Now we're in a position to use the machinery that we've developed earlier, to do an AC analysis of a simple circuit. Now, the one that we're going to look at is just an RC circuit like we looked at before. Now the difference here is we're not going to assume that there is a switch, but we're just going to assume that this circuit has a source of alternating current, voltage, AC voltage driving this circuit. Now again, keep in mind, this is any old frequency omega. It doesn't matter what frequency it is, it's just a general frequency omega. And so, if I can solve this circuit for the, a general frequency, I know the solution of this circuit for all frequencies. So, what we're looking for is the current as a function of frequency. Now, after we know the current, we can then find the voltage across this capacitor as a function of frequency. Now, we're going to use our phasor notation and write Kirchhoff's voltage law here, just like we did before. And we'll start at the ground and then go up through the AC source of voltage. So, you pick up a voltage V, and then when we pass through the resistor, we drop a voltage equal to I times R. And then when we go through the capacitor, there's a voltage drop, which is the impedance of the capacitor times I. So, we're treating the impedance here just like resistance. And so instead of this equation being, having derivatives in it like it did before, actually the derivative was over here. instead of having derivatives, these are just algebraic equations now. Now that is the great simplification in AC circuit analysis, is that these differential equations where we had the a function, like the current and derivatives of the current, and voltage and derivative of the voltage in one equation. Instead of having those differential equations in the frequency domain here. We have simple algebraic equation, and we can just solve these using the rules of algebra, but we have to keep the track of the complex numbers. So, I can solve this for I. So, I'll put the I terms on one side and the voltage terms on the other side, and then divide through by the factors, multiplying I and I get I is V over this factor. And then the voltage across the capacitor is the impedance of the capacitor times the current. Now, this current here, this is a current going through both elements. It's the same current going through R and C. If I want the voltage across the C, I just multiply this current times 1 over j omega, C. So I plug this in here and I get this expression, and now I'm going to take and just multiply this j omega C through the denominator. And so it cancels out this term and I get a 1, and then I get j omega C times R. So here's the voltage across the capacitor, is equal to this factor times the voltage driving the circuit. Now, what I want to do is compute the magnitude of VC. So, we introduced back when we were talking about complex numbers, we introduced the magnitude of a complex number. So, I don't care about the phase, I just want to know the magnitude of the voltage. And so, the magnitude of VC over V is just the magnitude of this factor here. And I just rewrote it as one plus j omega CR. So, I can write this magnitude of 1 over 1 plus j omega CR is, the factor times its complex conjugate. And I get the complex conjugate just by changing the sign in front of the j, then I take the square root of that entire expression. And so, if I multiply these two, then I get 1 plus omega squared, C squared, R squared. And the cross terms, here's plus j omega CR, and this is minus J omega CR, they cancel. So here is the expression for the voltage, magnitude of the voltage across the capacitor divided by the voltage driving the circuit. So if I plot that, VC over V. And now we're plotting this as a function of frequency. So remember it's very, it's very important to make a distinction between this and when we did the transient analysis of the circuit, we plotted the response of the circuit as a function of time. Now, we're plotting the response of the circuit as a function of frequency. Now, I'm taking and there's I'm expressing the frequency as the frequency divided by 1 over RC. Now RC, R times C is the so called time constant of the circuit, it has units of time. So if I multiply a frequency times a time, this is dimension less. so what I'm really doing is I'm taking my frequency and I'm normalizing it by 1 over that time, that characteristic time of the circuit. Now, this point when omega is 1 over RC, then that makes omega times RC equal to 1. And at that point, this factor, 1 over 1, becomes 1 over 1 plus 1. So that's 1 over the square root of 2 and that's 0.707, and that's this point right here. So this is the frequency when omega equals 1 over RC. The response of this circuit is down to 1 over the square root of two, times the response at DC. So this is an example of a low pass filter. What happens here, and it's much easier to see here in the frequency domain than it was when we were talking about this in the time domain. The low frequencies pass through with very little attenuation. The voltage across the capacitor is almost equal to the voltage driving this entire circuit. As I, the frequency goes up, the size of the voltage appearing here across the capacitor gets smaller. And when omega is 1 over RC, it's 1 over the square root of 2 times this driving voltage. when omega is much bigger than 1 over RC, the response goes down considerably. So high frequencies do not pass through this circuit very efficiently. Low frequencies are passed through this circuit very efficiently. And there is a very solid connection between this and the transient analysis we were talking about. The, at low frequencies, there's plenty of time for this capacitor to charge up and this voltage to become quite large. At high frequencies, there isn't enough time for the capacitor to charge up and for this voltage to become very significant. And so, high frequencies are attenuated by the circuit, low frequencies pass through this circuit.
