Chapter seven. So far we have discovered that if we take the magnitude of the current and multiply it by the magnitude of the voltage, we will come up with a term known as apparent power, which is short form with the letter S. And its units are volt amps, whose short form is VA. And if we make a phasor, whose magnitude is the volts times the amps and place it at an angle, such that that angle is the voltage angle minus the current angle, then we will have the p average or the real power average of that circuit, which is the volts times amps times the cosine of the angle between them and that's designated as p With the units of watts which has a short form of W. We also know or have learned that if we take the sine of the angle between the voltage and the current and multiply it by the apparent power, we will come up with reactive average power or q average and its units are volt amps reactive and its short formed var or V bars.

We also know that if we take the phasor of the power and add it to the phasor of the reactive power, we will come up with S or the apparent power and that is also known as complex power and complex power. Can Also be derived from multiplying the voltage times the current times the cosine of the angle between them, plus the voltage times the current times the sine of the angle between them. Or if you want it to designate the term of apparent power as a phaser itself, it is the voltage times the current at an angle of the voltage angle minus the current angle. Now if we wanted to, we could rearrange that mathematically keeping in mind the laws of multiplication for a phasor then we would have the voltage phasor which is the voltage magnitude times the angle of the voltage times the current magnitude at minus the current angle.

Or in mathematical terms, you can describe that as the voltage phasor times the current conjugate. And the conjugate of a current is described as this. And it's a definition in mathematics that the conjugate of a vector is just the mirror reflection of that, that phasor or vector across the horizontal axis such that if you have a phasor of I magnitude at plus theta, given us the current phaser, then its conjugate would be the minus angle, which would be described by the same magnitude, but minus the angle and that would be the current conjugate. Working with complex power requires a working knowledge of conjugate numbers or conjugate phasers. So I'm going to go through some exercises now that are built around complex power and conjugate numbers. And in the next couple of slides that we're going to be looking at actually the next two slides we're going to be looking at I'm going to denote any of the numbers in the subsequent slides, the red ones are going to be phasers or vectors and all the black numbers or letters are non vector quantities.

So if we were to look at this phasor here the blue phaser as magnitude x at plus theta, and it is the description of the phaser x The conjugate of x would be magnitude x at minus theta, and the phaser would be denoted as with a star in the upper right hand corner, that is the conjugate so that the star denotes a conjugate phaser or a vector. So now if we took two phasers together, actually one phaser and its conjugate and you multiply them together, we're just going to use our rules for multiplying phasers or vectors together. And x times x conjugate is the same as magnitude x plus theta times a magnitude x and minus theta. So you multiply the two x's together, and you add the two angles together while a plus theta and a minus theta will give you zero. And two x magnitudes together will give you an X magnitude squared at zero degrees.

And a lot of times we leave the zero degrees off a number when we're describing it, and that would just be magnitude x squared. So any any phaser times its conjugate will give you the magnitude of that phaser squared. Keeping in mind how we designate phasers in red and non phaser non vector quantities in black, the complex power is described by the voltage times the conjugate of the current. Now if we want to describe the voltage, current and impedance it would be done with the magnitudes of each times the angle of each. So the voltage would be magnitude v times theta the magnitude I for the current times the angle theta. I am The magnitude of Zed at some angle, thought, theta Zed.

Now, if that describes the phasers for the voltage, current and impedance, then the conjugate of those would be the same magnitudes. And the same angles, only the angles would be minus now, so I've described both the phasor notation for voltage, current and impedance, as well as their conjugates. And I'm going to go through a series of exercise here involving the complex power as well as ohms law. Now complex power again, I'm just repeating it is the voltage times the conjugate current. And we know from ohms law that the phasor V is given by the phaser i times the phaser Zed We can replace the VI in the formula for the complex power. Now we have AI, the phaser i times the phaser Zed times the conjugate VI and that is equal to from our previous in the upper right hand corner calculation, if you remember that any phaser times its conjugate is the square of that magnitude.

So, s can also be described as I magnitude squared times Zed, which is a phaser. Now ohms law is V over Zed, which can be described using our description of terms that we started with that magnitude or the voltage vector can be the magnitude v times theta v. All over magnitude said time. Theta said. And if we bring the calculation forward, in other words, we divide the numerator by the denominator, we can see that that's going to be magnitude V over manga to Zed at an angle of theta v minus theta Zed. And that's equal to the phaser I or the current phaser. Now, if I wanted to find the conjugate of that current phasor, using what we just developed there, I would have to take the minus of the angles same magnitude and leaving the magnitudes the same, and I'm just taking the minus of the angles, which would be minus v plus, sorry, minus theta v minus theta.

Or let's start again minus theta v plus theta Zed. And I can move that term into the denominator now and I now have magnitude v times minus V magnet over magnitude Zed times the angle minus data Zed, which is the conjugate over Zed conjugate. Now if we go back to our starting point for the description of complex power s is equal to v times conjugate I. And we replace conjugate I with conjugate V over conjugate Zed we get the times conjugate V over conjugate said, which is equal to in the numerator, v times conjugate, the all over conjugate said and we know it Any phaser times its conjugate gives us It gives us the magnitude squared, which is V magnitude squared, all over conjugate Zed. I'm going to take a closer look at the complex power this time in terms of a series circuit that you see in the upper right hand corner.

