The topic of this problem is Series and Parallel Independent Sources, and the problem is to determine the current in the circuit that’s drawn and the power associated with the 5 ohm resistor. In this problem, we have a series combination of Voltage sources and resistors. And we know that we can combine series voltage sources by adding them, and so that's what we're going to do in this problem. We first of all acknowledge that the 5V sources, all three of them, are a series of one another. And we combine those, and we get a 15 volt source as shown in the redrawn circuit. And the way that we get that is by going clockwise around the circuit with the negative polarity and the 5 volt source on the right hand side first. That's minus 5 volts. We could continue in the top part of the circuit. We hit the negative polarity of the 5 volt source at the top. So it's a minus 10 volts combining those two. And continually around the circuit in clockwise fashion, we hit the negative polarity of the 5V source on the right hand side. So that's how we ended up with a 15V source on the left hand side of our redrawn circuit. Similarly, we also know that the resistors are in series with one another, the 15 ohm, the 25 ohm, and the 5 ohm resistors. So we can combine those as well, and if you add those together, a series combination of resistors can be added, and we have a 45 volt equivalent resistance for those three resistors. Now we're going to try to find the current in the circuit and we can do that with our simplified circuit very easily. We notice that the current is going to be equal to the voltage divided by the resistance using Ohm's law. The voltage is going to be minus 15 volts because as we know through the passive sign convention a positive direction of the current in this circuit is clockwise and is hitting the negative polarity of the 15 volt source first. That's why we have amassed 15 volts in our ohm's law equation. We divide that by the resistance which is 45 ohms and that gives us a current of -0.333 amps. So, that's our current in the circuit. Now, if we want to find the power associated with the 5 ohm resistor we can use our well-known equation for power, where power is equal to I squared R. Our current which we just found, is -one-third amps. We squared that and we multiply it by the 5 ohm resistance to get our power associated with the 5 ohm resistor. We noticed that that power is going to be a positive power. So we have absorbed power for this resistor remembering using the passive sign convention that a positive power is absorbed power. And a negative power is a supplied power.
