# Lesson Video: Resonance in Alternating Current Circuits Physics

In this video, we will learn how to calculate the resonant frequency and Q value of simple resistive-capacitive-inductive circuits.

16:30

### Video Transcript

In this video, we’re going to learn about the phenomenon of resonance in alternating-current circuits. Resonance occurs because both the capacitive and inductive reactances depend on the frequency of the alternating voltage and current. Let’s start by reviewing reactance, which generalizes the current-opposing quality of resistance for resistors to include inductors and capacitors as well.

A resistor has the special property that its opposition to current is fixed. That is, the resistor’s resistance, usually given the symbol 𝑅, is unaffected by the strength, direction, or frequency of the voltage in the circuit. The same is not true of inductors and capacitors in alternating-current circuits. Although the capacitance 𝐶 and the inductance 𝐿 don’t depend on the voltage, the inductor and capacitor’s opposition to current does depend on the frequency of the voltage in the circuit.

For a capacitor, the more charged it is, the more it opposes current. The faster the electromotive force changes directions, that is, the higher its frequency, the less the capacitor charges before discharging again. So at higher frequencies, the capacitor has a smaller reactance. An inductor, on the other hand, builds up a magnetic field. And the stronger the magnetic field gets, the less the inductor opposes current. However, this magnetic field takes time to build up. So the higher the frequency of the electromotive force, the weaker the magnetic field will be before changing directions. As a result, the inductive reactance will be larger at higher frequencies.

As formulas, the capacitive reactance is one divided by the angular frequency of the voltage and current times the capacitance. The inductive reactance is the angular frequency of the voltage and current times the inductance. Note that both of these formulas give the correct qualitative relationship between reactance and frequency. The capacitive reactance is inversely proportional to frequency, while the inductive reactance is directly proportional to frequency.

So at higher frequencies, the inductive reactance is larger and the capacitive reactance is smaller. 𝜔, the angular frequency, is defined as two 𝜋 radians times the regular frequency, or cycles per second. We use 𝜔 because it helps us to write out these formulas simply without having to carry around factors of two 𝜋.

The last thing we need to recall is that for a circuit with both inductive and capacitive components, the total reactance is not simply the sum of the inductive and capacitive reactances. This is because inductors and capacitors also introduce a phase shift between the current and the emf. Capacitors cause the current to leave the emf, while inductors cause the current to lag the emf. The net effect of these different phase shifts is that the correct combination for total reactance is the difference between the two reactances, inductive reactance minus capacitive reactance.

Resonance will be possible in an alternating-current circuit precisely because total reactance is a difference instead of a sum. So let’s see how the difference between the inductive and capacitive reactances can lead to resonance. Let’s consider a simple circuit driven by an alternating-voltage source with an inductor and capacitor connected in series. Although we’ll limit our discussion to series circuits, the same principles still apply to parallel circuits.

At any rate, the total reactance in the circuit is the inductive reactance minus the capacitive reactance. Remember that the inductive and capacitive reactances depend on frequency in opposite ways. So if we vary the frequency from, say, very low value to very high values, the capacitive reactance will change from very large to very small. But the inductive reactance will change from very small to very large. This suggests that there may be some frequency in the middle of our range for which the inductive reactance and the capacitive reactance are equal.

If the inductive and capacitive reactances are equal, then their difference, the total reactance, is zero. So the inductor and capacitor combination provides no opposition to current at that frequency. To find the special frequency, we’ll start by equating the frequency-dependent formulas for inductive and capacitive reactance. To solve for 𝜔, we’ll multiply both sides by 𝜔 over 𝐿. On the left-hand side, the 𝐿 from our formula cancels the 𝐿 in the denominator. And on the right-hand side, the 𝜔 from our formula cancels the 𝜔 in the numerator. This leaves us with 𝜔 squared is equal to one divided by 𝐿𝐶.

If we now go ahead and take the square root of both sides of this equality, we get that the angular frequency versus the total reactance is zero is equal to one divided by the square root of the inductance of the inductor times the capacitance of the capacitor. We often write 𝜔 with a subscript of zero when referring specifically to this frequency.

The phenomenon of inductive and capacitive reactances exactly canceling at a particular frequency is known as resonance. And 𝜔 naught, the frequency at which resonance occurs, is called the resonant frequency. Because on resonance the total reactance is zero, the inductor and capacitor in our circuit behave like conducting wires. But this means that on resonance, the ideal circuit that we’ve drawn is effectively a short circuit. If this were a real circuit, a short circuit could cause serious damage to the various components and the AC voltage source. However, if this were a real circuit, there’ll be some resistance inherent in the components and in the wires.

So let’s model this real-life situation with an ideal circuit that consists of an inductor, a capacitor, and also a resistor. Here, we have a circuit with a resistor, inductor, and capacitor, all in series, driven by an alternating-voltage source. Since this circuit has both resistive and reactive elements, the total opposition to current is given by the combination of resistance and reactance, known as impedance. The size of the impedance is the square root of the sum of the squares of the resistance and the total reactance. We have to use this special combination because reactive components change the phase between emf and current, but resistive components don’t.

