How to calculate the resonance frequency of a parallel LC circuit with L having a DC resistance?

Question

Here is the parallel LC circuit with inductor having also r:

How can I calculate the resonance frequency?

Answer 1

How can I calculate the resonance frequency?

It's quite straightforward if you'll take the time to think about it.

By definition, the resonance frequency is, in this context, the frequency at which the impedance of the equivalent impedance is real.

The branch with the R and L has an impedance of: \$R + j\omega L\$

The branch with the C has an impedance of: \$\dfrac{1}{j \omega C} \$

Now, the equivalent impedance is just:

\$(R + j\omega L)||\dfrac{1}{j \omega C}\$

The resonance frequency is the frequency for which the above is purely real, i.e., set the imaginary part of the above equal to zero and solve for the frequency.

When you do this correctly, you should get

or equivalently, as the Wikipedia link Renan gives:

\$\omega_0 = \sqrt {\frac{1}{LC} - \left ( \frac{R}{L} \right )^2}\$

Answer 2

This is basic classroom or web material solution. f= sqrt (1/LC) / 2pi

Here is another way. http://www.testecvw.com/carl/images/ImpedanceNomograph.pdf Find the intersection of the same impedance for L and C.

Then look at the L/r ratio at resonance. This will be the Q or gain at resonance.

If you had a parallel R the R/C impedance ratio will another limiting factor on Q.

This is the exact answer and also listed in Wiki as same. The definition of resonance is when the impedance of both reactive elements are equal in scalar and inverse polarity in vector reactance. The phase shift is also zero at resonance. \$ f_0 = { \omega_0 \over 2 \pi } = {1 \over {2 \pi \sqrt{LC}}}. \$

R does not affect this result, only the Q factor.