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Exact Energy usage in an AC-R circuit

  1. Jun 8, 2009 #1
    Consider a circuit with an AC source (time varying emf) and a resistor. The voltage drop through the resistor is of course equal to the emf.

    Consider the region which bounds the resistor.

    The work done on a charge as it moves from one end of the resistor to the other (and in turn the change in kinetic energy) is equal to the work done by the electric field (at the rate VI) - the work done by the resistor (dissipated in the form of heat). Since the kinetic energy of the charge is changing (this is an AC circuit and so its velocity must change) it cannot be true that the energy dissipated by heat is = to the energy put in by the electric field, so why would the energy dissipated by heat by VI (as most physics textbooks say).

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    And if it is not VI then what is it? I presume a more complete treatment would involve looking at the change in the potential energy of the E-field, but I cannot figure out the best way to do this since the E field is not completely contained to within the wire.
     
  2. jcsd
  3. Jun 9, 2009 #2

    diazona

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    Why couldn't that be true?
     
  4. Jun 9, 2009 #3
    Well if the energy put in by the electric field is all lost to heat then no work is done on the charges and so there kinetic energy must remain constant.
     
  5. Jun 9, 2009 #4
    anyone??

    the only progress i have made is to assume that kinetic energy is equal to k*I^2 where k is some constant so that d Ke /dt = k*I*I' and from there I can derive the amount of energy lost to heat. Do you think this is the right approach?
     
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