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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

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  • \$\begingroup\$ What is significant about t = 0? Is there an event which happens, like the 5V source becoming connected? \$\endgroup\$ – Kaz Oct 14 '12 at 20:05
  • \$\begingroup\$ Furthermore, the assumption is that the capacitor is empty. \$\endgroup\$ – Kaz Oct 14 '12 at 20:09
  • \$\begingroup\$ @Kaz It looks like at t = 0 the switch opens - this wasn't entirely clear to me at first glance either! \$\endgroup\$ – Bitrex Oct 14 '12 at 20:10
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    \$\begingroup\$ The switch is closed for a long time before time 0, at t=0 the switch opens \$\endgroup\$ – 40Plot Oct 14 '12 at 20:18
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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.

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  • \$\begingroup\$ It will absolutely oscillate forever, if all the components are ideal, without any parasitic resistance to dampen the oscillations. \$\endgroup\$ – Kaz Oct 14 '12 at 20:16
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I'm not sure where this "behaves like a wire" and "behaves like an open circuit" stuff comes from, but you're not the first one to bring this up, so it must be how they're teaching this stuff these days. To an old-timer like me, it doesn't seem like the clearest way to go about presenting this material.

To solve this kind of circuit, first solve for the steady-state conditions at t = 0-. What is the voltage across the capacitor, and what is the current through the inductor? Keep in mind that a capacitor does not pass DC current, but an inductor does.

At t = 0+, you know two things: The voltage across a capacitor can't change in an infinitesimal amount of time, and the current through a coil can't change, either. These facts are implicit in the definitions of these components. The voltages and currents in the rest of the circuit will redistribute themselves in order to meet these conditions, and based on that, you can determine how they evolve over time from there.

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    \$\begingroup\$ I was taught acts like a wire and others in my circuit theory lectures. \$\endgroup\$ – Dean Oct 14 '12 at 20:16
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I wondering what are the behavior of the inductor, L, and capacitor, C, at t = 0+

(1) the capacitor and inductor are in parallel regardless of the switch position. Thus, at all times, \$v_C = v_L\$

(2) just before the switch opens, by virtue of the fact that the switch has been closed for a long time, the circuit is in DC steady state*, i.e., all transients have decayed. In DC steady state, \$i_C = 0\$ and \$v_L = 0\$.

(3) The voltage across a capacitor and the current through an inductor must be continuous.

By (1), (2) and (3) , \$v_C(0+) = v_C(0-) = v_L(0-) = 0V \$.

It follows that \$i_L(0+) = i_L(0-) = 5V / 10 \Omega = 0.5A \$

These are the initial conditions for the transient solution when \$t > 0\$

UPDATE: the capacitor current must discontinuously change from 0 to \$-i_L\$ across the switch opening time. To summarize the initial conditions:

\$v_C(0+) = v_C(0-) = 0V, i_C(0-) = 0A, i_C(0+) = -0.5A \$

\$i_L(0+) = i_L(0-) = 0.5A, v_L(0+) = 0V \$

*if it exists. Some circuits do not have a DC steady state solution.

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  • \$\begingroup\$ what about Ic(0+) ? \$\endgroup\$ – 40Plot Oct 15 '12 at 2:07
  • \$\begingroup\$ Ic(0+) = -Il(0+); there is no other path for the inductor current after the switch is opened. \$\endgroup\$ – Reinstate Monica Oct 15 '12 at 13:18
  • \$\begingroup\$ @Justin, I apologize, I typed my comment early this morning before coffee. You're absolutely correct. The capacitor voltage is continuous across the switch opening time, not the capacitor current. \$\endgroup\$ – Alfred Centauri Oct 15 '12 at 13:24
  • \$\begingroup\$ @40Plot, please see the update to my answer to address the capacitor current question. \$\endgroup\$ – Alfred Centauri Oct 15 '12 at 13:33

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