Power dissipation in inductors

[uzair_ha91]

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?

 

[Bob S]

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.

 

[astrokreng]

Thank you for bringing this to my attention. It does appear that there may be a mistake in the text. As you correctly pointed out, in the second quarter of the cycle, the voltage (V) is negative and the current (I) is positive, which would result in a positive power dissipation, not a negative one as stated in the text. This could potentially be a typo or an error in the figure. It is important to note that power dissipation in inductors can vary depending on the specific circuit and conditions, so it is always best to double check and verify any information provided in textbooks. I would recommend bringing this to the attention of the author or publisher so that it can be corrected for future editions. Thank you for bringing this to my attention and helping to ensure the accuracy of scientific information.