Chapter Five. We are now going to return to AC power with all the tools of our trig identities, as well as our review of inductors and capacitors and resistors. In an AC circuit, we are now going to get into some really interesting stuff using all those tools we've already reviewed. I'm going to start out by looking at a series AC RL circuit containing one resistor or one inductor. And I'm going to replace the resistor inductor combination with its equivalent impedance and we're going to call it Zed. So the current flowing in that circuit and the voltage that's across that impedance is described in two ways.

We can do it with a phasor diagram and in a tie domain and the phasor diagram I've written in the upper right hand corner, the time domain equations I've written just at the bottom of the slide and by time domain, we're going to plot these and have a look at them with one variable and that variable being the time t and the phasers in the upper right hand corner. They're rotating in a counterclockwise with an angular velocity of omega. The angle of the current is given by phi subscript i. And the angle of the voltage is given by phi subscript V and I've color coded these just to keep them separated and we can follow it a little bit easier. The angle between the current and the voltage is given by phi without a subscript and I've color coded it in black. Now, the time domain description of the current and voltages are very similar, the current is given by a maximum flow of current times sine omega t plus the phase angle of the current.

And the voltage similarly is described with the M, which is maximum value, and we'll see that in a plot in a few minutes times sine omega t plus the phase angle of the voltage. Now, just to make the mathematical manipulation that we're going to get into a little bit simpler, I'm going to move these phasers such that I'm going to use the current phaser as my reference, which means the phase angle of the current is now zero. And the phase angle of the voltage is going to be equal to the face angle of phi without a subscript That's describing the phasers with ISO them as zero phase angle. When I do that, the time domain equations become a little bit simpler the current is is just given by I am sine omega t, and the voltage is given by, similarly to what we had already but it's v m sine omega t plus phi without a subscript.

If we plot the current and the voltage in the time domain starting with the current itself, you can see that because the current is described as a maximum value times sine omega t, it will start at zero rate at the origin And he will rise to a maximum value, then return to zero and go to a minimum maximum value, and then return to zero and go to a positive maximum and then go through to a negative maximum. Again, that is a picture of the current in the time domain. Now I'm going to go through a series of time domain pictures here. They are really to help you visualize what's going on. Eventually, we will just be remembering the formulas, but the formulas will make a lot more sense if we can picture in our minds what is going on. So I'm going to spend a bit of time looking at the time domain plots of the currents and the voltages and the resultant power of the circuit itself.

The voltage is showing here in green We know that the voltage is a head of the current or is a phasers rotate, it means that the voltage leads the current. So you can see it's leading by an amount by which I've shown in the diagram here, and it also goes through a positive maximum, back through zero to a negative maximum, then up to a positive maximum down to a negative maximum, but at all times, it is leading the current so you can see that it gets to a peak value. Before the current gets to a peak value, it returns to zero just before the current returns to zero, and it gets to a maximum peak before the current gets to a maximum peak. And it's always indicated by that the phase difference in in the between the, the voltage and the current. The other thing to note and it goes without saying because they're both being multiplied by the same angular velocity omega and omega t. But the frequencies are the same, I'm indicating here the actual period of one cycle, which is related to the frequency, they are identical.

So the frequency of these oscillations are identical as well. You will recall that if we were to calculate the instantaneous power dissipated in the load or the impedance said it is given by multiplying the instantaneous voltage times the instantaneous current, and we can derive that using our time domain formulas. We just have multiply the two equations together, both the left hand sides and the right hand side, the left hand sides of the equations for voltage and current, when multiplied together give us the instantaneous power, of course, the right hand sides work out to be VM, I am sine omega t plus phi times sine omega t. And I'm going to plot that formula now for the instantaneous power on our time domain axis. And you'll see it in blue as it is plotted along that axis. Now, I used a computer generated graph in order to come up with my voltage and current as well as the power.

So you can see exactly visually how that how those graphs interact with each other. Starting with the fact that you'll notice the period of oscillation The the power is exactly half the period of oscillation of either the voltage or the current. And if the period of oscillation is one half, then the frequency of that power graph is twice that of the voltage or the current. The other thing you will note from our formula, we know that if either the voltage or the current are zero, then the power has to be zero and you can see it in our time domain graph there where I've indicated with the red arrows, that there's a zero crossing of the power plot every time either the voltage or the current is at zero. The other thing to note is that if both the voltage and the current are positive, then We're going to have a positive value for the power.

