Transient Analysis Lecture

Okay, good morning, everyone. Today, we'll look at Chapter 4. Chapter 4 is Transient Analysis. This is going to go by quickly. So today, I will cover the entire theory and the concepts for Chapter 4. And then Monday, we'll look at the problems. But the good thing about Chapter 4 is there's only one equation in that entire chapter that you need to know. So rather than taking down a lot of notes or looking at all the equations when I go through, I would rather students just take minimal notes only as much as they need to understand the concepts. Don't worry about writing down all the equations or the derivations in the middle. As I stated, at the end of the chapter, there is one single equation that works for everything. OK. All right. So let's go ahead and begin now. This chapter is known as transience and I will explain that in just a couple of slides. So, inductor and capacitors are energy storage devices. The energy transfer back and forth into the inductor or capacitor cannot take place in an instant. That transfer requires a certain amount of time. The time required to either receive the energy into the component or discharge it back into the circuit depends on the size of the component as well as the circuit conditions. Basically, the size of the container, the fluid flow, and the size of the pipe, right? So, the smaller the size, the faster it fills up or the faster it loses the energy. The bigger the size, the more time it requires. Now, because these are energy storage devices, they all are dependent on time. So a one energy storage device requires solving a first order ordinary differential equation. Because it's a first order differential equation, we refer to it as a first order circuit. Thankfully, greater minds have come before us and have already solved this equation, so we don't have to. All we have to do is basically recognize the circuit, set up the analysis conditions, extract the information, and then... plug it into that standard expression for the solutions. So that's why I said, we will look at the equations and look at the circuits, but at the end of it all, there's only one equation we need to understand. All right. Energy is of course being transferred back and forth from the inductance capacitor. When the energy is going into the components or the inductance capacitor receiving the energy, we say that they are being charged. Or if they are sending the energy out of the components, then we say they are being discharged. So energy flows into the inductor and capacitor, we say they are being charged. Energy flows out of those components back into the circuit, we say they are being discharged. Analysis of the circuit during these phases is referred to as transient analysis because the response of the capacitor in the... inductor is transient in nature. It is fleeting in nature because after a certain point in time, that response reaches a steady state. Steady state meaning the value remains constant. It does not change with respect to time. So that's the reason we call this as transient analysis. So during the transient analysis portion, the energy transfer portion, because it's constantly going back and forth, the response is fleeting. It's transient in nature. And this applies to both DC as well as AC. We won't look at AC in too much detail until we get to AC analysis. But AC, when you apply an AC signal to a capacitor inductor, there's also a point in time where it reaches steady state. That is a little bit different than, you know, AC, we say, you know, it's constantly changing with respect to time. That's different from this transient analysis. For this one, we're just looking at generic equations, especially for a DC applied source, DC voltage and DC current. Transient response for all circuits always follow the same pattern. It is exponential in nature. It's exponential in nature. Both of these waveforms are showing charging. This one is for voltage. this one is for current. Whether it's voltage or current, it doesn't really matter. Both of them follow an exponential form. A discharging waveform... also has an exponential relationship with time. So this is for discharging. It doesn't matter whether it's voltage, doesn't matter whether it's capacitor. If it's charging, it follows an exponential response like this. If it's discharging, it follows the exponential response that I drew on the right. And hence, all... we need is one single equation that applies to all circuits. Yes. Current is the rate of flow of charge, meaning the rate at which the electrons are moving and supplying the charge that is flowing through the circuit. Yes. So let's see, what was I doing? Oh, so yeah, the transient response is the one in the middle. So at the end on the far right, here we see that it's described as steady state. Steady state, if you remember from chapter three, is the point in time where the capacitor or inductor reaches full charge. If it reaches full charge, then the value does not change anymore. It remains constant with respect to time. When it remains constant... It's steady, hence steady state. If it's discharging, at some point in time, it reaches zero, right? It reaches zero. If it reaches zero, it still is constant with respect to time. So I like to call it steady state because, again, the value is steady with respect to time. It doesn't change, okay? But typically, steady state refers to the point. where the capacitor or inductor reaches full charge. When it reaches full charge, it has absorbed as much energy as it can, depending on the source, and then it just remains steady with respect to time. Okay, there are two kinds of transient responses, natural and forced. It is not as sinister as the word sounds, right? There's another word, another term that we use is known as the step response. but we still have to you know refer to this because this is the standard risk you know standard term that's been used for like decades okay so i still uh you know i would prefer not to use that word but again we have to make engineers aware of what is forced simply means you're connecting a source okay you're connecting a source which is what you've seen so far we see our resistive circuits where we connect a voltage source or a current source that voltage source or current source is external to the component, right? It's not part of the component. The energy is being supplied externally. That is what we have seen. All circuit analysis where we connect a source is forced response because the response is being forced from the component because of an externally applied energy source, okay? So again, it has a negative connotation, but it's not as sinister as it as the word sounds. It's basically charging from an external source, it's referred to as a forced or step response, and if it has received energy and there is no source connected in the circuit anymore, now it's simply losing the energy out into the circuit, then we call it a natural response because the voltage or current is simply being naturally drained out. And so discharging of the energy is referred to as the natural response so based on natural and forced we will look at four different circuits two each for the inductor and two each for the capacitor if you have a inductor capacitor in a circuit basically we cannot let's back up electric circuits will always have resistors resistors are a way of controlling the current right if current if resistor is not present current is infinite right so any electric circuit will always always always have resistors in the circuit now here you see the natural response circuit simply shows you one inductor one capacitor connected to one resistor each okay but note that it says eq that eq as you should be well aware by now stands for equivalent equivalent so Your circuit could have multiple inductors, it could have multiple resistors. But if it is possible to electrically reduce it all down to one single component, we refer to it as one single resistor and one single inductor. And so we refer to this as an RL circuit, RL circuit. Same thing with the capacitor. If we have multiple capacitors in the circuit, and it is possible to reduce those capacitors down to one single capacitor, we refer to it as a RC circuit, okay? One energy storage device. So it doesn't matter whether physically they might be multiple inductors or multiple capacitors. If they can be reduced to one single energy device, we refer to it as a RL or RC. These single energy storage devices will require a first single... the single energy storage device circuits would require a first-order differential equation. And because it's a first-order differential equation, we refer to these as first-order circuits. So one energy storage device, physical reduce two. one component okay so equal is just you reduce sorry not physically electrically reduced to one component yes that be a second order so that was going to be my next point yes so if you have more than one energy storage device in a circuit like two for instance then it would be a