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Parallel RC circuit

  1. Mar 20, 2012 #1
    Parallel RC circuit (Updated)

    I'm having trouble with a seemingly simple circuit. I have a voltage source (not alternating current) that increases exactly with time. In other words, if you graphed voltage as a function of time, the slope would be 1. The voltage equation is v(t)=t. The capacitor is 1/2 F and the resistor is 3Ω. Here's a picture of the circuit.
    http://img832.imageshack.us/img832/1351/circuite.png [Broken]
    I'm supposed to find the current i(t) in the left branch where the voltage source is connected.

    I applied the KCL law to the upper node where the resistor and capacitor branches meet. I labelled the current I'm solving for i(t). The current through the capacitor is ic(t) and the current through the resistor is ir(t). Using KCL, I get:
    i(t)=ic(t)+ir(t).

    The equation for current through a capacitor is:
    i(t)=C*(dv(t)/dt)
    Since the derivative of the voltage function is 1, and the capacitor is 1/2 F:
    ic(t)=1/2 A

    For the current through the resistor, I just applied Ohm's Law and got:
    ir(t)=v(t)/3Ω, which is just;
    ir(t)=t/3

    Therefore, my final answer is:
    i(t)=t/3 + 1/2 A

    The book's answer is:
    i(t)=t/3 + 5/6 A

    I might add that the function is only v(t)=t on the domain t={-1,0}. I didn't think this was too relevant since I'm only solving for the current on this domain. Also, the voltage source starts at 0V, so it peaks at 1V where t=0.
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Mar 20, 2012 #2

    gneill

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    Staff: Mentor

    Looks like the book is incorrect on the 5/6 A bit. Perhaps they originally designed the question with another V(t) function in mind and failed to change the answer when they changed the problem.
     
  4. Mar 20, 2012 #3
    Does my answer look correct? I felt a little unsure about applying Ohm's law directly to the right branch.
     
  5. Mar 20, 2012 #4

    gneill

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    Staff: Mentor

    Looks fine to me :smile: Ohm's law was the correct choice for the resistor.
     
  6. Mar 21, 2012 #5
    I got someone else to work it and he had the book's answer. The problem is that he did a lot of it in his head and I didn't have much time to write it down. I know that probably sounds strange, but I thought I would understand more if I looked at the small amount I wrote later. Anyway, I think I'm doing something wrong with the capacitor's current. I remember he added 1/3 A to the 1/2 A to get the 5/6 A. He applied the same differential equation that I used above, but on top of that he was able to add that extra 1/3 A. I vaguely remember him getting that 1/3 A from calculating a 1/6. I still can't figure out where that extra 1/3 A came from.
     
  7. Mar 21, 2012 #6

    gneill

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    Staff: Mentor

    I don't see where the extra 1/3 A could come from either, assuming that the applied voltage does indeed ramp up from 0 to 1V over 1 second.
     
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