The topic of this problem is Kirchhoff's Voltage Law. And what we're going to do in this problem is we're going to write the mesh equation or also known as the loop equation for this single loop circuit. Additionally we're going to solve for the voltage from node a to node e. And also from node E to node C and so we'll see what these voltages come out to be as we add up the voltage drops across various elements. So the first thing we want to do is write the mesh equation. So if we do that starting at the lower left hand corner which is node F and working our way around the circuit we would first encounter the 24 volt source and the nano polarity of that 24 volt source. Then as we continue we encounter R1 which has a 16 volt source, a 16 volt drop across it. We continue on to the next independent source which is 12 volts. We encounter the negative voltage polarity across it first on to R2, where we encounter a 4 volt drop Onto R3, we encounter a 6V drop. Onto R4, which is a 10V drop. And then ultimately back to the point at which we began, and that's equal to 0. So that is our mesh equation for that circuit. So it's just a practice problem with some relation to Kirchhoff's Voltage Law, that allows you to write that loop equation. But most importantly in this problem, we want to find V ae and V ec. When you see a voltage like Vae written as it's written. You of it as Vae plus two minus. So we're going to find the voltage form node a to node e with node a being positive voltage compared to relative to node d or node e. So first thing we do is we can draw that on a circuit if we want, this is what we're looking for, the voltage from a to e is called Vae. And so we disseminate that with an arrow with the head of the arrow on the positive A node and the tail of the arrow on the negative E node and we're looking for that voltage from one to the other. So, we can do that by adding up the voltages in either direction around our mesh just so we started A and we end at E. So if we started A, and we go around the left side of the circuit down from a working our way down, we end up adding a plus 24v for the independent source and then we continue. We had a negative 10v drop across resistor R4 before we get to e. So if we add those two together, we end up with a 14v drop Vae. We could have easily gone the other direction as well. So we could have found Vae by going, put or out here in front around the other way with the circuit. So we could start at a go through our one first and we have a 16 volt drop and then after that we have an encounter. The 12 volt, the negative polarity of it first, continuing around the loop we would hit the plus four volts across resistor two, continuing around the loop. We would hit a plus six volts across resister three and then we reach E. So if we add all those up we end up with 14 volts. So either way would have gotten us the same answer. The second part of this problem is to find Vec. Vec is the drop from plus and minus again from e to c. So we can draw some arrow, put an arrow end at the e, that's Vec. So we're going to start at E, we can go either direction again around the circuit to find the voltage drop. Let's take the direction where we take and go counter clockwise from E and reach up to C. If we do that, we first encounter a minus six volts across resistor three. And then we have a minus four volts across resistor two before we reach node C. So starting at e, going counterclockwise until we reach node c. And that gives us a minus ten volts for the voltage across eC. Similarly we could have gone the other direction. And if we do that we can find the C will come out to the same value. Starting at e, going to C but this time clockwise and looking at the voltage drop across R4 for our sets 10 volts and then we go through the independent source. Which is 24 volts R1 which has a 16 volt drop across it and the 12 volt source, we're hitting the negative polarity of that source first. And again, if we add those up we would get minus ten volts for Vec.