second order differential equation and so it would be known as a second order circuit okay second order circuits a little bit second order transient response is a bit more complicated um so you know for non-e major classes i only cover them if we like you know get to the end of the quarter and we have time left but for now it's just first order circuits charging and for discharging okay so these are the step response the charging circuits the charging circuits we have a source connected here on the left it's a voltage source so Thevenin equivalent and here you have a current source or a Norton equivalent. Let's look at a natural response circuit first. So you have RL circuit meaning you have one inductor one resistor in that circuit. If you look at this circuit, the first thing I'd like to point out is the switch. This is the chapter where we use the constant of switches flipping open and flipping close. So at time t equal to zero, the arrow is pointing up. It means the switch is being opened at time t equal to zero. This is an important point. Switch is being opened. opened at time t equal to zero. So your job as an engineer is to set up the circuit. Again, all of the circuits you see in this chapter, you only require one equation. And the equation, you're not solving the equation. Most of the analysis or most of the solution in this is basically understanding what the problem statement is giving you. and just extracting information, and that's it. That's like 90% of your solution. So it's theoretical concept that you apply. I wouldn't even call it electrical engineering circuit analysis. What you're doing here is you need to logically look at it, understand what it means, and extract information from the problem. So in this case, it tells you that at time t equal to zero, the switch is being flipped open. which means before time t equal to zero, the switch was closed. So time t less than zero, the switch was closed. And if the switch was closed, then the current source Is was connected to that inductor. Eventually, at some point in the past, that inductor had reached full charge. If the problem statement doesn't give you anything about the past, It simply states a very long time. Then you simply assume or infer that it has had sufficient time to reach steady state. It has had sufficient time to reach steady state. We don't care about what happened in the past. We don't. We are not analyzing the past. We are not analyzing the circuit before time t equal to zero. But we need to know that value so that we can begin our analysis. Does that make sense? Right? We are beginning our analysis at t equal to zero. So we need to know what the value of that current or voltage is at t equal to zero. For that, we look into the past. That's all. Any questions? Okay. So, inductor is a short because it has reached steady state. All of the current is flowing through the inductor. Nothing is flowing through those resistors at all. This allows us to assume that the maximum current is flowing through the inductor, and maximum current should have a value that is equal to that source, Is. And that's where we begin our analysis. Now, at t equal to zero, the switch flips open. flips open the circuit simply reduces to the inductor and the resistor on the right everything else is disconnected so if the source is disconnected and there is no source present in the circuit it is a natural response this is an inference or conclusion that you have to make by looking at the circuit again as i said the analysis in here you're not writing on equations it's pretty much You can of course draw out the circuits, but it's the conclusion that you make. What happens before the switch was flipped? What happens after the switch is flipped? After the switch is flipped, it's simply an inductor and a resistor. No source left in the circuit anymore, so it's a natural response. So then you should know which equation you need to use. Okay, we will go over the derivation. And then again, at the end, you will see how that one single equation applies to all. So the initial condition of the inductor depends on the energy that was present in the inductor right before the switch is flipped. This here is an important point. We refer to the beginning of analysis as time t equal to zero. Then there is the concept of t minus and t plus. This simply indicates the time right before. So minus is in the past, so it's right before. And t plus is the time right after. For this class, mostly it's just 0 minus and 0 plus. But E major classes, we look at problems where it's like, you know, t could have any value. two seconds, five milliseconds. You would say t five minus, five plus. It simply means the time right before and time right after. It does not have a fixed value. It's simply indicating an instantaneous moment, I mean instant in time, right before the switch is flipped and right after the switch is flipped. And the reason this is important is because the current through an inductor cannot change instantaneously. You cannot snap your fingers and transfer the energy in and out instantaneously. It doesn't matter if you have all five infinity stones. Snap your fingers, doesn't work that way. It takes a finite amount of time. So it doesn't change instantaneously. And this allows you to set up your initial conditions. So this is the most important concept. Switch flips. This is the circuit we are analyzing, but we have no idea what the stored energy is. So we have to look into the past to determine that. And the assumption we make is in the past, it was fully charged. Now in the present, it cannot change instantaneously. So it's still carrying that full charge. And we begin our analysis with that assumption. Let's take a pause. Let me know if you have questions. Any questions? Okay. So then we say that the current before and after is exactly the same. I0 is equal to Is. That 0 indicates... that zero indicates the initial period time t equal to zero sometimes it is also referred to as i i initial so i stands for initial both mean exactly the same thing that's the point in time that we begin our analysis okay so we begin our analysis the inductor starts discharging at this point we apply kvl at the top when you apply kvl we can write vl plus vr equals to zero You plug in the equations for both the inductor current as well as the resistor current. Resistor current is from IR, Ohm's law. Inductor current is LDI, I'm sorry, inductor voltage is LDI over DT. Resistor voltage is I times R. You have a first order ordinary differential equation. When you solve that equation, this is the form we get. I0 times E raised to minus T over tau. okay, I of t, the current through the inductor as a function of time can take on any value depending on which moment in time you're talking about. t is the variable, I0 would be the constant or has a constant fixed value because that's the initial value, the full charge, and then tau is the time constant L over R. Simply describes the speed at which the energy is being transferred. That's it. Those are the only two quantities in that equation, I0 and tau. Tau would be easy to calculate, figure out, not figure out. L would be given to you, R would be given to you. That's component values. Your job would be to understand the circuit and determine the value of I0. And once you have that, you simply plug it into this equation. As I said, 90% of the solution is simply understanding the problem statement, setting it up correctly, and then simply using the standard equation and plugging the values in. This is just the first of four equations, but you will see they all have a similar structure. And there will be one generic equation at the end. to rule them all. Were there five infinity stones or six? Okay, it's been a while. If we have the expression for the voltage or the current, we can find the other quantity by applying Ohm's law. In this case, we obtain an expression for current, right? I of t. Simply apply Ohm's law and we can obtain V of t. So V of t is simply the product V of t again, we are talking about the inductor, so the voltage across the inductor will also drop as a function of time. As it loses the energy, the current starts decreasing, the voltage will also start changing and that would depend on r times i. Instantaneous power p is a product of voltage and current in terms of the resistance is i squared times r. So you simply plug that into the equation and you obtain this. The total energy dissipated in a resistor would be the total energy that was stored in the inductor would be half Li square. That doesn't change. let's see uh rc oh any questions before i go to rc Same thing here, same concept. You have a capacitor, switch was flipped from A to B. So you analyze the circuit at A, then you analyze the circuit at B. If you look at the circuit at A, capacitor was connected to the voltage source VG. The G subscript simply stands for generator. So the generator voltage that is connected to C capacitor. We are not given anything about the past. simply states it's a