# Average power in series LC circuit

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1. Nov 20, 2014

### Rugile

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

3. 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!

2. Nov 21, 2014

### rude man

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.

3. Nov 21, 2014

### ehild

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.

4. Nov 21, 2014

### Staff: Mentor

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!