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

On Jan 12, 2:08 am, Roger Sparks wrote:
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.

In power engineering, the power can consist of two components, active
power, V*I*cos(A) = watts, and reactive power, V*I*sin(A) = VARS.
These terms are defined in The IEEE Dictionary. P=V*I results in
Volt*Amperes, part active and part reactive if the voltage and current
are not in phase.
--
73, Cecil, w5dxp.com

On Jan 12, 8:10 am, Keith Dysart wrote:
P(t) = V(t) * I(t) is simple, and always holds.

But one must remember the units of that result is Volt*Amps, not
watts.
--
73, Cecil, w5dxp.com

On Sat, 12 Jan 2008 03:49:18 -0800 (PST), Keith Dysart
wrote:

Take two flashlights.
I claim that the energy for the light produced by each
originates in the battery in each flashlight.
Is your claim otherwise?

Connect the negative terminals of the batteries together.
I claim that the energy for the light produced by each
still originates in the battery in each flashlight.
Is your claim otherwise?

Connect the positive terminals of the batteries together
through an ammeter. For the purposes of this experiment,
the batteries have exactly the same voltage so the ammeter
indicates that no current ever flows between the batteries.
I claim that the energy for the light produced by each
still originates in the battery in each flashlight.
Is your claim otherwise?

Hi Keith,

You have admitted that energy dissipated in the loads cannot be
identified to its source and you have used the above devices previous
to that statement. Your admission dismisses their applicability.
Separability is not demonstrated.

73's
Richard Clark, KB7QHC

On Sat, 12 Jan 2008 12:32:35 GMT, "Dave" wrote:

over 700 messages in this thread now, it would probalby be over 1000 already
if it hadn't had some contributions sucked away by the 'standing morphing

Chip,

This is not a terribly different posting from your own complaint.

73's
Richard Clark, KB7QHC

On Jan 12, 11:13*am, Cecil Moore wrote:
On Jan 12, 8:10 am, Keith Dysart wrote:

P(t) = V(t) * I(t) is simple, and always holds.

But one must remember the units of that result is Volt*Amps, not
watts.

You really should put a bit of effort into understanding
the different expressions for power.

P(t) = V(t) * I(t)

is saying that the function which describes power with
respect to time, P(t), is the product of the function
describing voltage with respect to time, V(t), and the
function describing current with respect to time, I(t).

As corollary, the power at an instant of time is equal
to the voltage at that instant times the current at
that instant.

And, of course the unit watts is the same as the units
volts times amperes. Just do the substitution.

V = J/C
A = C/s

V A = (J/C) x (C/s)
= J/s
= W

You are confused with the common use in power
engineering where VA is used to mean Vrms*Irms.
Whether VA is equal to watts depends on the
phase relationship of the voltage and current.

When the voltage and current are not in phase, then
the expression
Pavg = Vrms * Irms * cos(A)
can be used to derive the average power.

This is a special case of P(t) = V(t) * I(t).
It is applicable when V(t) is a sinusoid of
a single frequency, and I(t) is related to V(t)
using the expression V(t) = Z * I(t).

Pavg = Vrms * Irms * cos(A)
is completely derived from P(t) = V(t) * I(t)
by appropriately substituting for V(t) and I(t)
and then averaging P(t) over an appropriate
interval (typically one cycle).

Bottom line:
P(t) = V(t) * I(t)
is always correct.

The other variations only work when
applied under the appropriate conditions.

...Keith

On Jan 12, 1:22 pm, Keith Dysart wrote:
P(t) = V(t) * I(t)
As corollary, the power at an instant of time is equal
to the voltage at that instant times the current at
that instant.

You obviously mean the *real* instantaneous voltage times the *real*
instantaneous current. You should say that to avoid confusion.

Seems you need to add 'Re' to that equation to remove the obvious
ambiguity. P(t) = Re[V(t)]*Re[I(t)]

From "Fields and Waves", by Ramo & Whinnery:

W(t) = {Re[Vm*e^j(wt+A1)]}{Re[Im*e^j(wt+A2)]}
--
73, Cecil, w5dxp.com

On Jan 12, 2:05*pm, Cecil Moore wrote:
On Jan 12, 1:22 pm, Keith Dysart wrote:

P(t) = V(t) * I(t)
As corollary, the power at an instant of time is equal
to the voltage at that instant times the current at
that instant.

You obviously mean the *real* instantaneous voltage times the *real*
instantaneous current. You should say that to avoid confusion.

Seems you need to add 'Re' to that equation to remove the obvious
ambiguity. P(t) = Re[V(t)]*Re[I(t)]

From "Fields and Waves", by Ramo & Whinnery:

W(t) = {Re[Vm*e^j(wt+A1)]}{Re[Im*e^j(wt+A2)]}
--
73, Cecil, w5dxp.com

Good recovery.

But, of course, V(t) and I(t) are general functions
of time. In particular, the discusion was regarding
pulses.

Only when one constrains oneself to sinusoids and
chooses to use the complex exponential notation
is it necessary to use Re[].

But it was a good recovery.

...Keith

On Jan 12, 3:25 pm, Keith Dysart wrote:
Only when one constrains oneself to sinusoids and
chooses to use the complex exponential notation
is it necessary to use Re[].

I think that's a false statement. Do you have any proof? Why would the
imaginary part of a pulse produce any power?
--
73, Cecil, w5dxp.com

On Jan 12, 3:25 pm, Keith Dysart wrote:
But, of course, V(t) and I(t) are general functions
of time. In particular, the discusion was regarding
pulses.

Pulses can be analyzed as Fourier sinusoidal functions with multiple
frequencies. Point is that you could have saved a month of grief on
this newsgroup if you had initially said your power equation applied
only to real voltage and real current. If you had done that, nobody
would have argued with you.
--
73, Cecil, w5dxp.com