**
Tutorial 12 Extension - What Happens at Resonance?**

If we are dealing with a perfect inductor in parallel with a capacitor, there
comes a point where the reactance of the capacitor and the reactance of the
inductor are equal, *i.e.*:

We then derived the equation:

However, there is a fly in the ointment, the resistance of the inductor, which we model as a single series resistor. Energy will be lost in the resistor.

We know that the current lags the voltage (and the current in the resistor).
However it does not lag by 90 degrees, but by a phase angle, *
f*. This is shown in the diagram
below:

The total current through the *LR* branch of the circuit is given by *I*_{LR}.

The current through the resistor is given by:

The current through the inductor is given by:

We know that:

And:

At resonance, both *I*_{C} and *I*_{L} are the same,
so we can write:

So we can re-draw the phasor diagram in terms of resistance, reactance, and impedance, i.e., the triangle of impedance.

At **resonance**, we know that *X*_{C} = *X*_{L},
so we can write:

We also know that from the impedance triangle:

In terms of *I*_{C}:

And we can go one stage further to write:

Now, the *V* terms cancel out, and we can simplify the expression to give:

We can rearrange this to give:

We know that *X*_{C} = 1/(2*pfC*)
and *X*_{L} = 2*pfL*,
so we can write:

Since:

We can write:

We can expand this to:

Rearranging:

To get *f*, we rearrange:

This then gives us:

We can tidy this up to give us our **final result**:

This is a tedious derivation, but some of the other derivations are even more
horrendous, including the use of complex numbers (*j*^{2} = -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 1/4p^{2};

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

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