The topic of this problem is AC Steady State Power. The problem is to find the average power absorbed by the load in the circuit shown below. In the circuit we have a voltage source. Has an amplitude of 10 and a phase angle of 60 and we have a load, our load's comprised of a 2 ohm resistor and a J2 impedance inductor, and what we're looking for again is the average power absorbed by the load. So in order to work this problem, the first thing that we have to understand is that the average power is defined as ½ (V sub m) (I sub m) times the cosine of the angle between the voltage and the current. So in order to solve for the average power, we're going to need to find the current I sub m which is through our load 2 + j2. And we also need to find the current, so that we can find the angle of the current in order to solve for the average power. So let's start with finding the current I in the circuit. I, we know, is going to be equal to the voltage V, which is across a load and it is source voltage and it's 10 at an angle 60 degrees divided by the impedance z. So if we fill in for those, we have 10 at an angle of 60 for the voltage and again the impedance 2 + j2. And if we go through the complex math, we end up with a magnitude of 3.53 for the current and an angle of 15 degrees. And that's in amps. Now that we have the current, we can solve for the average power. The average power Through our definition is going to be one-half V sub m which is 10, times I sub m which is 3.53 and we take the cosine of the angle between the voltage and the current. So we have 60 degrees for the voltage, and 15 for the current. So we end up with an average power of 12.5 watts. Since the inductor in our circuit absorbs no real power, we can check the resistor power to see whether or not the resistor power equals our average power that we found from our calculation. In order to do that we need to find the voltage across the resistor. We know what the current is, we found that earlier in the problem. Now if we want to find the power associated with the resistor then we can first find the voltage associated with the resistor. And that voltage is going to be Through voltage division the applied voltage of 10 at an angle of 60 degrees across our load. And using voltage division we have the 2 ohm / 2 + j2 to find the amount of the voltage which is dropped across the resistor. And that comes out to 7. The angle of 7.07 at an angle of 15 degrees. So, its 7.07 and angle of 15 degrees. Therefore, the power associated the resistor, this is average power, is going to be again ½(V sub m) (I sub m) times the cosine of the angle between the voltage and the current. And you could see that first of all we're working with a resistor and if you look at the angle on the voltage and the angle on the current, they're equal. So we have cosine of 0 degrees which we know is 1 for the resistor. So it's really 1/2 (V sub m) (I sub m) for a resistor. And so if we plug in our numbers, ½ (7.07)(3.53) we indeed get at the same number. We end up with 12.5 watts that's absorbed by the resistor.