long time. So when it's a long time, we simply assume capacitor has reached full charge, steady state, capacitor voltage VC is equal to VG. That's an assumption that we have to make or a conclusion that we have to infer. At time t equal to zero, switch flips to the right. When it flips to the right, the left side of the circuit is disconnected. Capacitor is now connected to the R on the right. and it becomes like this. We look at the circuit and we realize there is no source connected in the circuit at all. If it's no source, meaning it's a natural response, capacitor is losing the energy. But what is the value of that energy? We look into the past. When we look into the past, we see that Vc is equal to Vg, the generator voltage, right? Full charge, sometime in the past. From that point on, capacitor behaves like an open circuit. It's steady with respect to time. So it has a value of Vg. Then, to analyze, we apply KCL at the top. We get capacitor current IC plus IR equals to zero. Plug in the capacitor current equation. IC is C dV over dt. Current is V over R. plug in those values you have a first order differential equation you solve that and you and you obtain the exact same expression except this time is for voltage v of t is v zero into e raised to minus t over tau again only two quantities v zero voltage that is the initial value of that capacitor voltage at the beginning of your analysis v zero or vi okay some books some authors and some instructors like to use zero some like to use i as an engineer you should understand that both mean exactly the same thing there is no difference so if you look at a different um you know you're sitting in a meeting you're working and somebody refers to as vi or ii that's what they're referring to tau in this case is r times c So for a RC circuit, tau is RC, product of the resistor and the capacitor. As straight, as easy as that. Tau is the same whether it's charging or whether it's discharging. So if it's a discharging, tau for a RC circuit is RC. If it's charging, it is still RC. Same thing for the inductor. L over R for charging as well as discharging. Same exact expression. v zero into e raised to minus t over tau no difference and that is the reason why we can take one single equation for all four circuits okay questions any questions okay let's look at oh sorry standard equations for voltage power and energy uh capacitor voltage uh We found the equation in the previous slide. So if we have the voltage, then we can find the current using Ohm's law. I is V over R. Power is still V times I. When you plug it into the equation, you can apply V squared divided by R. You obtain an expression for power. And the total energy dissipated in the resistor has to equal the total energy that was stored in the capacitor, half Cv half cv squared eventually all of that energy will be lost capacitor voltage will go to zero okay i'm going to slip over this summary slide you don't actually need it we look at the step response step response or the straightforward you have a voltage source or a current source and you connect it directly to the capacitor or the inductor inductor capacitor starts charging There may or may not be and initially stored energy in the capacitor or inductor. For instance, let's say initial energy for the inductor is zero. Then it begins charging. Now it starts discharging again. But before it can discharge all the way to zero, it starts charging again because of an externally applied source. So when it starts charging again, there is some energy that is stored in there. So unless the problem states that... the inductor capacitor has been discharged for a very long time excuse me you have to take that initial energy into account okay all right so apply kvl still the same thing vl plus vr should equal to some value because there is a source present in the circuit it's not zero it's equal to vs you get a first order equation you solve that and you obtain I of t is equal to I f plus I0 minus I f. Now we have an extra quantity in there. Previously, we only had the initial current, I0. Now we also have to account for the final value. Remember the charging expression, right? We start at some value and we reach, so this is 0 or I, this is f, final value. Initial value, final value. So you have to take both into account when you're doing a charging analysis. But the rest of the equation remains the same, E raised to minus T over tau. E raised to minus T over tau. If the component is completely discharged, if it did not have any stored energy, then I0 would be equal to 0, and the equation would simply be IF minus IF times e raised to minus t over tau okay so no initial energy then i zero goes to zero tau still remains the same l or r that does not change the time constant tau does not change so again the equation is exactly the same it follows the exponential function e raised to minus t over tau capacitor exactly the same thing you have a voltage that's being applied to the capacitor it begins charging you have some initial value of the energy storage you get a first order differential equation by applying kcl you solve it exactly the same structure v of t is vf plus v0 minus vf e raised to minus t over tau you have to take the initial energy into account if initial energy v0 is zero then the equation becomes Vf minus Vf into e raised to minus t over tau. Time constant still remains the same, r times c. Four circuits, exactly the same. Let's take a pause before I show you the one grand equation. Let me know if you have any questions. Any questions? Okay. One equation to rule them all. X of t is given by XF plus X0 minus XF into e raised to minus t over tau. X, let's take it one at a time. X is a variable could stand for inductor current or capacitor voltage. So it applies for both current as well as for voltage. Xf is the final value. X0 is the initial value. Tau is the time constant. That is, again, standard expression. So your job is to determine what is the value of tau, what is the value of X0, what is the value of... xf. That's it. You find these three quantities, you plug it into the equation, that's your final answer. But 90% of the work, as I said, is finding these three quantities. Tau, again, straightforward. You can find tau if you reduce the circuit. Tau is easy. Then you need to find XF and X0. X0, if it is discharging, sorry, if it is discharging, then XF, the final value is, what is the final value? If it's discharging, what is the final value? Zero, always zero. Doesn't matter if the discharge gets interrupted, right? As I said, for EE major classes, we solve examples. where we state, oh, this fully charged capacitor started discharging. And then before it could discharge fully, we apply external source and it starts charging again. So it never reaches zero. But when we analyze that first stage, XF will still be input as zero. So if it's discharging, XF is always zero. It doesn't matter whether it's capacitor or inductor. So as soon as you recognize... that it's a discharging problem, XF takes on the value of zero. That's it. Now, when it's charging, when it's charging, XF will always equal the voltage or current that's being applied. But there's a caveat, right? You need to find the equivalent circuit to find that final value. And again, I will look at that, you know, I will address that point number one. So XF while it is charging should be equal to the voltage or the current that's being applied. The initial value would depend if the problem statement says capacitor or inductor had some initial charge stored in it, then it would change. But the only standard one value that I would state is that XF is zero for discharging. So again, XF, X zero, and tau. Find those three values, plug into this equation, done. xf, x0, and tau. So the equation is x of t equals xf plus x0 minus xf e raised to minus t over tau. t is the variable. It can take on any value, and then tau is the time constant. Time constant, of course, is a standard expression, L over R, or R times C. Okay, let's address point number one. Does the equation make sense? As I said, all four circuits have the exact same structure. So this one equation rules them all. You don't have to do anything except understand this equation, understand your circuit, and know how to apply that equation, meaning calculate the values and then plug it into the equation. That's it. So point number one, find the equivalent circuit. What does it mean? you have to set up the circuit for time t less than zero and then time t greater than zero so you have to see how the circuit was before it changed and after it changes right before and after if this was your circuit you have a resistor and you apply it to a capacitor this is a random example so you don't have to draw this out but think through this with me. What is the voltage Vc? Meaning at some point in time, that capacitor is going to reach full charge. What is the value of that full charge? Vs, right? Eventually, that capacitor will reach full charge, Vs, and behave like an open circuit. So, your