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LRC circuit: netted time-averaged square of the magnetic field versus net charge flow

  1. Feb 16, 2008 #1
    Let's say for the:

    First 10 milliseconds: Current is 1 milliamp. The squared magnetic field resulting we will call "default".

    1 millisecond following, last millisecond: Current is -10 milliamps. The squared magnetic field is -100 * "default".

    The time average square of the magnetic field corresponds to: 10 * 1 - 1 * 100 = -90

    One could then imagine the ability for an electromagnetic field made this way to do net work with zero net charge flow, provided that the time constant of the wire is larger than 10 milliseconds. Capacitors could easily increase the time constant of the circuit. The capacitors could be lined up in circuit with a low voltage battery. The circuit could intermittently connect and disconnect a higher voltage, higher current battery with the opposite polarity for a fraction of the time. If timed precisely, the charges would travel back in forth through the wire in a linear way while producing time-average squared magnetic field that is positive.

    Something is wrong.
     
  2. jcsd
  3. Feb 16, 2008 #2
    One thing that is wrong is that the square of -10 is +100, not -100.

    The rest of your post is difficult to follow. I can't tell what you're trying to say.

    Does this have something to do with an earlier post where you were putting the primary and secondary of a transformer in series with a motor and commutator?

    It would help if you would provide a schematic (not a picture, but a regular schematic) and tell us what you're trying to do.
     
  4. Feb 16, 2008 #3
    Right. But this misses an important part of my point of flipping the sign. Let's go with the positive sign and see what happens:

    10 * 1 + 1 * 100 = 110

    To take the average we get:

    110/11 = 10

    So we get a net squared magnetic field without a net flow of charge.

    No

    Right now I'm not trying to do anything. I just need proof that a system that has 0 net charge flow can't do net work.
     
  5. Feb 17, 2008 #4
    You won't find any such proof because it isn't true. In fact, it would appear that you have just proved that it isn't true.

    If it were true, then AC powered induction motors wouldn't work.
     
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