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Consider the following circuit: enter image description here

Let both capacitors be charged to equal voltage. The circuit acts as LC oscillator circuit with resonance at $f= \frac{1}{2\pi\sqrt(LC)}$, where C is the resultant capacitance of C1 and C2. What frequency of oscillation will be experienced by both the individual capacitors? Will it be same as f, or different?

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    $\begingroup$ "Let both capacitors be charged to equal voltage." How could they ever have unequal voltage, since they are connected in parallel? $\endgroup$
    – alephzero
    Oct 23 '18 at 22:40
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Yes, the frequency remains the same, why ? Suppose you replaced the capacitors with an equivalent capacitor but the rest of circuit remains the same, so the current and voltage oscillate with frequency say $f$ across the inductor, now consider the original circuit, still you will find the same current and voltage across the inductor so nothing has changed, because this system is linear.

All the elements in this circuit are linear, so we can define the behaviour of the system with linear differential equations, we can rewrite the solution as a linear combination of imaginary exponential functions, the arguments of these exponential function have the common factor that we call angular frequency, is equal to $2\pi f$.

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  • $\begingroup$ so if both the capacitors are of different values, they will experience same oscillation frequency? You have assumed both the capacitors are of equal capacitance. $\endgroup$
    – edcvfr
    Oct 23 '18 at 14:46
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    $\begingroup$ @ujnmki Suppose the upper capacitor oscillate with different frequency, then as all the three are in parallel, so it means the voltage across the inductor oscillate with two different frequencies, clearly a contradiction. $\endgroup$ Oct 23 '18 at 14:49
  • $\begingroup$ your answer might be right but argument is not. there is no contradiction. suppose the two capacitors oscillate with two different frequencies, the inductor can oscillate with the resultant of the two frequencies. Let X be the function defining oscillation across the inductor, the fourier transform of X will give X1 and X2, which are the function defining oscillations across both the capacitors respectively. $\endgroup$
    – edcvfr
    Oct 24 '18 at 13:21
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    $\begingroup$ @ujnmki The three elements share the same node across their terminals, how on earth those can have different frequencies ? justify it please! The Fourier transform of voltage in function of time gives us the frequency response, the frequency response times the input is the out put, if they have different frequencies we expect different voltages at the same time at the output (output here is any node in the left or right hand site of any element ). $\endgroup$ Oct 24 '18 at 14:48
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They will see the same frequency.

Simple explanation: capacitors connected with no resistance or inductance anywhere in the loop from one to another will act as a single capacitor with capacitance equal to the sum of two capacitances. It is one capacitor for all practical purposes.

More complex explanation: there is a more nuanced way to calculate resonant frequency. It is when impedance reaches its peak. You can calculate the impedance for each 3 parts separately. After you've done this, you can take a 'view' of one capacitor, it will see that it is connected to a network with impedance of inductance-another capacitor. Second capacitor will see impedance of inductance-first capacitor. This two solutions will give exact the same answer, even with capacitors being different, that is two different equations will at the end give the same resonance frequency. This is general solution, that allows you to calculate resonance in much more complex circuits, and predict what will change if parts are changed.

https://en.m.wikipedia.org/wiki/Electrical_impedance -> device examples -> capacitor (and also check inductor)

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