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Common ground and supply voltages

  1. Jan 16, 2013 #1
    I have already asked this question at electronics.stackexchange but I still don't get it. I thought I understood it in terms of a water analogy but when I went back to electrons it still didn't make sense to me. My first reaction when encountering the idea of common grounding 2 separate power supplies was that it would alter the voltages of the supplies, but that doesn't seem to happen. So I thought of the simplest example I could (using capacitors instead of batteries etc.). Here are two capacitors acting as power supplies in circuits with very high resistance loads, one at twice the voltage of the other and with twice as much charge on its plates :

    http://i.imgur.com/46rPk.png

    if I common ground them :

    http://i.imgur.com/RGWKK.png

    I realise that this does not create an additional closed circuit. However the bottom plates were at different potentials yet they must now be at the same potential. That means the potential of one has increased and the potential of the other has decreased. Meanwhile the potentials of the top plates are unchanged so to my mind the voltages across both capacitors must have changed. Could someone please explain where I am going wrong in terms of potentials and / or electrons, before and after.
     
  2. jcsd
  3. Jan 18, 2013 #2
    Potential with respect to what? Where are you connecting the leads of your voltmeter?

    Again, the top plates are unchanged with respect to what point? Where are your voltmeter leads connected?
     
  4. Jan 24, 2013 #3
    The bottom plates may or may not be at different voltages before connection, the voltage difference is undefined. However that does not matter. In this circuit connecting the bottom ends has no effect on each circuit's energy. The right hand still has the same voltage diff across the capacitor, the left is equally unaffected. There is no closed circuit to carry current or provide a voltage reference, so each cct can be analysed completely independently even after connecting the bottom ends.

    Assume the original voltage across the right capacitor is 2v and the left is 1v. The potentials at the tops on each side are changed as the charge on the capacitor sets the right hand top as two volts above the bottom, and the left hand top as one volt above the bottom, so the difference of the tops is now defined as 1v.
     
    Last edited: Jan 24, 2013
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