The topic of this problem is series and parallel independent sources. In this problem we want to determine the power associated with the 80 volts source. As you can see from the circuit, it has three independent voltage sources associated with it, it also has four resistors. We're going to rearrange this circuit using our knowledge of series and parallel independent sources to make the circuit easier to solve. Ultimately, what we're looking for, again, is the power associated with the 80-volt source. We want to also determine whether that power is absorbed power or supply power. So the first step in our process is to redraw our circuit. We're going to take our circuit, we're going to redraw it and we're going to take it. And we're going to move the voltage sources so that they look a little easier as far as the combination of these voltage sources. The sources are in series of one another. These are voltage sources. And we know that when we have voltage sources in series with one another that we can add them. We also know, even though it's not relevant in this problem, that if we have current sources in parallel with one another they can be added as well. So we have our 80 volt source on the left hand side of the circuit. We have a 30 volt source at the top, a 20 volt source at the top of the circuit and then all of our resistances, the 10, the 7, the 5 and the 8 ohm resistors. So, we've redrawn the circuit so that it becomes more obvious that these voltage sources can be combined in series with one another. So we have an 80 volt source. So if we go around this circuit and we see that we hit the minus 80 so it's a minus 80 volts and then here we hit another minus 30 volts and then if we continue we hit a positive 20 volts. So the net result of that is going to be a 90 volt source. So we're going to redraw the circuit one more time with our 90-volt source in there. And we also know that we have this series combination of resistors as well, and we're going to combine those as a series combination of resistors, 10, 7, 5 and 8. So that gives us 30 ohms as a equivalent resistance for those four resistors. And we also have our current I, drawing it in both of our redrawn circuits. And so we've reduced our circuit to something that's very easy to find the current for. Ultimately if we want to find the power associated with the 80 volt source, we're going to need that current because we know that the power is equal to the voltage times the current. So we know our voltage and we're also going to find our current using Ohm's Law and our redrawn circuit. So if we go back to our redrawn circuit, and we're looking for the current, we know the current is equal to the voltage divided by the resistance. And so that's going to be equal to 90 volts divided by 30 ohms. And so that gives us 3 amps for our current source. We also can use that 3 amps using our power equation to find the power associated with the 80 volt source. So if we do that, we have a positive current 3 amps flowing up into the negative polarity of the 80 volt source. So we have a minus 80 volts using a passive sign convention where we have a positive current flowing into the element times the current, which is 3 amps. And so, in the end, we end up with a minus 240 watts of power associated with the 80 volts source. So since that's a negative power, we know that that is a supplied power. Remember that using the passive sign convention negative powers are supply powers and positive would be absorbed powers. So the 80 volt source is supplying 240 watts of power.