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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.
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.