Chapter Six. I want to now look at another series circuit, this time it's going to be RC or a resistance capacitive circuit. And it's going to be a series circuit. So the current through each component is equal. And I'm going to lump the resistant and capacitive impedance together to give us one lump impedance, and I'm going to look at the voltage across that impedance and the current through that impedance. Now that I'm going to use a sine wave for the voltage source, and the voltage source is going to be described and it's exactly the voltage across the impedance as well is going to be described by some maximum voltage V subscript m times the sine of the angle omega t, where t is the variable And it's going to be displaced by a lead in angle of phi subscript v. Now to look at that in the time domain graph, this is what it would look like.

The sine wave goes from zero to a positive maximum VM, and then back through zero to a negative maximum v m. And it's going to start a little bit ahead of the origin by a factor of phi subscript v. so this is the picture that describes that equation that we just wrote in green. Now the angular velocity is given by omega and that is degrees per second. So as we start to plug in the seconds, 1234 or wherever it is, along with horizontal axis will get a different value for the now the current will also be sinusoidal. And it will be at the same frequency as the generated voltage. However, it's going to be out of phase and a different magnitude. And the magnitude is I subscript M. and it describes the red graph that you see there, that rises up to a positive maximum, I am to a negative maximum negative I am and it starts also with the displacement of phi subscript i.

So they're both described almost the same way with a couple of different values in it, but they're also using the variable of time and the omega is the same for The for the current the angular velocity and it is also in degrees per second. So, if we plug in the seconds, we can plot the actual curve itself in the time domain graph that you see there. Now looking at it in phaser notation, we have a current phaser, which is the maximum 30 the magnitude is given by I subscript m, and it is at an angle of phi subscript i, and the voltage has a magnitude of v subscript M and it is lagging the current because we have a capacitive load on it, and it's going to be lagging by a certain amount of the current. We'll see that in a minute, but it's angle with the horizontal axis.

You is given by fi subscript v to in order to satisfy the two equations that we've already developed. Now with those two phasers on our graph, we can see that the angle between the current and the voltage is the difference between the current angle and the voltage angle in other words, phi subscript i minus phi subscript v. Now, the phasor diagram is a snapshot of the phasor at one particular moment and that those phasers are circulating at a speed of omega which is in our system 60 cycles a second so I can move those vectors anywhere I want in a circular motion because I can take a picture at any time that I want, as long as I maintain the phase angle between the current and the voltage and keep that the same. So, one thing is I want to use the current as my reference phasor.

So I'm going to put that at zero angle. And if I put that at zero angle, the angle of the current is going to equal zero or, or phi subscript i is equal to zero. So if I rewrite my time domain equation for the current, I'm going to replace phi subscript i with zero, that would make the formula now i is equal to i subscript m sine omega t. And if the angle of the current is zero Then phi is going to be equal to just minus phi subscript V or the angle of the voltage. And I'm going to rewrite that equation. Now, in terms of the voltage, I'm just going to switch the left and the right sides around. So, if the black phi is equal to minus phi subscript v then the phi subscript v is equal to minus phi.

So, now, I can rewrite that equation the time domain equation for the voltage and substitute minus phi for phi subscript v. And my formula now becomes the voltage is described as the subscript M which is the maximum voltage in a sinusoidal cycle. times sine of the angle omega t minus phi without a subscript. And as described by the phasers that you see in the upper right hand corner. I'm now going to have a relook at these formulas again and this time as they have been rewritten or recalculated. And the current is now going to be I am sine omega t now sine omega t, or sine of A variable angle goes from zero through to a positive one and then back through zero to a negative one and then repeats itself. So for multiplying sine omega t, by I am it'll go to a maximum of I am and down through zero into a negative I am, but it'll start at zero, as we've indicated here now If we plot the voltage, the voltage is similar in that it's repeating itself, and it will go to a maximum value of positive VM and to back through zero into a negative VM.

But because it's displaced by minus phi or the phi without a subscript colored in black, it's going to be delayed in starting up from its zero point as indicated here by an angle of minus phi. Now, we've seen these calculations before, where we're going to calculate the, the power that the instantaneous power that's dissipated by these two values, the current and the voltage. So I'm going to go through the Reader's Digest version of this a little bit fast because we've seen it before. For an inductor and it's very, very similar in in the case of a capacitor only the phase shift is a little bit different. But we're still going to get a instantaneous power whose frequency is double that of either the current current or the voltage. And whenever the current is zero, the power is zero whenever the voltage is zero, the power is zero.

