The topic of this problem is Operational Amplifier Circuits. And the problem is to determine V out and I out in the circuits shown below. It's an op amp circuit with a current source and a voltage source in it, as well as three resistors. Vout is the voltage measured across the load resistance which is the 6 kilo Ohm resistor. And Iout 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 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 know is the properties or 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 non-inverting input. And for an ideal op-amp, we know that the currents, both at the inverting input and the non-inverting input are equal to zero. We also know that the voltages at the two inputs are equal to each other, so V- = V+. We use these two properties which are not the only two properties by op amps but the two of the most important to this class for solving problems and that are associate 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 know is the voltage at this node at the bottom of our circuit, which is the ground node, the voltage is equal to zero. Which also tells us the voltage at the non inverting input is equal to zero. Using the properties of the ideal op amp we know the voltage at the inverting input is also zero which makes it it a zero volt at this node at the inverting input of the op amp. So, now we've used that first, or second property of the ideal op-amp, V- = V+. Now we're going to use nodal analysis, and we're going to sum the currents into this node which we've identified at zero volt potential. And if we do that we'll get something that involves Vout is 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 with the sum of voltage into, or sorry, we're going to sum the currents into, node 1. So if we start with the four kilo Ohm resistor and we want to sum that current it's going to be 12 volts- 0 volts / 4 kilo Ohms for the current through the 4 kilo Ohm resistor flowing left to right into node 1. We also have this 2 milliamps source which is flowing into node 1 as well. So we're going to add the 2 milliamp source. In addition to that, we have the current which is flowing through the 3 kilo Ohm resistor back into node 1. It'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 op-amp, ideal op-amp. want 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 Vout, we end up with Vout = -15 volts. Now, we want to find Iout. And Iout is our current through our current back in to the Op Amp. So we can use the node at the output, maybe we call this node 2. And we can use Kirchhoff's Current Law at node 2 to find that current Iout. So this is KCL at node 2 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 the current out through the 6 kilo Ohm resistor, we start with the voltage on one side minus the voltage on the other, which is 0 volts- Vout. We know Vout is -15 volts divided by 6 kilo Ohms. So it's 0- a minus 15 for Vout / 6 kilo Ohms. That's the current flowing up through the 6 kilo Ohm resistor. We also have the current flowing left to right through the 3k resistor, which is 0 volts- Vout, Vout- 15 volts / 3K and the other one we have is is we have Iout, and in fact it should be a -Iout since we're assuming the currents into it. We should have a -Io instead of a +1o for that current. And those are all the currents into node two and the sum is equal to zero. So as you can see we end up with equation which we only has one unknown, which is Io and Io comes out to 7.5 milliamps.
