The topic of this problem is AC steady state power. The problem is to determine the average power absorbed by the 0.6 kilo ohm resistor and the average power supplied by the current source in the circuit shown below. We have a current source which is 4 square root of 5 cosine of 10 to the 4th T milliamps. We have two resistors and we have a capacitor. So if we're looking for the average power absorbed by the 0.6k ohm resistor, then we can find that using our well known equation for average power and steady state power problems. The average power is equal to one-half V sub M, I sub M times cosign of the angle between the voltage and the current. So if we're looking for the average power absorbed by the 0.6 Kohm resistor, which is on the left hand side of our circuit, then if we can find the current through it, and we can find the voltage across it, then we're going to be able to find the average power. So if we're trying to find, first of all, the current through the 0.6 Kohm resistor, this current that'sin this direction, so we have using passive sign convention of positive to negative voltage drop in the direction shown. Then we can use current division of our current source because our current source is here, the center of our circuit, and it divides between two paths, between the path through the 6.6 Kohm resistor and the path through the 1.8 Kohm resistor and one twenty-fourths microhenry or microfarad capacitor. So we need to find out what that current is through the resistor. So in order to find that, we would use current division. So let's start with that. This current has caught I through the resistor, current I, the current which is flowing the opposite direction, so we're going to get a negative sign, four square roots of 5, and the angle on that current is 0 degrees. And again, we're going to divide that current between the two paths that it can take. And so through the current division, we would use 1.8k minus j2400 because this equivalent impedance for this capacitor is minus j 2400 using our well known expression, the impedance is equal to 1 divide by j omega C. Omega is ten to the fourth, C is 1 over 24 microfarands. That results in j2400 for the capacitor. It's actually a minus, I found that. So current division, if we're looking for the current through the 0.6 Kohm resistor, current division would be done in this way. We take the current that we are dividing, multiplied by the opposite path for the current to flow, divided by A combination of the two paths. So it's 1.8 Kohm minus j2400 path combined with the 0.6 Kohm. So that gives us our current I which is flowing through the 0.6 Kohm resistor. And so that comes out to be minus 5 square roots of 5 over 2, an angle of minus 8.1 degrees, and that's in milliamps. So if we know that, then we can then calculate the power for the 0.6 K Resistor. And this is again average power. We use our equation up here for average power. And we know that the current and the voltage across the resistor have the same angle. So this cosine term goes to 1, and we're left with, for the resistor, one-half V sub m times I sub m. We know what I sub m is, it's 5 square root of five-halves. So, put that in. 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. And again it's average power. And sowe are going to use the same approach. It's still one half V sub m I sub m cosine of theta V minus theta I. We know what I is and we know what V sub m is because it's the same voltage that was dropped across our resistor top to bottom. It's 0.6k times I sub m. So if we do that, then first of all l have our one-half term in the front, and then V sub m is going to be our .6K times 5 square root of 5 over 2. That's our V sub m. And then times our current, and our current in this case, I sub m is 4 square roots of 5. And then we take the cosine of the angle between them. And the cosine of the angle between them is going to be minus 8.1 degrees. And so the power associated with the source is 21 milliwatts.