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Natural response of an RC circuit with two capacitators.

  1. Oct 11, 2007 #1
    1. The problem statement, all variables and given/known data
    I'm supposed to find the voltage v0(t) for t > 0 for the following circuit:
    [​IMG]

    2. Relevant equations
    The first part of the excercise was to find the current io(t). I did this by using the node-voltage method to find the voltage across the 8k resistor and the 10k resistor to find the voltages across the two capacitators. Then I figured since, once the switches open, they'll be in series, I find R*C by finding the total C which was 100 nF.

    Then I found the voltage drop across Io which then became 120e^-5000t - 100e^-5000t = 20e^-5000t and divided this by the resistance to find the current, which would be 10e^-5000t mA.


    3. The attempt at a solution
    That's when things got ugly. I'm thinking that the voltage at v0 will have to be the voltage-drop between Io and the 300nF capacitator. Fair enough. If the current in I0 is 10e^-5000t mA I figure the voltage will have to be 20e^-5000t V. Then I'm thinking the voltage across the 300nF supply will have to be 100e^-5000t V (since the voltage at the 10k resistor was 100v before the switches were opened) which makes me believe the voltage at v0 should be -80e^-5000t v.

    However, they gave us the correct answers beforehand, and I know it's supposed to be:
    -6.67e^-5000t + 106.67 v.
    This has me somewhat stumped. If this is correct, that means even when t -> infinity there will be a voltage of 106.67 v over v0. But then i'm thinking, since these really are capacitators, won't the 150nF capacitator, being the one with the highest voltage, somehow charge the other one? Or what?

    I'm thinking some kind of equilibrium here is involved, but I'm having real trouble seeing where those numbers are coming from. A push in the right direction would truly be helpful!
     
  2. jcsd
  3. Oct 11, 2007 #2
    I've done some more thinking to the problem, should I treat this as some kind of step-response? Considering the 300nF capacitator goes from one voltage/current supply to a new one, that would explain how I'm supposed to get that extra voltage in the equation. I tried calculating as a step-response by using the foruma Vc = IsR + (V0 - IsR)e^(-t/(RC)) but to no appearant avail, maybe because according to my book i'd normally have to transform into a norton-equivalent before I could use this formula, and since I can't find anything on doing anything similar when another capacitator is the voltage supply, I should probably just drop that train of thoughts.
     
  4. Oct 11, 2007 #3

    CEL

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    You must treat it as a zero input response.
    [tex]v_c(t) = v_c(0).e^{-\frac{t}{RC}}[/tex]
    This means that the answer they gave you is wrong.
    Analyzing it physically: The two capacitors are previously charged. As time goes by they discharge into the resistor and the energy they had stored is lost as heat. After an infinite time all energy is spent and the capacitors are discharged.
     
  5. Oct 11, 2007 #4

    dlgoff

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    "As time goes by they discharge into the resistor..."
    Am I missing something? What resistor will the caps discharge through? Since there's switches on either side, ideal caps would come to some kind of equalibrium and remain there for infinity. Right?
     
  6. Oct 11, 2007 #5

    CEL

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    The 2k resistor through wich [tex]i_0[/tex] flows.
     
  7. Oct 12, 2007 #6

    dlgoff

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    Nah. There's no path to ground(-) inorder for them to discharge; once the switches are open.
     
  8. Oct 12, 2007 #7

    CEL

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    So, you are saying that the two capacitors and the 2k resistor don't make a closed loop?
     
  9. Oct 12, 2007 #8
    Everyone who has posted in this thread is wrong, but in different ways. Of course the resistor discharges some of the power; that's what resistors do. But they only do it when there is a voltage drop over them.

    The capacitors don't entirely discharge. They reach an equilibrium voltage of 106.67v, at which point there is no longer a current running through the circuit.

    The OP was right to combine the caps to find the tau, but you need to use 300e-9 F to find the equation for V(t). You need Io so that you can differentiate the equation Vf + Ae^-5000t and set it equal to Io/C and find A.

    The solution given by the book is absolutely correct.
     
    Last edited: Oct 12, 2007
  10. Oct 13, 2007 #9
    Aha, I was thinking something like this might happen if there's no voltage drop across the resistor. It was just so hard to analyze physically.
    Thanks loads for the help!
     
  11. Oct 13, 2007 #10

    dlgoff

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    No. They do make a closed loop. But like I said:
     
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