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Charging RC circuit

  1. May 2, 2013 #1
    1. The problem statement, all variables and given/known data
    Switch is initially opened with the capacitor uncharged. At t=0 the switch is clsoed. At time=t1 later, the potential difference between A and B is 2.00volts
    Find the charge on the capacitor at time t1.

    R=10,000ohms
    C=2.4nF
    Vb=12volts
    VA->B-=2.00volts


    2. Relevant equations
    I(t)=(V/R)*e^(-t/RC)
    q(t)=CV(1-e^(-t/RC))
    C=Q/V


    3. The attempt at a solution
    RC was correctly found to be 40microSeconds.So then the goal is to find the time at t1. The 2volts across A and B must be important to finding t1.
    So I assumed that the voltage drop from A-B must be equal to the voltage drop for R3. Thus 2Volt/R=I3.

    Then I used I3 to find t1: I3=(V/R)*e^(-t/RC) --> solving for t and then plugging into q(t).
    Does this seem right or is my logic wrong?
     

    Attached Files:

  2. jcsd
  3. May 2, 2013 #2

    mfb

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    I don't think this helps. You do not have a simple RC circuit, the charging will show a more complex time-dependence.
    You can use the usual circuit rules to get the charge, you don't even need the time.

    This would correspond to zero voltage drop at the capacitor...
     
  4. May 2, 2013 #3
    1) I mean RC as in the time constant= total resistance*capacitance = 5/3R*C

    "You can use the usual circuit rules to get the charge..."
    How do you do this?
    2volts = (q/c) - I3R3 and solve for q?
     
  5. May 2, 2013 #4

    mfb

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    The voltage drop at your RC part is not constant.

    Step by step. You can calculate the voltage drop at the right transistor. Based on this, you can calculate the current there, and so on.
     
  6. May 2, 2013 #5
    "The voltage drop at your RC part is not constant."
    yeah but isn't [itex]\tau[/itex] (time-constant) always constant from the time the switch is closed to a "long time after"?


    "Step by step. You can calculate the voltage drop at the right transistor. Based on this, you can calculate the current there, and so on."
    I assume you meant resistor not transistor?
    I'm confused. Is the above equation 2volts = (q/c) - I3R3 not correct. I don't see how you couldn't just find q from this.
     
  7. May 2, 2013 #6

    gneill

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    It looks to me like there's something wrong with the problem statement. I don't see how the magnitude of the potential difference between points A and B can ever be 2.00V after the switch is closed. It should only be able to range between the values afforded by two conditions: capacitor is a short (uncharged); capacitor is an open (charged).
     
  8. May 2, 2013 #7
    I just found an answer (without solution) and it said it is supposed to be 7.2nC. Using the first idea I had, I pretty much got the answer with t1=11.5microseconds :

    **"RC was correctly found to be 40microSeconds.So then the goal is to find the time at t1. The 2volts across A and B must be important to finding t1.
    So I assumed that the voltage drop from A-B must be equal to the voltage drop for R3. Thus 2Volt/R=I3.

    Then I used I3 to find t1: I3=(V/R)*e^(-t/RC) --> solving for t and then plugging into q(t).
    Does this seem right or is my logic wrong?"**

    but... I become completely confused by mfb's suggestions.


    gneill, why can't it ever be 2volts if the battery is 12volts?
     
  9. May 2, 2013 #8

    gneill

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    Calculate the potential difference when you replace the capacitor with a short. Calculate it again when you replace the capacitor with an open circuit. Does 2V lie between the values?
     
  10. May 2, 2013 #9

    ehild

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    That is true, UAB never can be 2 V.

    ehild
     
    Last edited: May 3, 2013
  11. May 2, 2013 #10

    mfb

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    At every moment, you can calculate τ, but you cannot use it in the exponential formula.
    Oh, right.
    It is, but you have two unknown values there. You have to find I3 first.

    Interesting, I did not test this. The capacitor would have to have a reverse charge.
     
  12. May 2, 2013 #11
    ***"Oh, right.
    I'm confused. Is the above equation 2volts = (q/c) - I3R3 not correct. I don't see how you couldn't just find q from this.
    It is, but you have two unknown values there. You have to find I3 first."***

    Right, I did that and I solved for q. This does not come out to the answer provided [7.2nC charge on the capacitor]


    BUT! When I the following:
    I3=I0(e^(-t/[itex]\tau[/itex])), solve for t.
    Solve for q(t)=q0(1-e^(-t/[itex]\tau[/itex]))
    I got 7.2nC...

    So now I'm just confused because you say this is not the correct use of the formula.
     
    Last edited: May 2, 2013
  13. May 2, 2013 #12
    Moving on for now
     
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