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Capacitor network finding potential

  1. Feb 8, 2012 #1
    For the capacitor network shown in the figure, the potential difference across ab is 12V. Find (a) the total energy stored in this network and (b) the energy stored in the 4.8 uF capacitor.

    figure: http://i.imgur.com/Wxbuu.png

    ATTEMPT:

    PART A:
    i first did the leftmost 2 capacitors and got C=3.1E-6.
    then i did the top 2 capacitors and got C=1.4E-6.
    then i used those top 2 as a parallel capacitor with the bottom one getting: 4.9E-6.
    i added them all together to get C=9.39E-6.
    using the forumla u=.5CV^2 i got U = 6.76E-4 J. did i do this right?
     
  2. jcsd
  3. Feb 8, 2012 #2

    gneill

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    Staff: Mentor

    Your value for the top two capacitors looks a bit low. Check your calculation.

    Otherwise, your method looks okay for finding the energy.
     
  4. Feb 8, 2012 #3
    i re-did it and got 4E-6. is this right?
     
  5. Feb 8, 2012 #4
    i then got a total capatance of 1.21E-5 F and a total energy of 0.0027 J
     
  6. Feb 8, 2012 #5

    gneill

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    That's better. 4.06E-6 would be a bit better... try to keep enough significant figures in intermediate values so that roundoff and truncation errors don't sneak into your results.

    The total capacitance looks a bit large. Can you show your steps?
     
  7. Feb 8, 2012 #6
    my bad. i dont know what i did wrong, but i redid it and got u=8.64E-4.
     
  8. Feb 8, 2012 #7

    gneill

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    That doesn't match what I'm seeing. Again, why not show your steps and the intermediate values?
     
  9. Feb 8, 2012 #8
    ok. adding all the Cs gives me 1.2E-5. i plugged this into the formula U=.5CV^2. for v i used 12 and solved for u
     
  10. Feb 8, 2012 #9

    gneill

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    That's not the total for C that I get. Show the steps that arrive at the total.
     
  11. Feb 8, 2012 #10
    3.1e-6 + 4e-6 + 4.9e-6
     
  12. Feb 8, 2012 #11

    gneill

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    You haven't specified what capacitor combinations arrive at those numbers, or why you think they should be added in that way. I recognize the first two from your earlier posts (again, you should keep more significant figures). But I don't see any justification for the last one or why they are being added together.
     
  13. Feb 8, 2012 #12
    i did this in three steps. i found C from the first two as a series. then i found c from the top two as a series. then i used my answer from the top two and the lower capacitator as a parallel network. so in the last calculation i did 4E-6. AH! there is my mistake. before i had my old answer in there. NOW it should be 4E-6 + 3.5E-6 = 7.6E-6. for the final answer i get 0.001 J
     
  14. Feb 8, 2012 #13

    gneill

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    Things are starting to get cleared up. What value did you get for the final equivalent capacitance, and what values did you combine (and how) to get it?
     
  15. Feb 8, 2012 #14
    final capacitance = 1.48E-5 F. i combined 3.1E-6, 4.1E-6 and 7.6E-6
     
  16. Feb 8, 2012 #15

    gneill

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    Okay, HOW were they combined? What is the configuration of the capacitors represented by those values? Didn't the 4.1μF value already get combined with another capacitance to arrive at the 7.6μF value?
     
  17. Feb 8, 2012 #16
    3.1E-6 is from series. so you add them by 1/C and the answer is 1/c. so for the first one i did 1/8.6E-6 + 1/4.8E-6 = 1/C where C is 3.1E-6

    4.1E-6 is also from series. i did 1/6.2E-6 + 1/11.8E-6 = 1/C where C is 4.1E-6.

    7.6E-6 is parallel so you just add them. i added 4.1E-6 to 3.5E-6
     
  18. Feb 8, 2012 #17

    gneill

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    Okay, and then what? What's left to be combined?
     
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