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RC circuit -- finding the charge on each capacitor

  1. Jul 12, 2017 #1
    1. The problem statement, all variables and given/known data
    I have to find the charge ##q## on the capacitors (all have capacity ##C= 4 \mu F##) and ##\mathscr{E} = 12 V##, ##R=50 \Omega##


    2. Relevant equations


    3. The attempt at a solution
    I wrote the nodes current equations
    ##\begin{cases}
    i_{1}+i_{2}=i_{3}\\ i_{3}=i_{4}+i_{5}+i_{6}\\ ...
    \end{cases}## etc
    But when I write the mesh equations I don't know if the current pass or not through the capacitors (because they are all charged). So how can I write the mesh equations?
     

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    Last edited by a moderator: Jul 13, 2017
  2. jcsd
  3. Jul 12, 2017 #2

    Merlin3189

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    Gold Member

    Since you say the capacitors are all charged, I think you are looking for the steady state solution. This means, in a DC circuit all the currents and voltages are constant.
     
  4. Jul 12, 2017 #3

    stevendaryl

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    Staff Emeritus
    Science Advisor

    If the capacitors are fully charged, then the current through the capacitor is zero. So in your drawing, [itex]i_4[/itex] and [itex]i_6[/itex] must be zero (after the capacitors are fully charged).

    The rules for solving these circuit problems are:

    1. The current into a node must be equal to the current coming out of the node.
    2. The sum of the voltage changes around any closed loop must be zero.
    You have to apply #2 to get the capacitor charges.
     
  5. Jul 12, 2017 #4
    So, for example the equation of the loop at the bottom right is ##3 \mathscr{E} - i_{3}R - \frac{q}{C} - i_{1}R=0## ?
    And ##i_{5} ## must not be ##0## as ##i_{4},i_{6}##?
     
  6. Jul 12, 2017 #5

    stevendaryl

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    Yes.

    No. [itex]i_5[/itex] is zero, also.
     
  7. Jul 12, 2017 #6

    gneill

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    Staff: Mentor

    Hi fedecolo,

    A handy trick when analyzing the steady-state of a circuit with L and C components is to suppress those components first. That means removing capacitors (open circuit) and replacing inductors with a wire (short circuit).
    With the resulting simplified circuit you should be able to determine the potentials at locations where the capacitors were removed and the currents where the inductors were located.

    Removing the capacitors from your circuit yields:
    upload_2017-7-12_9-32-47.png

    Pick a suitable reference node and determine the potentials at A, B, D, and E.
     
  8. Jul 13, 2017 #7
    Thank you so much!
    So the equation of the external loop is ##3 \mathscr{E}- \mathscr{E} -4i_{1} R=0##.
    The potential in D is ##18 V##
    in A is ##12 V##
    in B is ##6 V##
    in E is zero.
    To find the charges I only have to substitute the difference of potential in ##q= C \Delta V##
     
  9. Jul 13, 2017 #8

    gneill

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    Staff: Mentor

    That's the idea. Can you show the details of your calculations for the potentials? They don't all look correct to me.
     
  10. Jul 13, 2017 #9
    Ok.
    By the equation of the external loop I find ##i_{1}= \frac{\mathscr{E}}{2R}= 0,12 A##
    in D: ##3 \mathscr{E} - \mathscr{E} + i_{2}R =V_{D}##
    but ##i_{2}=-i_{1}##
    so ##V_{D}= 2 \mathscr{E}-i_{1}R= 2 \cdot 12 V - 0,12 A \cdot 50 \Omega = 18 V##

    in A: ##V_{D}-i_{1}R= 12 V##

    in B: ## V_{A} - i_{1}R = 6 V##

    Are they correct?
     
  11. Jul 13, 2017 #10

    gneill

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    Yes. You didn't mention that you chose node H as your reference node. That being the case, your calculations are good.

    What is the potential at E?
     
  12. Jul 13, 2017 #11
    Oh yes I forgot it, sorry

    The potential in E is zero, isn't it?
     
  13. Jul 13, 2017 #12

    gneill

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    How did you arrive at that conclusion? (hint: No, it isn't zero).
     
  14. Jul 13, 2017 #13
    The currents ##i_{4},i_{5},i_{6}## (in my first draw) are zero, and the current that goes in E must goes out equal. So the current that goes in E (##i_{3}##) must be zero, so the potential in E is zero?
     
  15. Jul 13, 2017 #14

    gneill

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    If the current that goes from E to G is zero, what is potential drop across the resistor in that path?
     
  16. Jul 13, 2017 #15
    So it's ## 4 \mathscr{E}##
     
  17. Jul 13, 2017 #16

    gneill

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    How do you arrive at that conclusion?
     
  18. Jul 13, 2017 #17
    Because of Kirchhoff's law the sum of currents that goes in E must goes out from E. So ##i_{3}=0##. So there is no voltage drop at the resistance.
    So E is at the same potential of G? If it's ok, the potential is ##3 \mathscr{E}##.
    Am I correct now?
     
  19. Jul 13, 2017 #18

    gneill

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    Staff: Mentor

    Yes; correct.
     
  20. Jul 13, 2017 #19
    Thank you!
     
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