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Solving a first order circuit

  1. Mar 26, 2015 #1
    1. The problem statement, all variables and given/known data
    Selection_026.png

    We are given K1 = 0. So the answer should be in the form of V(x) = K2e^(-t/z)


    2. Relevant equations
    V=IR
    V(x) = K1 + K2e^(-t/z)
    V(infinity) = K1
    V(0+) = K1 + K2
    z = RC

    3. The attempt at a solution

    I draw the circuit for t = 0- and find the voltage across the capacitor. Vc(0-) = 27/12*4 = 9V

    Now I draw the circuit for t = 0+ with the capacitor as a voltage source. I apply KVL to get the following:

    4i1 = 27
    i2 = 0
    -9 + 2i3 + 4i3 = 0

    Vo(0+) = -3V

    We know K1 = 0, so I don't draw the circuit for Vo(infinity)

    V(0+) = -3 = K2

    z = RC.

    Now I get the thevenin resistance across the capacitor; ignoring the short circuit I have: ( 1/(1/4+1/4)+2)=4k ohm

    So z = 4000 * 350 * 10^-6 = 1.4

    But my answer is wrong: V(t) = -3e^(-t/1.4)
     
  2. jcsd
  3. Mar 26, 2015 #2

    SammyS

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    Where does current, i3 flow?


    For t<0, no current flows through the capacitor, so no current flows through the 2kΩ resistor. That also means that for t < 0, you have the wrong voltage across the capacitor .

    Edit: I crossed out some bad info!
     
    Last edited: Mar 26, 2015
  4. Mar 26, 2015 #3

    gneill

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    No current flows through the 2 kΩ resistor once the capacitor is charged. That puts the right end of the capacitor at 0 V (assuming we take the bottom rail as the 0 V reference). The voltage at the left end of the capacitor is determined by the resistor voltage divider and the voltage source. 9 V looks good for that.

    @ x86, I think you're over complicating things by writing multiple loop equations. Once the switch closes the 27 V source and its attached 4 k resistor are effectively isolated by the switch short and become irrelevant to what's happening on the other side of the switch. You can simplify the remaining circuit by combining the two 4k resistors so you end up with a single loop (or a voltage divider if you wish).
     
  5. Mar 26, 2015 #4

    SammyS

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    Oh Snap !

    I misread what OP had!

    Thanks !!
     
  6. Mar 26, 2015 #5
    I believe doing that would give me a different answer (probably a correct one, which I will check at another time).

    I'm a bit confused about your method because what about the 27V power source? Surely all of the current from the 27V power source is going through the short. From left to right, we have three 4K resistors. The middle one is connected to the bottom node by this short. Wouldn't all that current flowing through the short affect the resistor a bit? Or perhaps this does not affect it at all because it doesn't pass through the middle resistor.
     
  7. Mar 26, 2015 #6

    gneill

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    Right.
    Nope. No effect at all (for a theoretically perfect short).
    Right.
     
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