The topic of this problem is operational amplifier circuit. 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 of 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 the current for the inverting input. We also have voltage associated with each one of these inputs as well, for the inverted and non-inverting inputs. For 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 zero. 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 two most important properties of the ideal op amp. And these 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 voltage is 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 ff 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 kilo ohm and the 2 kilo ohm, 4 kilo ohm and the 8 kilo ohm 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 a -2 volts. And if we do that, then we will come up with an equation which gives us V out is the unknown in the equation. So let's do that. Well, the Kirchhoff's Current Law, I'm going to call this node one. So we're going to do Kirchhoff's Current Law node one and we're going to sum the currents into node one. So we're going to sum the current through the four kilo ohm resistor into that node through the 8 kilo ohm resistor into that node and also this current which we know is zero flowing out of the op amp into that node. So let's do that. Starting with the current through the 4 kilo ohm resistor, it's going to be 12 volts- (-2 volts). Divided by 4 kilo ohms. That gives us the current through the 4 kilo ohm resistor. I'm going to add to that the current through the 8 kilo ohm resistor. It's going to be the voltage at this point which is V out minus a -2 volts Divided by 8 kilo ohms and of course we have the third current which you know is zero coming out of the inverting input of the op amp. I'm going to put it in there for completeness and the some of those currents is equal to zero. So now we have an equation that has V out is 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 one is I out, we want to find the current through the a kilo ohm resistor and we know what that's going to be because we're 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 one. 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 one, which is -2 volts minus this voltage which is V out. So it's -2- V out / 8 kilo ohms. And that will give us I out. We know what V out is, it's -30 volts. And so we can determine I out. So I out Is equal to 28 divided by 8k, which gives us 3.5 milliamps.
