CHAPTER THREE reactance and impedance our LC circuits if we were to plot the current and voltage for a simple AC circuit consisting of a AC voltage source acting on a resistor, it would look something like this. Because the resistor simply and directly resists the flow of electrons at all periods of time. The waveform for the voltage drop across the resistor is exactly in phase with the waveform for the current through it. We can look at any point in time along the her the horizontal axis of the plot and compare those values of current and voltage with each other. Any snapshot look at the values of the wave are referred to as instantaneous values meaning the values at an end In time, when the instantaneous values of current is zero the instantaneous value across the resistor for voltage is also zero. Likewise at the moment in time where the current through the resistor is at a positive peak, the voltage across the resistor is also at a pot positive peak and so on.

At any given point in time along the waves ohms law holds true for the instantaneous values of voltage and current and the voltage phasers are plotted in the upper right hand corner there and you can see that when the voltage source is at zero, so is the current. We can also calculate the power dissipated by the resistor and plot those values along with the voltages and current currents. Note that the power is never a negative value. When the current is positive above the line, the voltage is also positive, resulting in the power equaling i times R. Conversely, when the current is negative below the line the voltage is also negative, which results in a positive value for power, negative number multiplied by a negative number equals a positive number. This consistent polarity of power tells us that the resistor is always dissipating power, taking it from the source and releasing it in the form of heat energy.

Whether the current is positive or negative, the resistor still dissipates energy inductors do not behave the same as resistors whereas resistors simply oppose the flow of electrons through them by dropping the voltage directly proportional To the current inductors oppose changes in the current through them by dropping the voltage directly proportional to the rate of change of current. This is this induced voltage is always such a polarity as to maintain the current in its present value. That is if the current is increasing in magnitude, the induced voltage will push against the electron flow, if the current is decreasing the polarity or reverse and push with the electron flow to oppose this decrease this opposition to current changes called reactance rather than resistance expressed mathematically the relationship between voltage dropped across the inductor and the rate of change of current through the inductor is as this formula indicates the voltage is equal minus L di by dt, the minus sign indicates that the voltage is opposing the change of current flow.

The expression di by dt is one from calculus meaning the same thing as Delta II over delta t as delta goes to zero. So it's really the change of current over the change of time. So, this meaning the rate of change of instantaneous current over time, in amps per second, the inductance L is in Henry's and the instantaneous voltage of course, is in volts. If we plot the current and voltage for this simple circuit, it would look something like this. The circuit has An AC source of voltage et acting on an inductor with an inductance of L. Remember, the voltage drop across the inductor is a reaction against the change in current through it. Therefore, the instantaneous voltage is zero whenever the instantaneous current is at a peak zero change or level slope on the current sine wave and the instantaneous voltage is at a peak.

Whenever the instantaneous current is at a maximum change the point of steepest slopes on the current wave, not is where it crosses the zero line. This results in a voltage that is 90 degrees out of phase With the current wave, looking at the graph, the voltage wave seems to have a head start on the current wave, the voltage leads the current and the current lags behind the voltage. The phasers would look like this, where the voltage phasor would be leading the current phaser remember now that the phase rotation is counterclockwise and inductors opposition to change in current translates to an opposition to alternating current, in general, which is by definition, always changing in instantaneous magnitude and direction. This opposition to alternating current is similar to resistance but different in that it is always results in a phase shift. Between the current and the voltage and it dissipates zero power.

Because of the differences it is a it has a different name reactance reactance to AC as expressed in ohms, just like resistance is except that its mathematical symbol is x instead of R to be specific reactance associated with doctors is usually symbolized with a capital letter X with a subscript l as shown here. Since inductors drop voltage proportional to the rate of change of current, they will drop more voltage for faster changing currents and less voltage for slower changing currents. What this means is that reactance in ohms for an inductor is directly proportional to the frequency of the alternating current. The exact formula for determining reactance is as follows reactance for an inductor is equal to two pi f l where f is the frequency in hertz or cycles per second pi is the infamous number for pi 3.14159 and L is equal to the inductance in Henry's.

Let's consider now a series RL circuit with a supply voltage of 80 which is of course sinusoidal or AC voltage source. The resistor will offer resistance to the AC current while the inductor will offer resistance in the form Have reactance to the AC current. Because the resistors resistance is a real number in there and the inductors reactance is an imaginary number, the combined effect of the two components will be an opposition to current equal to the complex sum of the two numbers. This combined opposition will be a vector, it will be a vector combination of resistance and reactance. In order to express this opposition succinctly, we need a more comprehensive term for opposition to current than either resistance or reactance alone. This term is called impedance.

