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  1. #1
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    Capacitance question

    Hi. I have a question on Question 3b. My understanding is that the capacitors are in series, but the solution from the textbook seems to indicate a parallel arrangement assumption. See the attached images. I am puzzled!

    Help would be appreciated!

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    Re: Capacitance question

    Circuit is anti-parallel.



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    Re: Capacitance question

    Hi,

    two capacitors connected together.
    Are they now in parallel or in series? Its a chicken-and-egg problem.

    But one knows:
    * both voltages before connection
    * both charges before connection
    * both capacitances befroe connection
    * and how they are connected.

    --> No need to know whether they are in series or in parallel.

    Klaus

    - - - Updated - - -

    added:
    Circuit is anti-parallel.
    Why not "series", or "anti-series"? - I really don´t know.

    If we consider both capacitors as "power supply" or as "battery" and one connects "+" of the one with "-" of the other then we call it (straight) series connection.
    But what if we connect the still open "-" with the still open "+" will it now become "anti" and "parallel"?
    Couldn´t we call it "short circuited series connection"?

    But as said above: Maybe one should not care about the naming. The definition of the task is clear.

    Klaus
    Please don´t contact me via PM, because there is no time to respond to them. No friend requests. Thank you.



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    Re: Capacitance question

    You can of course interpret the circuit as series connection, if you like.

    I think that parallel (in the other thread started by the author) respectively anti-parallel (this thread) is closely related to the achieved charge balancing, which is sum in parallel and difference in anti-parallel case.



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    Re: Capacitance question

    Please draw the diagram yourself and you will see whether it is series or parallel.

    Consider the negative voltage as zero (it does not matter because what matters is the potential difference).

    Step 1: 0 V end of Cap A is connected to the 40V end of Cap B. Now the 40V end of Cap A will appear to be 80V with respect to the 0V end of Cap B. Just like connecting two cells in series.

    Step 2: The free end of Cap A (now at 80V) is next connected to the free end of Cap B (now at 0V). Because they are at different potentials, some current (or charge in fact) will flow. This is like shorting a series connection of two cells.

    Step 3: You now need to determine the new voltage across the two connections (the capacitors now look to be in parallel)- the two ends.

    The text book analysis is clear and correct.



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    Re: Capacitance question

    It is true that an ambiguity can occur if only two capacitors are involved. Your two capacitor circuit can be considered as connected either in series or parallel. If three or more capacitors are connected in series, then no amount of topological manipulation can make them into a parallel circuit. The following method works for any number of capacitors connected in series, with the higher voltage connected to the next capacitor's lower voltage. This is what your problem does. What is happening here? Well, you know that the sum of the voltages of the capacitor string before the head and tail are connected together is the sum of the voltages of each capacitor. After the final connection, the sum of the capacitor voltages in the string has to be zero. That means that some the the capacitor voltages in the string have the opposite polarity of their neighbors. You can conclude that the charge imbalance of each capacitor has changed. Some capacitors have lost a little their former charge imbalance, and some have lost so much imbalance that they passed the neutral point and are imbalanced the opposite way, so as to show a opposite voltage with respect to their former voltages. The following is the equation to calculate the charge imbalance change.

    (Q1-x)/C1 +(Q2-x)/C1...(Qn-x)/Cn = 0, where x is the charge imbalance change. x = 40.6 uC for your problem.

    In your problem the charge imbalance of the 0.4 uF capacitor changes from 24 uC to -16.6 uC. That corresponds to a -41.54 voltage. The 2.2 uF capacitor changes from 132 uC to 91.4 uC. That corresponds to a +41.54 voltage. The voltage totals are zero. Any questions?

    Ratch
    Hopelessly Pedantic



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