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Capacitance in a Closed Loop

  1. Feb 23, 2017 #1
    1. The problem statement, all variables and given/known data
    upload_2017-2-23_13-10-42.png

    2. Relevant equations
    C=Δq/V


    3. The attempt at a solution
    So part a:
    The capacitance of 3 should be the difference in charge of b and a divided by the voltage of the circuit?
    C3=(14μC-6μC)/12V
    =6.67*10-7 F
    This was wrong...
     
  2. jcsd
  3. Feb 23, 2017 #2

    cnh1995

    User Avatar
    Homework Helper

    No. You need to divide it by the voltage across C3, which is not same as total voltage in the circuit. Use voltage divider equations for capacitive circuits.
     
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