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Voltage drop across capacitor in RC circuit

  1. Mar 19, 2013 #1
    1. The problem statement, all variables and given/known data

    Please look at the attachment for the circuit.

    The question is this:

    The switch has been closed for a long time. It is then opened. How long does it take for the potential across the 4μF capacitor to fall to half of its initial value?

    2. Relevant equations

    V = V0e^(-t/RC)

    3. The attempt at a solution

    Once the switch is opened, the capacitor would begin to discharge creating a potential difference. The equation I tried to solve is

    0.5 = e^(-t/(4.0*R))

    The problem I'm having is that I don't know what R, the equivalent resistance is for this circuit.

    The answer is that it takes 10.2 μs for the potential to drop to half of its initial value. Working backwards, I get the equivalent resistance is 3.68 ∏. How do I get that?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Mar 19, 2013 #2

    gneill

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    Staff: Mentor

    What components comprise the circuit of interest when the switch has opened?
     
  4. Mar 19, 2013 #3
    When the circuit is opened, the left loop should still has current because there is a battery in that loop.
     
  5. Mar 19, 2013 #4

    gneill

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    Staff: Mentor

    Is the current in that loop of interest? What part of the overall circuit is the question asking about?
     
  6. Mar 19, 2013 #5
    No ... it's the right loop. When the capacitor discharges, it creates a current that flows around the right loop. I3 becomes zero. Is that correct?
     
  7. Mar 19, 2013 #6

    gneill

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    Staff: Mentor

    Yes, that is correct. Look at the components in that loop. How might you determine the time constant for it?
     
  8. Mar 19, 2013 #7
    The time constant = RC.

    The only resistor is the 5 ohm resistor. Is that R?

    And for C, do I use the equivalent capacitance or just the 4 microfarad?
     
  9. Mar 19, 2013 #8

    gneill

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    Staff: Mentor

    Yes. (being the only resistor, it must be!)
    The loop comprises a single circuit, so all the components participate... Find the total equivalent capacitance.
     
  10. Mar 19, 2013 #9
    C = (1/4+1/11)^-1 = 2.9333

    Time constant = RC = 5*2.93333 = 14.6667 microseconds

    0.5 = e^(-t/14.6667)
    t = 10.2 microseconds

    Thanks for all the help!

    btw ... does it also take 10.2 microseconds for the potential across the 6 microfarad resistor to drop to half of its initial value?
     
  11. Mar 19, 2013 #10

    gneill

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    Staff: Mentor

    You're welcome!
    Yup.
     
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