On Tue, 15 Apr 2008 14:40:28 -0700 (PDT)
Keith Dysart wrote:

On Apr 15, 12:47*pm, Cecil Moore wrote:
Keith Dysart wrote:
Is there some reason why you think that it is only
necessary to account for the energy removed from the
circuit by heating?

Heating is the only way that the energy can be removed
from the ideal closed system.

And that you can ignore energy being removed in other ways?

No energy is removed in other ways. There is no other
way for energy to leave the ideal closed system.

And how do you know the ideal source does not dispose
of the energy it receives by getting warm?

An ideal source has zero source impedance, by definition.
All of the source impedance is contained in Rs, the
ideal source resistor.

I see we need to back up a bit further.

Consider a 10 VDC ideal voltage source.

When 2 amps are flowing out of the positive terminal, the
ideal voltage source is delivering 20 joules per second
to the circuit.

Q1. Where does this energy come from?

When 1.5 amps is flowing into the positive terminal, the
ideal voltage source is absorbing 15 joules per second
from the circuit.

Q2. Where does this energy go?

Answer to both. We do not know and we do not care. An
ideal voltage source can deliver energy to a circuit
and it can remove energy from a circuit; that is part
of the definition of an ideal voltage source. It does
not matter how it does it.

But just as easily as it can supply energy, it can
remove it.

Without understanding these basics of the ideal voltage
source, it will be impossible to correctly analyze
circuits that include them.

Perhaps this has been the root cause of the misunderstandings.

...Keith

This WIKI article mentions the ability of an ideal voltage source to absorb power.

http://en.wikipedia.org/wiki/Voltage_source
--
73, Roger, W7WKB