It is symbolized as with a letter Zed and it is also expressed in units of homes just like resistance and reactance. So, the term or the impedance is also a phasor. Let's see how we calculated the voltage around the loop. According to current child's Voltage Law is the supply voltage is equal to the voltage drop across the resistor plus the voltage drop across the inductor. And they are phasor values and must be treated as such. The current in the loop must be the same for all components because it is a series circuit and the current remains the same at all times.

If we take the voltage equation in divided by the current i which is the same in each one of the components we are left with an accent equation of reactance is an impedance is because e over i is a reactance or an impedance or a resistance the term e L over AI is the reactance XL, the term e r, which is the resistance over the current is just the reactance of the resistor or just the resistor itself are added together using phaser addition because they are at 90 degrees to each other, they provide the resultant impedance Zed T for the total impedance of xR plus XL So, We have a way of calculating the impedance of two components, an inductor and a resistor. The inductor is at 90 degrees to the real component of resistance. So you can use the Pythagorean Theorem to calculate what Zed is using the real value, R, or the resistance and the reactance of the inductor x out.

Okay, let's try and cement some of that theory with an example. Let's put some real numbers and into a series RL circuit and just see how it works. In our circuit here, the resistor is five ohms and the inductor is 10 milli Henry's and the frequency that we're working with is a standard North American frequent You have 60 hertz or 60 cycles per second. In this example, Excel is given by two pi times the frequency times the inductance which is two times 3.14159 times the frequency which is 60 times 10 milli henries and 10 milli Henry's is 10 over 1000 Henry's, so we multiply those out and end up with a reactance for our inductor of 3.7699 ohms. Therefore, the resistance or sorry the impedance a total impedance of our l in series is the five ohm resistance added to 3.7619 ohms inductive reasoning Now remember these are both phasers.

So they have to be treated as phasers. So they are the resistance is five ohms along the real axis which is at zero degrees, so it's five and zero degrees. The reactance is 90 degrees, so it is 3.7699 at 90 degrees and we want to add these two together. So we're going to convert them to to a rectangular notation, which is five plus j zero ohms plus zero plus j 3.7699 ohms. Which when added together will give us five plus j 3.7699. And if you want to convert that to a polar notation, it's a magnitude of 6.262 ohms at 37.016 degrees.

As with the purely inductive circuit, the current wave lags behind the voltage wave of the source. Although this time the lag is not as great only 37.016 degrees as opposed to a full 90 degrees as was the case in the purely inductive circuit. In this case, the current is modified by The fact that we put a resistor in series with the with the reactance or the inductor and that has a tendency to straighten out the lag for the resistor and the inductor individually and their voltages the phase relationship between the voltage and current hasn't changed. voltage across the resistor is in phase zero degrees phase shift with the current through it and the voltage across the inductor is still 90 degrees out of phase with a current going through it. Let's revisit a DC Circuit momentarily and have a look at how a capacitor reacts to being charged.

We know that the voltage across a capacitor is dependent on its charge that is the voltage across the capacitor is given by the Q over c, where q is the charge and C is the capacitance and BC is the voltage across the capacitor. That equation can be rewritten as Q is equal to c times VC, which are have the same variables. Now, at time if we measured the charge and the voltage at time one, we would have this equation q one is equal to the same capacitor of with the voltage at the voltage time equals one. If we measured the voltage again at time two as things are charging, we would have a second value of charge which would be equal to see the capacitance times the voltage at that time as well. So if we essentially subtract a question One Two from equation one, we could say that q2 minus q1 all over T two minus T one, which is equal to delta Q over delta t is equal to the same voltage difference.

In other words, as the charge changes over a period of time, the voltage given by the equation Q is equal to c v sub c would give us delta V over delta T the same Delta time. But, delta Q over delta t by definition is equal to the current of the charging capacitor, which is still equal to c delta VC over the delta t. Or in terms of calculus, we have this term current is equal To the capacitance d dV all over dt and and the D is just the change in voltage or the change in time. What this formula tells us is that the rate of flow of electrons through a capacitor is directly proportional to the rate of change of voltage across the capacitor. This opposition to voltage changes another form of reactance, but one that is precisely opposite to the kind exhibited by inductors.

