Tutorial 12 B - Parallel RLC Circuits and Resonance

Learning Objective

To explain the conditions for resonance in parallel circuits.

To recognise and use the equations.

To explain the uses for electrical resonance.

Key Questions

How does resonance happen in a pure LC circuit?

What happens if we put in a resistor in series with the LC circuit?

How do we explain electrical resonance in C-LR systems?

What is the Q factor?

Parallel CLR Circuit

Consider this circuit, consisting of a capacitor of capacitance C in parallel with a pure (perfect) inductor of inductance L.  It is connected to a sinusoidally alternating voltage, V which has a frequency, 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 directions.

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.


Question 12

Click HERE to see Question 12.

Question 13

What will IC and IL be like if the frequency is low?

Question 14

Sketch the phasor diagram for this situation, and show the resulting current.



Remember the formulae for the reactances for the capacitor and the inductor:



And we need to remember how reactance relates to voltage and current:


Now we will put in the numbers in the next example.


Question 15

A parallel circuit consists of a capacitor of 400 mF and a 0.0045 H inductor in parallel.  The components are connected to a supply of 6 V at a frequency of 300 Hz.

a.       What is the reactance of the capacitor?

b.      What is the reactance of the inductor?

c.       What is the resultant current?  Does it lag or lead the voltage?

d.       What is the impedance of the circuit?


LC Parallel Resonance

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 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 now derive an expression to work out the resonant frequency, f0.  We know that:



Question 16

A parallel circuit consists of a capacitor of 400 mF and a 0.0045 H inductor in parallel.  What is the resonant frequency?



For angular velocity fans, we can rewrite the equation as:

The impedance, Z, is infinite at resonance.


Question 17

Explain why Z = ∞



The graph below shows the current against the resistance for a parallel combination of a 2 mF capacitor in parallel with a 4.5 mH inductor with a resistance of zero.  At the resonant frequency, the current = 0.


Question 18

Calculate the resonant frequency for this parallel combination.



The answer to the question above is consistent with where the current = 0.


Adding a resistor in series with the parallel LC Circuit


In the circuit above, we have added a resistor outside the LC parallel combination.  The voltage will always be in phase with the resistor current.  Will it affect the resonant frequency?  The answer is no.  Although current has to pass through the resistor, at resonance the current is zero, so there will be zero current flowing through R.

If we have a frequency that is not the resonant frequency, then R has a significant part to play.  Let us see how this happens.

Worked example

A parallel circuit consists of a capacitor of 4 mF and a 4.5 mH inductor in parallel.  A resistor of resistance 20 ohms is place in series with the parallel combination as shown.


The components are connected to a supply of 6 V at a frequency of 1000 Hz.

a.       What is the reactance of the capacitor?

b.      What is the reactance of the inductor?

c.       What is the overall impedance?

d.      What is voltage across the resistor?




c.       We need to work out the parallel reactances to give a single equivalent reactance, in exactly the same way as we do parallel resistors.


1/X = 1/39.8 + 1/ 28.3   X = 16.5 W


We then work out the impedance by Z2 = X2 + R2

Z2 = 16.52 + 202 = 672.25

Z = 25.9 W

d.      To work out the voltage, we need the current across the resistor:

I = V/Z = 6 ÷ 29.5 = 0.231 A

V = IR  = 0.231 × 20 = 4.63 V

Let’s go on to work out the voltage across the capacitor and resistor.

The answer is NOT 6.0 – 4.63 = 1.37 V.

Question 19

Why is it not 1.37 V?



Look at the triangle of impedances:

We have with the reactive side of the circuit a single reactance equivalent, in series with a resistor.  The voltages add up vectorially:

So we can do the calculation:

VX2 = 6.02 – 4.632 = 14.6

 VX = 3.82 V


So the voltage across the inductor and the capacitor is nearly 4 V, rather than 1.4 V.  We could also simply find out the voltage by multiplying the reactance by the current.  It gives us the same result.


Now we can use the fact that the voltage is the same across each parallel branch to work out the current in each component

For the inductor:

I = V/XL = 3.82 ÷ 28.3 = 0.135 A


For the capacitor:

I = V/XC = 3.82 ÷ 39.8 = 0.0959 A


Add the two together, we get 0.231 A.  This is consistent with our previous calculation of the current through the resistor.


