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Power in AC Circuit

  1. Nov 15, 2012 #1
    In an AC circuit, the average power dissipated is given by [itex]P=VIcos(\phi)[/itex]. Does that mean that in a highly inductive, or highly capacitave, circuit where [itex]\phi[/itex] approaches [itex]\pm \pi/2 [/itex], the power can be made arbitrarily small? Even if a resistor were present? Does that mean it wouldn't heat up at all?
  2. jcsd
  3. Nov 15, 2012 #2


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    Staff: Mentor

    As a fraction of apparent power, real power can be small, but adding a capacitor doesn't reduce the actual value of the real power.
  4. Nov 15, 2012 #3


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    Staff: Mentor

    In a highly inductive element, there is only a very small component of current that is in phase with the voltage (leaving most to be in phase quadrature). But if resistance is added, then ɸ will no longer be close to Pi/2.

    If a current I (RMS) passes through a resistance R, the power loss is I²R. ALWAYS.
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