final value in this case, Xf, would be the capacitor voltage which would equal the source voltage Vs. But what if, what if, so this is R1, this is R1 here, and there is another resistor R2. Now, what is the value of the voltage across that capacitor? Is that equal to Vs? Let me ask that. No, right? It's not equal to Vs because it's not being applied directly. If I redraw that circuit in a way that makes sense, the circuit would look like this. This is R1, this is R2, capacitor C is in parallel with R2, and then you have your voltage source here. If you remove the capacitor, you know that the two resistors are in series so you have some voltage across R1 and some voltage across R2 which would eventually be the case when C reaches full charge because it behaves like an open circuit but the capacitor voltage in parallel with the R2 resistor will remain exactly the same but that voltage is not equal to Vs so you have to find the equivalent circuit by removing the capacitor and analyzing between those open terminals. That's what we mean by finding Thevenin or Norton equivalent circuit, because we have to reduce that circuit down to one single voltage source and one single resistance. So we have to find the Thevenin or the Norton equivalent circuit as seen by the capacitor. You remove the capacitor from your circuit and you analyze it between those open terminals. And that's why Thevenin and Norton equivalent circuits are a very important concept, very important part of circuit analysis. So you find the Thevenin circuit between terminals A and B. Capacitor is removed. Now, for non-E major classes, you're not going to do some large circuit solving and then I ask you to find this. But this is a concept. I like to point this out to students because one of the questions is, when am I ever going to use this? When we will ever apply Thevenin? You see right here in chapter four, how and why it can be important, right? You have to know the voltage that is being applied to that capacitor. We can't do that without finding the equivalent circuit. and that's what i mean by step number one when it says find the equivalent circuit we reduce the components down to one single resistor one single voltage source does that make sense all right questions yes So the question being asked is, let's say the switch flips and voltage source Vs now gets connected to R1, R2, and Vc. Is that what you're stating, the circuit conditions? Yes, because the current flowing through R1, R2, and C would depend on the capacitor charging. So the answer is yes. If it's in series, then it would be the same current. If it's in parallel like R2, then again, the voltage... across R2 is exactly the same as the voltage across the capacitor. So the capacitor equation applies to R2 also. Yes. Yes. So the question being asked is, would the Thevenin voltage that we find in this example, hypothetical example, would that be equal to XF? The answer is yes. Correct. So Vth that we find from here would be equal to the final value. The final value is basically the source voltage that is being applied to the capacitor. And in this case, it would be the Thevenin voltage. Yes. So the question being asked is, in general terms, I think what the question that's being asked is, how can we use a capacitor, right? Applications of a capacitor in real life. Am I understanding the question correctly, right? Applications. So one of the most common examples I like to give was in a camera flash. You know, the big ones that are attached on top of a professional camera. If you've ever, have you ever heard like a whining noise? In that flash, what is happening is there is a capacitor that's being charged from the batteries. Usually there are like four AA batteries, so six volts of energy, six volts of charge that's supplied to the capacitor. And when you click the camera shutter button, all of that energy stored in the capacitor is released all at once. And so there's a flashlight, there's a light bulb that goes off like a flash, hence the name flash. Does that make sense? So there's a capacitor. And when you click on the shutter button, that energy is released all at once. Even closer, everybody uses a capacitor every time you touch your touchscreens, because these screens are now capacitive touchscreens. When you touch it with a bare skin, it causes a capacitance. I mean, you know, the human skin along with the screen causes a capacitance. That change in the capacitance is detected and is registered as a touch, converted into an electrical signal. Previously, it used to be resistive touchscreens. Resistive touchscreens, you had to apply pressure. So when you applied pressure on it, it would change the resistance value. And depending on the spot where it changes the resistance, that is transferred into an electrical signal. Resistive touchscreens do not work with skin. It requires like a sharp point or a sharp object. Hence, we used to have those styluses to use those, you know, to use those screens. So as I was saying, capacitors, one of the common example is where you store, you know, store energy that is released all at once. And then of course, your capacitive touchscreens. Also, there is there are multiple applications we look at another application in chapter six which is frequency response which is when we use it for electronic filtering because the response of a capacitor to an ac circuit is different well not different but that's where we'll apply okay let's look at the next slide here where i show a universal time constant chart Remember, I stated all the examples are exactly this. All the equations are exactly the same, right? You have e raised to minus t over tau, right? e raised to minus t over tau. Whether it's charging or discharging, this is the only equation you need to understand. Because it's e raised to minus t over tau. It's exponential relationship. This here shows a charging waveform. And this here shows a discharging waveform. Doesn't matter what kind of circuit it is. Doesn't matter what kind of value the capacitor inductor has. Doesn't matter the value of the resistors. It will always have this exact waveform. That does not change. That's one of the, for me, it's like one of the mysteries of nature. When we are going into more like philosophy in that sense. How is it that everything behaves exactly the same way? That's just the property. It has an exponential relationship with time, whether it's charging, whether it's discharging, it remains exactly the same. In terms of pure numbers, we can plug in values of time to any value we want, 0.01, 0.005, 0.0003, and we can find the value of the world's current at any given point in time. But if we plug in the value of time constant as multiples of tau, the time constant. we obtain that universal time constant chart. But look at the first time constant. When it's charging, e raised to minus t, e raised to minus 1, e raised to 0, of course, is 1. e raised to minus 1 would be 0.368. So when it's charging, what happens is it acquires more than 50%, 63.8%. or 63.2% of its charge after the first time constant. So if you look at this chart, you can see that when it's charging within a very short period of time, meaning the first time constant, it acquires more than half of its charge, exponential, right? There's an initial burst, and then it kind of slowly flattens out. Discharging, exactly the same thing happens. it loses more than 60% of its charge after the first time constant. So it's a drastic drop, and then it flattens out towards zero. And that's what that equation signifies. So whether it's charging or whether it's discharging, it acquires or it loses more than 60% of the charge or the stored energy. But it's a mathematical expression. So if you plug in a value of T, as like 100 seconds, 1,000 seconds, a million seconds, or a million days, mathematically, you will still obtain some small value of that voltage or current. right? But realistically speaking, after a five-time constant, it acquires more than 99% of a charge, or it loses more than 99% of its energy. So if it's charging, it's more than 99%. If it's discharging, it's less than 1%. So after five-time constants, we put an end to it by standard accepted practice, by standard accepted convention. So after a five-time constant, We say it is either fully charged or fully discharged. Okay, fully charged or fully discharged. All right, so Monday, we will look at the example problems. Quiz on Monday is based on chapter three. Again, I will post a announcement later this afternoon. It will be on piecewise linear analysis. I'm letting you know right now. Piecewise linear analysis. So you know what you have to study for. And there are a couple of extra solved examples on the website. It's been there since the beginning of the quarter. There is a PDF with solved examples. Go through those examples. Go through the examples that are solved in class. And then prepare accordingly. So piecewise linear analysis quiz on Monday. All right. Have a good weekend. I will see you all on Monday. Yes.

