How a LC in parallel fed with DC supply works?
It is the classical analog of the spark gap transmitter.
If the switch is pressed too long, the battery is shorted. But during the time current builds up in the inductor and the capacitor is getting charged, oscillations will be produced.
When the switch is turned off, the oscillations start again.
if the switch is turned on and off at the right frequency and at the right time, we shall get a conventional oscillator.
Consider the simple Hartley oscillator: the job of the transistor is to turn on and off the power at the right time and it needs a feedback from the tank.
Consider your circuit in post #2. Take a tap from the inductor and put a transistor switch and you will have the Hartley oscillator.
If you put the L and C in series the circuit is not complete; to complete the circuit, you need to connect the free ends of the L and C and you will make them parallel. But if you connect the free ends of L and C to a battery, you will again get the same circuit but the potential is applied at different points. This still will not produce continuous oscillations because the timing is not given. We need a feedback from the tank and apply potential at right time and at the right place.
I hope I have made it clear. Sorry if I appear a bit vague. I am just trying to explain the process.
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All it does is to quickly drain the battery. The very low internal resistance of the battery will prevent the LC from ringing.
It will still ring during the period the current is building up in the coil; it will again ring when the battery is disconnected. The oscillations will be damped because of internal resistance of the battery, coil and the capacitor.
I do not understand the simulation you have given in post #4.