So the power average for the 10 ohm resistor is going to be one-half,

V sub m, I sub m cosine of the angle between them.

We know that since it's the resistor, the angle between the voltage and

current is 0, so we end up with a 1 for this cosine term.

So it's equal to one half V sub M,

I sub M and the cosine term is 1.

And so for the resistor, we have our V determined,

it's 50 square root of 5 at angle of minus 126.6 degrees.

So it's one-half, 50 square root of 5 for

our V sub m and then our I sub M is going to be 50 square roots of 5,

for the voltage drop, divided by 10 ohms.

And that's the current flowing downward in this case, through this 10 ohm resistor.

And as for the direction that we want to choose it based on our passive sign

convention where we've chosen the positive polarity of the voltage drop across

the 10 ohm resistor at the top of the resistor.

So through the passive sign convention we know that the current enters

the positive of the voltage drop across resistive or absorbing elements.

So defining this current is going to be V1 divided by ten.

And so V1 is 50 square root of 5, so

I'm going to square this term because we have two V sub ms and

we divide that by our R, which is 10.

So we end up with an average power absorbed by the resistor

which is 625 watts again, this is absorbed.