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Capacitor discharge

  1. Apr 18, 2004 #1
    I know that in a parallel connection, two capacitors that are charged and connected across a resistor, the switch is open, the capacitors discharge the same as if there was only one by itself? I need a physical argument as why this is? Any suggestions?

    I know that in a series connection, two capacitors that are charged and connected across a resistor, the switch is open, the capacitors discharge slower than if there was only one by itself? I need a physical argument as why this is too? Any suggestions?
     
    Last edited: Apr 18, 2004
  2. jcsd
  3. Apr 19, 2004 #2

    chroot

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    I'm not sure what "discharge the same" means exactly, but it doesn't sound correct.

    A capacitor is basically just two metal plates separated by a small distance. Two capacitors wired in parallel are equivalent to one larger capacitor with the combined plate area of the two smaller capacitors. If you have two 1 F capacitors, wiring them in parallel results in a 2 F capacitor. A 1 F capacitor can discharge 1 ampere for 1 second per volt of charge. A 2 F capacitor doubles this to 2 amperes for 1 second per volt of charge. That doesn't sound like "the same" to me.

    Two capacitors wired in series, however, results in a smaller capacitance. Why? Because by connecting the positive plate of one to the negative plate on the other, you're forcing both plates to have the same potential. (Every point in a conductor has the same potential.) Wiring them in series essentially eliminates half the plate area, and half the capacitance. If you halve the capacitance, you halve the time to discharge.

    - Warren
     
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