What happened in a LC circuit when I connect two ends of the Inductor by a wire with very small resistance, at the time the current in the Inductor is $$0 < I < I_{max}$$
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2 Answers
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Zero crossings of voltage happen where there are current maximums. That is because V = L dI/dT. The derivative is zero at maximums, whether positive or negative.
Thus, when you add that wire and 0
The wire, being non-zero inductance and resistance. will do two things. It will (dramatically) change the resonant frequency AND it will dissipate a LOT of the stored energy, causing the amplitude to decay VERY rapidly (possibly less than one cycle).
Precisely what the waveform will look like will depend on the value of the capacitor, the value of the inductor, and the resistance AND inductance of the "wire". Practically (in the real world), it will also depend on the inductor's parasitic resistance and the capacitors "effective series resistance".
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\$\begingroup\$ So, in ideal situation, all energy stored in capacitor vanished at that time, and the current in the inductor still remained in "inductor + wire" circuit. \$\endgroup\$– HLongDec 9, 2014 at 5:33
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Just model your situation to a circuit's below, there are initial voltage on C1 and initial current on L1. And do transient analysis on it. If possible, you can transform the circuit from time domain to the s-domain.
simulate this circuit – Schematic created using CircuitLab
