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Two capacitors

  1. Nov 28, 2003 #1
    62 on GRE. Two capacitors of capacitances 1.0 microfarad and 2.0 microfarads are each charged by being connected across a 5.0-volt battery. They are disconnected from the battery and then connected to each other with resistive wires so that plates of opposite charge are connected together. What will be the magnitude of the final voltage across the 2.0-microfarad capacitor?
    a. 0 V
    b. 0.6 V
    c. 1.7 V
    d. 3.3 V
    e. 5.0 V

    The answer is C. Can someone outline the process to arrive at this?

    By the way, does anyone know a good website that explains the answers to the practice GRE for physics posted on gre.org?
     
  2. jcsd
  3. Nov 28, 2003 #2
    Capacitance adds backwards of how resistance adds; Therefore when the two capacitors are put together in series the capacitance adds as if the two capacitors were resistors in parallel. This yeilds that the series capacitance on either capacitor when the two are connected in series is equal to

    1/C1+1/C2=1/Cs or....
    Cs=C1*C2/(C1+C2)

    plugging in numbers you get that Cs=2/3 microfarads. Comparing this with the initial capacitance it is easy to see that the series capacitance on the 2 uF capacitor is exactly 1/3 of the initial. Following from this if 5V are on the 2uF capacitor initially then 1/3 of 5V will be on the 2uF capacitor when in series with the 1 uF capacitor. 5V/3 =1.6667 or about 1.7V.

    Bored CSMPhysicist.
     
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