Tutorial 11 B - Series Resonance
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.
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?
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 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.
We know the following about this circuit:
The voltages add up vectorially. Since there is no resistive element, there is no resistive voltage. Therefore the phasor diagram looks like this.
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.
The graph of impedance against frequency looks like this:
The resonant frequency occurs when the impedance is zero.
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.
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:
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.
We can work out the voltages across the inductor and the capacitor quite easily:
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.
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.
∑ 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.
∑ 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.
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:
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:
Now have a strong cup of coffee.