Where does it go? (mismatched power)
On Jun 16, 11:10*pm, Cecil Moore wrote:
On Jun 16, 5:58*pm, K1TTT wrote:
no sines, no cosines, no square roots, no Pfor/Pref... so much simpler
using voltage or current and simple transformations.
Yes, but determining where the reflected energy goes is the title of
this thread. My method shows where the reflected energy goes and yours
does not! That's the entire point.
--
73, Cecil, w5dxp.com
ah, but it does... it gives the exact same results but without all the
power manipulations. the point is, in sinusoidal steady state you can
calculate the impedance seen by the source and get all those same
results much simpler than trying to track forward and reflected
powers. and if you really want to know the source and load powers
they are easy enough to get as i showed.
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