On Apr 2, 9:17*am, Roger Sparks wrote:
On Wed, 2 Apr 2008 02:41:24 -0700 (PDT)

Keith Dysart wrote:

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But your phraseology suggests another way to approach
the problem. Let us consider what happens just after
the arrival of the reflected wave. For computational
convenience, we'll use a frequency of 1/360 hertz so
that 1 degree of the wave is one second long.

The reflected wave arrives at 90 seconds (or 90 degrees)
after the source is turned on. If the line is
terminated in 50 ohms, then no reflection arrives and
during the one second between degree 90 and degree 91,
the source resistor dissipates 0.01523 J of energy. The
line also receives 0.01523 J of energy and the source
provides 0.03046 J.
* Esource.50[90..91] = 0.03046 J
* Ers.50[90..91] * * = 0.01523 J
* Eline.50[90..91] * = 0.01523 J
* Esource = Ers + Eline
as expected.
* Efor = Eline
since Eref is 0.

I come up with 141.4v across 50 plus 50 ohms. *The current should be 1.414a. *Power to each 50 ohm resistor would be *50*1.4142^2 = 200w. *For the 1 degree interval of 1 second, that would be 200 joules.

Right? *Peak current flows at 90 degrees?
--
73, Roger, W7WKB- Hide quoted text -

- Show quoted text -

We may be using different sources. My Vs is 141.4cos(wt)
so that between 90 degrees and 91 degrees, my source
voltage is going from 0 to -2.468 V.

And an ooopppps. I actually did the calculations for
a shorted load rather than 12.5 ohms as stated.
With the reflection coefficient of -1 and a 90 degree
delay, the reflected voltage between 90 and 91 degrees
changes from -70.711 V to -70.700 V.

Hoping these details resolve the disparity,

...Keith