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RC Circuit and current from a discharging capacitor

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  1. Dec 10, 2016 #1
    1. The problem statement, all variables and given/known data
    In the circuit of the figure below, the switch S has been open for a long time. It is then suddenly closed. Take emf.gif = 10.0 V, R1 = 56.0 kΩ, R2 = 200 kΩ, and C = 14.5 μF.
    28-p-039.gif
    (a) Determine the time constant before the switch is closed.
    (b) Determine the time constant after the switch is closed.
    (c) Let the switch be closed at t = 0. Determine the current in the switch as a function of time. (Assume I is in A and t is in s. Do not enter units in your expression. Use the following as necessary: t.)
    2. Relevant equations
    V=IR
    The solved diff eqs for current over time
    Kirchoff's rules

    3. The attempt at a solution
    (a) and (b) aren't problems for me but I do want to make sure my reasoning is accurate.

    (a) Is pretty easy I guess. I just combined the two resistors in series and then multiplied by C, the reasoning being that the electrons that go through C also have to go through R1 and R2.
    (b) The switch closing means that there is a short circuit, so all of the current from the voltage source goes through the middle branch and the right branch is ignored basically. So I guess you would multiply R2 by C because the discharging electrons have to go through R2?

    (c) Is the problem here for me. At first it's kind of easy. The current from the left branch would be I1, the current in the middle branch can be Iswitch, and the current in the right branch can be I2. Using Kirchoff's junction rule, I1+I2=Iswitch.

    The basic solution to the diff eq would be I(t)=I0e-t/RC. However, getting the initial current is kind of a problem for me. The only value for any sort of voltage we have is the ε, but how exactly can I use that?

    I mean, what I'm thinking is that I could use Kirchoff's loop rule for before the switch is closed, but we don't have the charge on the capacitor, do we? Also, I believe the solution doesn't do it that way either. We could also use V=IR but it seems like that wouldn't work. Am I wrong?
     
  2. jcsd
  3. Dec 10, 2016 #2

    gneill

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    Staff: Mentor

    To find the initial current from the capacitor (at t = 0+) you'll need the potential on the capacitor at that instant in time. That's where you need to know what the circuit was doing just before the switch closed.

    According to the problem statement, "...the switch S has been open for a long time. It is then suddenly closed." What does the switch being open for a long time tell you about the circuit conditions? What current do you expect to see flowing in the circuit? Can you determine the potential on the capacitor?
     
  4. Dec 10, 2016 #3
    This is kind of what I'm struggling with. Maybe I'm overthinking this. I'd consider the situation right as the switch is closed. That would give me the initial current, but I'm not really sure what to do. Maybe I'm just lacking context of what's going on.

    Now that I think about it, after a long time with the switch being open, wouldn't the current stop completely in everything because the capacitor is charged? So then apply V=IR? I'm not really sure. If the current is stopped that wouldn't work would it?
     
  5. Dec 10, 2016 #4

    gneill

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    Yes, at steady state no current will flow and the capacitor will have some fixed potential. What must that potential be? Remember, this is before the switch is closed. Don't even think about the switch yet.
     
  6. Dec 10, 2016 #5
    So the potential of the capacitor has to be equal to ε right? What would I do with that, though? Find the charge? I'm not sure how that would be useful.

    Also, I think I have some misunderstanding here that might be causing my problem. Immediately after the switch is turned on, wouldn't it still be at steady state and thus no current would exist? How exactly would there be an initial current?
     
  7. Dec 10, 2016 #6

    gneill

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    Right. The voltage on the capacitor will be equal to ε. You can find the charge if you wish, it won't be particularly useful as you say. What is important is the voltage across the capacitor immediately before the switch is closed.
    No, immediately after the switch is closed current will begin to flow. However, the voltage on the capacitor will still be whatever it was just prior to the switch closing. This is an important concept! Capacitors cannot change their potential difference instantaneously. It takes time and current flow to change the potential difference of a capacitor. So if you know the potential difference on a capacitor at some time t = 0-, and if something happens in the circuit (such as a switch changing) at time t = 0, then at time t = 0+ the capacitor will still have the same potential difference that it had at time t = 0-.

    You can use this capacitor potential difference to determine the initial current that the capacitor will drive through the switch.
     
    Last edited by a moderator: Dec 10, 2016
  8. Dec 10, 2016 #7

    NascentOxygen

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    No. Current doesn't have inertia.

    Current can be instantaneously changed through any [closed] path that includes no inductance. The current can be changed at the flick of a switch. Literally. :smile:
     
  9. Dec 10, 2016 #8
    So what should I use to find the current using that voltage? The capacitor would start discharging. Is it the electrons we care about here rather than the actual "current"? Because the electrons would go through R2. The "current" doesn't go through anything.

    The weird thing about these problems is that I've never actually had to apply the concept that the current is just lack of electrons. I usually just follow the current when it's just resistors or just capacitors. Is my reasoning right for all this stuff, even for (a) and (b) in my original post?

    Also, why does the potential on the capacitor have to end up being equal to the potential from the voltage source? Is it because of the fact that eventually the current stops?
     
  10. Dec 10, 2016 #9

    NascentOxygen

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    If an electron moves, that constitutes a current.

    Maybe you are stumbling over the fact that when electrons flow from A to B that constitutes current in the opposite direction. This can certainly be cumbersome, at times. But in some situations if it is going to make things easier it is okay to think in terms of electron flow. Though in the circuit under discussion, I don't see this being needed.

    You can instead picture positive charges flowing, because this would be in the same direction as current. To wit, "In attaining the steady state, current from the + terminal flows into the capacitor's upper plate, charging that plate increasingly + . When the switch is closed, some of these positive charges move ...."

    I guess it's one of those rare cases where two wrongs do make a right. :biggrin:
     
  11. Dec 10, 2016 #10

    NascentOxygen

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    Your reasoning in (a) and (b) is right.
     
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