On Jan 12, 2:08*am, Roger Sparks wrote:
On Fri, 11 Jan 2008 18:49:20 -0800 (PST)

Keith Dysart wrote:
On Jan 11, 7:30*pm, Roger Sparks wrote:
On Fri, 11 Jan 2008 13:59:47 -0800 (PST)

Keith Dysart wrote:
On Jan 11, 11:15*am, Roger Sparks wrote:
On Thu, 10 Jan 2008 18:52:28 -0800 (PST)Keith Dysart wrote:
P(t) = V(t) * I(t)
by substitution from V = I * R
P(t) = V(t) * V(t) / R
and
P(t) = I(t) * I(t) * R

Correct.

But recall that power is the flow of energy per unit time.

The energy present in any section of line is the energy
present in the electric field of the capacitance plus
the energy present in the magnetic field of the inductance.

When the capacitance is in phase with the magnetic field, both are measured to peak at the same time. *This "resistive" condition is found in a transmission line when there are no standing waves. *Under resistive conditions, the electric field and inductive field merge into one electromagnetic field that carries only one energy. *The identical charges carry both electric and magnetic fields, with the fields in a ratio defined as impedance. Each field is a separate way of looking at the same energy.

I am not quite sure what you are attempting to say here and
whether you are disagreeing with my assertion that the total
energy in a section of line is equal to the sum of the energy
present in the capacitance and the energy present in the
inductance.

So from your expressions above
U = (CV^2)/2 + (LI^2)/2

You may have been reading about traveling waves. *They offer an explaination for how magnetic and electric fields may get out of phase. *When the two fields are out of phase, the U = (CV^2)/2 + (LI^2)/2 calculation will ALWAYS sum to more energy than can be found in the resistive wave. *This occurs because more energy actually is resident on the line with reflections, due to waves traveling in two directions. *The reflection effectively doubles the length of the transmission line available for energy storage. *

This seems like a complicated way to think about things. So
complicated that I will
not comment on correctness.

If P = (v^2)/r = (i^2)r is correct, what is the precondition that MUST be true?
Answer: The i and v measurements must be across a pure resistance, without reactance.

This is why it is often better just to use the original
equation:
P(t) = V(t) * I(t)
for it is always true and requires no preconditions.

I learned long ago that it DOES have a precondition. *The power measured will only be resistive if measured across a pure resistance. *If the measurement is across resistance plus reactance, then reactive power is also measured so P(t) = V(t) * I(t) becomes the sum of an unknown mix of two or more waves. * *

This too seems like a complicated way to think about things.

P(t) = V(t) * I(t) is simple, and always holds. One does not need to
consider
the wave content in any way. Or impedances. Measure the instaneous
voltage and
the instanteous current; multiplying will always yield the
instantaneous power.
Integrating instantaneous power will always yield the net energy
transfer over
the interval of integration.

...Keith