And that series circuit is made up of a resistance and reactance portion that are in series as far as a load is concerned with a source voltage called V and that v is sinusoidal. And the current in the circuit is designated as AI. And it is equal to the current through the reactance portion of the circuit, which is I subscript x, and it's also equal to the current through the resistor Of course, because it's a series circuit, which is designated I subscript r And the components themselves as far as impedance elements, we have a resistor and a reactance. Now, we don't know exactly what the reactance is for now, and it doesn't matter, but we will look that a little bit closer Later, we'll just say that it is either capacitive or inductance. But we'll just call it x for now. And the voltage drop across those components are given by the subscript x and v subscript R. Now the power triangle is shown in the upper left hand corner, and we can see that the apparent power is given by the sum of the real power and the reactive power.

And we have also learned that we can also describe the apparent power as the square of the magnitude of the current times the impedance or we can take This square of the magnitude of the voltage all over the complex conjugate of the impedance. And that also would give us the complex power. Now as we go through a series of these manipulations and equations, I'm going to color all of the phaser elements in red to differentiate it from the scalar portions which will be in black, of course, and the conjugates are going to have a an asterisk in the upper right hand portion of that part of the equation. So for the moment, let's assume that we're going to reduce the reactance portion of this series circuit from what it is now to make it zero and if we Did that then the apparent power would equal the real power.

In other words, s would equal P. And now the voltage of the source will equal the voltage drop across the resistor v subscript R. And the current course is still I subscript R, or AI in the circuit because it's still a series circuit even though there's only one impedance element, and that impedance is now given by r which is the resistance of the resistor. So now if we took a look at the complex power s is equal to v times I conjugate. The s now can be replaced by P, the v can be replaced by V subscript Are because they are equal and the AI conjugate, which is equal to AI subscript r can replace the AI conjugate of the complex power equation. So, if we have only a resistive element, then the the real power or the p can be found by taking the voltage times the complex conjugate of the current.

Also. The next equation that we had for complex power can now be rewritten such that P will be equal to the magnitude of the current, which is through the resistor, which is squared times the resistance, which is the total impedance of the circuit now, and the V magnitude squared all over the complex conjugate of Zed now becomes the subscript r squared all over the resistance conjugate. And that is equal to the real. One thing worth noting in these equations is that we're taking the conjugate of the impedance of a resistor. And the impedance of a resistor is really a magnitude at zero degrees. So the the conjugate of some phasor at zero degrees is really, there isn't one it's it's the same thing.

So really, the conjugate symbol is really superfluous in this case, so you could really drop that from the equation, because as I said, the impedance is at zero degrees, and the conjugate is the same as the actual phasers. Now, let's go back to our original diagram with the current flowing through both the reactants and the resistor. You can see the formulas that we derived using the real power p, that we are only considering the voltage drop across the resistor, as well as the current through the resistor and the impedance of the resistor in the three formulas for real power. So regardless of what the value of the reactance is, if we're only looking at the resistor and its voltages and current, then the three equations that we just derived, hold true. Now let's assume that we're going to drop the resistance to zero And we're only going to have the reactants in the circuit.

So the apparent power is now going to be equal to the reactive power in that circuit. And the voltage of the across the reactance is going to be the same voltage as the source voltage. And the current, of course through the reactance is going to be the rate the current through the circuit. And the impedance is going to be the impedance of the reactance, which is J x the reactance is x and the reactance is J x where j is the operator that moves through 90 degrees. And as we've just seen, the apparent power is going to equal the reactive power In the circuit. So now we can look at our equations for complex power, and S is equal to v times the conjugate good, I now becomes Q is equal to the voltage across the inductance, sorry, the voltage across the reactance times the current conjugate through the reactance.

And s is equal to the magnitude of the current squared times the impedance now becomes the square of the magnitude of the current through the reactance times j times the x, which is the impedance of the reactance and the final formula s. Now let's again go back to the original circuit And we can see that the three formulas that we derived for the reactive power in this circuit are only looking at the voltages across the reactance and the current through the reactance and the impedance of the reactance itself. So regardless of whether we reduce the resistance to zero or not, because we're only looking at the reactance and the current through it and the voltage through it, then the three equations that we've derived for reactive power will still hold true just as the three equations we develop for the real power hold true. You will recall that while we are having a look at the functionality of this circuit in regard to complex power That we didn't really worry about whether the reactive component X was capacitive or inductive, but let's take a look at that right now.

Let's assume for the moment that the reactive component is an inductor or has an ducted part of it, it is going to be described as x subscript L, which we know is given by the frequency in the inductance of that inductor and it is given by two pi FL or f is the frequency and cycles per second pi is the infamous 3.14159 and L is the inductance in Henry's. Now, if we were to take the conjugate of that element, it would just be the minus which is minus two pi f L. And if that inductive element were a capacitor, it would be given by the reactance of minus one all over two pi f c, where F and pi are the same as for the inductor, but C is the capacitance in ferrets. And if we were to look at the conjugate of that capacitance, it would be one all over two pi fc.

In other words, we would drop the minus sign, because when we pick up these inductances in our circuit, we apply the J operator, but once we apply the J operator by taking the complex conjugate of them, we reverse the sign of that essentially of that j operator This ends the chapter