Okay, so let’s see what happens to the impedance when we drive the circuit at the resonant frequency. Recall that on resonance, the inductive and capacitive reactances are equal. So the total reactance is zero. So the impedance on resonance is the square root of 𝑅 squared plus zero, which is the square root of 𝑅 squared, which is just 𝑅. So on resonance, the impedance of the circuit is identically the resistance.

This tells us several important things. First, since the total reactive contribution to impedance is zero on resonance, there is no phase shift introduced between the current and the emf. Secondly, for a general alternating-current circuit, Ohm’s law tells us that voltage is equal to current times impedance. On resonance, this becomes voltage is equal to current times resistance, which is just Ohm’s law for purely resistive alternating-current circuits.

Furthermore, looking back at our formula for the size of impedance, total reactance squared is always positive unless it’s zero. So on resonance, when total reactance is zero, the impedance is minimized. Looking back at Ohm’s law, if the peak voltage doesn’t change, then as impedance gets smaller, peak current gets larger. So a minimum impedance implies a maximum peak current.

Now that we’ve seen what happens when we drive the circuit at the resonant frequency, let’s see what happens when we drive the circuit at frequencies other than the resonant frequency. To see how the current behaves at frequencies other than the resonant frequency, we use a graph with angular frequency on the horizontal axis and relative amplitude of the current on the vertical axis. The relative amplitude for the current at a particular angular frequency is found by dividing the peak current at that frequency by the peak current at the resonant frequency.

So, by definition, the relative amplitude for the current at the resonant frequency is one. If at some other frequency the current had a peak value that was half the value of the current at the resonant frequency, then the relative amplitude of the current at that frequency would be one-half. Using relative instead of absolute amplitude allows this discussion to be very general so it applies to a wide variety of resonant phenomena.

Turning back in particular to our electronic circuits at frequencies increasingly large relative to the resonant frequency, the inductive reactance is larger and larger. And so the relative amplitude of the current is smaller and smaller. Similarly, at frequencies increasingly small relative to the resonant frequency, the capacitive reactance is larger and larger. And so, again, the relative amplitude of the current is smaller and smaller.

This graph that we’ve drawn actually has a shape that’s typical for a wide variety of resonant systems. One of the most striking features of this graph is the sharp peak at the resonant frequency. We say that the peak is sharp because it is much narrower than it is tall. What this means physically is that the current in our circuit will be much larger when driven at the resonant frequency than when driven at frequencies not more smaller or larger than the resonant frequency.

For many applications, from measurement equipment to radio communications, it’s useful to quantify the sharpness of the resonant peak. This is because the sharper the resident peak, the more selectively our system responds strongly at a particular frequency. Equivalently, the sharper the peak, the more our systems respond to changes for smaller shifts away from the resonant frequency.

The number that we use to quantify the sharpness of the peak is called the 𝑄- or quality factor of the resonance. For series circuits like the kind we’ve been considering, the 𝑄-factor is equal to the angular frequency of the resonance times the inductance of the inductor divided by the resistance of the resistor. There are a number of other ways we could define the 𝑄-factor. No matter how we define it though, larger 𝑄-factors correspond to graphs that are more sharply peaked around the resonant frequency. And smaller 𝑄-factors correspond to graphs that are more broadly spread out around the resonant frequency.

In fact, it turns out that the width of the peak at about half the maximum value is approximately the resonant frequency divided by the quality factor. This provides a pretty good way to determine 𝑄 since just by looking at the graph we can determine the width of the peak and also its frequency, which is the resonant frequency. Furthermore, if we know any three of the quantities appearing in our full formula, we can use that formula to find the fourth.

Alright, now that we’ve learned about the resonant frequency and the 𝑄-factor, let’s work through some examples.

A circuit consists of a resistor, a capacitor, and an inductor all of which are in series. An alternating-voltage source is connected to the circuit, and an alternating current is generated. How does the resonant frequency of the circuit change if the inductance of the inductor is increased? (a) The resonant frequency decreases. (b) The resonant frequency increases. (c) The resonant frequency does not change.

The question is asking us about the resonant frequency of an alternating-current circuit. Specifically, for a series circuit with a resistor, capacitor, and inductor, the question is asking us what will happen if the inductance of the inductor is increased. Here’s a diagram of our circuit. We have the alternating-voltage source, a resistor of resistance 𝑅, an inductor of inductance 𝐿, and a capacitor of capacitance 𝐶. We’ll use the symbol 𝜔 for the angular frequency of the voltage source.

Recall that resonance will occur in this circuit when the difference between inductive and capacitive reactances, that is, the total reactance, is zero. In other words, resonance is when the inductive and capacitive reactances are equal. We also have the formulas that relate angular frequency to reactance as inductive reactance is angular frequency times inductance and capacitive reactance is one divided by angular frequency times capacitance. If we equate these expressions, as will be true at the resonant frequency, we get that 𝜔 naught 𝐿 is equal to one divided by 𝜔 naught 𝐶, where 𝜔 naught is the resonant angular frequency.

