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Circuit help!

  1. Jan 30, 2005 #1
    1. For this picture.. Calculate the potential at point D.

    2. I 15-V battery is connected to three capacitors in series. The capacitors have the following capacitances: 4.5microFarads, 12.0microFarads, and 32microFarads. Find the voltage across the 32microFarad capacitor.

    I am so lost. :frown: :confused:
     
  2. jcsd
  3. Jan 31, 2005 #2
    i can't see the circuit
     
  4. Jan 31, 2005 #3

    s_a

    User Avatar

    1. Can't load the circuit diagram.

    2. The absolute value of the charge collected on each capacitor plate will be equal (conservation law of charge), suppose it is Q. The overall capacitance of the three capacitors in series is: (4.5^-1 + 12^-1 + 32^-1)^-1 uF = 2.97 uF (uF = microfarad)

    Q = CV
    = 2.97 * 10^-6 * 15 C
    = 4.45 * 10^-5 C

    So the voltage across the 32 uF capacitor is:
    V = Q/C
    = (4.45 * 10^-5) / (32 * 10^-6) V
    = 1.39 V

    NB: You could have also used voltage division by treating the capacitances as impedances with values -j/(2pi*fC) ohms, since the 15V DC source can be thought of as a sinusoidal voltage source with infinitesimal frequency.
     
    Last edited: Jan 31, 2005
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