The topic of this problem is operational amplifier circuits. And the problem is to determine V out in terms of the input voltages. In this circuit, we have two op amps, an upper and lower op amp. We also have input voltages V sub 1 associated with the non-inverting input of the upper op-amp, and V sub 2, associated with the non-inverting input of the lower op-amp. We have this resistor network, at the output of the op-amp where V out is measured across that series of resistors and parallel combinations of resistors. So we're going to use our properties of the ideal op-amp to solve the circuit, namely that the currents into the op amp are equal to 0. And the voltages at the inverting and non-inverting input are equal to one another. So if we use that property, then we know that V sub 1, which is tied to the non inverting input of the upper op amp is the same voltage down below at the inverting input and we have the same voltage out at this node at the output of our op-amp circuit. So that node is at V sub 1. Similarly, if we look at the lower circuit of the lower op-amp which is part of our circuit, we see that the voltage at the node between R2 and R1 is equal to V sub 2 using the same approach. So, now we have our input voltages conveyed over to the output side of our circuit. Now if we use Kirchhoff's Current Law to sum the currents into the node where the voltage is V1 and also into the node where the voltage is V2, we could find some equations which relate those input voltages to the output voltage. So let's start this upper node where the voltage is V1. And so we're going to sum the currents into that node. We the current down through R2. We have the current out of the inverting input of the op-amp. We have the current up through R1 and we have a current up through R sub G, there's four currents. So let's start with the current through R2 at the top coming down. That current is going to be V out minus V1 divided by R2. The current out of the op-amp is 0, because there's no current flowing out of the op-amp. The current of through R1 is going to be V sub A, minus V1, divided by R1, plus the current through R sub G which is V2 minus V1 divided by R sub G. So there's our four currents and they're equal to 0. We're going to call that equation one. Equation two, for the node down below, sum into currents into that node. First of all, the current up through R2 is going to be 0 minus V sub 2 divided by R2. A current pop-up is 0. The curent down through R sub G is going to be V1 minus V2 over R sub G. And then we have the current through R sub 1 from node A down to the node where the value of the voltage is V2, so it's going to be VA minus V2 divided by R1. That's our last current and some of these currents is equal to 0. So we have two equations which have the input voltages V1 and V2, the output voltage V sub out and this V sub A for node A in our circuit. So, if we take for example, equation two and we solve equation two for V sub A. And then we take V sub A expression and plug into equation one, then we'd have an equation that related V out and V1 and V2. So if we indeed do that, then we end up with a V out which is equal to V1 minus V2, so it's a difference amplifier, times this factor which is a function of the resistors that are in our circuit the R1 and R2 values and the R sub G values as well. So it's a difference amplifier where the voltage is equal to the difference in the input values for the voltages times this factor, which is a function of the resistors.