And whenever the voltage and current have a light sign, in other words, they're both positive, then we have a positive voltage. And whenever we have the current and the voltage have a different sign than the power will be negative. It's shifted a little bit because of the fact that we're dealing with capacitors now instead of inductors. But the characteristics are the same as you see there. I'm going to take a closer Look at this equation. And we did this before for the series RL circuit.

But now we're dealing with a capacitor. And it's slightly different. If you can remember, we had a positive black phi in the first equation with the inductor, this time we have a minus phi that we're dealing with. But we're going to use the same trig identities and just follow through mathematically to the bottom and see what we come up with. So the first trig identity that we used was the sine x times sine y. And we're going to let x equal omega t minus phi, and we're going to let y equal omega t. And the portion that we're dealing with in that equation, I just recovered in blue as I did before, so we can see it a little bit better.

And the equation now looks like this. And the two omegas in the first part of that right hand side, we'll cancel out because we've got a positive and a negative of the same thing, they will cancel out. And we have two omega T's. So we can just rewrite that as two omega t in the last portion of that equation. We can rewrite the equation now and it'll look like this. And now I want to look at just the last portion of the equation.

And again, use the same trig identity we did before, the coax x minus y, which is equal to cosine x cosine y plus sine x sine y. This time, we're going to let x equal to omega t, and y is going to equal phi. Now we've got to be careful here because we have a minus phi, but we're using the trig identity coasts x minus y. So I'm only going to I don't have to worry about the minus sign. I'm just going to make a substitute for phi. And I'm going to rewrite the equation And again, I'm going to color the portion that I'm dealing with in blue, so we can follow it better.

And our equation now will look like this. And if we collect like terms and rewrite the equation, and the other thing to keep in mind is that coasts minus x is equal to cosine plus x. That's another trig identity. And we can rewrite the equation getting rid of the minus sign for the phi term in the first part of the equation. And the rest of the equation I'm just rewriting here. Now if I collect like terms, this is what I come up with.

VM I am over two times the quantity one minus coast tote to omega t times coast phi minus VM, I am over two times sine two omega t times sine phi. Now this, if you recall, and if you don't, you can go back and have a look at it. This is almost the same equation that we had for the inductor with the exception of this minus sign. And we're going to talk about that right away. I'm going to place the phasor diagram in the upper right hand corner just for our reference as we go along. And keep in mind that this equation that we see here, is describing the instantaneous power dissipated in the RC circuit.

And for the moment, I'm going to put the time domain current and voltages on the page, and I'm going to superimpose the computer generated instantaneous power graph in the time domain. Now, if we look at just the left hand portion of the equation, VM, I am over to one minus cosine two omega t times cosine phi. It's the orange curve that you see on the diagram here. And it is twice the frequency of course of the each of the currents and or voltages. And the variable in the time domain, of course, is the key. The rest of the values of that portion of the equation are constant as long as we've got a constant AC voltage in a constant AC current the variable portion of this equation is the one minus cosine to omega t, which is represented by the orange graph, any change in the VMI M over two or the coasts phi will just change the magnitude of that curve.

The second portion of the equation VM, I am over two sine two omega t times sine phi is described with the black curve that I've drawn in the slide here. And the variable portion is the time domain and I'm just bringing out the actual variable portion of that part of the equation which is signed to omega t, the rest of the stuff around that variable portion. The curve is a constant, the VMI, M is a constant, two, of course, is a constant, and sine phi is a constant. So I'm just bringing out the two variable portions of the equation. And because it gets kind of confusing looking at all the rest of the curves that are there, I'm only going to plot the two portions of the equation, which is the VMI M over two, one minus cosine omega t times cos phi, which is in orange, and the M or I should say the minus VM is over two times sine two omega t times sine phi, which is in black.

