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
and through the source. The point is that the ideal
source doesn't dissipate that energy, i.e. it doesn't
heat up. 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.
--
73, Cecil http://www.w5dxp.com