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Wheatstone bridge with capacitances

  1. Aug 1, 2016 #1
    1. The problem statement, all variables and given/known data
    https://www.physicsforums.com/attachments/circuit-gif.64003/
    Basically I am asked to find the equivalent capacitor.

    2. Relevant equations
    C = q/v
    Conservation of energy.

    3. The attempt at a solution
    I have read a lot about this topic and how if they were for example resistor and R4 * R3 = R1 * R5, the bridge doesn't go through it a current and I derived that. But I couldn't do the same with capacitors, however I found that this also happens with capacitors. In the question, The equation above applies ( replace R with C)

    So here is my question, Is there is a simple highschool derivation for this? and what if the wheatstone wasnt balanced how would I calculate it?
     
  2. jcsd
  3. Aug 1, 2016 #2

    cnh1995

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    Yes. Are you familiar with KVL and KCL?
    Using star-delta transformation technique which may not be there in your highschool syllabus.
     
  4. Aug 1, 2016 #3
    for the first one Yes,
    Second Nope
     
  5. Aug 1, 2016 #4

    SammyS

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    You answered this as if you were OP.

    Is that the case ?
     
  6. Aug 1, 2016 #5
    Me and him are friends irl. We are solving this together. So you could consider us both the OP
     
  7. Aug 2, 2016 #6

    SammyS

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    If you know how to solve the similar circuit but with resistors, rewrite that relevant equation as follows.

    ##\displaystyle \ V=q\cdot \frac 1 C \ ##

    Compare that with Ohm's Law.
     
  8. Aug 2, 2016 #7

    gneill

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    To expand on what SammyS stated above: When you have a network where no current flows at steady state as in this example where the network consists of capacitors only, then if a voltage source is connected across the network some finite amount of charge will flow into the network over some limited time and then all current flow ceases when equilibrium is achieved. This leaves charges on each of the capacitors, hence potential differences.

    For the short time that current does flow in the circuit all of the usual circuit laws are obeyed (KVL, KCL). More importantly, the steady state condition when current ceases to flow will also satisfy KVL. So the potentials left on the capacitors will satisfy KVL.

    What this means is that you can write KVL loop equations for the "static" condition where charge replaces current and ##V = q/C## is used in place of Ohm's law. Rather than net currents in branches you have net charge moved in those branches.

    You could even use mesh analysis to write loop equations, where you assume mesh charges rather than mesh currents:

    upload_2016-8-2_6-38-42.png
    In the above diagram the assumed mesh charges are q1 through q3, representing the total charge moved by analogous mesh currents during the short time that current flows when reaching steady state. Mesh equations can be written using ##q##'s in place of ##I##'s and ##1/C##'s in place of ##R##'s.
     
  9. Aug 2, 2016 #8
    Thanks a lot for the answer, I am going to try it and see what I will come up with.

    Thanks again.
     
  10. Aug 2, 2016 #9
    Got the right answer, Thanks everyone!
     
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