What we are looking at here, just as a quick reminder is the instantaneous power consumption of the RC circuit. And I've drawn it in the time domain here in the blue curve that I You've just seen superimposed over the other two curves. But most of the time, as I have said previously, we are looking for average power, that's what your house is, is being built on as far as the power utility is concerned. So the meter on your house is actually reading average power. So if we look at the average power of that equation, and we take the variable portions of it, which is the coast to omega t and assign to omega t, they will average to zero because any sinusoidal function will average to zero over a long period of time and a long period of time is longer than a second. And if we average close to omega t, over a long period of time, it will average out to zero, leaving just the one inside that square bracket.

If we average out the sine to omega t, it will average out to zero, meaning the second portion of the equation is zero. So if we now look at what is left, after averaging everything we now have the average power flow have are the average power consumption of the RC circuit. And those portion This is what's left after we average everything out. So the average power consumption is the M, Im over two and remember that voltage V m is the maximum voltage of the sinusoidal wave and the i m is the maximum current of the sinusoidal current. And that is all over two and it's times cosine phi, which is the phase angle between the current and the voltage and everything in that equation is constant. As long as the voltage and the current, the AC quantities are constant, we will have a constant average power flow, which basically looks like it might even be equivalent to a DC quantity.

However, the just it suffice to say that the average power consumption is given by VMI M over two times the cosine of the phase angle between the current and the voltage. In actual fact, if we take a close look at what we are left with the left hand portion of this equation is actually on average equal to the power average that we just written there VMI M over two times cosine phi because the right hand portion which we're going to have a closer look at in the next slide. is actually averages out to zero. So the p average is the average of the sinusoidal. Orange Line that's there. And as I said, it looks pretty constant.

And it might not even be equivalent to a DC value if you want to look at it that way. And we'll talk more about that later. For the moment, I'm going to replace the orange and black curves that are computer generated with the equation for instantaneous power just so we can have a closer look at that. What we're looking at is two formulas if you would that are added together or in this case, one is subtracted from the other because of the minus sign. The left hand portion of the of the equation is denoted by the orange sinusoidal curve that you see there, and the right hand portion which is actually a minus a variable minus formula, and that's denoted by the black curve line in the time domain. And those two functions together will add up to the instantaneous power or in the case of the second portion, it's subtracted.

But if we're adding a subtraction if you want and if you combine the two which is a better term, the orange line and the black line you will give the instantaneous power. So I just want to have a look at that and look at a couple of points just to verify it in our own mind and it it helps to visualize it as as we move on. So the times where the black line and the orange line are both to zero, that indicates that the instantaneous power has to be zero. And that is indeed the case you can see it there. Every time that both the orange line and the black line or zero, the blue line is zero. Now the other time that the power is zero is when there's an equal portion of the orange line above the line above the zero line and an equal portion of the black line below the zero line.

And I'm going to indicate that with this dotted line here and the arrows. You can see that the power at that point is zero. But there is still a positive quantity for the orange line and a negative quantity for the black line. But they are equal and opposite and when added together We'll give us zero. And that is in again indeed the case that the instantaneous power, the blue line crosses the zero line at that point. So those are the two zero points.

And they verify the fact that while at least they satisfied the equation that we have evolved here. Now, I want to look at the second part or the right hand portion of the equation. And that is given by minus VM, I am all over two sine two omega t times sine phi. I would like to concentrate on the second half of this equation, the minus V VMI M over two times sine two omega t times sine phi, which is the black curve in the time domain graph that we have been looking at. And I want to get a better understanding of what is going on here. This black curved line is sinusoidal and its frequency is two omega t. And what it is telling me is that the power the instantaneous power to our load for this part of the equation is for half the cycle is flowing into the load, but the other half of the cycle is negative which means the power is now flowing from the load back into the system.

So, what is happening, the actual thing that is happening and this time domain curve describes is the power that's flowing into the capacitor to build the electrostatic field. And once it builds the electrostatic field, the field will then because of the AC voltage is being applied to the circuit. And the current is flowing in at once the the electrostatic field reaches a maximum which takes which the actual instantaneous power isn't a maximum, then the field starts to collapse and the field will start to collapse and then go in the other direction when it goes into the other direction. It is actually pushing power back into the circuit and then it the field will collapse after that negative field is built and it'll go to zero, the power will go to zero, then it'll start to build in the other direction. So you get this cyclical flow of energy.