And that yes is indeed the case you can see it here and I've indicated the that area in with the red arrows. Now, if either the voltage or the current have a different sign value, in other words, if voltage is negative and current is positive, or voltage is positive and current is negative, the signs of their values are unlike then we're going to have a negative value for the power and you can see it the first red arrow there that the the voltages negative and the current is positive. So we have a negative value for the power and the second arrow, you can see that the current is negative and the the voltage is positive. So we again have unlike signs, so we're going to have a negative value for the power. And you can see that all along the graph there where I've indicated with a red arrow.

In order to see that positive and negative power flow more visually, I'm going to fill in the areas for the power positive and the power negative. So you can see where the power is positive by the burgundy color and where it's negative by the yellow color. And you can indeed see very, very plainly that whenever the voltages and current are both positive or both negative, then we have a positive power flow. And the reverse is also true. When the sign is different between the voltage and the current, we're going to have a negative value For power. I'm now going to take a closer look at the formula for instantaneous power.

And I'm going to use the trig identities that we developed in further developing this power equation. And what you have to know is that we are going to change the right hand side of that equation. But in doing so we are not changing the fact that that equation is the equation for instantaneous power. So firstly, I'm going to use the trig identity. sine x times sine y is equal to one half the quantity cosine x minus y minus cosine x plus y and I've used x&y as the angles by They represent the angles of the power equation such that x is going to be equal to omega t plus phi, and y is going to be equal to omega t. So rewriting the equation using our trig identity now, I'm going to have you focus on the part that I'm applying the trig identity to by changing it to a blue color.

And we are now going to have the instantaneous power equal to VM, I am one half the quantity cosine omega t plus phi minus omega t minus cosine omega t plus phi plus omega t. And in the first cosine quantity, the two omega T's cancel each other out because one is paused division one is negative, and that just leaves the angle phi in that cosine value in the second cosine value, we have two omega T's, and they add together to give us two omega t. And that's plus phi. I can now rewrite that equation using those values, and I am left with the M, I am one half times the quantity cosine phi minus close to omega t plus phi. Now, I want to use one more trig identity, and that trig identity is coasts x plus y is equal to cosine x times cosine y minus sine x times sine y and this time I'm going to concentrate on only the last portion of that equation that contains the quantities close to omega t plus phi.

And I'm going to let two omega t be the x value of my trigger identity, and I'm going to let phi be the Y portion of that identity. So now I can rewrite the equation again. And this time I have VM, I am one half cosine phi minus cosine two omega t times cosine phi plus sine two omega t times sine phi. I'm now collecting like terms and readjusting the equation and I get an equation that has the sum of two values. And that is the First value is VM, I am all over two times the quantity one minus cosine two omega t times cos phi. The second quantity in that equation is VM, I am all over two times sine two omega t times sine phi.

Now, we're going to do some further investigation of this formula. But keep in mind, as I said at the beginning, this is still the formula for the instantaneous power. All I've done is manipulate the the right hand side of that equation, so that it will define how that power equation acts and we're going to have a look at that right now. And look at it in terms of the time domain I'm going to bring back the computer generated graphs for the voltage, the current and the power that we saw before. And I'm going to want you to keep in mind that we are looking at an inductive series circuit here. So the voltage will lead the current and remember the current is our reference phaser.

And the angle phi is the phase angle between the voltage and the current. Now, as I said, the formula for instantaneous power is made up of the sum of two functions now as we've read and re manipulated that equation, and I'm going to look at the first function first of the sum of the functions v I am over two times one minus close to omega t times cosine phi. And if I was to plot just that portion of the of the equation, it would look like this orange line that's plotted on our graph. And the you can see that the orange line is sinusoidal, but it is always above the zero line. And that is because the term is one minus cosine two omega t. And I'm going to bring that down just so that that is the only part of the function that is changing, because the only variable in this time domain graph is T. All of the rest of the values are set values.

