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Conceptual question regarding inductors and capacitors

  1. Mar 20, 2012 #1
    1. The problem statement, all variables and given/known data

    http://img207.imageshack.us/img207/3756/24134024.jpg [Broken]

    N.B: (0+) means just after switch is open, and (0-) means just before switch is open.


    I have the solutions for this, but there are some problems I have with understanding how inductors and capacitors work in circuits like this.


    1) Before that switch is open (t < 0), the inductor is essentially a short, and the capacitor is an open circuit

    2) When that switch is open, the currennt I[itex]_{R}[/itex] at time (0+) is going to be different to (0-) because resistors unlike capacitors and inductors respond quickly to changes in current.

    3) The current through the inductor i[itex]_{L}[/itex] at time (0+) is considered to be the same as (0-) because inductors oppose sudden changes in current.

    What I do not understand is when the switch is open, current i[itex]_{C}[/itex] at time (0+) - the current through the capacitor has a value of 0.278mA, I thought the capacitor is acting like an open circuit and like the inductor doesn't respond suddenly, how is there a current flow through it?

    The solutions also say V[itex]_{C}[/itex](0+) = V[itex]_{C}[/itex](0-) = 0

    I understand that just before the switch is open, the capacitor is acting like a short circuit, and there is no current through it, but how can there be 0 voltage at time (0+) and have a current i[itex]^{C}[/itex](0+) = 0.278mA?

    i = Cdv/dt, there is no change in voltage, so current should be zero.
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Mar 20, 2012 #2

    tiny-tim

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    Hi NewtonianAlch! :smile:
    as SGT said seven years ago (https://www.physicsforums.com/showthread.php?t=89729) …
    The current in a capacitor is
    i = Cdv/dt​
    and the voltage in an inductor is
    v = Ldi/dt​

    So, for an instantaneous change of voltage (dt = 0) in a capacitor, you need an infinite current. In the same way, for an instantaneous change of current in an inductor, you need an infinite voltage.

    Notice that there is an infinite mean power, but a finite energy, since dt = 0.​
    no, it's like when you let go of something …

    its acceleration is g, but its speed is 0

    here, there's no instantaneous change in voltage (v), but there is an instantaneous change in dv/dt :wink:
     
  4. Mar 20, 2012 #3
    Instantaneous change in [itex]\frac{dv}{dt}[/itex]? You mean [itex]\frac{d}{dt}[/itex]([itex]\frac{dv}{dt}[/itex])?

    What do you mean by an instantaneous change in [itex]\frac{dv}{dt}[/itex]?

    Although if the capacitor is still acting as an open-circuit at t = (0+), how is there going to be a current through it anyway?

    P.S - Nearly 20,000 posts, congrats!
     
  5. Mar 20, 2012 #4

    gneill

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    No, an instantaneous change in V. That is, some substantial dV while dt = 0. For a capacitor that would require an infinite current.
    The capacitor behaves as an open circuit simply because it has matched the potential being presented to its terminals -- there is no potential difference available to drive current one way or the other, so the current is zero just as for an open circuit. If you change the conditions though (such as by closing a switch and presenting the capacitor with a new 'equilibrium target') it will take in or release charge accordingly to reach a new equilibrium condition.
    Agreed! :smile:
     
  6. Mar 20, 2012 #5

    tiny-tim

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    20,000th post!

    oooh, this is my 20,000th post!! thanks for pointing it out! :smile:

    in a graph of v against t, an instantaneous change in v (from 0) would look like a flat-line to the left of 0, then a discontinuity, a jump up the vertical (v) axis, and then a continuous curve to the right

    but an instantaneous change in dv/dt (from 0) would look like a flat-line to the left of 0, then a "corner" as the graph starts rising to the right …

    not a smooth change at 0, but a sharp corner :wink:

    for a capacitor, the former (a jump) isn't allowed, but the latter (a corner) is allowed
     
  7. Mar 29, 2012 #6
    I was just doing this problem again and ran in to the same kind of issue, although a bit different this time.

    I still do not understand how ic(0+) can be non-zero because iL(0-) = iL(0+) has a value, because it's essentially a short-circuit, so if in the inductor is short-circuiting the capacitor at 0+, why is there a current through the capacitor at 0+ ?
     
  8. Mar 29, 2012 #7

    gneill

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    The short-circuit and open-circuit behaviors of inductors and capacitors apply to their DC steady state conditions. That is, when things have settled down and the voltages and currents in the circuit stop changing.

    When you create a change (by, say, opening or closing a switch), then the circuit is no longer in steady state -- A change in voltage is presented to the inductor or capacitor, or a previous current path for the inductor's current is altered, which sets new "target" currents and voltages for a new steady state.

    To find out why there might be a given current or voltage, determine what the new steady state conditions might be. Do they differ from the current state? If yes, how will the circuit get from here to there?
     
  9. Mar 29, 2012 #8
    So iL(0+) has the same current as iL(0-) because the inductor doesn't respond immediately to changes in current, BUT at 0+ the inductor is no longer a short-circuit but now starting to function as an inductor is that what this means?
     
  10. Mar 29, 2012 #9

    gneill

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    In truth, inductors always behave as inductors and capacitors always behave as capacitors. They're not schizophrenic components :smile: We say that an inductor "behaves like" a short circuit at steady state because at that time all the transient AC components have died away and a steady current flows through it as though it were nothing more than a piece of wire. BUT! add some new transient voltages or currents and you quickly find out that that inductor is still an inductor after all.

    The same goes for the capacitor. At steady state its potential difference matches the potential presented to it from the surrounding circuit. With no differences, no current flows. No current flowing even though there's a potential difference across its terminals makes it "look like" an open circuit. BUT! Change the circuit a bit so that there's a new current flowing into the capcitor and voila! You find out it's still a capacitor.
     
  11. Mar 29, 2012 #10
    Thanks gneill, you've always been a big help. Much appreciated!
     
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