Tutorial 11 B - Series Resonance

Learning Objective

To describe electrical resonance

To explain how it occurs when reactance of the inductor is the same as the reactance of the capacitor.

Explain the Q-factor.

Key Question

What is resonance?

How do we explain electrical resonance in LC systems?

How do we explain electrical resonance in LCR systems?

What is the Q factor?

Electrical Resonance


The physics phenomenon of resonance can be easily demonstrated in a mechanical system.  One example is pushing a child on a swing.  You have to push the child at the right point, so that the child swings higher and higher (they like it).  If you donít push at exactly the right time, the swings get less.  We can demonstrate mechanical resonance with a vibration generator acting on a mass on a spring.



If we alter the frequency we see that the mass bounces with varying amplitude.  However at the resonant frequency, the amplitude gets very large.  It is not unknown for the masses to fly off!  Typically the resonant frequency of this kind of system is about 1.5 Hz.


The condition for mechanical resonance is for the natural frequency (the frequency at which the spring bounces freely) to equal the forcing frequency.


Mechanical resonance can be very useful, but can also be a nuisance, or even destructive.


Electrical Resonance

Electrical circuits can be made to resonate.  If we have a circuit with an inductor and a capacitor, we find that at a certain frequency, the current goes to a very high value.  This is electrical resonance.  This phenomenon can be used in:

         Electrical filters;

         Tuned circuits, such as those found in a radio receiver.


We are now going to study this in more detail.  Consider a pure LC circuit, i.e. one with no resistance at all.

Question 11

Explain what is meant by a pure inductor.




We know the following about this circuit:

  • The current is the same throughout;

  • VL = IXL;

  • VC = IXC.

 The voltages add up vectorially.  Since there is no resistive element, there is no resistive voltage.  Therefore the phasor diagram looks like this.


Question 12

Explain what the phasor diagram is showing



Now consider what happens as we change the frequency:

         The reactance of the capacitor decreases as the frequency goes up. 

         The reactance of the inductor increases as the frequency goes up.


This stands to reason as the reactance of the capacitor is given by:


And the reactance of the inductor is given by:

Consider a circuit with a pure inductance of 0.15 mH and a capacitor of 47 mF.


If we use the values in the equations to generate data for a graph, we see:




At a certain point, about 1900 Hz in this case, we find that the reactance of the capacitor = reactance of the inductor.  If we look at the voltage phase vectors, we see that they are of equal magnitude and opposite directions.  So they add up to 0.


Since the potential difference is 0, and a current is flowing, we can say that the impedance (the vector sum of the reactances) is 0.


Question 13

Why is resistance not mentioned?



The graph of impedance against frequency looks like this:


The resonant frequency occurs when the impedance is zero.


Question 14

 What frequency does this occur at?


You can see there is a spike in the current at resonance, somewhere at about 1900 Hz.  The peak current is about 710 A.  The spike is shown in more detail in this graph:


In theory the current should be infinite at resonance.  In practice this would not occur, as there is resistance in the inductor and the wires.


Letís look at how we can find the precise frequency at which resonance occurs.  We know that:

         The reactance of the capacitor is given by:

         And the reactance of the inductor is given by:

We know that in resonance:


We can rearrange this to give:

This will give us an expression to give us the resonant frequency:

This can be square-rooted to give:

Letís substitute the values from our example to get the resonant frequency. The  circuit has a pure inductance of 0.15 mH and a capacitor of 47 mF.

This is consistent with our observation on the graph.


Question 15

A 330 mF capacitor is in series with a perfect inductor of inductance 50 mH.  Both are connected to an alternating supply voltage of 6 V.

a.       Which data item is irrelevant to the question?

b.       Work out the resonant frequency.

Question 16

In a circuit, there is an overall pure inductance of 0.141 mH.  It is found to resonate at a frequency of 25000 Hz.  Calculate the capacitance of the circuit.


Resonance in real LC circuits

In real circuits, there is always resistance.  The wires have a certain (small) resistance, while the inductor, being a coil of wire, has a measurable resistance.  The coil has a resistance that we can easily measure with an ohm-meter.


