Standing-Wave Current vs Traveling-Wave Current
On Dec 24, 9:39*am, "Dave" wrote:
"Keith Dysart" wrote in message
...
On Dec 23, 11:33 am, Roger wrote:
You are basing this conclusion on the observation that Power = V*I, and
because we can not detect V or I at some points in the standing wave,
then V*I is zero at these points. Correct math, but wrong conclusion.
Are you really saying that if I measure the instantaneous
voltage and the instantaneous current then I can NOT multiply
them together to obtain the instantaneous power?
It certainly works some of the time.
If I can not do it all the time, when can I do it?
you can do it when it makes physical sense. *
Can you kindly articulate the rules you use to know
when it is appropriate to use P = V * I?
At a minimum, the rules should cover DC circuits,
AC circuits, and transmission lines with various
excitations and terminations. Be sure that the
rules cover 60 Hz circuits and transmission lines
since this is a common application of P = V * I.
it does not make sense in
standing waves for all the obvious reasons that i have pointed out. *it does
make sense in the individual traveling waves. *just accept what your little
swr meter tells you, it shows the forward power and reflected power, that is
all you need and the only powers that make sense.
And for a challenging use case, please consider two
circuits connected together. The circuits are in black
boxes so you do not know their details, but the voltage
on the connection between the circuits is measured as
10 V RMS at 4 MHz. The current is measured as 0.
How much energy is being transferred between the
circuits?
1) P = VI, so 0.
2) P = VI some of the time, so there is insufficient
detail to answer the question.
Do you choose 1), 2) or perhaps some third answer?
...Keith
PS. The connection between the circuits is very small
so that it is possible that it is part of a
transmission line that continues into each box,
but you can not be sure.
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