Okay, good morning, everyone. Today, we'll look at Chapter 4. Chapter 4 is Transient Analysis. This is going to go by quickly.

So today, I will cover the entire theory and the concepts for Chapter 4. And then Monday, we'll look at the problems. But the good thing about Chapter 4 is there's only one equation in that entire chapter that you need to know. So rather than taking down a lot of notes or looking at all the equations when I go through, I would rather students just take minimal notes only as much as they need to understand the concepts.

Don't worry about writing down all the equations or the derivations in the middle. As I stated, at the end of the chapter, there is one single equation that works for everything. OK. All right. So let's go ahead and begin now.

This chapter is known as transience and I will explain that in just a couple of slides. So, inductor and capacitors are energy storage devices. The energy transfer back and forth into the inductor or capacitor cannot take place in an instant.

That transfer requires a certain amount of time. The time required to either receive the energy into the component or discharge it back into the circuit depends on the size of the component as well as the circuit conditions. Basically, the size of the container, the fluid flow, and the size of the pipe, right?

So, the smaller the size, the faster it fills up or the faster it loses the energy. The bigger the size, the more time it requires. Now, because these are energy storage devices, they all are dependent on time. So a one energy storage device requires solving a first order ordinary differential equation.

Because it's a first order differential equation, we refer to it as a first order circuit. Thankfully, greater minds have come before us and have already solved this equation, so we don't have to. All we have to do is basically recognize the circuit, set up the analysis conditions, extract the information, and then...

plug it into that standard expression for the solutions. So that's why I said, we will look at the equations and look at the circuits, but at the end of it all, there's only one equation we need to understand. All right.

Energy is of course being transferred back and forth from the inductance capacitor. When the energy is going into the components or the inductance capacitor receiving the energy, we say that they are being charged. Or if they are sending the energy out of the components, then we say they are being discharged. So energy flows into the inductor and capacitor, we say they are being charged.

Energy flows out of those components back into the circuit, we say they are being discharged. Analysis of the circuit during these phases is referred to as transient analysis because the response of the capacitor in the... inductor is transient in nature. It is fleeting in nature because after a certain point in time, that response reaches a steady state. Steady state meaning the value remains constant.

It does not change with respect to time. So that's the reason we call this as transient analysis. So during the transient analysis portion, the energy transfer portion, because it's constantly going back and forth, the response is fleeting.

It's transient in nature. And this applies to both DC as well as AC. We won't look at AC in too much detail until we get to AC analysis.

But AC, when you apply an AC signal to a capacitor inductor, there's also a point in time where it reaches steady state. That is a little bit different than, you know, AC, we say, you know, it's constantly changing with respect to time. That's different from this transient analysis. For this one, we're just looking at generic equations, especially for a DC applied source, DC voltage and DC current.

Transient response for all circuits always follow the same pattern. It is exponential in nature. It's exponential in nature.

Both of these waveforms are showing charging. This one is for voltage. this one is for current. Whether it's voltage or current, it doesn't really matter. Both of them follow an exponential form.

A discharging waveform... also has an exponential relationship with time. So this is for discharging.

It doesn't matter whether it's voltage, doesn't matter whether it's capacitor. If it's charging, it follows an exponential response like this. If it's discharging, it follows the exponential response that I drew on the right.

And hence, all... we need is one single equation that applies to all circuits. Yes. Current is the rate of flow of charge, meaning the rate at which the electrons are moving and supplying the charge that is flowing through the circuit.

Yes. So let's see, what was I doing? Oh, so yeah, the transient response is the one in the middle.

So at the end on the far right, here we see that it's described as steady state. Steady state, if you remember from chapter three, is the point in time where the capacitor or inductor reaches full charge. If it reaches full charge, then the value does not change anymore.

It remains constant with respect to time. When it remains constant... It's steady, hence steady state. If it's discharging, at some point in time, it reaches zero, right? It reaches zero.

If it reaches zero, it still is constant with respect to time. So I like to call it steady state because, again, the value is steady with respect to time. It doesn't change, okay?

