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Series vs Parallel

  1. May 3, 2015 #1
    1. The problem statement, all variables and given/known data
    We connect a capacitor C1 = 8 microfarad to a power supply, charge it to a potential difference V = 120V, and disconnect the power supply (Fig. 24.12).

    http://imgur.com/7VQDEKo [Broken]

    2. Relevant equations


    3. The attempt at a solution
    How are these capacitors in parallel after S is closed?
    I'm confused on how I'm supposed to differentiate conductors in a series and parallel. I thought they are in a series if the electrons can only travel 1 route between the capacitors and in parallel if they can branch off into multiple routes. In this case, when the switch is closed, the electrons can only travel either on the top or bottom path.
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. May 3, 2015 #2
    http://www.allaboutcircuits.com/vol_1/chpt_5/1.html

    For example, on this website, the first picture represents a series. Even though these are not capacitors, R1 and R3 are in a series (despite being parallel to one another, if you removed the wire). It seems like the image I posted mirrors this situation, except that R1 and R3 are on the left and right side, instead of the top and bottom and the wire is bent near the switch.
     
  4. May 3, 2015 #3

    gneill

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    When there are only two two-lead components in a circuit, and both components share two distinct nodes, then you can interpret their connection as being either series or parallel. Indeed, the topology satisfies the basic definition of both. This an be quite handy for the analysis of particular situations (such as capacitor circuits where conservation of charge is a helpful approach).

    Note that the orientation of components on the page, or the bending or slant of wiring is completely irrelevant. What matters is the topology of the circuit: what is connected to what. You are free to bend or twist or stretch the drawing in any fashion you like, provided that all the component connections remain the same. You can reposition any component or component connections so long as it remains connected to the same nodes.
     
  5. May 3, 2015 #4
    I'm having a hard time understanding how it can interpreted as both series and parallel. In a series, the charges on both are the same, while in parallel, the potential difference is the same. According to my book, the charges differ in this situation, which means that they can't be in a series.
     
  6. May 3, 2015 #5

    gneill

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    The charges are both the same if they are charged at the same time with the same current. If they are charged separately when they are not in the same circuit then all bets are off! If they are pre-charged before they are connected the you need to follow the charges...
     
  7. May 3, 2015 #6

    vela

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    When elements are in series, the current through them is the same.

    I think you're thinking of charging uncharged capacitors connected in series. In that case, all of the capacitors end up with the same amount of charge. This is a different situation.
     
  8. May 3, 2015 #7
    That makes sense. I have one more question about solving this problem though. How would I know to treat this problem as a parallel or series? Because I need to calculate the potential difference to solve for Q (Q = CV). When a battery is disconnected, do I assume that the potential difference is the same, then calculate the charge from there? I guess I'm mainly confused because you said that the rules for the distribution of charge between the two conductors differ when they are not charged at the same time. So does the arrangement no longer matter, and I'm supposed to know that the potential difference is the same between capacitors when the source is disconnected?
     
  9. May 3, 2015 #8
    My book doesn't mention current until the next chapter. I just finished reading up on charging uncharged capacitors. I'm not sure what happens when the two capacitors are connected by the wire after the battery is removed. I'm assuming the excess electrons from the charges capacitors flow to the other capacitor that is uncharged. I don't understand what makes the potential difference of the two capacitors the same.
     
  10. May 3, 2015 #9
    Here is the problem and solution:
    http://imgur.com/q6adfLv [Broken]

    They mention that because the capacitors are parallel, they have the same potential difference. I can't seem to figure out how that conclusion can be drawn from the information given and from the information that was provided in this post.
     
    Last edited by a moderator: May 7, 2017
  11. May 3, 2015 #10

    gneill

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    When the battery is removed from the capacitor, the capacitor remains charged to the same potential difference that the battery imposed upon it. This is so because there is no path for current to flow for the charges on the capacitor when the circuit is "broken" from the battery.
    Yup. If no current can flow then the charges stay put.
     
  12. May 3, 2015 #11
    Sorry, I didn't mean right when the battery is disconnected. I meant to say after the battery is disconnected and the wires are connected. At this point the potential difference decreases from 120V of the original capacitor to 80V for both capacitors. Is potential difference always the same on both capacitors when capacitors are connected with a wire without a battery?
     
  13. May 3, 2015 #12

    gneill

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    What drives charges to move? Potential difference. Charges will move until an equilibrium is established.

    Batteries impose a "permanent" potential difference which is what drives current though a circuit. A device like a capacitor can also have a potential difference that is capable of driving a current, but its supply of charge is limited and the potential difference on the capacitor depends upon that charge. As charge leaves the capacitor its potential difference drops. So when you connect a charged capacitor to an uncharged one, the potential difference on the charged capacitor will drive a current which will charge the uncharged capacitor. This will continue until the potential differences on the two capacitors equalizes. When current stops flowing the potential differences on the capacitors must be the same (else current would continue to flow).
     
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