Average power in series LC circuit

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In a series LC circuit connected to a resonant voltage source, the average power is zero if there is no active resistance. The discussion highlights that with a finite input voltage, achieving 1V across the capacitor or inductor leads to infinite current, making the circuit non-realizable. Introducing resistance complicates the calculations, but the average power can be expressed as V^2/R. Participants emphasize the importance of demonstrating understanding by exploring different scenarios, even if the question is poorly framed. Ultimately, the consensus is that without resistance, average power remains zero.
Rugile
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Homework Statement


A voltage source is connected to a series LC circuit. The frequency of the source is resonant. The voltage amplitude of capacitor is 1V. Find the average power in the circuit.

Homework Equations

The Attempt at a Solution


I realize that if there is no active resistance the power is 0. I tried assuming that there is active resistance of the voltage source, but it leads me nowhere since there are so many unknowns left: I_{max} = \frac{U_{max}}{R}
$$P=\frac{U_{max}I_{max} }{2}$$
$$U_{maxC} = \frac{I_{max}}{\omega C} = \frac{U_{max}}{R \omega C} = \frac{U_{max} \sqrt{LC}}{RC}$$
I really don't know how to work around this... Or should I just assume that there is no active resistance? Any help appreciated!
 
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Leave it at that: average power is zero.
For ANY finite input voltage V it is impossible for a pure L-C circuit to have 1V across the capacitor (or inductor for that matter). For ANY finite input voltage the voltage across either component would be infinite, as would be the current.
Of course, this circuit is non-realizable.
If you stuck a resistance R in series with the L-C section, the power would simply be (V^2)/R, again getting you nowhere.
 
As the maximum voltage of the capacitor is finite, there is some resistance in the circuit. Assume that t the generator voltage and L, C are given. Determine the average power in terms of them.
 
Rugile said:
I really don't know how to work around this... Or should I just assume that there is no active resistance? Any help appreciated!
In situations like this, try to cover all bases. You've considered that R may be zero. So if R is non-zero write the power loss in a couple of ways, e.g., (wC)2.R, V2/R, while explaining these terms.

By making the best of a bad situation (a badly contrived question) you at least demonstrate competency in the topic, and that is basically what any marker is looking for so he/she can justify awarding you marks. Make it easy for the marker---show what you know!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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