Find frequency such that two components have same average power

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SUMMARY

The discussion focuses on determining the frequency at which two resistors in an RLC circuit have the same average power. The user correctly identifies that the average power in each resistor is half the total power, leading to the conclusion that the resonance frequency can be expressed as $$\omega = \frac{1}{\sqrt{LC}}$$. The user also explores the implications of Kirchhoff's Loop and Current Rule, ultimately concluding that equal reactances of the inductor and capacitor result in equal power distribution across the resistors. The solution emphasizes the importance of understanding voltage dividers and reactance in AC circuits.

PREREQUISITES
  • Understanding of RLC circuits and resonance frequency
  • Familiarity with Kirchhoff's Loop and Current Rules
  • Knowledge of AC circuit analysis and power calculations
  • Basic concepts of reactance in inductors and capacitors
NEXT STEPS
  • Study the derivation of resonance frequency in RLC circuits
  • Learn about the implications of Kirchhoff's Laws in AC circuits
  • Explore the concept of voltage dividers in electrical engineering
  • Investigate the relationship between reactance and impedance in AC analysis
USEFUL FOR

Electrical engineering students, circuit designers, and anyone involved in AC circuit analysis will benefit from this discussion, particularly those focusing on resonance and power distribution in RLC circuits.

Tobias Holm
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Homework Statement
At what frequency is the average power in the two resistors the same? (Hint: Use a symmetry argument as calculations are tough). Use inductance ##L=2CR^2##.
Circuit: https://drive.google.com/file/d/17yHMDfH5zY7J42wewqI4P4nLTkpsXhQa/view?usp=sharing
Relevant Equations
Kirchhoff's loop law:
$$ \sum_j \tilde{V}_j = \sum_j \tilde{I}_j Z_{L, j} + \sum_j \tilde{I}_j Z_{C, j} + \sum_j \tilde{I}_j Z_{R, j} $$

Component impedance:
##Z_L = -i\omega L##, ##Z_C = i/(\omega C)## and ##Z_R = R##.

AC current:
$$ \tilde{I}(t) = |\tilde{I}_{max}| \cos (\omega t + \phi) $$

Average power:
$$P_{ave} = \frac{1}{2} |\tilde{I}_{max}|^2 R = \frac{V_0^2}{|Z|} \cos (\phi) $$
Where ##|\tilde{I}_{max}|## is the amplitude of the complex current and ##V_0## the amplitude of the AC voltage.
(This is my first time posting here, sorry in advance for any difficulties. )

All componenets of same type has same magnitude, so e.g. the two resistors both have $R$ resistance.
Given the difficulty of the previous exercises, I believe I'm over complicating the problem. However, here is what I've tried:

1) Use that the power in each resistor is half the total power.
I know the effective impedance is simply ##Z_{eff} = R## (calculated in earlier exercise). Thus the total average power is: $$P_{avg}^{tot} = \frac{1}{2} |\tilde{I}_{max}|^2 R$$ and each of the resistors must experience half of this power:
$$ P_{avg} = \frac{1}{2} |\tilde{I}_{max, R}|^2 R $$ $$ = \frac{1}{2}P_{avg}^{tot} = \frac{1}{4} |\tilde{I}_{max}|^2 R$$
I thought I might be able to find the angular frequency by using ##|\tilde{I}_{max, R}| =\frac{\tilde{I}(t)}{\cos (\omega t + \phi)}## but this seems rather complicated.

2) Set the two average power expressions equal to each other
This one I'm less certain about (I will leave out the tildes for brevity)
$$ P_{avg, 1} = P_{avg, 2} $$ $$ \frac{1}{2} I_{max, 1}^2 R_1 \cos(\alpha) = \frac{1}{2} I_{max, 2}^2 R_2 \cos (\beta)$$
I believe the trick is to say that ##\omega## is the resonance frequency, such that ##\alpha = \beta = 0##:
$$ I_{max, 1} = I_{max, 2} $$
But that does not give me an expression for ##\omega##, nor does it prove that the two currents actually are equal.

3) Use Kirchhoff's Loop- and Current Rule.
Since each resistor has same resistance, I should find an ##\omega## that makes the current flowing in the resistors equal to each other. If I write all the necessary equations for the loops and the condition:
$$ |\tilde{I}_{R1}| =|\tilde{I}_{R2}| $$
I get the result:
$$ \omega = \frac{1}{\sqrt{2} CR} = \frac{1}{\sqrt{LC}}$$
Which is the resonance frequency for a simple RLC-circuit. I believe this is the right answer, however it requires immense amount of algebra, and I have to do this exercise by hand, thus I'm inclined to believe this is not the correct method of obtaining the result.

Thanks in advance.

EDIT: Formatting
 
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If you ignore the resistors for a moment, you'll see that you have a pair of voltage dividers consisting of series-connected L and C. What can you say about the reactances of L and C if both arrangements are to produce identical potential drops at their junctions?

1639323366636.png
 
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If we ignore the resistors, we will see that for the two paths:
$$ V = I_1 (Z_1+Z_2) = I_2 (Z_2 +Z_1) $$
$$ I_1 =I_2 $$
So the current that runs through the pair is equal to each other.
However, I cannot see how we can just ignore the resistors? I get that we can conclude that their reactans is identical, but is that necessarily true for the original case, where the last one is in parallel with a resistor?
gneill said:
If you ignore the resistors for a moment, you'll see that you have a pair of voltage dividers consisting of series-connected L and C. What can you say about the reactances of L and C if both arrangements are to produce identical potential drops at their junctions?

View attachment 294034
 
If the two voltage dividers behave identically at the given frequency, then adding equal resistors in equivalent locations will have identical results for each path.
 
gneill said:
If the two voltage dividers behave identically at the given frequency, then adding equal resistors in equivalent locations will have identical results for each path.
I'm not sure exactly where you are getting at - Wouldn't adding the resistors at "equivalent locations" be on the same component, so e.g. we put the resistors on both capacitors not a capacitor and an inductor?
However, your comments made me realize the solution:
If I can make ##Z_L = Z_C## then the two voltage dividers is equal to each other, and thus the power in the resistors must be the same.

You have my thanks! The many hours I have spend on such a simple problem.
 
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