Electronics Tutorial 6 - LC Resonance

To make sense of this tutorial, you may want to review the section on AC Theory.


Mechanical 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 don’t 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.

The capacitor is analogous to the spring, while the inductor is analogous to the mass in the mechanical resonance system.


Series Resonance

Consider a pure LC circuit, i.e. one with no resistance at all.  It is connected to a supply voltage V at a frequency of f Hz.


The current is the same throughout.  The voltage VL is related to the reactance in the inductor by the equation:



The voltage VC is related to the reactance in the capacitor by the equation:  




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




Question 1

Explain what the phasor diagram is showing.



Now consider what happens as we change the frequency:
• The reactance of the capacitor gets less as the frequency goes up.
• The reactance of the inductor goes up 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 F.  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 2

Why is resistance not mentioned?



The graph of impedance against frequency looks like this:




The resonant frequency occurs when the impedance is zero.  In this case it is 1900 Hz.  If we plot the current against the frequency, we get:



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:


The value of the peak shown in the graph was produced by data modelling, not by direct observation in the lab.  Signal generators that can give out such a current are not available, certainly not in the school or college physics lab.  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.  The maximum current would be determined by the resistance of the inductor. 


Parallel Resonance

Consider this circuit that consists of an inductor of inductance L, and capacitor of capacitance C. It is connected to an alternating voltage V that has a frequency of f.

We know:
• The voltage across
L and C is the same (V);
• The currents add up vectorially;
• The current
IC leads the voltage by 90 degrees;
• The current
IL lags the voltage by 90 degrees.

We can show this using the phase vector diagram:


We can see that the total current, IT is the difference between IC and IL.  This is because the current vectors are pointing in the opposite direction.  Therefore:


We know that:


We know that if X is bigger, I will be smaller. We can explain this in terms of the reactances and the frequency. As frequency goes up, the reactance of a capacitor goes down, while the reactance of the inductor goes up. Therefore as frequency goes up, IC gets bigger, while IL gets smaller.


If we change the frequency, we have seen that the currents change with the reactances of the inductor and the capacitor. There comes a point where the reactance of the capacitor and the reactance of the inductor are equal, i.e.:

It doesn’t take a genius to see that if the voltages and reactances are the same, the currents are the same. However the directions are opposite, so the currents cancel out to 0.

We have achieved parallel resonance.  We can see this in the graph below:


At the resonant frequency, the current falls to (almost) zero.  The corresponding impedance graph looks like this:


Here we see that the impedance goes up to a high value at the same frequency as the current drops to a very low value.  Although these data were obtained using data-modelling, a school frequency generator can quite easily produce a very low current, so parallel resonance can be studied easily in the school or college Physics lab.


I have included both series and parallel resonance here, although the AQA syllabus requires only parallel resonance.  Other syllabuses require series resonance as well.  To sum up:


Resonant Frequency of the LC Circuit

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


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:


Let’s substitute the values to get the true resonant frequency for the case we looked at above:


This is consistent with what we saw on the graph.


You are not expected to know this derivation for the exam.  The same argument can be applied for both series and parallel resonance.


Question 3

A 330 F 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 4

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.



If there is resistance in the LC circuit, the resonance is damped.  The resistance is analogous to the card you stick onto the slotted mass when you are investigating damped resonance.  If you want to know more about damped electrical resonance, please click HERE.  Go down to about half way down the page.


The voltage magnification at resonance is called the
Q factor, which is defined as:

The ratio between the inductor voltage and the supply voltage

So we write:



It's a ratio, so has no units.


Q-factor can also be defined in energy terms:


The ratio between the stored energy per cycle and the supplied energy per cycle


This is entirely consistent with the voltage definition as voltage is energy per unit charge.  Therefore:


We know that:

E = Pt


For one cycle, we can rewrite this as:

E = PT


Since for the period of one cycle, T


we can write an expression for the energy per cycle:




It can be shown that the Q factor can be related to the bandwidth of the system.  The bandwidth is defined as:

the range of values in which the energy is above 50 % of the maximum energy per cycle. 


Since P = I2XL the current would be 0.707 of the maximum value.  This is shown on the graph:


The Q factor is given by the bandwidth relationship:


where f0 is the resonant frequency and the fB is the frequency range of the bandwidth:


fB = Highest frequency - lowest frequency


Question 5

Show that the Q factor in the graph above is about 19.



In the Exam

  The concept of resonance is quite likely to have interpretation of graphical data, especially when the Q-factor is considered.

  Find the lowest and the highest frequency that gives 71 % (1/Ö2) of the maximum voltage or current.

  To get the magnified voltage, multiply the supply voltage by the Q factor.