductive reactance increases when the frequency goes up. Additionally, the capacitative reactance and the inductive reactance have opposite signs—negative for capacitative reactance and positive for inductive reactance.

What is interesting is that if you have a combination of inductors and capacitors, there is always some frequency at which their reactances exactly cancel each other out. This point is known as the resonant frequency of the circuit.

When a capacitor and an inductor are in series with each other, it is termed an LC series circuit. Because the inductor inhibits high frequencies and the capacitor inhibits low frequencies, LC circuits can be used to let through a very specific frequency range. The center of this range is known as the resonant frequency of the circuit.

Now, if you are good with algebra, you can combine Equation 22.1 and Equation 22.2 to figure out the resonant frequency of a circuit (i.e., set them to add up to zero and then solve for the frequency f ). However, to spare you the trouble, there is a formula that you can use to find the resonant frequency of a circuit:
$$ f=frac{1}{2pi sqrt{Lcdot C}} $$
(22.7)
At this frequency, there is no total reactance to the circuit—the only impedance comes from the resistance.
  • Example 22.27 Let’s say that you have a 20 μF capacitor in series with a 10 mH inductor. What is the resonant frequency of this circuit?

  • To find the resonant frequency, we only need to employ Equation 22.7:

$$ {displaystyle egin{array}{l}f=frac{1}{2pi sqrt{Lcdot C}}\ {}kern0.6em =frac{1}{2pi;sqrt{0.01cdot 0.00002}}\ {}kern0.6em =frac{1}{2pi;sqrt{0.0000002}}\ {}kern0.6em approx frac{1}{2pi cdot 0.000447}\ {}kern0.6em approx frac{1}{0.00279}\ {}kern0.6em approx 358;mathrm{Hz}end{array}} $$
  • Therefore, this circuit has a resonant frequency of 358 Hz. This means that, at this frequency, this circuit has no reactive impedance.

Resonance frequencies are important in signal processing. They can be used in audio equipment to boost the sound of a specific frequency (since all other frequencies will have resistance). They can be used to select radio stations in radio equipment (since it will be the only frequency allowed through without resistance). You can also remove a specific frequency by taking a resonant frequency circuit to ground, thereby having a specific frequency short-circuited to ground with no resistance.

22.6 Low-Pass Filters

As mentioned, capacitors allow AC and block DC. If you wanted to remove the AC component of a signal, a capacitor can be connected to take the AC portion of your signal to ground. This is known as a low-pass filter, because, since the high-frequency (AC) component of the signal is being sent to ground, the low-frequency (DC) component of the signal is being allowed to pass.

How do you know what size of capacitor to use? Well, a rule of thumb is to have a capacitor whose reactance to the signal that you want to keep is greater than the resistance coming in to the capacitor (we want a larger resistance because we don’t want that signal coming through the capacitor).

Therefore, we can modify Equation 22.1 to say
$$ f=frac{1}{2pi cdot Rcdot C} $$
(22.8)
or
$$ C=frac{1}{2pi cdot Rcdot f} $$
(22.9)

If you know the incoming resistance, you can use that last equation to just solve for the size of capacitor you will need (if you don’t have any incoming resistance, then you can simply add a resistor of your choosing and calculate for C).

22.7 Converting a PWM Signal into a Voltage

In Chapter 15, we talked about how the Arduino microcontroller can’t emit an analog voltage, but can mimic it with a PWM signal, where it turns its 5 V signal on and off to mimic in-between voltages. We can use a low-pass filter to transform a PWM signal into a voltage.

The circuit to do so is fairly straightforward and is given in Figure 22-2. The resistor and capacitor serve to tune the lower end of the frequency of the signal that gets removed.

According to the specs, the PWM frequency is about 490 Hz. In theory, we could find the reactance of the capacitor using Equation 22.1. However, in practice, because the PWM is a square wave (it has very sharp edges), it requires a larger capacitor. In any case, this filter removes the AC component of the PWM and just leaves the residual voltage that it represents.
../images/488495_1_En_22_Chapter/488495_1_En_22_Fig2_HTML.jpg
Figure 22-2

Converting PWM to Analog with a Capacitor

Review

In this chapter, we learned the following:
  1. 1.

    Reactance (X) is a property of some electronic components that is similar to resistance, but it prevents the flow of current instead of dissipating the flow (i.e., converting it to heat).

     
  2. 2.

