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. So we have a circuit that has one voltage source in it as a five volt source and I out is measured as the current in the bottom leg of our circuit in this configuration. V out is the voltage measured at the output of our op amp, so to solve this circuit we're going to use the properties of an ideal op amp. First of all, we have the symbol for an ideal op amp, which has two inputs, an inverting input and a non-inverting input. Each one of those inputs has a voltage and a current associated with it. We know that the ideal op amp has these properties. First of all, the currents into the op amp are equal to zero. There's no current flowing into the input of an ideal op amp. And that the voltage is that the inputs are equal. V minus is equal to V plus. So, these are the properties that we're going to use to solve our circuit that is given in this problem. So, the first thing we're going to use is this first property of the circuit that there's no current flowing into the ideal op amp. So if there's no current flowing into this non inverting input of the op amp, that means that I out is a current which is confined to this loop that we have at the bottom of our circuit. So if we want to find I out, we can use Kirchhoff's Voltage Law to do that and we can use Kirchhoff's Voltage Law around that mesh. And in this case I out is measured in a counter clock wise direction. So we're going to sum our voltages up counter clock wise around this circuit. So if we start at the bottom right hand corner and start around the circuit we first encounter the negative polarity of the minus 5 volts source. So it's minus 5 volts. And then we have the volts that drop across 20 kilohm resistor which is 20k(I out) plus the voltage drop across 10k as we continue around this loop. And the voltage around the 10k is going to be 10k(I out), and then we return back to the starting point. And so we know through cross voltage law the the sum of those voltages is equal to 0. So if we use that equation and we solve for I out, then we end up with an I out which is equal one-sixth milliamps. Now that we know I out, we can go back and perhaps find V out. And the way we're going to do that is using the second property of the ideal op amp. If we know what the voltage is at the non-inverting input it's going to be equal to the voltage at the inverting input. So can we find the voltage at the non-inverting input, which is the voltage at this point in our lower loop. It's going to be equal to 10k times I out so the voltage at the noninverting input is equal to 10 kilohms times I out and I out is 1/6 milliamp. So, our voltage at the non-inverting input is equal to 10 6th of a volt. We also know that that is equal to the voltage at the inverting input. So, the voltage at this node, it's a non-inverting input. I'm sorry, at the inverting input, has a voltage level of 10, 6 volts. Then, we can use Kirchhoff's Current Law at that node. Let's call it, node 1, and we're going to sum the currents into that node, in order to find an equation, which has the V out as the unknown. So, let's sum the currents into that node. We have the current through the 10 resistor flowing from left to right. It's going to be 0 volts minus the voltage here, which is 10 6th volts, divided by 10k. 0 minus 10 6ths. Divided by 10k. We also have the current which is flowing from right to left through the 20 kilohm resistor. It's going to be Vout minus 10/6 volts Divided by 20 K. So that's the current flowing right to left into node one through 20 kilom resistor. And we also have the current which is flowing out of the inverting input of the op amp. We know that that is equal to zero. We're going to add it in for completeness. That's all of our currents, they're summing into node one. So this is an equation which just has V out as the unknown. It's the only unknown in the equation. So we can solve for V out and if we do, we end up with a V out equal to 5 volts. So now we I out, which we've been looking for, and also V out.
