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Capacitor circuit

  1. Apr 29, 2017 #1
    1. The problem statement, all variables and given/known data
    c143bde5-6559-4291-ae49-b2de8e4feb5e.png
    2. Relevant equations


    No equations needed
    3. The attempt at a solution
    I know that capacitors are used to store charge but what exactly is the 40 micro-farad capacitor doing? No emf/voltage is shown so is the 40 micro-farad capacitor acting like the emf/voltage? or is it doing something else? Just trying to better understand this circuit
     
    Last edited: Apr 29, 2017
  2. jcsd
  3. Apr 29, 2017 #2

    mfb

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    The image doesn't work, I guess you have to be registered at that website to see it.
     
  4. Apr 29, 2017 #3
    fixed
     
  5. Apr 29, 2017 #4

    phinds

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    It still doesn't "work" in the sense that it makes no sense. You show a voltage source labeled as a 40uf cap.
     
  6. Apr 29, 2017 #5
    Thats how the circuit was given to me in the question lol
     
  7. Apr 29, 2017 #6

    gneill

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    No doubt the author just re-purposed an existing circuit drawing, thinking it "good enough".

    The intention is that the capacitor has some initial charge. If the problem doesn't specify an initial value for that charge or the voltage on the capacitor you'll have to assign it a variable name and solve the problem symbolically.
     
  8. Apr 29, 2017 #7
    Yea this was the initial question-
    In the circuit shown in the figure, all the capacitors are air-filled. With the switch S open, the 40.0-µF capacitor has an initial charge of 5.00 µC while the other three capacitors are uncharged. The switch is then closed and left closed for a long time. Calculate the initial and final values of the total electrical energy stored in these four capacitors.

    I already figured out the question was just trying to understand the circuit more. No current would be flowing from the 40 µF right?
     
  9. Apr 29, 2017 #8

    gneill

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    No current would flow until the switch closed. Then current would flow very briefly until the new steady-state is reached.
     
  10. Apr 29, 2017 #9
    Ok so where would the current flow from? since i guess no voltage or emf is present, the current would initially flow from the 40µF?
     
  11. Apr 29, 2017 #10

    gneill

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    Yes, the 40 µF capacitor has an initial charge so it has an initial emf associated with it.
     
  12. Apr 29, 2017 #11
    Ok thanks a lot.

    Dont want to make another thread so i thought id ask this question here to see if im right

    An uncharged 30.0-µF capacitor is connected in series with a 25.0-Ω resistor, a DC battery, and an open switch. The battery has an internal resistance of 10.0 Ω and the open-circuit voltage across its terminals is 50.0 V. The leads have no appreciable resistance. At time t = 0, the switch is suddenly closed.
    (a) What is the maximum current through the 25.0-Ω resistor and when does it occur (immediately after closing the switch or after the switch has been closed for a long time)?

    Already know the answer but trying to figure out when it occurs. It would occur immediately after closing the switch right? I figure that if the switch is left closed after a long time the capacitors will eventually be fully charged meaning there will be no more current running through the circuit? So wouldnt the maximum current occur one the switch is closed? Im trying to find a more detailed explanation
     
  13. Apr 29, 2017 #12

    gneill

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    Your thinking is correct. Just remember that at steady state capacitor current goes to zero; the circuit reaches an equilibrium where there are no longer potential differences to drive current into or out of the capacitors. In a series circuit that means all current ceases.
     
  14. Apr 29, 2017 #13
    Ok thanks a lot
     
  15. May 8, 2017 #14
    Didn't want to create another thread but for my first question with the picture of the capacitor circuit I calculated the initial energy with q^2/2C and got 0.313micro joules then calculated the final with C=60 then used the same equation for the final energy q^2/2C=0.208 but I want to know how the energy dropped?
     
  16. May 8, 2017 #15

    mfb

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    A real circuit will have some resistances in it, dissipating the difference between these energies. And if you make the whole circuit out of superconductors to avoid that, it still emits electromagnetic radiation (it acts like a resonant circuit then).
     
  17. May 8, 2017 #16

    gneill

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    In a circuit made of real components the energy would be largely dissipated by the small resistance in the wires, the rest lost to a radiating electromagnetic field (which we're not accounting for when we do circuit problems in this way). The bottom line is that when charges are moved around from capacitor to capacitor some potential energy gets lost. After all, the thing that drives the movement of charge is that it is seeking a lower energy state.

    edit: Ah! mfb beat me to the punch!
     
  18. May 8, 2017 #17
    So in this circuit potential energy is dropped because the charges are moving from capacitor to capacitor. And ultimstely the charges want to go to a lower energy state. Is this some how related to like the potential drop from I guess in this case high energy to low energy?
     
  19. May 8, 2017 #18

    gneill

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    Yes. A mechanical analogy might be water flowing from one reservoir to to another when a path is opened between them. As the water level lowers in one the other rises, but ultimately the center of gravity of the system as a whole has been lowered somewhat so there has been a net loss in potential energy.
     
  20. May 9, 2017 #19

    epenguin

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    Yes the first question sounds like a wrong question. There is no way you could have the 40 μFcapacitor charged and others not charged in a resistanceless circuit. Even if you have resistances and this is an initial situation it seems rather artificial when you think about how you could get that, but then there would be current flows, redistribution of charge, without closing the switch. I wouldn't worry about it until you know what the right question was.
     
  21. May 9, 2017 #20

    gneill

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    With the switch open, there's no reason why such a circuit could not be assembled using one initially charged capacitor and three uncharged ones. No current would flow until the switch is closed. The question is fine as posed.
     
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