Engineering Resonate Frequency of Parallel RLC Circuit

[p75213]

Homework Statement

Determine the resonant frequency of the circuit in Fig 14.28 (See attached)

Homework Equations

\begin{array}{l}<br /> {\rm{How is }} \to {\omega _0}0.1 - \frac{{2{\omega _0}}}{{4 + 4\omega _0^2}} \\ <br /> {\rm{derived from }} \to j{\omega _0}0.1 + \frac{{2 - j{\omega _0}}}{{4 + 4{\omega ^2}}} \\ <br /> \end{array}

I'm sure this is not difficult but I just can't see it.

The Attempt at a Solution

 

[NascentOxygen]

You consider just the two terms with a "j" in them. I've corrected where you omitted a "2".

\begin{array}{l}<br /> {\rm{How is }} \to {\omega _0}0.1 - \frac{{2{\omega _0}}}{{4 + 4\omega _0^2}} \\ <br /> {\rm{derived from }} \to j{\omega}0.1 + \frac{{2 - j{2{\omega} }}}{{4 + 4{\omega ^2}}} \\ <br /> \end{array}

 

[p75213]

Still can't see it.

 

[NascentOxygen]

One of the terms with a j in it is jω0.1

What is the other term with a j in it?

 

[p75213]

One of the terms with a j in it is jω0.1

What is the other term with a j in it?

The other term is -j2w. I still can't see how this helps.

 

[NascentOxygen]

The other term is -j2w. I still can't see how this helps.

That's only the numerator; what is its denominator?

 

[p75213]

How's this? I new it was simple and the answer was already sitting there. It felt a bit like pulling teeth.

Thanks for that.

\begin{array}{l}<br /> \begin{array}{*{20}{c}}<br /> {{\rm{How is}} \to {\omega _0}0.1 - \frac{{2{\omega _0}}}{{4 + 4\omega _0^2}}} \\<br /> {{\rm{derived from}} \to j\omega 0.1 + \frac{{2 - 2j\omega }}{{4 + 4{\omega ^2}}}} \\<br /> \end{array} \\ <br /> {\rm{Answer:}} \\ <br /> {\rm{At resonance }}Im(Y){\rm{ = 0:}} \\ <br /> j{\omega _0}0.1 - \frac{{j{\omega _0}2}}{{4 + 4\omega _0^2}} = 0 \to j\left( {{\omega _0}0.1 - \frac{{{\omega _0}2}}{{4 + 4\omega _0^2}}} \right) = 0 \\ <br /> {\omega _0}0.1 - \frac{{{\omega _0}2}}{{4 + 4\omega _0^2}} = 0 \\ <br /> \end{array}