### 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.