The topic of this problem is Operational Amplifiers, and we're going to work with a circuit that is an inverting amplifier. The problem is to find V sub out in terms of the input voltages. In our case, we have one input voltage, it's V sub n, that's attached to a branch with Rs of 1, and ultimately, to the inverting input of our op-amp. We know that, for op-amps, we have a number of properties, which help us solve our circuits. So, if we look at, just a general op-amp configuration, and this is the circuit symbol for it, we have currents associated with the inverting input and with the non-inverting input. We also have voltages associated with those inverting and non-inverting inputs as well. And we know that some of the properties of ideal op amps are that the current into the op amp are equal to 0. They're equal to each other and they're equal 0. We also know that another important property for us when we're solving for these op amps is that the voltage at the inverting input of the op am is equal to the voltage at the non inverting input of the op amp. So V minus is equal to V plus. So we're going to use these two properties when we solve circuits with op amp in them for linear circuit applications. So let's look and see how we can solve the problem that we have in this example, using our the two properties of the op amp. If we use those properties, one thing that we know is that the voltage at this point is equal to the voltage at ground, which is zero volts. So we're at 0 volts at this non-inverting input. So if it's 0 volts at the non-inverting input, using our second characteristic of op amps, then we're also at 0 volts at the inverting inputs. So that's 0 volts as well at that point. So once we have that, we can use tools that we've already learned in the past to solve this problem. Again, we're looking for V out in terms of the input voltages. In our case the one input voltage which is tied to our inverting input. So if we use Kirchhoff's current law at that node, we'll call it node one. This is Kirchhoff's current law at node one, then we can sum the currents into that node. We have a current through R1, we have a current through R sub F and we have the zero current associated with the non-inverting or the inverting input. So let's first look at the current through R sub 1. It's going to be the N minus 0 over R1. It's voltage at this point minus the voltage on the other side of the resistor divided by the resistor value. If we look at the feedback path in the current through R sub f, is going to be V out, minus 0 volts, divided by Rf. And we know that the other current, which is associated with the inverting input, is equal to 0. And the sum of those currents using Kirchhoff's current law is equal to 0. So we're summing currents into node 1. So we have an equation which relates V out to Vn along with our resistor values which we've associated with our circuit. So if we solve that equation for the V out, we end up with a V out which is equal to minus R sub F divided by R1 times V sub n. So if we look at that relationship, the output voltage is equal to an inversion of our input voltage, as amplified by the factor Rf/R1. So we can use our external components to control properties of our op amps. In this case, the amplification factor for an input voltage, with respect to its output. So the negative sign gives us this inverting. Characteristic that we call inverting operational amplifiers. And so V out is equal to V in times a factor which is a negative number. It could be a number greater than one, it can be a number less than one, or it can be a number equal to one where R sub f is equal to R1. Depends on the application.
