Parallel RLC circuit: find resonant frequency and Input at that frequency

  • Thread starter VinnyCee
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Homework Statement



In the circuit below, find the resonant frequency ([itex]\omega_0[/itex]) and [itex]Z_{IN}\left(\omega_0\right)[/itex].

http://img143.imageshack.us/img143/9845/problem1444dx4.jpg [Broken]



Homework Equations



[tex]\omega_0\,=\,\frac{1}{\sqrt{LC}}[/tex]



The Attempt at a Solution



The resonant frequency is easy to find:

[tex]\omega_0\,=\,\frac{1}{\sqrt{20mH\,9\mu F}}\,=\,2357\,\frac{rad}{s}[/tex]

Now I reconfigure the circuit a little to combine the resistor and capacitor into the element [itex]Z_1[/itex]:

http://img156.imageshack.us/img156/4173/problem1444part2hm4.jpg [Broken]

Does that seem right?

Now I get this:

[tex]Z_{IN}\,=\,\frac{1}{1\Omega}\,+\,\frac{1}{j\omega\,0.02}\,+\,9X10^{-6}j\omega\,+\,10[/tex]

[tex]Z_{IN}\,=\,\frac{1}{j\omega\,0.02}\,+\,9X10^{-6}j\omega\,+11[/tex]

Now what do I do?
 
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Answers and Replies

  • #2
mjsd
Homework Helper
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Now I get this:

[tex]Z_{IN}\,=\,\frac{1}{1\Omega}\,+\,\frac{1}{j\omega\,0.02}\,+\,9X10^{-6}j\omega\,+\,10[/tex]

[tex]Z_{IN}\,=\,\frac{1}{j\omega\,0.02}\,+\,9X10^{-6}j\omega\,+11[/tex]

Now what do I do?


what does the question ask you to find? [tex]Z_{in}(\omega_0)[/tex]
 
  • #3
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Yep, I could just substitute the [itex]\omega_0[/itex] value into the last equation. Would that then be the final answer for [itex]Z_{IN}(\omega_0)[/itex]? Is there any simplification I can do for the j's?

[tex]Z_{IN}\left(\omega_0\right)\,=\,\frac{1}{j(2357)\,0.02}\,+\,9X10^{-6}j(2357)\,+11[/tex]

[tex]Z_{IN}\left(\omega_0\right)\,=\,11\,+\,\frac{1}{47.14\,j}\,+\,0.0212\,j[/tex]
 
Last edited:
  • #4
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[tex]Z_{IN}\left(\omega_0\right)\,=\,518.5j\,+\,0.0006[/tex]

Can someone double-check that the above is correct?
 
  • #5
SGT
[tex]Z_{IN}\left(\omega_0\right)\,=\,518.5j\,+\,0.0006[/tex]

Can someone double-check that the above is correct?
At resonance the circuit should be purely resistive. So, your answer is wrong.
Your mistake is in using [tex]\omega_0\,=\,\frac{1}{\sqrt{LC}}[/tex] to calculate the resonant frequency. This is valid only for series or parallel RLC circuits. For different connections you must calculate the complex impedance [tex]Z_{IN}[/tex] and make the imaginary part zero.
 
  • #6
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At resonance the circuit should be purely resistive. So, your answer is wrong.
Your mistake is in using [tex]\omega_0\,=\,\frac{1}{\sqrt{LC}}[/tex] to calculate the resonant frequency. This is valid only for series or parallel RLC circuits. For different connections you must calculate the complex impedance [tex]Z_{IN}[/tex] and make the imaginary part zero.
yes, your answer is right but I have a question in my mind. Why this is the case? Why at resonance the imaginary part should be zero? I am taking the complex analysis course but have not seen any theorem yet which suggest that the minimum of a complex function occurs where the imaginary part is zero. So I did not get exactly the reasoning behind this approach. Could you please explain it a little?
 
  • #7
To actually figure out the resonant impedance you use w0 for XL and XC or ZL and ZC depending on nomenclature u maybe using it’s the same thing. Any way when the circuit is at resonance it is to be completely resistive like previously stated. But what that means is XL – XC = 0. To find XL the equation is (w0*L) and the equation for XC is (1/(w0*C)). Now once you find XL and XC you put those in their respective places in the circuit as though they were resistors. Now you can solve for the resonant impedance of the circuit like any parallel or series circuit.
 
  • #8
Defennder
Homework Helper
2,591
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Did you really have to resurrect a 10-month old homework thread just to answer it? The OP has probably long lost interest in the thread.
 

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