....
Regarding your last point :
General notice if I may: in circuits that include capacitors and coils that oscillate at a certain (mostly at resonant) frequency, the instantaneous peak power levels may exceed the instantaneous peak input power levels taken from a voltage or current source.
My question is as follows:
Where did this apparent current and voltage gain come from? Even if it is at peak level at nS or mS time ? 
I really appreciate your answers, as well as Itsu's. My thanks
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Hi Nelson,
The instanteneous currents and voltages via or across capacitors and coils cannot really be termed as "gain" because they come from normal operation of such reactive components.
Consider a 10 mH coil with 2 Ohm DC resistance. Suppose we connect a 22 nF capacitor in series with it and at their resonant frequency, 10.73 kHz we drive this series LC circuit from a function generator that has only 0.1 Ohm internal resistance and the output voltage is set to say 1 V.
So the current flowing in this circuit at resonance will be I = 1 V / 2.1 Ohm = 0.476 A.
Now consider this current establishes a magnetic field in this coil, the stored energy in this field would be E = 0.01 H x 0.476 A x 0.476 A / 2 = 0.00113288 J i.e. 1.13288 mJ.
When the field changes to zero as per the AC input current dictates, the energy in the diminishing field drives a charging current into the 22 nF capacitor. Now we can calculate the voltage this capacitor will be charged up to.
Assuming a nearly lossless capacitor, quasi all the 1.13288 mJ energy would manifest as stored voltage and we can reverse calculate this voltage level: 0.00113288 J = 22 nF x V x V / 2. This gives V = 320.9 V, this will manifest across the 22 nF capacitor periodically at particular moments.
And also this 320.9 V will appear across the coil but just in opposite phase with respect to the capacitor voltage because the discharging capacitor current will build up the magnetic field in the coil and of course the coil will fight against the increase of this current by developing an increasing counter voltage. Normal induction: a changing coil current builds up a changing magnetic field and this latter induces voltage across the coil.
Now we can also consider the 0.476 A through the coil causes a voltage drop across the wire resistance this calculates as
2 Ohm x 0.476 A = 0.952 V and yet there is also the 320.9 V across the same coil but not across the 2 Ohm DC resistance but across the coil inductive reactance (the coil is a series RL circuit in this respect).
The inductive reactance is 674.18 Ohm for this coil at 10.73 kHz and the Q quality factor is 674.18 Ohm / 2 Ohm = 337. And this same 337 is given by the ratio of the voltage across the 2 Ohm to the voltage across the coil i.e. 320.9V / 0.952V = 337.
Now what could be said is that the coil in a series resonant LC circuit keeps maintaning a voltage across itself which is proportional to the voltage drop across the coil wire resistance (neglecting any other series resistance in the circuit).
And the proportionality constant is the Q quality factor, the ratio of the coil inductive reactance to its resistance.
Regarding a parallel LC circuit at resonance, the input current to the circuit increases Q times inside the LC circuit i.e. if we drive an LC circuit with 10 mA current at the resonant frequency, then inside the LC circuit the current will be 1 A if the loaded Q of this resonant LC circuit would be 100.
Hopefully these 'ramblings' answer your above question.
Good night,
Gyula
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