If we solve this equality for 𝜔 naught, we find that the resonant angular frequency is equal to one divided by the square root of the inductance of the inductor times the capacitance of the capacitor. This formula relates resonant frequency to inductance, so let’s use it to answer our question. As inductance is increased, the square root of inductance times capacitance is increased. So the denominator of our fraction is getting larger, which means the value of the overall fraction is getting smaller. But the value of this fraction is just the resonant frequency. So as the inductance of the inductor increases, the resonant frequency decreases. Interestingly, we can see from our formula that the resonant frequency would also decrease if we increase the capacitance of the capacitor. But if we change the resistance of the resistor, the resonant frequency wouldn’t change.

Let’s now see another example that deals with resonance in a more quantitative way.

What is the resonant frequency of the circuit shown in the diagram?

The circuit consists of an alternating-voltage source connected to a series combination of a 35-Ω resistor, a 7.5-henry inductor, and a 350-microfarad capacitor. And we’re asked to find the resonant frequency of this circuit. Recall that the inductive reactance in a circuit is the angular frequency of the voltage source times the inductance. And the capacitive reactance is one divided by the angular frequency of the voltage source times the capacitance. On resonance, these two reactances are equal.

If we call the resonant angular frequency 𝜔 naught, then we have that 𝜔 naught 𝐿 is equal to one over 𝜔 naught 𝐶, which we can solve for 𝜔 naught. When we solve this equation for 𝜔 naught, we find that the resonant angular frequency is equal to one divided by the square root of the inductance of the inductor times the capacitance of the capacitor. Now, this is a formula for angular frequency, but we’re looking for just regular frequency. So we need to use the relationship that angular frequency is two 𝜋 times the regular frequency.

Alright, so let’s plug our definition for angular frequency into our equation for the resonant angular frequency. We have two times 𝜋 times the resonant frequency is equal to one divided by the square root of the inductance times the capacitance. To get this expression into the final form we need, we simply divide both sides by two 𝜋. On the left-hand side, two 𝜋 divided by two 𝜋 is one, and we’re just left with 𝑓 naught. On the right-hand side, the two 𝜋 just becomes part of the denominator of our fraction. This leaves us with the final formula we need. Resonant frequency is equal to one divided by two 𝜋 times the square root of the inductance, in henries, times the capacitance, in farads.

So now we just need to plug in values. We have an inductance in henries. It’s 7.5 henries. However, our capacitance is given in microfarads instead of farads. To convert to farads, recall that there are one million microfarads per farad. In other words, one microfarad is equivalent to 10 to the negative sixth farads. Since we have 350 microfarads, our capacitance is equivalent to 350 times 10 to the negative sixth farads.

Plugging our inductance and capacitance into our formula for resonant frequency, this gives us one divided by two 𝜋 times the square root of 350 times 10 to the negative sixth farads times 7.5 henries. It turns out that the square root of one farad times one henry is one second. So we can rewrite the denominator with units of seconds.

Now, one divided by seconds is the unit hertz, which is used for frequency. So now we have an expression for the resonant frequency. That’s a number times a unit hertz, which is the right unit for frequency. So now all we have to do is evaluate this number with a calculator. When we do this evaluation, we find that the entire numerical expression is approximately equal to 3.1. So the resonant frequency of this circuit is 3.1 hertz. It’s worth noting that the 35-Ω resistor played no role in our calculation of the resonant frequency.

Alright, now that we’ve seen some examples, let’s review some of the key points that we’ve learned in this lesson. In this video, we considered a circuit consisting of an alternating-voltage source driving a resistor, inductor, and capacitor, all connected in series. Because inductive and capacitive reactance both depend on the frequency of the alternating-voltage source, we saw that it was possible to find a frequency where the total reactance, inductive minus capacitive reactance, is zero.

By equating the frequency-dependent formulas for inductive and capacitive reactances, we were able to find that the resonant angular frequency is equal to one divided by the square root of the inductance of the inductor times the capacitance of the capacitor. When the frequency of the voltage is equal to the resonant frequency, the net effect of the inductors and the capacitors is to provide no opposition to current in the circuit. This means that the only opposition to current is from the resistor, and so the impedance is identically the resistance. This is also the minimum possible value for the impedance because off resonance, the reactive contribution to the impedance is greater than zero. Correspondingly then, by Ohm’s law, if the impedance is minimum, then the current amplitude is maximum.

Finally, we looked at how the relative amplitude of the current depends on the angular frequency. We saw that a graph with relative amplitude on the vertical axis and angular frequency on the horizontal axis shows a sharp peak at the resonant frequency. This corresponds to the same magnitude of drive voltage resulting in a much larger current at the resonant frequency than at lower and higher frequencies. To quantify the sharpness of the resonant peak, we define the 𝑄- or quality factor as the resonant angular frequency times the inductance of the inductor divided by the resistance of the resistor. Larger values of 𝑄 correspond to resonances with sharper peaks, and smaller values of 𝑄 correspond to resonances with broader peaks.

Finally, we stated but didn’t prove that we can determine the value of 𝑄 from a graph like this one by measuring the width of the peak and also the peak’s location, which is the resonant frequency. So by knowing any three of the values in our formula for the quality factor, we can determine the value of the fourth quantity.