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Question about a paralell-like RC circuit

  1. May 21, 2010 #1
    1. The problem statement, all variables and given/known data

    I have the circuit that's attached as a .bmp image. I'm asked to find the initial current that passes trough the battery (V=60 V) knowing that the capacitor is initially charged with charge q_{0}=0.1 C. The resistances are R_1=R_2=R_3=1 Ohm and C=1/1000 F.

    2. Relevant equations

    The relevant equations are given by Kirchoff's Laws:

    [tex]
    \oint \vec{E}\cdot d\vec{l}=0
    [/tex]

    [tex]
    I_3=I_1+I_2
    [/tex]

    3. The attempt at a solution

    My attempt to the solution was to use Kirchoffs first law. Making two closed loops: One passing trough V, then resistance R_1, C and finally trough resistance R_3 and the other loop passing trough V, resistance R_2, and finally resistance R_3. With this we get the two equations:

    [tex]
    V-R_{1}I_{1}-q_{0}/C-R_{3}I_{3}=0
    [/tex]

    [tex]
    V-R_{2}I_{2}-R_{3}I_{3}=0
    [/tex]

    From here we get:

    [tex]
    I_{1}=\frac{V-q_{0}/C-R_{3}I_{3}}{R_{1}}
    [/tex]

    [tex]
    I_{2}=\frac{V-R_{3}I_{3}}{R_{2}}
    [/tex]

    And we finally use Kirchoffs second law, I_{1}+I_{2}=I_{3} and get:

    [tex]
    I_3=\frac{V-q_{0}/C-R_{3}I_{3}}{R_{1}}+\frac{V-R_{3}I_{3}}{R_{2}}
    [/tex]
    And you finally solve for I_3, wich gives I_3=6.7 Amperes. My question is that if you replace the data on I_1 you get I_1=-46.7 A. I'm unsure about the current going in the opposite way of what i tought. Anyways, I tought it would have some phyical meaning like all current flowing into the capacitor, but i'm not sure. What do you think?
     

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    Last edited: May 21, 2010
  2. jcsd
  3. May 21, 2010 #2

    vela

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    The letter I stands for current, not intensity.

    The problem is ambiguous because it doesn't specify how the voltage across the capacitor is oriented. The end with the higher potential could be connected to R1 or it could be connected to R3.

    Calculate the initial voltage across the capacitor and you'll see why the current flows in the direction it does.
     
  4. May 21, 2010 #3
    Oh, sorry, my bad. I'm a spanish native speaker and i confused terms.

    In my work I assumed that the initial positive charge was on the left side of the capacitor thus leaving the negative 0.1 C on the right. If that's the case, then it DOES make sense that current flows in the opposite direction that i was expecting, right?
     
    Last edited: May 21, 2010
  5. May 21, 2010 #4

    vela

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    It depends on the initial voltage across the capacitor. Did you calculate it?
     
  6. May 21, 2010 #5
    The initial voltage accross the capacitor is:

    [tex]

    \Delta V=\frac{Q}{C}=\frac{0.1}{10^{-3}} V=100 V

    [\tex]

    So if there is an initial potential difference of 100 V, and positive charge is distributed on the left side, then current will flow to the left on R1, because of the electric field produced by the capacitor, right?
     
  7. May 21, 2010 #6

    vela

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    Not exactly. If the battery weren't present in the circuit, what you said would be correct. But say the battery was 1000 V, not 60 V, then the current through R1 would flow to the right. It's the difference between the battery and the capacitor that's important.
     
  8. May 21, 2010 #7
    Right! Thanks for you reply :)!
     
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