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Energy stored in Capacitor Network

  1. Oct 1, 2016 #1
    1. The problem statement, all variables and given/known data
    A potential difference Vab = 46.0 V is applied across the capacitor network of the following figure.
    If C1=C2=4.00μF and C4=8.00μF, what must the capacitance C3 be if the network is to store 2.80×10−3 J of electrical energy?

    2. Relevant equations
    (1/C) = (1/c1) + (1/c2) + ... capacitors in series
    C = C1 + C2 + ... capacitors in parallel

    3. The attempt at a solution
    U = (1/2)CV^2
    2.80×10−3 = (1/2)C(46^2)
    C = 2.6 μF

    Capacitance of C1 and C2 = 1/(1/4 + 1/4) = 2 μF
    Total capacitance:
    1/2.6 = 1/(2+C3) + 1/8
    C3 = 1.85 μF

    I can't figure out why this answer is incorrect. Help much appreciated!
  2. jcsd
  3. Oct 1, 2016 #2


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    Science Advisor

    I think your method is correct. Perhaps it's just round off error.? If I keep more digits, I get C = 2.6465 μF, which gives C3 = 1.955 μF. Is that the problem?
  4. Oct 2, 2016 #3
    That was it! Thank you so much, I didn't even consider that!
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