Minimize Impedance - (eletromagnetism)

AI Thread Summary
To minimize impedance in an RLC circuit with a resistance of 16kΩ, capacitance of 8.0μF, and inductance of 38.0H, the resonant frequency must be calculated. The impedance formula Z = √(R² + (X_C - X_L)²) indicates that minimum impedance occurs at resonance when Z equals R. The resonant frequency is derived as f = 1/(2π√(LC)), resulting in a calculated frequency of 9.12Hz. This frequency ensures that the reactances of the inductor and capacitor cancel each other out, leaving only the resistance. Double-checking calculations and units is advised before the test.
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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 16k\ohm, a capacitance of 8.0\mu F and an inductance of 38.0H what frequency is needed to minimize the impedance?

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

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

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

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

Solving this for \omega yields:

\omega = \frac{1}{\sqrt{LC}}

And frequency is given by: f = \frac{\omega}{2\pi}

So solving f for \omega and substituting into the equation above gives:

f 2\pi = \frac{1}{\sqrt{LC}}

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

And finally plugging in L,\,C from above gives:

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

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.
 
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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!
 
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