The topic of this problem is Operational Amplifier Circuits, and the problem is to find V out in the circuit shown below. It's a circuit with multiple resistors in it, one voltage source, an independent 12-volt source, two op amps, and the output is measured across the load resistance, which is 40 kilo-ohms at the output of the second op amp. In order to solve this problem, we have to use the properties of an ideal op amp, so we start with the symbol for an ideal op amp. It has an inverting input with a current associated with it, I-, and a volt associated with it, V-. It also has a non-inverting input with a current, I+, and a voltage, V+, associated with it. It has an output and a ground, we know that the properties of an ideal op amp are that the currents at both of the inputs are 0. We also know that the voltage is at the inverting and the non-inverting input are equal. Those are two of the properties of an ideal op amp that we use to solve linear circuits. So, when we use along with our knowledge of nodal analysis and measure analysis in order to solve this two op amp circuit that we have in this problem. So, let's start with finding voltages at the input. So, if we can find the voltage at this non-inverting input of the first op amp, then we know the voltage at the inverting input of the op amp, and that will allow us to work further back into our circuit, ultimately getting back to our voltage drop across our 40 kilo-ohm resistor. So, let's sum our currents into this node, into the V+ node, into our non-inverting input node. Let's call this Node 1, so if you sum currents into Node 1, then we have first of all the current through the 40 kilo-ohm resistor, which is minus 12 minus V+. Divided by 40 kilo-ohms. We also have the current which is flowing through the 20 kiloohm resister. And that current is going to be 0 volts because this is a ground point in our circuit, the bottom of the circuit. 0 volts- V+ Over 20k. And we have the current which is flowing out of the non-inverting, we know that through ideal op-amp that that current is equal to 0. And so the sum of those using Kirchhoff's current law At node one is equal to zero and so it's an equation which just has the one variable that is the voltage at the inverting or the non inverting input of our first op amp. If we solve for that voltage, V plus We get minus 4 volts. So it's minus 4 volts at node 1. And it also tells us that it's minus 4 volts at the inverting input as well. So we need to work further back into our circuit. And so we rely on the first property of an op-amp to get us a little bit further into the circuit. That is, that the current into each one of those inputs of the op-amp is always equal to zero. If that's the case, then there's no current flowing through this 10kOhm resistor, because it is the current flowing into the Inverting input of our first op-amp there's no current flow in through the 10 kilo ohm resistor therefore there's no voltage drop across the 10 kilo ohm resistor.So what that tells us is, that this voltage between the 10 and 20 kilo ohm resistor at the top of our circuit is also a minus 4 volts, at that point. Now we can use that to work a little bit further into our circuit. Let's look at our circuit a little bit closer. If we look at V out, V out is measured across the 40 kilohm resistor. It's the voltage at this output node of our second We also notice that there's no voltage drops between these two points along the feedback path of our second op amp. So that also tells us that V out is the voltage at this point in our circuit. At the top of our circuit is equal to the voltage at the inverting input of our second op amp. We know that that voltage is the voltage at the non-inverting input of our second opamp as well as using our second properties of an opamp. So we have a V out which can be measured at this point in front of our second opamp. So that helps us a little bit. because what we can do is we can start summing voltages, or summing currents in this case, into nodes and come up with equations which allow us to solve for V out. Let's see if we can do that. Let's look at the top node which is at minus four volts first and let's call that node one. Sorry if I had node one let's call this node two. So let's look at node two and we're going to use Kirchhoff's current law at node two and we're going to sum the current into node two. So it's a current first of all through the 10 kilohm resistor which we know is zero, it's a current through the 20 kilohm resistor Up through the 20 kilo ohms resistor. Let's call this node 3 down below. So in this case the current up through the 20 kilo ohms resistor is going to be V sub 3 -(- 4) volts. Dividend by. 20 kilohms that's a current up through our 20 kilohm resistor we also have the current flowing through the 20 kilohm resistor at the top of the circuit. It's going to V out minus a -4 divided by 20 kilohms for the current flowing into node two. So V out Minus a minus four volts divided by 20 kilohms. V sub 0 minus a minus four volts divided by 20k. And that's all the currents that are flowing into node 2. Because we know that the current flowing through the 10-kilo ohm resistor is 0 when I add it in there just for completeness, and that's equal to 0. So that's an equation which has two unknowns in it. It has V3 as an unknown and also has V out as an unknown. Let's look at this node where we have VOut measured, let's call this node 4. So let's sum the currents into node 4. So all these are summing currents into the nodes. So if we sum the current into node four, we first of all have the current up through the 10 kilohm resistor is going to be zero volts minus Vout over 40k. We have the current flowing left to right from node three To node 4 through the 20 kilohm resister, so that's going to be V3 minus V out divided by 20 k And we have the current which is flowing out of the non-inverting input of the second opamp, and we know that current = 0 through the properties of an ideal opamp. Those are all the currents flowing into node 4, and the sum = 0. So now we have in the second equation, which is independent of our Other equation that if we used these two equations simultaneously to solve for V sub 3 and V out, we can do that. And so V out can be determined by finding these voltage levels. So if we use this two equations to solve for V out simultaneously, then we end up with a V out which is equal to two thirds V sub f, V sub 3. And that's equal to minus sixteen-fifths of a volt.
