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Homework Help: Capacitor networks

  1. Aug 28, 2008 #1
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Aug 28, 2008 #2


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    Your diagram is rather unclear. I don't see any voltmeter. Where is it supposed to be?
  4. Aug 28, 2008 #3


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    The statement of the problem is incorrect and I will assume it is sloppy wording, because after equilibrium is reached, there is no current flowing anywhere in the circuit with no restive path between the terminals of the battery.

    But taking the awkward wording as meaning they want to know the total flow of charge as a result of shorting the switch, I have to ask where it is you got the figures for the charges on the capacitors before or after the switch is shorted?

    Please show your work as to how you arrived at those figures at the points x and y.
  5. Aug 29, 2008 #4
    Actually there are the figures deduced from my friend's tutor.
    But I know that first we can treat the network as two parallel networks, and the two sides of the capacitors can simply computed by using the formula Q=VC.

    I just doubt how <+60>and <-60> can be found out. And how to deduce in what way the charge will flow.
  6. Aug 29, 2008 #5


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    Why don't you start out then by calculating the equivalent capacitance of each leg. Then you will know the first part of what you need - the charge on each capacitor - and hence the voltage at each node.

    Then you proceed to calculate the equivalent capacitance of the new network after the short is affected. Armed with that you can calculate the effective new charges and then you can describe the difference.
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