On Apr 15, 6:16*am, Cecil Moore wrote:
Keith Dysart wrote:
Or is it accepting energy from the circuit, the negative
P representing an energy flow into the source?

Of course there can be an energy flow into the source

Good.

and through the source.

It is not clear what you mean by "through the source".
The source provides or absorbs energy. It does not have
a "through", since it only has a single port.

The point is that the ideal
source doesn't dissipate that energy, i.e. it doesn't
heat up.

It is not obvious why you want to draw a distinction
between elements that remove energy from a circuit by
heating and those that do so in some other manner.

Could you expand?

Is there some reason why you think that it is only
necessary to account for the energy removed from the
circuit by heating?
And that you can ignore energy being removed in other ways?

And how do you know the ideal source does not dispose
of the energy it receives by getting warm? Nowhere
do I find in the specification of an ideal source any
hint of how it disposes of its excess energy. It could
be by heat, could it not?

And the resistor, could it not also radiate some of
the energy it receives? Perhaps even as visible light?
Would that make it less of a resistor because it was
not 'dissipating' the energy?

All of the heating (power dissipation) in the
entire example occurs in Rs and RL because they are
the only resistive components in the entire system.
Any additional heating in the ideal source would
violate the conservation of energy principle.

This is quite a leap. The energy flows into the source.
We have accounted for that energy. We don't know where
it goes from there.

How would it violate conservation of energy if it was
dissipated rather than going somewhere else?

In your model, what things could be done with the energy
that would not violate conservation of energy? What other
things (besides heating) would violate conservation of
energy?

...Keith