Expressed mathematically the relationship between current through a capacitor and the rate of change of voltage across the capacitor. In the way of review is given by AI is equal to d over d t or d is the change of voltage over the time period dt. The expression d over dt is one of calculus meaning the same thing as Delta II over delta t and just as a way of a review, and as delta the Delta goes smaller and smaller it becomes more significant. voltage E over time is in volts per second capacitance C as in fare ads and the instantaneous current I have courses in amps. Remember, the current through a capacitor is a reaction against change in voltage across it. Therefore, the instantaneous current is zero whenever the instantaneous voltage is at a peak Zero change or level slope on the voltage sine wave and the instantaneous current is at a peak, whenever the instantaneous voltage is at a maximum change the points of steepest slope on the voltage wave where it crosses the zero line.

The results are similar for the negative side of the wave. This means that the voltage wave is 90 degrees out of phase with the current wave. Looking at the graph and the current wave seem to have a head sorry looking at the graph, the current wave seems to have a head start on the voltage way, the current leads the voltage or the voltage lags behind the current. If we draw the phasers for the voltage and current current, it becomes obvious capacitors opposition to the change in voltage translates to an opposition to alternating voltage in general, which is by definition always changing in instantaneous magnitude and direction. For any given magnitude of AC voltage at any given frequency a capacitor of given size will conduct a certain magnitude of AC current. Just as the current through a resistor is a function of voltage across the resistor and the resistance offered by the resistor.

The AC current through a capacitor is a function of the AC voltage across it. And the reactance offered by that capacitor has within Drs. The reactance of the capacitors is expressed in ohms and is symbolized By the letter x or x subscript C to be more specific, since capacitors conduct current in proportion to the rate of change of voltage, they will pass more current for faster changing voltages as the charge and discharge to the same voltage peaks in less time and less current for slow slower changing voltages. What this means is that reactance in ohms for any capacitor is inversely proportional to the frequency of the alternating current. In other words, the reactance XC is equal to one over two pi f c where f is in frequency or hertz. sec cycles per second pi is our infamous 3.14159 and capacitance is in ferrets.

Note that the relationship of capacitance or capacitive reactance to frequency is exactly opposite from the inductive reactance capacitive reactance in ohms decreases with increased AC frequency. Conversely, inductive reactance in ohms increases with increased AC frequency inductors oppose faster changing currents by producing greater voltage drops. capacitors oppose faster changing voltage drops by allowing greater and greater currents. Let's consider a series RC circuit now with a supply voltage at generating an alternating voltage. The resistor will offer resistance and we'll call it x R, which is equal to the resistance of the resistor y capacitor offer offers resistance in the form of reactance which we will call XC and we just saw that that was equal to one over two pi f c, because the resistors resistance is a real number and the capacitors reactance is an imaginary number, the combined effect of the two components will be an opposition to current equal to the complex sum the complex sum of the two numbers, which is impedance.

And impedance is given as Zed subscript T for the total impedance, and that's equal to the phaser sum or the complex sum of xR plus x C. Now the voltage around the loop according to her chops Voltage Law is equal to the supply voltage is equal to the voltage drop of the resistor plus the voltage drop of the capacitor. Given that the voltage drops are complex numbers, they're phasers and they have to be added that Coralie or as complex numbers, the current in the loop is the same because again it is a series circuit. So the current at any one time in a series circuit is always equal. So, the total current is equal to the current in the resistor is equal to the current in the capacitor. If we take the voltage equation and divide each of the voltages by the current, we would be left with this equation.

Et over i is equal to er all over i plus EC Oliver etc overall Is the reactance of the capacitor. Er all over i is the resistance of the resistor or the reactance of the resistor which is just resistance. When added together give us the impedance and they have to be added vectorial Li and that would give us the total impedance for the circuit is xR plus x C, xR being along the real axis, x c being at minus 270 degrees. Let's look at an example. In our example here we have a resistor of five ohms in series with a capacitor of 100 microfarads. Being energized in a state system have a frequency of 60 hertz or 60 cycles per second.

In this example, since the reactance XC is given by one over two pi fc, the reactance is going to be one over two times 3.14159 times 60 times 100 over 1 million seems we're dealing in microfarads and that calculates out to 26.5258 ohms when added together that will give us a total impedance. So we total impedance is made up of five ohms of resistance and 26.5 to five eight ohms of capacitive reactance or that is the impedance is made up of five ohms at zero degrees plus 26.5258 ohms at minus 90 degrees or five plus J zero ohms plus zero minus j 26.5258 ohms. That's just given it in terms of a rectangular coordinates. And that can be either written as a five and minus j 26.5258 ohms or 26.993 ohms at minus 79.3 to five degrees. So impedances are related to voltage and current, just as you might expect, in a manner similar to resistance in ohms law.