In theory, the electrical oscillations should continue for ever.  In reality they die away rapidly, as there is a resistance.  This is the same as in mechanical systems, where the oscillations decrease in amplitude due to friction and other factors.  We can increase the damping for mechanical systems by putting on dampers.

The picture here shows damped oscillations.


Consider this LCR circuit:

Question 17

The resistance of the inductor is 5 ohms.  The wires have a total resistance of 0.5 ohms.  There is an external resistance of 10 ohms.  What value would you put in for the R term in the circuit above?



The phasor diagram for the LCR circuit at resonance looks like this:

We know that impedance, Z, is the vector sum of the resistance and the total reactance.  We have seen how we can write a relationship:

Since XL and XR are the same at resonance, it doesnít take a genius to see that Z2 = R2; in other words, Z = R.


Letís go back to our circuit we had before.  The circuit has an inductance of 0.15 mH and a capacitor of 47 mF.  Now it has a resistance of 5 W.

The graph of impedance against frequency looks like this:

If we look at the current against the frequency, we see:

The peak current happens at the resonant frequency, 1900 Hz.  The peak gets sharper as we reduce the resistance.

We see a very similar thing with mechanical resonance: the less the damping, the sharper the peak.  So we can say that the resistance in an electrical resonance circuit is the equivalent of the damping in a mechanical resonance circuit.  Earlier we saw that the resonant frequency of the circuit was 1896 Hz.


Question 18

Will the resonant frequency be affected by adding a resistor?

Question 19

Calculate the maximum current in the circuit at resonance, assuming the resistance is 5 W and the voltage is 6 V.

Question 20

Show that the reactance of the inductor is about 2 ohms

Question 21

What is the reactance of the capacitor at resonant frequency?  Explain your answer.



We can work out the voltages across the inductor and the capacitor quite easily:



Question 22

Work out the voltages across the inductor and the capacitor.



This calculation showed that the voltage across capacitor and the inductor were smaller than the supply voltage.  The graph of voltage against frequency is shown here for a resistance of 5 ohms.

Note the voltage scale: 2.00E+00 means 2.00 volts.


Question 23

Calculate the peak current in this case



Suppose we reduced the resistance to a much lower value, say 1 ohm, we see a much different picture as shown on the graph below.


Notice that:

         The voltages are equal at the resonant frequency;

         The capacitor voltage peaks just below the resonant frequency;

         The inductor voltage peaks just after the resonant frequency.


Now letís make the resistance even less, about 0.1 ohm.

Note that:

         There is a much sharper peak. 

         Both the inductor voltage and the capacitor voltage have a peak at the resonant frequency.

         The value of the peak voltage is about 120 V, much higher than the 6 V input.


Question 24

Calculate how many times bigger the inductor voltage is than the input voltage.



The voltage has been magnified.  The magnification factor is often called the Q-factor or quality factor.



At resonance, if R is small compared to the reactances XL (and XC), the inductor voltage can be much bigger than the supply voltage.  The voltage magnification is called the Q factor, which is defined as:


The ratio between the inductor voltage and the supply voltage


So we write:

As the Q factor is a ratio (volts over volts) itís just a number; there are no units.  From the equation above, we can write some further expressions.  We will do inductance first.


Since V = IXL, we can write:

And it doesnít take a genius to see that the current terms cancel to give:

We can go on to write:

Using a very similar argument, we can write expressions using the capacitance:

This gives us:


Question 25

A capacitor of 2 mF is in an LCR circuit that is found to resonate at 5600 Hz.  The total resistance is 0.8 ohms.  Show that the Q-factor is about 18.


Question 26

Use the Q-factor to work out the value of the inductance.



We can write an expression that allows us to predict the Q factor without knowing the resonant frequency.  We know that:


We also know:

So we can substitute:

This tidies up to give:



Question 27

A resonant circuit has a capacitance of 3.3 mF, an inductance of 3.6 mH, and a resistance of 0.5 ohms.  What is its Q-factor?


Home Principles Extension Tut 11 A Self Test Answers

Now have a strong cup of coffee.