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RC Circuit current through capacitor question

  1. Mar 21, 2014 #1
    1. The problem statement, all variables and given/known data


    Specifically #13 only.

    2. Relevant equations

    Same image as above.

    3. The attempt at a solution

    Loop equations
    10 - 500,000* i1 - 500,000 * i2 - 500,000 * i3 = 0
    10 - 500,000* i1 - q/ 5 * 10^-6 - 500,000 * i3 = 0
    Junction equations
    i1 = i2 + dq/dt
    i1 = i3

    Solving for i1 in the second loop equation, you get

    i1 = 1 * 10^-5 - 0.2q

    Substituting second junction and above i1 into the first loop equation, you get

    i2 = 2 * 10^-5 - 2*10^-5 + .4q = .4q

    Substituting into the first junction,

    1*10^-5 - .6q - dq/dt = 0

    Using the equation i = E/R * exp(-t/RC) to solve for time, I used

    (2/3)(8) uA = (1*10^-5 / 1)(exp(-t/1.6667))

    However, the correct answer is using 4 uA with the above equation. My main question is why do we use 4 uA as the current going through the capacitor instead of 16/3 uA. The overall resistance of the path through the switch would be 1,000,000Ω while the resistance of the capacitor path is 500,000Ω. Doesn't this mean that 2/3 of the current would flow through the capacitor, so 2/3 of 8 uA?


    (PS. I realized afterwards that you don't need to use the current equation, but can use the given charge equation once you figure out the charge on the capacitors when 8 uA goes through the capacitors, which is 10 uC. However, I still think my way should work, or want to understand why it doesn't.)
  2. jcsd
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