The voltage is equal to the current times the impedance and that that equation can be converted or manipulated such that the current is related to the voltage over the impedance. In other words, i is equal to E all over said, and Zed can be found from the voltage drop across the impedance over the current through the impedance. These are similar to what we are doing with resistances. The only thing is now we're dealing with complex numbers. And all quantities are phasers, or they're complex numbers. And you have to use the rules of addition, subtraction, subtraction, multiplication and division using complex numbers and using phasers.

Let's work through an example just to see how impedance is related to the current and the voltages as we just described. Here's an example of an RLC series circuit, we can now determine the equivalent impedance of the circuit. The first step is to determine the reactance is in ohms for the inductor, and we know that we're dealing with a 60 hertz system. The XML is given by the equation two pi f L. and if we convert two pi FL to pure numbers, that's two times pi times 60 times 650 megahertz go through the math and that works out to 245.04 ohms. The reactance of the capacitor is given by one over two pi f c. So xe is going to be one over two times pi times 60. times 1.5 microfarads. And if you do the math that works out to 1.7684 k ohms.

The impedance of the resistors is going to be given by 250 ohms plus j zero because there's no reactive component, just a real component, the impedance of the inductor, there is no real component. So the first part of the impedance number is zero, and it's plus j times zero reactance, which is 245.04, ohms. And Zed c is. Again, there's no real component, just a reactive component. And this time it is a minus j because it's a capacitor, and it's minus j times 1.7684 k ohms. Now, those three impedances can also be written in polar notation.

The impedance for the resistor is 250 ohms. at zero degrees for the for the inductor is 245.04 ohms at 90 degrees, and for the capacitor it's 1.7684 k ohms at minus 90 degrees. Now with all the quantities of opposition to the electric current expressed in complex number format, they can be handled the same way as we handled plane resistances In a series DC circuit, we can draw up an analysis table for this circuit and insert all the given figures total voltage and the impedances of the resistors inductors capacitors. unless otherwise specified the source voltage will be our reference for phase shift and so will be written at an angle of zero degrees. Remember that there is no such thing as an absolute angle of phase shift for a voltage or current since it always it's always a quantity relative to another waveform. phase angles for impedances however, like those of resistors inductors and capacitors are known absolutely because the phase relationship between voltage and current at each component are absolutely defined.

Notice that we are assuming a perfectly reactive inductor and capacitor with impedances phase angles of exactly plus 90 and minus 90 degrees respectively. Although real components won't be perfect in this regard, they should be fairly close. For simplicity, we'll assume perfectly reactive inductors and capacitors from now on in my example calculations except were noted otherwise. Since the above example circuit is a series circuit, we know that the total circuit impedance is equal to the sum of the individuals. So the total impedance is going to be given by the sum of the resistance impedance the reactor or sorry, the inductor impedance and the capacitor impedance. So in terms of numbers, that's 250 plus j 245.04 minus j 1.7684 k ohms.

Doing the math for those numbers, the total impedances written in rectangular form 250 ohms minus j 1.5233 k ohms or 1.5437 k ohms at minus 80.680 degrees. We can now plug that into our chart. We can now apply ohms law is equal to E over Zed in the total column to find the total current in this series circuit Now the the M designates milliamps since the impedance was in K ohms. Now being a series circuit current must be equal through all components. Thus we can take the figure obtained for the total current and distributed to each of the other columns. Now, we were prepared to apply ohms law, again this time solving for the voltage drop across the individual impedances we use e is equal to i times Zed and we can fill in across the top of the table for the voltage drop across the resistor across the inductor and across the capacitor.

So, we have successfully used ohms law and complex numbers or phasers and impedances to calculate and solve for the current and the voltage drops in a series RL circuit. No notice something strange here though. Although our supply voltage is only 120 volts, the voltage across the capacitor is 137.46 volts. So how can this be? Well, the answer lies in the fact that when you connect impedances up and those impedances include capacitor inductance, there is a resonant frequency that will allow the voltages to essentially your Back and forth building and discharging to mount to a voltage level. That's, in this case 137.46.

Now, the theory and the way this happens is the subject of another lesson. However, you don't have to worry about it if all you're wanting to do is calculate the voltage drops and the currents. The math looks after that for you. And so this ends another chapter chapter three.