Re: The Rest of the Story

On Apr 14, 5:28 pm, Cecil Moore <nos...@xxxxxxxxx> wrote:
Keith Dysart wrote:
What is your explanation for this phenomenon?

You first! :-)

I suppose. If you are happy with energy equations that
don't balance.

My energy equations balance perfectly. Yours are the
energy equations that don't balance in violation of
the conservation of energy principle.

"It depends" was no answer - that was just mealy-mouthing.

Except, that it does depend.

Sounds more like religion than anything else.

You have to read more carefully. I did not use the word
dissipation.

You have to read more carefully. I used the word
"dissipation" and you disagreed with me. Please
read it again to verify that fact.

Since we don't know the internals of the source, we do not
know if it is dissipating or not.

Sorry, the source is, by definition, lossless. All of
the source dissipation is lumped in the source resistance
drawn separately on the diagram as Rs.

You really should rethink this a bit. When current flows into
a voltage source, the voltage source is absorbing energy.

But the source is NOT *DISSIPATING* energy because
Rs is not inside the source. Rs is clearly drawn outside
the lossless source so the source generates *ZERO* heat.
I am amazed at the lengths to which you will go to try
to obfuscate the discussion.

You do need to back up a bit and review voltage sources.

When conventional current is flowing out of the positive
terminal of a voltage source, it is usually agreed that
the voltage source is providing energy to the circuit.
This comes from
P(t) = V(t) * I(t)
when V and I are the same sign, P is positive representing
a flow of energy from the source to the other elements
of the circuit.

What do you explain is happening when conventional current
is flowing into the positive terminal of the source?
Is the source still providing energy to the circuit now
that P is negative?
Or is it accepting energy from the circuit, the negative
P representing an energy flow into the source?

...Keith
.

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