Power dissipation in inductors

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SUMMARY

The discussion addresses a common misconception regarding power dissipation in inductors, specifically in the context of alternating current (AC) circuits. The author highlights an error in a textbook that states no power is dissipated in a pure inductor, asserting that during the second quarter of the cycle, the voltage (V) is negative while the current (I) is positive. This contradicts the textbook's claim, as the correct interpretation shows that the integrated area per cycle is zero, confirming that no power is dissipated. The integration of sin(wt) * cos(wt) over 2π radians supports this conclusion.

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  • Familiarity with inductors and their behavior in circuits
  • Knowledge of trigonometric functions and their integration
  • Ability to interpret graphical representations of waveforms
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  • Study the principles of energy storage in inductors
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uzair_ha91
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I have found a possible mistake in my book...here's the text which contains it,


"Referring to figure, it can be seen that no power is disspated in a pure inductor. In the first quarter of cycle, both V and I are positive so the power is positive, which means that energy is supplied to the inductor. In the second quarter, V is positive but I is negative. Now power is negative which implies that the energy is returned by the inductor..."
(The figure looks "something" like this) ::
http://img25.imageshack.us/img25/485/grapha.png

Isn't V negative and I positive in the 2nd quarter according to the figure?
 
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The author is incorrect; I is positive and V negative in second quadrant. If the integrated area per cycle is zero, then there shouldn't be any dissipated power. Integrate sin(wt) * cos(wt) over 2 pi radians, and determine what the area is.
 

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