Question 20

A 10 mF capacitor is in parallel with a 5 mH inductor.  They are connected to a 6 V supply which has a frequency of 800 Hz.  What is the overall reactance?




While the voltage and current have directions which are important, the reactance is like resistance.  It is the ratio of the voltage to the current, and this is NOT directional.  Series reactances add up; parallel reactances add up by their reciprocals.  This does not affect the impedance triangle.


Adding the resistor in series with the inductor

Consider this circuit.

This is a much more common circuit, as every inductor has a certain amount of resistance.  We need to work out the impedance of the series part of the circuit first of all, before we analyse the parallel part.

The impedance of the LR part of the circuit is:


Z2 = XL2 + R2


The reactance of the C part is simply XC.


To work out the total impedance we would write:

This doesn’t look very nice, so we need to work out a problem-solving strategy:

·         Work out XL2 + R2

·         Take the square root of this.

·         Take the reciprocal (x-1) and add it to the reciprocal of XC.

·         Then take the reciprocal of the answer to get ZT.


Question 21

What can we say about the voltage and current in the series part of the circuit?

Question 22

The circuit below is connected to a sinusoidally alternating voltage of 6 V at a frequency of 500 Hz.

a.       Work out the reactance of the capacitor;

b.      Show that the reactance of the inductor is about 16 ohms;

c.       Work out the impedance of the of the inductor and the resistor together;

d.      Work out the total impedance of the whole circuit.


At resonance in a circuit that has an inductor with a resistance the resonant frequency is given by this equation:

The derivation of this is long and tedious.  Click HERE to see how it's done.

This is a tedious derivation, but some of the other derivations are even more horrendous, including the use of complex numbers (j2 = -1).  We will not attempt these.


If R is small, then we can ignore it and the relationship we can use is simpler:

Problem solving strategy:

·         Square the equation to get rid of the square root bit;

·         Evaluate the bit in brackets;

·         Then multiply that result by 1/4p2;

·         Then square-root the answer to give you f.


Now we will do a worked example:

Worked example


 What is the resonant frequency of this circuit?


Square the equation:

Evaluate the bit in brackets:

Now multiply by 1/4p2:



Now square root it to get:

f  = 1460 Hz



At resonance, the current in the resistor can be shown to be:


Question 23

What is the current that flows at resonance?


Q factor

In a parallel circuit, the Q-factor is a measure of the current magnification.   Let’s go back to the phasor diagram:

There is a current circulating in the parallel portion.  The inductor is giving the current to the capacitor, and vice versa.


Q = circulating current ÷ supply current

So we can substitute the formula for inductive reactance into the formula above to get:

At low frequencies, the Q factor tends to be quite low, less than 10, but at very high frequencies, the Q factor can be high.


Question 24

The resonant frequency of the circuit below is 1460 Hz. 

What is the Q factor?  What is the circulating current?



Mechanical Analogy

If we have a mass bouncing on a spring, it will bounce up and down at a natural frequency for some time.  There is constant interchange between kinetic (movement) energy and potential energy in the spring.  Physicists describe the movement as simple harmonic

motion.  The movement is sinusoidal.


However the amplitude of the oscillations gradually decreases until the mass simply hangs there.  This is because there are energy losses, from air resistance, or within the spring itself.  This is called damping.


To keep the spring bouncing, we have to put in energy at exactly the right time, by using a forcing oscillation.  If the forced oscillations are the same as the natural frequency, then resonance happens, and the amplitude can get very large indeed.


In our electrical system, the signal generator provides energy at exactly the right time.  The resistor acts as a damping system.  If the resistor is small, the damping is light.  The bigger the resistor is, the heavier the damping becomes.


Applications of Resonance

Resonance is not just a curious electrical effect.  It can be used to:

·         Produce a stable AC signal;

·         Act as an electrical filter to block (or enhance) particular frequencies;

·         Tune a radio to pick up a particular frequency from all the others that are available.


As the impedance is very high at resonance, the resultant current is at a minimum.  Therefore the circuit can be described as a rejector circuit.


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