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Circuit with capacitors in series and parallel

  1. Jul 5, 2010 #1
    1. The problem statement, all variables and given/known data
    Problem taken from : Fundamentals of Physics , Volume 2 ( 7th edition) Halliday , Resnick , Walker

    In the figure, a 20.0V battery is connected across capacitors with capacitances C1=C6=3.00mF and
    C3=C5=2.00C2=2.00C4=4.00mF

    What is the equivalent capacitance Ceq of the capacitors?


    2. Relevant equations

    Capacitors in series:
    Q = Q1 = Q2 = ...
    V = v1 + v2 + ...
    1/Ceq = 1/C1 + 1/C2 + ...

    Capacitors in parallel:
    Q = Q1 + Q2 + ...
    V = V1 = V2 = ...
    Ceq = C1 + C2 + ...

    3. The attempt at a solution

    Given data : C1=C6=3.00mF
    C3=C5=4.00mF
    C2=C4=2.00mF

    I tried combining C3 and C5 since they were connected in series:

    So 1/C35= 1/4+1/4= C35=2mF

    Then I tried combining C35 and C4 as they were connected in parallel :
    C345 = 2+2=4mF

    Then I'm stuck on how to solve the network of C345,C1,C6 and C2.

    Did I do any of the above steps wrong and can anyone give me some hints on how to continue ?? Thanks !!


    PS: I am sorry that the image in the pdf file may be quite small. Please magnify it to see the numbers .Thanks again !
     

    Attached Files:

  2. jcsd
  3. Jul 5, 2010 #2
    You're doing great. I really don't know the rules as to only help you, or give the complete solution, so I'll just go for the hint first:
    You say C35 and C4 are connected parallel. That's true. But are there any other ones that are connected parallel?

    What I usually do in problems with circuits that are hard to see is to start drawing the hole thing again. Start from the battery, and draw the circuit again, making sure to connect points that are equipotetial(sorry i don't know the english word for that), and have them meet in one and only one point.
    I basically got (counting C35 as one capacitor) a series of two packs of capacitors, one pack having 3 connected parallel, the other only 2. That's what you should get as well.
     
  4. Jul 5, 2010 #3

    Filip Larsen

    User Avatar
    Gold Member

    Welcome to PF!

    You are probably letting the layout of the drawing confuse you. Try redraw the diagram such that common wires between three elements or more are replaced by a node (draw a dot on the diagram) and then a wire from the node to each element. For instance, the negative of the battery (the top wire) is connected to C4, C5 and C2, so replace that wire by with a node(say, draw a dot over C4) and the connect the node to the anode, C4, C5 and C2.

    Once you have done so for the whole diagram it should be easier to spot which element is in parallel and which is in series. If you can trace a path from one node to another without meeting any other nodes on the way, the elements you passed on the path is in series which you can reduce to one equivalent element. If you from one node to another have multiple paths (each only one element) those elements are in parallel and you can replace the paths with a single path with the equivalent element.
     
  5. Jul 5, 2010 #4
    I tried redrawing and saw that C35, C4 and C2 are connected parallel ; C6 and C1 are connected parallel .

    Then I combined them and C2345= 6mF C16= 6mF

    So the answer is 1/Ceq=1/6+1/6

    Ceq=3mF?
     
  6. Jul 5, 2010 #5

    Filip Larsen

    User Avatar
    Gold Member

    Looks good.
     
  7. Jul 5, 2010 #6
    Thanks for all the help !!
     
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