The topic of this problem is Operational Amplifier Circuits. And the problem is to determine V out and I out in the circuit shown below. The circuit has an op amp, it has two voltage sources, a 12 volt and a 2 volt source, and three resistors associated with it. What we want to do is we want to use the properties of an ideal op amp in order to help us solve this circuit. So if we look at an ideal op amp, first of all, the symbol for the ideal op amp has two inputs. It has an inverting input and a non-inverting input. And each one of those inputs have a current associated with it. So we have a current for the non-inverting input and a current for the inverting input. We also have voltage associated with each one of these inputs as well, for the inverting and non-inverting inputs. If we want an ideal operational amplifier, we know that the current going into the inverting input is equal to the current going into the non-inverting input, and that both of them are equal to 0. We also know that the voltage at the inverting input is equal to the voltage at the non-inverting input. So there are other properties of the ideal op amp that are of interest, but for this class, these are the two most important properties of the ideal op amp. And those are the ones we need in order to solve our problem. So if we go back to our problem here, and we use these properties, it'll allow us to work the problem knowing voltages at certain points in the circuit. For instance, we know what the voltage is at this point. It's -2 volts at the non-inverting input of the op amp, which means it's also -2 volts at the inverting input of the op amp, which means that it's -2 volts at this node between the 4 kiloohm and the 2 kiloohm, 4 kiloohm and the 8 kiloohm resistor. So, having known that, we can use that in order to solve the problem. The way we do that is we use Kirchhoff's current laws, and we'll sum the currents into this top node that we've identified as a node at -2 volts. And if we do that, then we will come up with an equation, which gives us V out as the unknown in the equation. So let's do that, we're going to do Kirchhoff's current law, and we're going to call this node 1. So we're going to do Kirchhoff's current law at node 1, and we're going to sum the currents into node 1. So we're going to sum the current through the 4 kiloohm resistor into that node, through the 8 kiloohm resistor into that node, and also this current, which we know is 0, flowing out of the op amp into that node. So let's do that. Starting with the current through the 4 kiloohm resistor, it's going to be 12 volts- (-2) volts. /4 kiloohms, that gives us the current through the 4 kiloohm resistor. We're going to add to that the current through the 8 killoohm resistor. It's going to be the voltage at this point, which is V out,- (-2) volts /8 kiloohms, and of course, we have the third current, which we know is 0 coming out of the inverting input of the op amp. I'm going to put it in there for completeness, and the sum of those currents is equal to 0. So we now we have an equation that has V out as the only unknown in the equation. If we solve for V out, we end up with a V out, which is equal to -30 volts. So now that's one of our variables that we want to determine in this problem. The other is I out, we want to find the current through the 8 kiloohm resistor. And we know what that's going to be because we've already investigated that, but as it's described and defined in this problem, I out is in the opposite direction of the current that we were summing into node 1. So I out, if we're going to find that current, we know I out. It's going to be equal to the voltage at this point, node 1, which is -2 volts,- this voltage, which is V out. So it's -2- V out / 8 kiloohms. And that will give us I out. We know what V out is, it's a -30 volts. And so we can determine I out. So I out, Is equal to 28/8k, which gives us 3.5 mA.
