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#11
August 20th 03, 06:38 PM
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W5DXP wrote:
wrote: In reality there is not zero volts in the incident wave or in the reflected wave. There`s full voltage coming and going. The volts just happen to be out-of-phase at this point. Yes, indeed. But there is no power. Power is the same as irradiance in optics. When total V=0, it is simply the result of destructive interference. Perhaps this quote from _Optics_, by Hecht, will enlighten you. "The principle of conservation of energy makes it clear that if there is constructive interference at one point, the "extra" energy at that location must have come from elsewhere. There must therefore be destructive interference somewhere else." My knowledge of optics is insufficient to comment on any analogies you choose to draw. Fortunately, a knowledge of optics is unnecessary to understand circuits and transmission lines. The voltage goes to zero because two voltage waves are engaged in destructive interference. The current goes to maximum because two current waves are engaged in constructive interference. The momentum in the voltage waves simply transfers to the current waves and they just keep on rolling along. There is no mechanism of physics existing at that point to change the momentum of the waves. Believing that no energy crosses a superposed V=0 boundary is just a wet dream. This puts you in group a) P(t) is not always equal to V(t) * I(t); or group c) "double think". Care to think about which and comment? The current is at an absolute maximum point so plenty of charge carriers are crossing that boundary. Yes indeed, but current by itself is not energy. Remember P(t) = V(t) * I(t) [unless you choose option a)] Both volts and amps are simultaneously necessary for power. ....Keith |
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