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RC circuit with current source

  1. May 20, 2016 #1
    1. The problem statement, all variables and given/known data
    FUHimw2.jpg

    2. Relevant equations


    3. The attempt at a solution
    so when t=0, the capacitor is a closed circuit.

    The Req of the circuit is (20k)(30k)/(20k+30k)= 12000 ohms.

    U = I Req = (7.5x10-3A) (12000 ohms) = 90 Volts

    so there is a voltage drop of 90 volts across the 20k resistor (R1), when t=0.



    RC= Req C = (12000 ohms) (0.1 x10-6F) = 0.0012 sec
    ∴ UR1= Uo (e-t/RC) = UR1= 90V (e-t/0.0012)


    Ok, so then the current flowing through the capacitor would be the same current flowing through the 30k resistor when t=0.

    Using current division, 7.5x10-3A [20 / (20+30)] = 0.003 A

    iC1 = io (e-t/RC) = (0.003A)(e-t/0.0012)

    This is my final answer, I've erased my initial attempt as it was wrong. Could someone verify if this is correct?
     
    Last edited: May 20, 2016
  2. jcsd
  3. May 20, 2016 #2

    SammyS

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    The capacitor is initially uncharged. The voltage across it is zero at t=0. That's like a short circuit, not an open circuit.
     
  4. May 20, 2016 #3
    If it is a short circuit, then my drawing is wrong? :(

    So a capacitor only behaves as a open circuit when it is fully charged?
     
  5. May 20, 2016 #4

    SammyS

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    Yes. This is when current ceases to flow through the capacitor, like the case for an open circuit.
     
  6. May 20, 2016 #5
    Ok, so then the current flowing through the capacitor would be the same current flowing through the 30k resistor when t=0.

    Using current division, 7.5x10-3A ( 20 / (20+30)) = 0.003 A

    RC = (30,000)(0.1 x 10-6) = 0.003 seconds

    iC1 = io (e-t/RC) = (0.003A)(e-t/0.003)

    or is RC not 0.003 because I had to use Req of the circuit before capacitor became fully charged?
     
  7. May 20, 2016 #6
    so when t=0, the capacitor is a closed circuit.

    The Req of the circuit is (20k)(30k)/(20k+30k)= 12000 ohms.

    U = I Req = (7.5x10-3A) (12000 ohms) = 90 Volts

    so there is a voltage drop of 90 volts across the 20k resistor (R1), when t=0.



    RC= Req C = (12000 ohms) (0.1 x10-6F) = 0.0012 sec
    ∴ UR1= Uo (e-t/RC) = UR1= 90V (e-t/0.0012)

    assuming this is correct, my answer to the previous post should be:
    iC1 = io (e-t/RC) = (0.003A)(e-t/0.0012)
     
  8. May 21, 2016 #7

    cnh1995

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    This means voltage across R1 is exponentially decreasing. Is it? Think about the steady state of this circuit. Which quantities become zero in the steady state here?
     
  9. May 21, 2016 #8
    so the capacitor has increasing voltage across it over time, and decreasing current across it, so when current decreases, R2 will have less current passing through it... and since we have a current source, we have a constant flow of current flowing through it, therefore all the current (or most) overtime will have to flow through R1 once the capacitor is fully charged (where it becomes open circuit)

    applying ohms law, the voltage drop across R1 should be increasing.

    Yeah... I'm having a harder time understanding/remembering the equations than understanding the concepts. :H

    The voltage drop across R1 should be (7.5 mA) (20k ohms) = 150 V.
    but i'm having trouble obtaining an equation that equates to 150V when t → ∞
     
    Last edited: May 21, 2016
  10. May 21, 2016 #9

    cnh1995

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    Right. Have you studied source transformation? It would be easier.
     
  11. May 21, 2016 #10
    No, I'm in an intro class, and we haven't even (or just started) AC circuits.
     
  12. May 21, 2016 #11

    cnh1995

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    Ok but source transformation means converting current source into voltage source.
    You can replace a current source of magnitude I with parallel resistance R with a voltage source of magnitude I*R with the same resistance R in series. This way, you'll end up with a simple RC circuit with two resistors in series with a capacitor and a voltage source.
    images.png
     
  13. May 21, 2016 #12

    cnh1995

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    Initial current through the capacitor will be 30mA and that through R1 is 20mA, as you calculated earlier. Now, voltage across R2 decays to 0 exponentially and voltage across R1 increases to 150V exponentially. Also, capacitor voltage increases exponentially to 150V. Using the source transformation, you'll get the expression for ic(t). You can use it in the original circuit to find i1(t) and U1(t).
     
  14. May 21, 2016 #13
    I'm trying to answer it without using source transformation, I know it'd be easier but we haven't covered it yet in class.
     
  15. May 21, 2016 #14

    cnh1995

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    Ok. How about Thevenin's theorem? Have you covered it yet?
     
  16. May 21, 2016 #15
    UR1(t) = [Is - IC1(t)]/ 20kΩ

    Plugging in IC1(t)] from what I found before:


    UR1(t) = [Is - io (e-t/RC) ]/ 20kΩ = [Is - (0.003A) (e-t/0.0012) ]/ 20kΩ
     
  17. May 21, 2016 #16
    Nope. we are low on options haha, sorry.
     
  18. May 21, 2016 #17

    cnh1995

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    This is the right expression.
    Edit: Oh dear! How did I miss that..:-p V=IR and not I/R...
    The time constant is incorrect.
     
    Last edited: May 21, 2016
  19. May 21, 2016 #18
    I didn't derive this equation, I just thought for a bit and this equation made sense.
    How would one derive this?


    I wasn't sure if I had to use the resistors of R2 (the resistor it was in series with, so they have the same current) or Req of the circuit when the capacitor was short.

    would I be wrong if I said Time constant for a circuit = (Req) (Ceq)? (assuming there were more than 1 capacitor)
     
  20. May 21, 2016 #19
     
  21. May 21, 2016 #20

    cnh1995

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    No. The time constant considers both R1 and R2. How are they connected when you open the current source?
     
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