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Initial Voltage in an RC circuit with a Current Source and no Voltage Source

  1. Feb 19, 2017 #1
    1. The problem statement, all variables and given/known data

    Derive an expression for vc(t) in the circuit of Figure P4.13 and sketch vc(t) to scale versus. I attached a screen shot of the circuit for this problem.

    Screen Shot 2017-02-19 at 12.04.57 AM.png time.

    2. Relevant equations

    V = IR
    V = V0exp(-t/tau)

    3. The attempt at a solution

    I am trying to find the initial conditions to this problem. To do this, I have said that at t < 0 the capacitor will act as a short circuit for a dc current. we would then have that the current that is going through the resistor R is 10mA. This would yield to Vc(0) = (.01)(2000) = 20V. When I checked a solution of this problem, they get that Vc(0) = 0V. I am not really sure why that is! Thank you very much for your help!
     
  2. jcsd
  3. Feb 19, 2017 #2

    ehild

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    Short circuit means zero resistance. Would the current flow through the resistor, when it can choose a parallel zero-resistance path?
     
  4. Feb 19, 2017 #3
    Thanks you for the quick reply! Oh! I I did not think of it this way. My answer would be "No, it would not." That raises another question that might be related to a fundamental piece of information that I don't know. Current would also not flow if there was no change in voltage, which the resistor offers, right?
     
  5. Feb 19, 2017 #4

    ehild

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    Yes, no current through the resistor implies zero voltage across it.
    But you can think in the point of view of the capacitor. At t=0, there is zero charge on it. The voltage across a capacitor is V=Q/C. So zero charge means zero voltage across the capacitor, and the resistor and capacitor are parallel connected, so V(t)=0 at t=0.
    Is your "relevant equation" V = V0exp(-t/tau) correct then?
     
  6. Feb 19, 2017 #5
    So, I think that I am misunderstanding the problem. The arrow on the switch in the diagram means that the switch was closed for a long time and then it was opened at t=0s. If this is the case, then would that not mean that the capacitor was charged at t <= 0?
     
  7. Feb 19, 2017 #6

    ehild

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    If the switch is closed, it means short-circuit for the current source. Why should the current choose a different path?
     
  8. Feb 19, 2017 #7
    I think I am starting to understand where you are coming from. Initially, I thought that current would not go through the parallel piece of wire because there is no voltage drop there, but it would go to the resistor because there there is a voltage drop and the current could flow. I have always thought that current would not go throw a wire that did not have a voltage drop. I think this is where my misunderstanding is coming from.
     
  9. Feb 19, 2017 #8

    gneill

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    Moderator's note: Changed the thread title to make it less generic and more specific to the problem being addressed.
     
  10. Feb 19, 2017 #9
    Thank you for the note. I changed the title of the thread.
     
  11. Feb 19, 2017 #10
    Would the current not choose the path that has a voltage drop, which is the one with the resistor?
     
  12. Feb 19, 2017 #11

    gneill

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    Current *causes* a voltage drop when it flows through a resistor. If it's flowing elsewhere, there will be no voltage drop.
     
  13. Feb 19, 2017 #12

    ehild

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    How do you know that there is a voltage drop across the resistor? The voltage across a resistor is U=IR. The resistor is not connected to a voltage source, the voltage across it is determined by the current. Zero current, zero voltage.
    The branch including the closed switch has zero resistance. So the voltage across the current source is zero. So is the voltage across both the resistor and capacitor.
     
  14. Feb 19, 2017 #13
    Thank you very much for this explanation! I really appreciate it!
     
  15. Feb 19, 2017 #14
    Thank you!
     
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