The VM I am over two is a set value and the coast phi is a set value. The only thing that is changing in the time domain graph is the time. Looking at the second part of the equation, that is a function, and that is VM Iam over two times sine two omega t times sine phi. And if we graph that equation, or that function using a computer generated graph, it will look like that black line that you see there. And I'm going to bring down the only variable portion of that part of the function. And that is the sine to omega t because T will change with time.

The rest of that function that I've circled in red doesn't change the VM I am over two is not changing and the sine phi is not changing. So we are only lucky At sine two omega t, now that our curves are getting rather confusing, so I'm going to remove just the, the power, the voltage and the current graphs are now. So we can just look at the two functions that are sum together and have a can see them a little bit planar, you can see that the first function, the orange line, is always above zero. That is because the variable portion is one minus coast to omega t. So that forces the graph or the sinusoidal graph to be above zero, the part that's sine to omega t, the black line is also sinusoidal. It is the same frequency. However, it is exactly through the zero line.

So there's an equal portion above zero as there is below zero. Now I'm going to put back in the instantaneous power curve, just so you could see how they relate to each other we're going to go through is just a bit of a comparison here just to visually prove the fact that the instantaneous power is given by the arithmetic sum of two functions that you see there. One function being the orange line and the other function being the black line. And if you look at where both orange and black lines are zero, which means the sum would be zero, yes, indeed, you can see that the blue line or the blue graph, which is the instantaneous power is also zero. And something that's not as as obvious but if you look very close, you can see that where the power is zero again, there's an equal portion of orange line above the zero and an equal portion of the black line below the zero, those so the sum of the orange and the black line net out to zero.

And that is where the instantaneous power, the blue line is zero. So the takeaway here is that you now split, we have now split out the formula for instantaneous power into the sum of two functions, which you see here in a time domain, one being the orange curve, and the other being the black curve. Now, there was a reason why we went through the huge exercise of splitting out the instantaneous power equation into the sum of two functions and you're going to see that right Now, the formula that we have devised, and the one we started with gave us instantaneous power. And there's no doubt about that. But there's very few times that we want to look at instantaneous power. In fact, most of the time, we're going to be dealing with average power and average values.

In fact, the kilowatt hour meter that's on your house that the power company builds you. They bill you on average power and average energy that's coming into your house. And industry is all based on the consumption of average power. So we need a way of calculating what that average power is. And the way we've broken out the equation allows us to do that. What is the average?

Well, the average is is the function over a long period of time. And if we take the average of the four of that equation, over a long period of time, what happens, the changing portions that oscillate back and forth, add out to zero. In other words, cosine to omega T will add out to zero and sine to omega t goes to zero. So the second term of our instantaneous power equation will disappear because that goes to zero. The first term, the close to omega t goes to zero but that leaves one inside those square brackets, leaving us with the value of VM. I am all over two times cosine phi and that is the average of the orange line.

That you see above the zero portion of our graph. So now we know that the average or if we want to calculate the average power being delivered to our inductive load, it is the VM, I am over two times cosine phi. Now, let's look at the second part of our two function equation. The VM i m over two times sine two omega t times sine phi, you can see that it is the power is being delivered to and from, in other words as a positive side, an equal positive side and an equal negative side to that black line as it progresses through time. And that is hauled the reactive power, and it's given for the letter Q. And it is exactly the amount of power that's being delivered to the inductor to build the magnetic field in the inductor.

And then in the second half cycle, that magnetic field collapses, and sends power back into the system. But the net power delivered and the net power delivered back to the system is zero, because it goes to and from, but we'd like to be able to calculate what the reactive power is. In other words, we like to give a quantity to it. But if we took the average of the black line over time, as we saw, it will be zero, and it will be meaningless to us. So we'd like to try to find the average value of it similarly to the way we found The average power the real power that was delivered in a circuit, well, we can do that by moving the black line above zero, just so it's just above zero like the orange line was, and no one and the way you do that is introducing the one into the formula such that it's VM Iam over to one minus sine two omega t times sine phi.