But typically, steady state refers to the point. where the capacitor or inductor reaches full charge. When it reaches full charge, it has absorbed as much energy as it can, depending on the source, and then it just remains steady with respect to time.

Okay, there are two kinds of transient responses, natural and forced. It is not as sinister as the word sounds, right? There's another word, another term that we use is known as the step response. but we still have to you know refer to this because this is the standard risk you know standard term that's been used for like decades okay so i still uh you know i would prefer not to use that word but again we have to make engineers aware of what is forced simply means you're connecting a source okay you're connecting a source which is what you've seen so far we see our resistive circuits where we connect a voltage source or a current source that voltage source or current source is external to the component, right?

It's not part of the component. The energy is being supplied externally. That is what we have seen.

All circuit analysis where we connect a source is forced response because the response is being forced from the component because of an externally applied energy source, okay? So again, it has a negative connotation, but it's not as sinister as it as the word sounds. It's basically charging from an external source, it's referred to as a forced or step response, and if it has received energy and there is no source connected in the circuit anymore, now it's simply losing the energy out into the circuit, then we call it a natural response because the voltage or current is simply being naturally drained out. And so discharging of the energy is referred to as the natural response so based on natural and forced we will look at four different circuits two each for the inductor and two each for the capacitor if you have a inductor capacitor in a circuit basically we cannot let's back up electric circuits will always have resistors resistors are a way of controlling the current right if current if resistor is not present current is infinite right so any electric circuit will always always always have resistors in the circuit now here you see the natural response circuit simply shows you one inductor one capacitor connected to one resistor each okay but note that it says eq that eq as you should be well aware by now stands for equivalent equivalent so Your circuit could have multiple inductors, it could have multiple resistors. But if it is possible to electrically reduce it all down to one single component, we refer to it as one single resistor and one single inductor.

And so we refer to this as an RL circuit, RL circuit. Same thing with the capacitor. If we have multiple capacitors in the circuit, and it is possible to reduce those capacitors down to one single capacitor, we refer to it as a RC circuit, okay?

One energy storage device. So it doesn't matter whether physically they might be multiple inductors or multiple capacitors. If they can be reduced to one single energy device, we refer to it as a RL or RC. These single energy storage devices will require a first single... the single energy storage device circuits would require a first-order differential equation.

And because it's a first-order differential equation, we refer to these as first-order circuits. So one energy storage device, physical reduce two. one component okay so equal is just you reduce sorry not physically electrically reduced to one component yes that be a second order so that was going to be my next point yes so if you have more than one energy storage device in a circuit like two for instance then it would be a second order differential equation and so it would be known as a second order circuit okay second order circuits a little bit second order transient response is a bit more complicated um so you know for non-e major classes i only cover them if we like you know get to the end of the quarter and we have time left but for now it's just first order circuits charging and for discharging okay so these are the step response the charging circuits the charging circuits we have a source connected here on the left it's a voltage source so Thevenin equivalent and here you have a current source or a Norton equivalent. Let's look at a natural response circuit first. So you have RL circuit meaning you have one inductor one resistor in that circuit.

If you look at this circuit, the first thing I'd like to point out is the switch. This is the chapter where we use the constant of switches flipping open and flipping close. So at time t equal to zero, the arrow is pointing up. It means the switch is being opened at time t equal to zero. This is an important point.

Switch is being opened. opened at time t equal to zero. So your job as an engineer is to set up the circuit.

Again, all of the circuits you see in this chapter, you only require one equation. And the equation, you're not solving the equation. Most of the analysis or most of the solution in this is basically understanding what the problem statement is giving you. and just extracting information, and that's it. That's like 90% of your solution.

So it's theoretical concept that you apply. I wouldn't even call it electrical engineering circuit analysis. What you're doing here is you need to logically look at it, understand what it means, and extract information from the problem.

So in this case, it tells you that at time t equal to zero, the switch is being flipped open. which means before time t equal to zero, the switch was closed. So time t less than zero, the switch was closed. And if the switch was closed, then the current source Is was connected to that inductor.

Eventually, at some point in the past, that inductor had reached full charge. If the problem statement doesn't give you anything about the past, It simply states a very long time. Then you simply assume or infer that it has had sufficient time to reach steady state. It has had sufficient time to reach steady state.

We don't care about what happened in the past. We don't. We are not analyzing the past. We are not analyzing the circuit before time t equal to zero.

But we need to know that value so that we can begin our analysis. Does that make sense? Right?

We are beginning our analysis at t equal to zero. So we need to know what the value of that current or voltage is at t equal to zero. For that, we look into the past. That's all. Any questions?

Okay. So, inductor is a short because it has reached steady state. All of the current is flowing through the inductor.

Nothing is flowing through those resistors at all. This allows us to assume that the maximum current is flowing through the inductor, and maximum current should have a value that is equal to that source, Is. And that's where we begin our analysis.

Now, at t equal to zero, the switch flips open. flips open the circuit simply reduces to the inductor and the resistor on the right everything else is disconnected so if the source is disconnected and there is no source present in the circuit it is a natural response this is an inference or conclusion that you have to make by looking at the circuit again as i said the analysis in here you're not writing on equations it's pretty much You can of course draw out the circuits, but it's the conclusion that you make. What happens before the switch was flipped?

What happens after the switch is flipped? After the switch is flipped, it's simply an inductor and a resistor. No source left in the circuit anymore, so it's a natural response.

So then you should know which equation you need to use. Okay, we will go over the derivation. And then again, at the end, you will see how that one single equation applies to all.

So the initial condition of the inductor depends on the energy that was present in the inductor right before the switch is flipped. This here is an important point. We refer to the beginning of analysis as time t equal to zero. Then there is the concept of t minus and t plus. This simply indicates the time right before.

So minus is in the past, so it's right before. And t plus is the time right after. For this class, mostly it's just 0 minus and 0 plus. But E major classes, we look at problems where it's like, you know, t could have any value. two seconds, five milliseconds.

You would say t five minus, five plus. It simply means the time right before and time right after. It does not have a fixed value. It's simply indicating an instantaneous moment, I mean instant in time, right before the switch is flipped and right after the switch is flipped.

