R1 is not in parallel with R2, it is in series with R2. However, once the capacitor becomes resistive, it is then in parallel with R2 which means the "R" variable in the equation is no longer a constant.
No, R1 is not in parallel with R2 and it's not in series with R2. But the RC constant can be calculated with an equivalent resistance Req, that equals R1//R2. This is a consequence of Thevenin's theorem. See the attachment and use your formula with the second circuit. Veq=Es*R2/(R1+R2)=18/2=9 V. Req=500 ohms.
I guess the inductor, in practical cases has some losses and needs soem energy to be fed into the tank circuit to compensate these losses. So I believe that there is no way to create an osicllator without an amplifier. Tell me if I am wrong.
Do you mean sustained oscillation when you say resonance? or do you count transient responses as well?
Now that 3iMaJ, mentioned the Laplace Transform, I am recollecting a few things.
In an ideal circuit (like a test problem) capacitors and inductors have NO losses - they are ideal.
Remember for oscillatory behavior, there needs to be both inductance and capacitance.. if the circuit contains only 1 type of energy storage element, then it is First order and oscillation CANNOT occur.
I would not go as far to say resonance since that requires a particular condition setup in the circuit (C & L values & FREQ of operation)
Things that oscillate are not necessarily resonant.
I guess the inductor, in practical cases has some losses and needs soem energy to be fed into the tank circuit to compensate these losses. So I believe that there is no way to create an osicllator without an amplifier. Tell me if I am wrong.