Now, this is just a method of coming out with the average value, similar to the what the power was. And if we take the average over time, we know that sine to omega T will again go to zero, but this time, it's going to leave us with the value VM, I am over two times sine phi, which will give us the average value of the reactive power. And we're going to have a little bit further look at that in a couple more slides. Going back to our formula for average power, that average power, which is the real power that's delivered to the circuit is, and if you'll remember what we started with it was a series RL circuit, the real power is all delivered to the resistor, and it's dissipated in the resistor. So the average real power is still all dissipated into the resistor and if we had a kilowatt hour meter on that circuit, and if a utility was billing us for that average power, it would be all delivered to the resistor because the inductor doesn't dissipate any real power.

It takes power for half a cycle and delivers it back to the system for the other half cycle as the magnetic field builds and collapses. Now looking at the formula, we can split out the current and voltage values by dividing each one of them by root two because if we multiply them together, we'll get VMI M over two, but I'm going to split them out right now, because if you remember the root mean square of a sinusoidal voltage is given by the M all over root two. And the sinusoidal root mean square or RMS value for current is given by the maximum value of i am all over root two. So we can now rewrite the average power equation as long as we measure the voltage in RMS terms and the current in RMS term and multiply by the phase angle between them. Now, most Most volt meters and most amateurs and most meters in an AC system record values in RMS value for this very reason that we can now take a direct reading of the voltage and current and come up with the average power as long as we know the phase angle between them.

And similarly, we can also say that the average reactive power is given by the RMS voltage times the rms current times the sine of the phase angle between them. You remember when we started this process out, we are looking at the phasers for the voltage and current and I've have the map pictured on the bottom of the slide here. And we used our current as a reference phaser. So we set it at zero However, the phasers are rotating at phase at a rotation OF 60 CYCLES a second in our system could be 50, if you're in Europe or any other angular velocity for the phasers, so they're not always going to be in a position where the current is at zero, they could be anywhere along this plane. So, but that doesn't matter because our formulas don't involve the phase angle of the current or the phase angle of the voltage.

It only involves the phase angle between the current and the voltage. And that never changes as long as the passive components of the circuit doesn't change. In other words, as long as we have a set impedance in a circuit At the phase angle will always remain constant. So you can always calculate the average power and the our average reactive power of the system. So, we can now replace the fly with the phase angle difference. In other words, the phase angle of the voltage minus the phase angle of the current.

That is what phi is and we can use the difference of the angles in order to generate the average power and the average reactive power. So far we've been looking at a series RL AC circuit. And if we were to take and measure the rms current flowing into that circuit and the RMS voltage across the load of that service Get the phasor diagram would look like this, the voltage as we've seen is leaving the current. And there's a phase angle between the voltage and the current of phi such that phi is equal to the phase angle of the voltage minus the phase angle of the current. Now, if we were to take the magnitude of the voltage and the magnitude of the current and multiply them together, that would give us a term known as apparent power. And if we plotted that as a phasor, with the same angle, as the angle between the current and the voltage, it would look like this.

Now if we took the projection of the apparent Power phasor onto the x axis. It would provide us with the RMS I RMS Coase phi, which is real power. And it's equal to the average power says same formula that we use to calculate average power. And if we took the projection of the apparent power onto the vertical axis, that would give us a phasor of RMS v times RMS i sine phi which is reactive power, which is equal to q average. Now if we move the vector of reactive power over to the right by adding the The reactive component to the phasor of the real component, it would give us the apparent power. So we're adding the real power and reactive power together.

And the vector sum or the phasor sum of those two quantities give us the apparent power and that forms a triangle which is noted as the power triangle. And this power triangle is very useful, as you will see later on in doing some analysis on our AC circuit. Before leaving the power triangle, there are a couple of terms that are used often in the analysis of power in AC system that you'll often hear and you should be aware of, and that is, in the real power, we have v rms RMS cosine phi cosine phi is what is known as the power factor and the power factor is always less than one and usually zero or greater than zero. Now, the angle associated with the power factor is known as the power factor angle and it is the angle between the voltage and the current and it is also the contained angle of the power triangle.

You should know the difference between these two terms because sometimes they are mixed up when they're referred to and that is the power factor is cosine of an angle and the power factor angle is that angle. This ends the chapter