Electronic – Behavior of RLC circuit elements at t = 0

capacitorinductorpassive-networksresistors

Circuit Diagram

I'm wondering what are the behaviors of the inductor, L, and capacitor, C, at t = 0+

Please correct my understanding of the above circuit if I am wrong.

  • At t = 0, the capacitor acts like a wire. So I think that Vc must = 0 since a theoretical wire has no voltage across it.
  • However, if that is the case wouldn't Infinite current be flowing through the capacitor at t = 0 ?
  • What would the current flowing into the inductor be?
  • Shouldn't inductor act like a open (broken) circuit at t = 0, which mean no current?

Clarification: The switch is closed for a long time before time 0, at t=0 the switch opens

Best Answer

You confuse the role of capacitor and inductor.

  • The inductor acts as a wire and the capacitor is an open circuit in static conditions
  • At t = 0- you have iC = 0, iL = 0.5A and vL = vC = 0
  • At t > 0 you have iL = -iC, vL = vC and Ic = C*vC', vL = L*iL'
  • As a consequence of vL = -L*C*vL'' (and iL = -L*C*iL'') you get an oscillation of infinite duration for t > 0. I.e. the voltage at and current through inductor and capacitor are sinusoids with frequency of 1/(2pi*sqrt(LC)) = 1/12.6 Hz.

This is also easily understood: The energy stored in the inductor (supplied at times t<0) is at times t>0 continuously oscillating between the capacitor and the inductor at a frequency corresponding to the resonance frequency of the L-C circuit.

Related Topic