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Net capacitance

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  1. Feb 28, 2016 #1
    1. The problem statement, all variables and given/known data
    Find the potential difference between points a & b. Diagrams attached below. I have doubts in part (b) and (c).

    2. Relevant equations
    Q = CV, kirchhoff's laws

    3. The attempt at a solution
    I honestly don't get how the current can flow in opposite directions in part(b). if we choose the conventional notation, then the current flows out of both positive terminals and both intersect?? If not, how can we let the current flow out of the positive terminal in one battery and out of the negative terminal in the other?

    In part (c), I'm confused as to how current could flow through that bridge in the middle. If we look at current coming out of the battery on the left, it will flow up the bridge. If we look at current flowing out of the right battery, it will flow down the bridge. This is contradictory, and if I change the direction of the current, I'm messing up the conventions again.
     

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  2. jcsd
  3. Feb 29, 2016 #2
    It appears that each of the various networks involve capacitors in a steady state -- there would be no current flow. It seems you are being asked to determine the voltage potential that exists between points a & b in each network. You need to apply Kirchhoff's voltage laws.
     
  4. Feb 29, 2016 #3
    so the capacitors are charged already? how can there be potential difference without current?
     
  5. Feb 29, 2016 #4
    It is the electric field in between the capacitor plates that is responsible for a potential difference; the p.d. across a capacitor is not related to current.
     
  6. Feb 29, 2016 #5
    okay but i will have to assume charges to use kirchoff's laws. how can the charges exist if the batteries don't give a current....unless the capacitors are charged already?
    To clarify, the circuit is not physically possible, right? what exactly is the purpose of the extra batteries?
     
  7. Feb 29, 2016 #6

    cnh1995

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    The capacitors are charged by the batteries and hence, no current flows in the circuit in the steady state. This circuit is totally possible. Now, do you know how KVL is used for this type of capacitor circuits?
     
  8. Feb 29, 2016 #7
    yes, i think:
    For (b):

    VA +12 - q/4 -12 = VB
    VA - VB = q/4

    Also VA - VB = -q/2 if I go the other way
    but this gives q = 0...
    The answer is given as difference in potential = -8 volts
     
  9. Feb 29, 2016 #8

    cnh1995

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    I believe you are right. There is no net voltage in the circuit to charge the capacitors, hence, if you connect a voltmeter between a and b, it will read 0V.
     
  10. Feb 29, 2016 #9

    epenguin

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    If it helps you can see or justify this by a way of thinking I recently found useful. Going down, each voltage on the left is double the last. But each capacitance is half the last. So CV is equal for each. And they are all directly connected on the right hand side. So OK it's not V but (V - vb), but they are directly connected so vb is the same for each, so (V - vb) same for each. So the charges must be equal for each. But now for the plates on the right hand side the charge can't be the same for each, because if there is any positive charge there has to be a negative one. So the charges can only be 0.
     
  11. Mar 1, 2016 #10
    but the voltages are 12 volts each, so how can the voltage on the left be double of the last?
     
  12. Mar 1, 2016 #11

    epenguin

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    I am talking about problem b where the voltages are 6, 12, 24; second double of first, third double of second.
     
  13. Mar 1, 2016 #12
    Oh, alright.
    But I don't think I understand. I get that q= CV is the same for each, but if all the charges are zero, then there is no potential change at the capacitors. The answer is given as -10.3 volts
     
  14. Mar 1, 2016 #13

    cnh1995

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    Yes. It is -10.3V and not 0.
     
  15. Mar 1, 2016 #14
    You mean problem d, right? @Penguin

    I believe 0V was your answer to problem b, no? @cnh1995

    Didn't you already know how to solve that question? @Epiclightning

    Trying to get us back on track - I got a little mixed up. I apologise for any confusion I may have caused.
     
  16. Mar 1, 2016 #15

    cnh1995

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    The circular circuit with two
    12V sources...
     
    Last edited: Mar 1, 2016
  17. Mar 1, 2016 #16

    cnh1995

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    True. But I believe that doesn't mean charges should be 0 on each capacitor. According to KCL, sum of charges of all the capacitors should be 0. It turns out that RHS plates of 1uF and 2uF have a -ve charge and RHS plate of 4uF has a +ve charge such that
    total charge on RHS plates=total charge on LHS plates=0.
     
    Last edited: Mar 1, 2016
  18. Mar 1, 2016 #17

    epenguin

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    Problem d

    Yes this will teach me to practice what I was just preaching to someone else here and do the easy check calculations, and also check qualitative arguments by calculations..

    In fact having done it I agree Vb ≈ 10.3 V.

    It is better to adopt the Wild idea :oldwink: giving it as a rational number, Vb = 3×24/7 so then we are able to see that the charges add up exactly, not just approximately and so plausibly exactly, to 0.
    I get

    Q1 = -24×5/7, Q2 = 24/7, Q3 = 24×4/7
     
  19. Mar 1, 2016 #18

    epenguin

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    I think so:confused: , now labelled my last post so. Got confused I think because they were presented out of order. #9 is wrong.
     
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