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2 Capacitors in Parallel

  1. Mar 2, 2009 #1
    1. The problem statement, all variables and given/known data
    Two identical capacitors are connected in parallel to an ac generator that has a frequency of 660 Hz and produces a voltage of 27 V. The current in the circuit is 0.14 A. What is the capacitance of each capacitor? Note: The ac current and voltage are rms values and power is an average value unless indicated otherwise.


    2. Relevant equations
    Z = x of c = V/I
    x of c = 1/(2pi*fc)


    3. The attempt at a solution

    x of c = 27V /.14A = 192.86 ohms

    192.86 ohms = 1 / (2pi*fc) = 1 / (6.28*(660)c) = 1/(4144.8 x c)
    c=1.25 uF

    1.25uF equals the capacitance of the two identical capacitors connected in parallel.
    The capacitance of each capacitor equals 1.2uF/2 = .625 uF

    This answer is incorrect, but I have absolutely no idea what I'm doing wrong.. Thanks for any help!!!
     
  2. jcsd
  3. Mar 2, 2009 #2

    Delphi51

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    Looks okay to me!
     
  4. Mar 2, 2009 #3
    They're asking me to submit my answer in units of F, and I think I may just have my decimal point in the wrong place... (I submitted .625 F). And now I'm confusing myself a lot as to where the decimal place really should be. Should my final answer actually be 6.25x10^-13 F?

    6.25x10^-7 uF x (1x10^-6 F / 1 uF) = 6.25x10^-13 F ???
     
  5. Mar 2, 2009 #4

    Delphi51

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    micro means 10^-6
    .625 uF = .625 x 10^-6 F = 6.25 x 10^-7 F
     
  6. Mar 2, 2009 #5
    Whew, dumb mistake on my part. Thanks!
     
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