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. It's an op-amp circuit with a current source and a voltage source in it as well as three resistors. V out is the voltage measured across the load resistance, which is the 6 kiloohm resistor. And I out is the current which is flowing back into the op-amp from the output side of the circuit. So what we want to do is we we want to use the ideal properties of op-amps along with our tools such as nodal analysis and mesh analysis to solve the problem. So the first thing we want to note are the properties of ideal op-amps. First of all, we have the symbol for the ideal op-amp, it has an inverting and a non-inverting input. Each one of those has a current associated with it. And each one of them has a voltage associated with it as well. So there's a voltage at the inverting input and a voltage at the noninverting input. And for an ideal op-amp, we know that the currents both at the inverting input and the noninverting input are equal to 0. We also know that the voltages at the two inputs are equal to each other. So V minus is equal to V plus. We use this two properties, which are not the only two properties of ideal op-amps, but the two that are most important to this class for solving problems and that are associated with linear circuits. We'll use these two properties to solve our original circuit. So let's do that. The first thing we want to note is the voltage at this node at the bottom of our circuit, which is the ground node, the voltage is equal to 0. Which also tells us the voltage at the noninverting input is equal to 0. Using the properties of an ideal op-amp, we know the voltage at the inverting input is also 0, which makes it a 0 volts at this node at the inverting input of the op-amp. So now we've used that second property of the ideal op-amp, V minus is equal to V plus. Now we're going to use nodal analysis and we're going to sum the currents into the node, which we've identified at zero volt potential. And if we do that, we'll get something that involves V out as the unknown variable in the equation along with other known voltages in the circuit. So we're going to do Kirchhoff's current law at node one, and we're going to sum the currents into node one. So if we start with a 4 kiloohm resistor and we want to sum that current, it's going to be 12 volts minus 0 volts, divided by 4 kiloohms for the current through the 4 kiloohm resistor flowing left to right into node one. We also have this 2 milliamp source which is flowing into node one as well, so we're going to add the 2 milliamp source. In addition to that, we have a current which is flowing through the 3 kiloohm resistor back into node oen. That's V out minus 0 over 3k. And we have the current which is flowing out of the inverting input, which we know is zero, our first property of the ideal op-amp. I'm going to add it in there for completeness. And there's our sum. So this is Kirchhoff's current law at node one. So we can see from this that we have an equation which has one unknown, V out, and we can solve for it. If we solve for V out, we end up with V out equal to minus 15 volts. Now we want to find I out. And I out is our current back into the op-amp. So we can use the node at the output, maybe we call this node two, and we can use Kirchhoff's current law at node two to find that current I out. So this is KCL at node two. And again, we're going to sum the currents into that node. We could choose to sum them out of the nodes, but for consistency, we're going to sum the currents into the nodes. So starting with a current through the 6 kiloohm resistor, we start with the voltage on one side minus the voltage on the other, which is 0 volts, minus V out, we know V out is minus 15 volts, divided by 6 kiloohms. So it's 0 minus the minus 15 for V out divided by 6 kiloohms. That's the current flowing up through the 6 kiloohm resistor. We also have the current flowing left to right through the 3k resistor, which is 0 volts minus V out. V out being a minus 15 volts. Divided by 3 kiloohms. And the other one we have is we have I out. And in fact, it should be a minus I out, since we're summing the currents into it, we should have a minus I out instead of a plus I out for that current. And those are all the currents into node two, and the sum is equal to 0. So as you can see, we end up with an equation which only has one unknown, which is I out, and I out comes out to 7.5 milliamps.
