The topic of this problem is Energy Storage Elements and Operational Amplifiers. The problem is to find V out in terms of V in in the circuit shown below. We have a circuit that has an op amp in it with resistors and capacitor. As well as a input voltage, V sub in, which is tied to the inverting input of the op amp. The output is measured at output of the op amp for this problem. So we want to use the properties of the ideal op amp, as well as our knowledge of nodal analysis to solve this problem. So we'll start with the circuit symbol for the ideal op amp. The ideal op amp has two inputs. It has an inverting input with the current associated with it. It also has a non-inverting input a current associated with it as well. It also has voltages associated with these inputs. And we know that there's properties of the ideal op-amp that help us solve our circuits. The first of which is that the currents and the inputs of the op amp are equal to 0. There's no current that flows into the input of the op amp. And the second one is that the voltages at the inverting and non-inverting inputs are equal. So we'll use these two properties along with our knowledge of nodal analysis, to solve this problem, and what we're looking for again is V out in terms of V in. So let's start with our knowledge of op-amps and see where it leads. So we know that this voltage at the non-inverting input of the op-amp is 0 volts because it's tied to ground. We know that the voltage at the inverting input is the same as the non-inverting input so the voltage at this node is also 0 volts. And so that gets us a long ways toward solving the problem. It allows us to then take advantage of Kirchhoff's current law and nodal analysis to write an equation around this node for the currents that are flowing into it or out of it depending our choice on which one we use. So if we use Kirchhoff's current law and we call this node one and we're going to use Kirchhoff's Current Law on node one and we're going to sum the currents that are flowing into node one. The current through the resistor flowing left to right is going to be V sub n minus 0 divided by R. A current flowing right to left through the capacitor is going to be using the relationship for capacitance and the IV characteristics of our capacitor, it's going to be equal to C dv dt. And our V in this case is V out which is the voltage that the rightmost node and then minus 0 divided by dt because we know that the current through a capacitor is C dv dt. And we also have the current which is flowing into this node out of the op amp and we know that current is zero through our properties of the ideal op amp. We're going to add it in just for completeness and the sum of those is equal to zero through Kirkoff's Current Law. So what we end up with is V sub in over R is equal to minus C d v0, dt. So if we want to solve for V 0t, then we're going to have to take our equation and rearrange it. So V out of t is going to be equal to minus 1 over RC times the integral from minus infinity to t of, Our input voltage, And so if we solve that we end up with V out of t, Which is equal to minus 1 over RC and we're going to break our integral up into two separate integrals. We are going to have integral from t0 to t, Again, Vin of x dx, And our other integral is going to be from minus infinity up to t0. V of n of x, dx, so this second part of our integral relationship is what we call the initial condition. So four time t is equal to t0. A lot of times t0 is equal to 0. It's when the circuit is turned on. This would be the initial condition for the output voltage. So if the output voltage were at some non zero value before we turned on the circuit, in this case, it wouldn't be, then you would have to replace this by whatever that voltage level was. And then we'd have this integral of v in, of t, dt. So what this represents is a voltage output which is a integral of the input voltage, and because of this this circuit is known as an integrator. It's an integrator circuit because it takes the input voltage and takes the integral of the input voltage, to obtain the output voltage.