And the reason this is important is because the current through an inductor cannot change instantaneously. You cannot snap your fingers and transfer the energy in and out instantaneously. It doesn't matter if you have all five infinity stones.

Snap your fingers, doesn't work that way. It takes a finite amount of time. So it doesn't change instantaneously.

And this allows you to set up your initial conditions. So this is the most important concept. Switch flips. This is the circuit we are analyzing, but we have no idea what the stored energy is. So we have to look into the past to determine that.

And the assumption we make is in the past, it was fully charged. Now in the present, it cannot change instantaneously. So it's still carrying that full charge. And we begin our analysis with that assumption.

Let's take a pause. Let me know if you have questions. Any questions? Okay. So then we say that the current before and after is exactly the same.

I0 is equal to Is. That 0 indicates... that zero indicates the initial period time t equal to zero sometimes it is also referred to as i i initial so i stands for initial both mean exactly the same thing that's the point in time that we begin our analysis okay so we begin our analysis the inductor starts discharging at this point we apply kvl at the top when you apply kvl we can write vl plus vr equals to zero You plug in the equations for both the inductor current as well as the resistor current.

Resistor current is from IR, Ohm's law. Inductor current is LDI, I'm sorry, inductor voltage is LDI over DT. Resistor voltage is I times R. You have a first order ordinary differential equation.

When you solve that equation, this is the form we get. I0 times E raised to minus T over tau. okay, I of t, the current through the inductor as a function of time can take on any value depending on which moment in time you're talking about. t is the variable, I0 would be the constant or has a constant fixed value because that's the initial value, the full charge, and then tau is the time constant L over R. Simply describes the speed at which the energy is being transferred.

That's it. Those are the only two quantities in that equation, I0 and tau. Tau would be easy to calculate, figure out, not figure out. L would be given to you, R would be given to you. That's component values.

Your job would be to understand the circuit and determine the value of I0. And once you have that, you simply plug it into this equation. As I said, 90% of the solution is simply understanding the problem statement, setting it up correctly, and then simply using the standard equation and plugging the values in.

This is just the first of four equations, but you will see they all have a similar structure. And there will be one generic equation at the end. to rule them all. Were there five infinity stones or six? Okay, it's been a while.

If we have the expression for the voltage or the current, we can find the other quantity by applying Ohm's law. In this case, we obtain an expression for current, right? I of t. Simply apply Ohm's law and we can obtain V of t.

So V of t is simply the product V of t again, we are talking about the inductor, so the voltage across the inductor will also drop as a function of time. As it loses the energy, the current starts decreasing, the voltage will also start changing and that would depend on r times i. Instantaneous power p is a product of voltage and current in terms of the resistance is i squared times r. So you simply plug that into the equation and you obtain this.

The total energy dissipated in a resistor would be the total energy that was stored in the inductor would be half Li square. That doesn't change. let's see uh rc oh any questions before i go to rc Same thing here, same concept. You have a capacitor, switch was flipped from A to B. So you analyze the circuit at A, then you analyze the circuit at B.

If you look at the circuit at A, capacitor was connected to the voltage source VG. The G subscript simply stands for generator. So the generator voltage that is connected to C capacitor.

We are not given anything about the past. simply states it's a long time. So when it's a long time, we simply assume capacitor has reached full charge, steady state, capacitor voltage VC is equal to VG.

That's an assumption that we have to make or a conclusion that we have to infer. At time t equal to zero, switch flips to the right. When it flips to the right, the left side of the circuit is disconnected. Capacitor is now connected to the R on the right. and it becomes like this.

We look at the circuit and we realize there is no source connected in the circuit at all. If it's no source, meaning it's a natural response, capacitor is losing the energy. But what is the value of that energy? We look into the past.

When we look into the past, we see that Vc is equal to Vg, the generator voltage, right? Full charge, sometime in the past. From that point on, capacitor behaves like an open circuit.

It's steady with respect to time. So it has a value of Vg. Then, to analyze, we apply KCL at the top.

We get capacitor current IC plus IR equals to zero. Plug in the capacitor current equation. IC is C dV over dt. Current is V over R.

plug in those values you have a first order differential equation you solve that and you and you obtain the exact same expression except this time is for voltage v of t is v zero into e raised to minus t over tau again only two quantities v zero voltage that is the initial value of that capacitor voltage at the beginning of your analysis v zero or vi okay some books some authors and some instructors like to use zero some like to use i as an engineer you should understand that both mean exactly the same thing there is no difference so if you look at a different um you know you're sitting in a meeting you're working and somebody refers to as vi or ii that's what they're referring to tau in this case is r times c So for a RC circuit, tau is RC, product of the resistor and the capacitor. As straight, as easy as that. Tau is the same whether it's charging or whether it's discharging.

So if it's a discharging, tau for a RC circuit is RC. If it's charging, it is still RC. Same thing for the inductor.

L over R for charging as well as discharging. Same exact expression. v zero into e raised to minus t over tau no difference and that is the reason why we can take one single equation for all four circuits okay questions any questions okay let's look at oh sorry standard equations for voltage power and energy uh capacitor voltage uh We found the equation in the previous slide.

So if we have the voltage, then we can find the current using Ohm's law. I is V over R. Power is still V times I. When you plug it into the equation, you can apply V squared divided by R. You obtain an expression for power.

And the total energy dissipated in the resistor has to equal the total energy that was stored in the capacitor, half Cv half cv squared eventually all of that energy will be lost capacitor voltage will go to zero okay i'm going to slip over this summary slide you don't actually need it we look at the step response step response or the straightforward you have a voltage source or a current source and you connect it directly to the capacitor or the inductor inductor capacitor starts charging There may or may not be and initially stored energy in the capacitor or inductor. For instance, let's say initial energy for the inductor is zero. Then it begins charging.

Now it starts discharging again. But before it can discharge all the way to zero, it starts charging again because of an externally applied source. So when it starts charging again, there is some energy that is stored in there. So unless the problem states that... the inductor capacitor has been discharged for a very long time excuse me you have to take that initial energy into account okay all right so apply kvl still the same thing vl plus vr should equal to some value because there is a source present in the circuit it's not zero it's equal to vs you get a first order equation you solve that and you obtain I of t is equal to I f plus I0 minus I f.

