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Quick question about make before break switches and capacitors

  1. May 19, 2013 #1
    1. The problem statement, all variables and given/known data

    A capacitor is initially connected to a voltage source in series for a long time. At time t0 the make before break switch switches to another series circuit which includes the capacitor. This means that the capacitor will have an initial voltage v(0-) = voltage source, correct? And if the switch was a break before make switch, the initial voltage v(0-) across the capacitor would be zero, correct?

    2. Relevant equations

    3. The attempt at a solution
    Last edited: May 19, 2013
  2. jcsd
  3. May 19, 2013 #2


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    Staff: Mentor

    It would depend upon the details of the circuit and what the switch is accomplishing when it commutes. A diagram would help.

    It might help you to know that if a charged ideal capacitor is removed from a circuit (so that one or both of its leads are no longer connected so that there's no closed circuit for current to flow into or out of the capacitor), then it will retain the charge it had at the moment of disconnection indefinitely.
  4. May 19, 2013 #3

    rude man

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    Homework Helper
    Gold Member

    The answer depends on the circuit the capacitor is switched to. Assuming it's not switched to a second hard voltage source, the capacitor voltage would be the same before and after the switching regardless of whether the switch is mbb or bbm: v(0+) = v(0-). That's because a capacitor can't instantly change its voltage unless switched to another hard voltage.

    If it's switched to a second hard voltage source a mbb switch would cause instant chaos since you'd be connecting two hard voltage sources to each other. If it's bbm then v(0-) would be the first hard voltage and v(0+) = second hard voltage.
  5. May 19, 2013 #4
    Thanks for the replies. I took my electric circuits course a long time ago and I need a refresher. I think I got it now.
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