First of all, we have one-half, and then we have for

our I sub m term 5 square roots of five halves.

And for the V sub m term, it's going to be VI,

the current I times 0.6k.

So it's going to be 5 square roots of 5 over 2, so

we're going to square that times the 0.6k.

And so if we do that calculation,

we end up with 18 .75 milliwatts of

power that's absorbed by the 0.6K resistor.

Again it's average power or

also the real power associated with our steady state AC circuit.

The next thing we're looking for is the power from the current source.