I do not know if this is an appropriate place to ask this question here but this is the only website that I am a member of so I hope that it is okay. If not, I can delete it.
This circuit question was brought to me while tutoring yesterday that I was not able to answer physically. Initially the capacitor had been charged to 2V so $v_C(0^−) = 2V$. Before the switch is thrown there is no current through either the resistor or inductor, so $i_L(0^−) = 0A$.
Since the inductor and capacitor cannot have an instantaneous change in their values just before and after the switch is thrown, these values must remain the same:
$i_L(0^−) = i_L(0^+−) = 0A$, $v_C(0^−) = v_C(0^+) = 2V$
This implies that at the moment the switch is closed, the inductor shorts out the capacitor and resistor, and no current occurs along any branches of the circuit. That is, the voltage drop across all three elements is zero.
What is happening to the capacitor’s and inductor’s current and voltage in the circuit? How can the capacitor have a voltage drop of 2V and yet the inductor appears to short it out simultaneously?