The topic of this problem is AC steady state power. The problem is to determine the average power delivered to each of the elements in the circuit shown below. The circuit has two sources, it has a current source which is a independent current source. And it has a voltage source, which is a voltage controlled voltage source. It's a dependent source where the voltage V sub 1, which is the voltage across the 10 ohm resistor controls the overall voltage of this dependent voltage source. We also have two resistors, a 10 and a 15 ohm resistor and a capacitor with an impedance value of minus J20. So if we're looking for the average power delivered to the elements, we need to find the currents and we need to find voltages so we can ultimately find the average power. Remember that our equation for average power is one-half V sub m, I sub m times a cosine of the angle between those voltages and current levels. So perhaps one way to solve this problem is to find a nodal voltage at this point. So use a nodal analysis summing currents into that node to find the voltage at this point. Once we know the voltage at this point, we can find the current through the 10 ohm resistor. We already know the current associated with the source so we have our Vs and Is that we need in order to solve for our average power. And we can then solve for the current and using the current over, across and through the capacitor and resistor, we can find the average power associated with those as well. So let's start with a nodal equation, let's call this node 1 at the top of the circuit. And we're going to write a nodal equation summing the currents into node 1. So we're going to sum the currents that are flowing into this node 1. So first of all, we have the current which is from the source, 20 at angle of 0 degrees is flowing into node 1. We also have the voltage drop across a 10 ohm resistor which leads to our current flowing into node 1. And voltage drop is 0 for the ground node at the bottom minus V1 for the voltage drop across the resistor divided by the resistance which is 10. We also have this current which is flowing through the resistor and capacitor back into node 1. And that current is first of all the voltage drop from one side which is minus three halves V1. On the right hand side of this series combination of resistor and capacitor and then the voltage on the other side is V1. So it's minus three-halves of V1 minus V1 divided by 15 minus J20 for the impedance. And the sum of those three currents at node 1 equal to 0 using Kirkoff's Current Law. If we look at this equation we notice that there's only one unknown in this equation, it's V sub 1. So the voltage V sub 1 from our nodal equation above is equal to 50 square roots of five. At an angle of minus 26.6 degrees. That's our voltage at the nodal point one. So if we have that voltage V sub 1, then we can find everything else that we need in the current. So if we're looking for, perhaps, this current I, that's going through the capacitor and resistor on the top right side of our circuit, then that current I is going to be V1 minus whatever the voltage is here, which is three-halves, minus three-halves in V1 divided by the impedance which is 15 minus j20. So the current I is going to be equal to V1. Again, the voltage at node .1 minus a minus three-halves V1, and minus three-halves V1 is the voltage at the node on the right-hand side of our circuit, divided by our Z value, 15 minus j20. And so the current comes up to be five, square-roots of five at an angle of 26.6 degrees. That's for the current I that we've designated in our circuit. So now we can use these voltages and currents to find everything else we want to in the circuit. So let's start with the average power associated with the 10 ohm resistor in the center of our circuit. 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. So now let's look at the power associated with the current source. Again, this is an average power so it's real power that's being absorbed or supplied by the current source. We know that we're going to use the same equation that we had before for the average power. It's one half and the V sub m is the same voltage across the current source as across the resistor and it's equal to our nodal voltage, V sub 1. So our V sub m is 50 square roots of 5. Our I sub m is the current which is flowing down through current source in this direction, since again, we've chosen the voltage as plus to minus across the element. So that current is minus 20, because we have a current source of 20 at an angle of 0 going up through the left-hand leg of our circuit. And then, we have the cosine of the angle between the voltage and the current, the cosine of minus 26.6 degrees, and the current is 0 degrees. So it's minus 26.6 minus 0 degrees for our angles. And so we end up, for the current source, we end up with a minus 1,000 watts power absorbed. In other words, the current source is supplying 1,000 watts. Now let's look at the 15 ohm resistor on the upper right-hand side of our circuit. And we look at the average power associated with it, again using our same equation. So if we want to find that, then we have to use our current. We know what the current I is. The current I is 5 square roots of 5 at an angle of 26 degrees and so we have our l sub n value. We need to have a V sub m value as well. Our V sub m value is going to be equal to the I times the resistor because V is equal to IR. And so for our voltage, we have 5 square roots of 5 times 15, that's our voltage and our current is 5 square roots of 5. So the power associated with the 15 ohm resistor and we know that these angles are the same, the angle between the current and the voltage are the same, it's 26.6 for the voltage and 26.6 for the current. So for resistor again, that cosine is always equal to 1. And so we end up with this 15 ohm resistor calculating an average power absorbed by the resistor at 937.5 watts again, absorbed power. Now let's look at the capacitor, the minus j20 capacitor and we look at the average power absorbed by it. Again we use our equation that we have above and we have a value for our current. It's 5 square root of 5 and angle of 26.6 degrees and we use the same approach as we did for the 15 ohm resistor. We would take that current multiply it by our impedance value to get the voltage. We see that when we multiply the current, 5 square root 5 at an angle of 26.6 degrees times this impedance which is 20 to angle minus 90 degrees. Then we end up with a cosine which is between these two angles theta V minus theta I of 90 degrees. The difference in these two across associated with this capacitor is 90 degrees, so the cosine is 0. So the average power associated with the capacitor is 0 watts. And that makes sense because surely inductive or capacitive components or elements do not absorb real power, they only absorb reactive power. There's no real power loss associated with those elements. So we have 0 watts of real power also known as the average power for the capacitor. The last one that we might want to find the average power for is the dependent source. And so it's the three-halves V1 source that we're finding the average power for. And the average power for that source is going to be one-half for our term out front. V sub m is going to be minus three-halves and then we know what V1 is, 50 square root of 5. So that's our VM term and then we have the current and we also have our current term which is 5 square root of 5. And we have the angle between those two, cosine theta v which is minus 26.6 degrees minus theta sub i which is 26.6 degrees. So our average power associated with the dependent source is minus 562.5 watts. And that's minus 562.5 watts absorbed power, which means that the source is supplying 562.5 watts.
