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RC circuit differential equations

  1. Oct 16, 2013 #1
    1. The problem statement, all variables and given/known data
    0ogitdsnch30040rzerr.png
    Task is to write differential equation for this circuit.

    2. Relevant equations


    3. The attempt at a solution
    I'll try to solve the task, but now I want to know, is it possible to use voltage source instead of current source. For example, I can calculate ekvivalent resistance R_1, R_2 an R_3 parallel. Then current source * ekvivalent (total) resistance = voltage source?
    Like this:
    8e3b70c65r5vaq7fb947.png
    Don't look at numbers, they are random.. I just want to know about idea.
    If it's possible, then I can use V(Thevenin) and R(Thevenin).

    OR:
    I can calculate R(Thevenin) = (R2*R3)/(R2+R3), I guess.
    But I don't know, how to calculate I(Thevenin). Or I have to use Northon method, when we've got capacitor?

    Can you suggest something?
     
  2. jcsd
  3. Oct 16, 2013 #2

    gneill

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    Staff: Mentor

    In your original circuit the current source and first resistor form a Norton equivalent when the switch is closed. No reason why you couldn't turn that into a Thevenin equivalent.

    Be sure to take into account the circuit properties affected by the switch if you need to determine the circuit operation if the switch is re-opened at some future time.
     
  4. Oct 16, 2013 #3
    Do you mean like this:
    svwxr4sksduu1c6ofb1a.png ?

    And then again I need to change circuit to Thevenin equivalent as I did before with RL circuits?
     
    Last edited: Oct 16, 2013
  5. Oct 16, 2013 #4

    gneill

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    Staff: Mentor

    Yes and yes, in your figure you can absorb R2 and R3 into the Thevenin model.
     
  6. Oct 16, 2013 #5
    Then:
    R(Thevenin) = ((R2*R3)/(R2+R3)) + R1
    E(Thevenin) = E = I*R1

    (((R2*R3)/(R2+R3)) + R1)*C*dUc/dt + Uc = E

    Is it correct?
     
  7. Oct 16, 2013 #6

    gneill

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    Yup. Looks good.
     
  8. Oct 16, 2013 #7
    Easier than I thought. But which initial condition I have to use? From picture in my first post?
    But there is Ic(0), but here I need Uc(0). Seems to me, that I don't understand, how to find intial condition.
     
  9. Oct 16, 2013 #8

    gneill

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    You can use the initial conditions for your "new" circuit. If the capacitor begins with no charge then its initial potential is zero. It will "look like" a short at time t = 0, so you can find the initial current, too.
     
  10. Oct 16, 2013 #9
    Ic(0) = 0 and Uc(0)= E ?
    Anyway, I'll try to finish all calculations tomorrow. It's to late today in my country.
     
  11. Oct 16, 2013 #10

    gneill

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    With the Thevenin circuit, when the switch closes there will be a closed loop. The capacitor will look like a short, so just the Thevenin resistance is seen as the load at time t = 0.
     
  12. Oct 17, 2013 #11
    It means that there is no voltage through the capacitor, so Uc(0)=0.
     
  13. Oct 17, 2013 #12

    gneill

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    Staff: Mentor

    "Across". Voltage is "across" a component, current goes "through". But yes, that's the right idea.
     
  14. Oct 18, 2013 #13
    gh6opbjjtb9zd988yy.png

    And it is same as solution for this:
    038c1138tl22lgdareup.png
     
  15. Oct 18, 2013 #14

    gneill

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    Staff: Mentor

    One difference between the two is that the current will be flowing in opposite directions and the potential across the capacitor will have the opposite polarity. To make them identical, change the polarity of either the current or voltage source.
     
  16. Oct 18, 2013 #15
    phvy3951mzd0o1wsshi.png

    How to redraw circuit here, when current source and resistor are in series? I guess, here we can't use Norton because resistor and current source not parallel.
     
  17. Oct 18, 2013 #16

    gneill

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    Big tip: A resistor in series with an ideal current source has no effect on the circuit! The current source will produce its specified current no matter what is in its path. It won't even contribute to Norton/Thevenin resistances for the circuit.
     
  18. Oct 18, 2013 #17
    Tried to solve second one (from post #15) with two methods:
    e5d7hily8tx7235icjr.png
    uubughflwtf7we1iuvmm.png

    And method 2:
    3o2fplrt8cr6p9vdw.png
    Then like before I used Norton/Thevenin equivalent:
    r4rj1wxqr3c80qv6kpm6.png
    E = R3 * J
    s7yy7w5yvkqvheeslg0v.png
    E(Thevenin) = E = R3*J
    R(Thevenin) = R2 + R3

    E = L*dIL + (R2+R3)*IL

    Actually the first circuit (from post #15) is simular, but why there are switch without anything? Is there something tricky in the solution?
     
    Last edited: Oct 18, 2013
  19. Oct 18, 2013 #18

    gneill

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    Staff: Mentor

    The switch changes the circuit by eliminating R3 when it closes (it shorts it out). So there will be a different steady-state situation for each position of the switch.
     
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