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Homework Help: Charging/discharging of capacitor. (sticky wicket)

  1. Mar 3, 2014 #1
    Charging/discharging of capacitor.

    1. The problem statement, all variables and given/known data

    Attached to this post, I have two circuits. The left, circuit 1. The right, circuit 2.

    For circuit 1 I'm supposed to compute T (tau). Find the expression for Uc1(t) and IR1(t). This is regarding discharging of a capacitor.

    For circuit 2, I'm supposed to compute T, Uc1(t) and Uc1(∞)

    2. Relevant equations

    3. The attempt at a solution

    For circuit 1:

    Doing this for a series circuit is straight forward. For parallel circuits, from whatever I can see in my notes and my book, I'm supposed to convert this circuit into a thevenin equivalent. When I do that, using C1 as load, pressing in the switch and other wise use my knowledge about thevenin equivalent circuits, I get R to be 0. When V1 is short circuited, the "current" will flow through the resistance free path, avoiding R1.

    How to I solve this?

    And my mathematical expressions for Uc1 and IR1, would they differ any from what I have from a series circuit (identical to circuit 1 here, only R1 and C1 are in series)

    There I have:

    T = RC = 4700Ω * 680*10-6F = 3.196S

    UC1 = V1*(1-e-t/Tau) = 8v*(1-e-t/3.196s)

    IR1 = V1*e-t/Tau = 8v*e-t/3.196s

    Attached Files:

    Last edited: Mar 3, 2014
  2. jcsd
  3. Mar 3, 2014 #2


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    Staff: Mentor

    Circuit #1 is strange. Connecting an ideal voltage source (with zero series resistance) generates infinite current to charge up the capacitor in zero time when the switch is closed. Is that really the circuit? Does the problem statement give any source resistance that is built into that voltage source?

    EDIT -- Wait, is the switch opened or closed at t=0? I see now that you mention "discharging" with respect to circuit #1. That would maybe imply that the switch is opened at time t=0?
  4. Mar 4, 2014 #3
    I'm supposed to do all calculations considering the voltage source an ideal one. Therefore, there's no source resistance.

    The problem statement says nothing about this. All it says is "discharging capacitor". Discharging is the best word I can translate it to.

    "the circuit discharges C1 through R1 when the switch is opened. When the switch is closed, C1 is charged".

    That's absolute everything the problem says, except for the computing parts.
  5. Mar 4, 2014 #4


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    Staff: Mentor

    For the first circuit I think you'll just have to assume that the voltage source is a real battery with some finite but negligible internal resistance and that the part of the circuit action of interest is when the switch opens allowing the capacitor to discharge. So at some time in the past the switch was closed and the capacitor charged over negligible time, then at time to the switch opens.


    Attached Files:

    • Fig1.gif
      File size:
      14.4 KB
  6. Mar 4, 2014 #5
    Ok, I understand that now. There were three problems, I just asked for help for the two last ones. The first one was the attached circuit, and I just realized that it was regarding charging, and the 2nd was regarding discharging. The third is labeled "analysis".

    On the first (now, the first of three, sorry for any confusion) circuit, I got these results:

    T = RC = 4700Ω * 680*10-6F = 3.196S

    UC1(t) = V1*(1-e-t/Tau) = 8v*(1-e-t/3.196s)

    IR1(t) = IR1 = IC1 = [itex]\frac{V1}{R}[/itex] * e-t/T = [itex]\frac{8V}{4700Ω}[/itex] * e-t/3.196s = 1.7mA * e-t/3.196s

    (in my OP, the IR1(t) has unfortunately been mixed up with, well, something wrong.)

    Now, for my 2nd circuid (of three), I have very similar answers!

    My T (RC) is the same: T = RC = 4700Ω * 680*10-6F = 3.196S

    UC1(t) = V1*e-t/Tau = 8v*e-t/3.196s

    IR1(t) = IR1 = IC1 = -[itex]\frac{V1}{R}[/itex] * e-t/T = - [itex]\frac{8V}{4700Ω}[/itex] * e-t/3.196s = -1.7mA * e-t/3.196s

    Is this even remotely correct?

    And by the way, gneill, how did you draw that graph?

    As always, thanks a lot for your help.

    Attached Files:

  7. Mar 4, 2014 #6


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    Staff: Mentor

    You'll want to review the form of the exponential functions. In the (first) circuit, the voltage on the capacitor is decaying from a starting value of 8V down to zero when the switch opens. That is, the discharging case. That would be form
    $$U(t) = A e^{-t/\tau}$$
    The other form,
    $$U(t) = A (1 - e^{-t/\tau})$$
    applies to the case where the voltage starts at zero and increases towards a final value of A.

    The problem statement should have some text describing what time intervals to consider and what to take as the initial conditions.
    For the second circuit, when the switch is in the initial position the capacitor "sees" the output of a voltage divider, so it can't charge to the full value of V1. I suspect that the problem would say something about the switch having been in the initial position for "a long time", so you won't have to write an expression for any initial charging, just determine the steady state voltage on C1 for the switch in that position.

    The interesting bit occurs when the switch is thrown to its other position.

    I used Microsoft Visio to draw it out.
  8. Mar 4, 2014 #7
    Considering the misunderstanding, I think - at least according to my book - that my values are correct now.

    To clear things up, from post #5: First circuit = series circuit. Second circuit = parallel circuit. Third circuit = the "analysis" circuit.

    My first circuit is regarding charging of capacitor. Thus using the form V1(1 - e-t/tau) to express UC1.

    For the current IR1 I'm using the positive value for [itex]\frac{V1}{R1}[/itex] * e-t/tau.

    My second circuit is regarding discharging of capacitor. Therefore I am using the form V1 e-t/tau to express UC1.

    For the current IR1 I'm using the negative value for [itex]\frac{V1}{R1}[/itex] * e-t/tau.

    My RC (tau) value for both circuits is the same - 3.196s.

    I hope this clears up any confusion.
  9. Mar 4, 2014 #8


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    Staff: Mentor

    Okay. In future it might be best to post a single question per thread. This has been an excellent example of why that's a good idea :smile:
  10. Mar 4, 2014 #9
    Well, at the very least, the moron OP should avoid asking two questions in the first post, and then drag in a third question in the fifth post that is actually a new first question.

    On a less self-sarcastic note, though, I just want to avoid clogging up the forum. I kind of feel I ask too many questions and contribute too little.

    If/when I need help for my third problem, I will post a new thread though. I hate being the cause of time wasting misunderstandings.
  11. Mar 4, 2014 #10


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    Staff: Mentor


    Post as many separate threads as you need and can keep track of diligently. Multitasking is a wonderful thing so long as it doesn't affect quality :smile:
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