In and out of the capacitor to build a magnetic field in one direction. And then in the other direction. The average power flow as you can see, is going to be zero because there's an equal amount of power going into the capacitor as coming back out of the capacitor. This power is actually known as reactive power, and it's given the designation as Q, and the instantaneous reactive power is given by the equation minus VMI M over two times sine two omega t times sine phi. But we'd like to know something that's a little bit more meaningful. In other words, we'd like to get a handle on the actual power that is flowing and if we took the average of this of this quantity Would average to zero, so that's not going to be very useful.

So we would like to move the curve down just so the crest of the curve is touching the horizontal axis. And then we can calculate the average reactive power that's flowing in and out of the capacitor, by analyzing it the same way as we did with the real power or the orange curved line that we did in a few slides back. But in order to do that, we have to readjust our formula because the formula that we see here does not describe the curve that we've just moved down. But it will, if I put a one in the equation and subtract sign tool me get T, that's just changing the, the formulas slightly and that will cause the curve to move down. Now I can animate lies the average mount by the same reasoning that we used when we did the orange curve and that is the sinusoidal parsh portion of this curved line goes to zero because it will average to zero so sine two omega t is zero, the VM i m over two is constant as is sine phi, which is the constant phase angle between the current and the voltage and what is left is the average reactive power, which is equal to minus VM I am over two sine phi.

Before going on, I want to take a nother look at visually what's going on with this RC circuit. And we've analyzed the equation here and broken it out into two parts. That are arithmetically added together to give us the instantaneous power flow. The first part is the actual power that's flowing into the resistor. And that's given by VMI M over two times one minus cos two omega t times cos phi. And that is denoted as the real power is really the power that's flowing into the circuit into the resistor.

The resistor is taking electricity and converting it into heat. And if it was a light bulb, it would be converted into heat and light. But let's just say it's a resistor for now and it converts the power or consumes the power in terms of real power, which is shown by the orange curve in the time domain curves that we've shown here. And what happens is the power flows into the resistor returns to zero Returning flows into the resistor then returns to zero. So it's actually pulsating 60 times a second. Now, the black line is the actual power that's flowing into and out of the capacitor.

And the average flow of power here as we've already studied is zero. But it is actually taking power in to build an electrostatic field in the capacitor and then flowing out in the other direction as it builds the electrostatic field in the other direction. And this is known as reactive power. So there's two portions and two things that are going on simultaneously in this circuit at a very fast speed. In other words, 60 cycles a second. Those two curves, keep in mind still add up to the instantaneous power flow into the whole circuit and In other words, it's the amount of power flowing out of the source voltage.

And it's made up of the arithmetic sum of the real power and the reactive power. So looking at this series RC circuit, we've developed a formula for calculating the average power, and that's given by VMI M over two times cosine phi, where the angle phi is the phase angle separating the current and the voltage. And just to have a quick review of what the current and the voltage looks like in that time domain, it is actually two sinusoidal waves, both of them have the same frequency, but separated by a phase angle and that phase angle is phi that you see in our formula and the formula Then can be also written as VM over root two times I am over root two. All we're an all I'm doing is mathematically splitting out the denominator two into two factors root two and root two, because we know from previous slide presentation that the RMS voltage can be VM all over root two and the rms current can be I am all over root two.

So now, the formula can be rewritten as the average power can be the RMS voltage times the rms current times the phase angle between them. And since most equipment in the AC world is measures voltage and current, then power and watts in RMS terms This is the famous formula for calculating average power using RMS voltage and rms current and the cosine of the angle between them and this is real power. We know that the reactive power is given by a similar formula only using sine phi and it is we have discovered for a capacitive circuit minus VM I am all over two times sine phi which also can be rewritten the RMS times I RMS times sine phi and that gives us the reactive power and the reactive power in a capacitive circuit is minus Looking at the phasers with respect to the voltage and the current, when we started the analysis, we place the current phaser on the horizontal axis, which meant that the phase angle of the current was equal to zero, which meant the phase angle for the voltage was a minus angle, it was actually lagging behind the current.