    Reactance, like resistance, is measured in ohms. A j is placed in front of the reactance to specify that it is a reactance value.

     
  3. 3.

    Reactance is frequency dependent—the amount of reactance depends on the frequency of the signal.

     
  4. 4.

    Capacitors and inductors each have formulas that can be used to calculate the reactance of the components.

     
  5. 5.

    Capacitors yield negative reactance and inductors yield positive reactance. This means that their reactances will oppose and cancel each other out to some degree.

     
  6. 6.

    Impedance (Z) is the total inhibition of the flow of current, combining both resistive and reactive elements.

     
  7. 7.

    Reactance and resistance are combined into impedance in the same way that walking two different directions can be combined into a total distance from your originating point—using the Pythagorean theorem.

     
  8. 8.

    RMS voltage (VRMS) is the average voltage of an AC circuit, regardless of the direction (positive or negative) of the voltage. This can be used to summarize the effects of an AC voltage.

     
  9. 9.

    RMS current (IRMS) is the average current of an AC circuit, regardless of the direction (positive or negative) of the voltage. This can be used to summarize the effects of an AC current.

     
  10. 10.

    Ohm’s law for AC circuits yields the summary relationship between RMS voltage, RMS current, and impedance in an AC circuit. It is identical to the previous Ohm’s law, but uses the summary values for the circuit at a particular frequency, rather than the values at a particular point in time.

     
  11. 11.

    The resonant frequency of a circuit is the frequency at which inductive reactance and capacitative reactance cancel each other out.

     
  12. 12.

    Resonant frequencies can be used in any application where isolating a frequency is important, because the resonant frequency will be the only frequency not encountering resistance.

     
  13. 13.

    Low-pass filters remove AC signals and allow lower-frequency signals to pass.

     
  14. 14.

    A low-pass filter can be used to convert an Arduino PWM signal into a voltage.

     

Exercises

  1. 1.

    As the frequency of a signal goes up, how does that affect the reactance from a capacitor? What about with an inductor?

     
  2. 2.

    As the frequency of a signal goes down, how does that affect the reactance from a capacitor? What about with an inductor?

     
  3. 3.

    What is true about the relationship between the capacitative reactance and the inductive reactance at the resonant frequency?

     
  4. 4.

    Why is power not used up with reactance?

     
  5. 5.

    How are reactance and resistance combined to yield impedance?

     
  6. 6.

    Calculate the capacitative reactance of a 3 F capacitor at 5 Hz.

     
  7. 7.

    Calculate the capacitative reactance of a 20 μF capacitor at 200 Hz.

     
  8. 8.

    Calculate the inductive reactance of a 7 H inductor at 10 Hz.

     
  9. 9.

    Calculate the inductive reactance of a 8 mH inductor at 152 Hz.

     
  10. 10.

    Calculate the impedance of a circuit with a 200 Ω resistor in series with a 75 μF capacitor with a signal of 345 Hz.

     
  11. 11.

    Calculate the impedance of a circuit with a 310 Ω resistor in series with a 90 nF capacitor with a signal of 800 Hz.

     
  12. 12.

    Calculate the impedance of a circuit with no resistor and a 60 mH inductor with a signal of 89 Hz.

     
  13. 13.

    Calculate the impedance of a circuit with a 50 Ω resistor in series with a 75 μH inductor with a signal of 255 Hz.

     
  14. 14.

    If I have an AC circuit with an RMS voltage of 6 V and an impedance of 1 kΩ, what is the average (RMS) current of this circuit?

     
  15. 15.

    If I have an AC circuit and I measure the AC voltage as 10 V RMS and I measure the AC at 2 mA RMS, what is the impedance of this circuit?

     
  16. 16.

    If I have an 80 Hz AC circuit that has an 8 V RMS voltage source in series with a 500 Ω resistor, a 5 H inductor, and a 200 nF capacitor, what is the RMS current flowing in this circuit?

     
  17. 17.

    Calculate the impedance of a circuit with a 250 Ω resistor in series with a 87 μH inductor and a 104 μF capacitor with a signal of 745 Hz.

     
  18. 18.

    What is the resonant frequency of the circuit in the previous question?

     
  19. 19.

    What is the reactance of a circuit at its resonant frequency?

     
  20. 20.

    If I have a 10 μF capacitor, what size inductor do I need to have a resonant frequency of 250 Hz?

     
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