Minimize Impedance - (eletromagnetism)

[FrogPad]

OK, I think I am doing this question right, but I'm not exactly sure. The question is as follows:

For an RLC circuit with a resistance of [tex]16k\ohm [/tex], a capacitance of [tex]8.0\mu F [/tex] and an inductance of [tex]38.0H[/tex] what frequency is needed to minimize the impedance?

Well impedance is give by:
[tex] Z = \sqrt{R^2 + (X_C - X_L)^2} [/tex]

Putting [tex] X_C [/tex] and [tex] X_L [/tex] in terms of [tex] L, C, \omega [/tex] we then have:

[tex] Z =\sqrt{R^2 + \left(\omega L - \frac{1}{\omega C}\right)^2} [/tex]
Minimum impedance is acheived at resonance, so [tex] Z = R [/tex]

Thus we have:
[tex] R =\sqrt{R^2 + \left(\omega L - \frac{1}{\omega C}\right)^2} [/tex]

Solving this for [tex] \omega [/tex] yields:

[tex] \omega = \frac{1}{\sqrt{LC}} [/tex]

And frequency is given by: [tex] f = \frac{\omega}{2\pi} [/tex]

So solving [tex] f [/tex] for [tex] \omega [/tex] and substituting into the equation above gives:

[tex] f 2\pi = \frac{1}{\sqrt{LC}} [/tex]

Now solving for [tex] f [/tex] yields:
[tex] f = \frac{1}{2\pi\sqrt{LC}} [/tex]

And finally plugging in [tex] L,\,C[/tex] from above gives:

[tex] f = \frac{1}{2\pi\sqrt{(38.0H)(8.0\mu F)}} = 9.12Hz = 0.009kHz[/tex]

So I'm pretty sure there are going to be a few questions like this on my test tomorrow, so I just want to make sure I'm doing this correctly. Thank you.

 

[OlderDan]

Looks good.

 

[thepatientmental]

Your calculation and approach seem correct. In order to minimize impedance, you need to find the frequency at which the reactance of the inductor and capacitor cancel each other out, leaving only the resistance. This is known as resonance. Your calculation for the resonant frequency, 9.12Hz, is correct. Just make sure to double check your units and calculations before your test tomorrow. Best of luck!