However, when we went through the mathematics, we found out that all we needed to know was the phase angle between the voltage and the current, we did not have to know what the actual phase angle of the voltage was or the phase angle of the current. All we needed to know, though was the phase angle between them, because that's what our formula tells us. So in actual fact, and the phasers are rotating 60 times a second. So that the phase angles could be stopped at any position. And there's still the same phase angle between the current and the voltage and that is given by phi which is written in black there, it has nothing to do with the absolute values of each of them it has to do with a difference of the two. So, what you have to note is that phi is the voltage phase angle minus the current phase angle, but the voltage phase angle is less than the current phase angle.

So, the difference is going to be less than zero. In other words, phi is going to be negative. Hence, the Q average is going to be negative. However, the p average is Positive because we're taking a cosine of the angle and a cosine of any angle is the same whether that angle is positive or negative. If we were to take voltage and current readings of the RC load in this series circuit and we took those readings in RMS values, the multiplication of the magnitude of the voltage in RMS and current in RMS would give us a factor known as the apparent power of the load. And that is given by multiplying the RMS voltage times the rms current regardless of any the phase angle between them.

Now if we plotted that apparent power as a face whose magnitude was the RMS times I RMS and we plotted that phasor at the same angle as which is between the current and the voltage. Then we took the projection of the phaser on to the horizontal axis that would give us the RMS I RMS cosine phi, which is real power, real average power. Now, if we took the projection on the vertical axis of that phasor that would give us minus v rms times I RMS times sine phi, which is the formula for reactive power which is q average. It's the average reactive power of course. Now, if we took that vertical phasor of reactive power and added it to the horizontal phasor of the real power that would give us apparent power. So, if we add real power to reactive power it gives us the apparent power and that right angle triangle forms what is known as the power triangle.

Similar to the inductive circuit, the term coasts phi is known as the power factor and the power factor of any type of load is always a positive number And the angle associated with the power factor is called of course the power factor angle. And in this case I look capacitor of a vo load that has a leading current that is the currents leading the voltage or the voltage is lagging the current. In this particular case, the angle phi will be a negative angle because the voltage lags the current. So, far we have looked at and studied two series type AC circuits one with an inductor one with a capacitor and in the case of the inductor the current will lay The voltage and in the case of the capacitor or the capacitive circuit, the current leads the voltage. So now I'd like to just talk generically about these circuits now and talk about just the impedance instead of the fact that we have an inductor and a capacitor which actually gives us the characteristic lead like that we are looking at.

And I'm only going to refer to currents that lag and lead voltage from here on in. And in the case of the current legs the voltage, the angle of the voltage is greater than the angle of the current. And in the case of the current leads the voltage, the angle of the voltage is less than the angle The current So, the power factor angle which is given by the voltage angle minus the current angle is always positive for current lags the voltage and in the case of current leads the voltage, the power factor angle of the voltage angle minus the current angle, the power factor angle is always negative, these give rise to the power triangles that look like this. And in the case of the power train goes for the current legs the voltage the Q average or the reactive power is positive and in the case of the current leads the voltage the queue average or the reactive power is Negative the p average is positive, where current lags the voltage and current leads the voltage the power average is still positive.

One thing that I did not emphasize, but in the case of the power triangle, the real power is given by the letter P and the reactive power is given by the letter Q. And the VA or apparent power is referred to as S. There is a third case that we haven't really looked at in detail, but that is when the voltage and the current are in phase with each other. That means that the Voltage angle is equal to the current angle in the circuit or the power factor angle is equal to zero or the angle of the voltage minus the angle of the current is zero. That would give rise to a power triangle that would look like this. And cosine of the power factor angle is equal to the cosine of zero which is equal to one and the sine of the power factor angle is equal to the sine of zero which is equal to zero.

This means that the real power equation v rms I RMS times cos phi is going to be equal to simply the apparent power which is the voltage times the current in RMS values, which is equal to s. And as I said, this is the real power all of the apparent power is the real power or the average power. And of course, that would mean that the v rms I RMS sine phi is going to be equal to zero. So all of the reactive power is zero. In fact, there isn't any in that circuit. And I think you already realized that this can only happen when the impedance is purely resistive. In other words, there are no reactive loads present in the circuit or if there is they counterbalance each other and there that leaves only an impedance equivalent to whatever is the resistance value of the circuit.

Therefore there current and the voltage will be in phase and all of the current flowing will be supplying real power. This ends the chapter