Resonance in AC circuits

Question

Derive the frequency for resonance condition in the following circuit.

My perception is that I need to nullify the j part of equivalent impedence, is that correct?

Question:

My approach:

Answer 1

The first step is finding the impedance: \$(R+j\omega L)\hspace{0.2cm}||\hspace{0.2cm}\frac{1}{j\omega C}\$: -

$$Z \hspace{1cm}=\hspace{1cm} \dfrac{(R+j\omega L)\cdot\frac{1}{j\omega C}}{R+j\omega L + \frac{1}{j\omega C}}\hspace{1cm}= \hspace{1cm}\dfrac{R+j\omega L}{1-\omega^2LC +j\omega RC}$$

Then, take the complex conjugate of the denominator and multiply top and bottom of the equation. The denominator becomes purely real but, the numerator is complex. Numerator: -

$$R-RLC\omega^2 -j\omega R^2C+j\omega L-j\omega^3L^2C+\omega^2RLC$$

And the solution is when the imaginary parts are zero i.e.: -

$$=R^2C+L-\omega^2L^2C = 0$$

My perception is that I need to nullify the j part of equivalent impedance, is that correct?

Yes, that is what I did on the previous line.

Drilling down a bit and you'll find the frequency where the impedance is real: -

$$\omega = \sqrt{\dfrac{1}{LC}-\dfrac{R^2}{L^2}}$$

That's the same as your equation (if you moved \$\frac{1}{L}\$ into the square root part).