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

In summary, a parallel RLC circuit is an electrical circuit consisting of a resistor, inductor, and capacitor connected in parallel, commonly used for filtering and tuning purposes. Its resonant frequency can be calculated using the formula f<sub>r</sub> = 1 / (2π√(LC)), and it is the frequency at which the circuit has the highest impedance. The input at resonance can be calculated using Ohm's Law, V = IZ, and the resonant frequency can be adjusted by changing the values of the components.
  • #1
VinnyCee
489
0

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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  • #2
VinnyCee said:
1.

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
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
[tex]Z_{IN}\left(\omega_0\right)\,=\,518.5j\,+\,0.0006[/tex]

Can someone double-check that the above is correct?
 
  • #5
VinnyCee said:
[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
SGT said:
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
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.
 

1. What is a parallel RLC circuit?

A parallel RLC (resistor-inductor-capacitor) circuit is an electrical circuit consisting of a resistor, inductor, and capacitor connected in parallel. This type of circuit is commonly used in electronic devices for filtering and tuning purposes.

2. How do you find the resonant frequency of a parallel RLC circuit?

The resonant frequency of a parallel RLC circuit can be calculated using the formula fr = 1 / (2π√(LC)), where fr is the resonant frequency, L is the inductance in henries, and C is the capacitance in farads.

3. What is the significance of the resonant frequency in a parallel RLC circuit?

The resonant frequency is the frequency at which the circuit has the highest impedance, meaning it resists the flow of current the most. This frequency is important in filtering applications as it allows for precise control of which frequencies are allowed to pass through the circuit.

4. How do you calculate the input at the resonant frequency in a parallel RLC circuit?

The input at the resonant frequency can be calculated using Ohm's Law, V = IZ, where V is the input voltage, I is the current, and Z is the impedance of the circuit. At resonance, the impedance is equal to the resistance of the circuit, so the input voltage can be calculated by multiplying the current by the resistance.

5. How can the resonant frequency of a parallel RLC circuit be adjusted?

The resonant frequency of a parallel RLC circuit can be adjusted by changing the values of the inductor and/or capacitor. Increasing the inductance will decrease the resonant frequency, while increasing the capacitance will increase the resonant frequency. The resistance can also be adjusted to fine-tune the resonant frequency.

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