Now we have an extra quantity in there. Previously, we only had the initial current, I0. Now we also have to account for the final value.

Remember the charging expression, right? We start at some value and we reach, so this is 0 or I, this is f, final value. Initial value, final value. So you have to take both into account when you're doing a charging analysis. But the rest of the equation remains the same, E raised to minus T over tau.

E raised to minus T over tau. If the component is completely discharged, if it did not have any stored energy, then I0 would be equal to 0, and the equation would simply be IF minus IF times e raised to minus t over tau okay so no initial energy then i zero goes to zero tau still remains the same l or r that does not change the time constant tau does not change so again the equation is exactly the same it follows the exponential function e raised to minus t over tau capacitor exactly the same thing you have a voltage that's being applied to the capacitor it begins charging you have some initial value of the energy storage you get a first order differential equation by applying kcl you solve it exactly the same structure v of t is vf plus v0 minus vf e raised to minus t over tau you have to take the initial energy into account if initial energy v0 is zero then the equation becomes Vf minus Vf into e raised to minus t over tau. Time constant still remains the same, r times c. Four circuits, exactly the same. Let's take a pause before I show you the one grand equation.

Let me know if you have any questions. Any questions? Okay.

One equation to rule them all. X of t is given by XF plus X0 minus XF into e raised to minus t over tau. X, let's take it one at a time. X is a variable could stand for inductor current or capacitor voltage.

So it applies for both current as well as for voltage. Xf is the final value. X0 is the initial value.

Tau is the time constant. That is, again, standard expression. So your job is to determine what is the value of tau, what is the value of X0, what is the value of... xf.

That's it. You find these three quantities, you plug it into the equation, that's your final answer. But 90% of the work, as I said, is finding these three quantities. Tau, again, straightforward. You can find tau if you reduce the circuit.

Tau is easy. Then you need to find XF and X0. X0, if it is discharging, sorry, if it is discharging, then XF, the final value is, what is the final value?

If it's discharging, what is the final value? Zero, always zero. Doesn't matter if the discharge gets interrupted, right? As I said, for EE major classes, we solve examples.

where we state, oh, this fully charged capacitor started discharging. And then before it could discharge fully, we apply external source and it starts charging again. So it never reaches zero.

But when we analyze that first stage, XF will still be input as zero. So if it's discharging, XF is always zero. It doesn't matter whether it's capacitor or inductor.

So as soon as you recognize... that it's a discharging problem, XF takes on the value of zero. That's it.

Now, when it's charging, when it's charging, XF will always equal the voltage or current that's being applied. But there's a caveat, right? You need to find the equivalent circuit to find that final value. And again, I will look at that, you know, I will address that point number one.

So XF while it is charging should be equal to the voltage or the current that's being applied. The initial value would depend if the problem statement says capacitor or inductor had some initial charge stored in it, then it would change. But the only standard one value that I would state is that XF is zero for discharging.

So again, XF, X zero, and tau. Find those three values, plug into this equation, done. xf, x0, and tau. So the equation is x of t equals xf plus x0 minus xf e raised to minus t over tau. t is the variable.

It can take on any value, and then tau is the time constant. Time constant, of course, is a standard expression, L over R, or R times C. Okay, let's address point number one. Does the equation make sense? As I said, all four circuits have the exact same structure.

So this one equation rules them all. You don't have to do anything except understand this equation, understand your circuit, and know how to apply that equation, meaning calculate the values and then plug it into the equation. That's it.

So point number one, find the equivalent circuit. What does it mean? you have to set up the circuit for time t less than zero and then time t greater than zero so you have to see how the circuit was before it changed and after it changes right before and after if this was your circuit you have a resistor and you apply it to a capacitor this is a random example so you don't have to draw this out but think through this with me. What is the voltage Vc? Meaning at some point in time, that capacitor is going to reach full charge.

What is the value of that full charge? Vs, right? Eventually, that capacitor will reach full charge, Vs, and behave like an open circuit.

So, your final value in this case, Xf, would be the capacitor voltage which would equal the source voltage Vs. But what if, what if, so this is R1, this is R1 here, and there is another resistor R2. Now, what is the value of the voltage across that capacitor? Is that equal to Vs? Let me ask that.

No, right? It's not equal to Vs because it's not being applied directly. If I redraw that circuit in a way that makes sense, the circuit would look like this. This is R1, this is R2, capacitor C is in parallel with R2, and then you have your voltage source here.

If you remove the capacitor, you know that the two resistors are in series so you have some voltage across R1 and some voltage across R2 which would eventually be the case when C reaches full charge because it behaves like an open circuit but the capacitor voltage in parallel with the R2 resistor will remain exactly the same but that voltage is not equal to Vs so you have to find the equivalent circuit by removing the capacitor and analyzing between those open terminals. That's what we mean by finding Thevenin or Norton equivalent circuit, because we have to reduce that circuit down to one single voltage source and one single resistance. So we have to find the Thevenin or the Norton equivalent circuit as seen by the capacitor.

You remove the capacitor from your circuit and you analyze it between those open terminals. And that's why Thevenin and Norton equivalent circuits are a very important concept, very important part of circuit analysis. So you find the Thevenin circuit between terminals A and B. Capacitor is removed.

Now, for non-E major classes, you're not going to do some large circuit solving and then I ask you to find this. But this is a concept. I like to point this out to students because one of the questions is, when am I ever going to use this? When we will ever apply Thevenin?

You see right here in chapter four, how and why it can be important, right? You have to know the voltage that is being applied to that capacitor. We can't do that without finding the equivalent circuit.

and that's what i mean by step number one when it says find the equivalent circuit we reduce the components down to one single resistor one single voltage source does that make sense all right questions yes So the question being asked is, let's say the switch flips and voltage source Vs now gets connected to R1, R2, and Vc. Is that what you're stating, the circuit conditions? Yes, because the current flowing through R1, R2, and C would depend on the capacitor charging.

So the answer is yes. If it's in series, then it would be the same current. If it's in parallel like R2, then again, the voltage... across R2 is exactly the same as the voltage across the capacitor.

So the capacitor equation applies to R2 also. Yes. Yes.

So the question being asked is, would the Thevenin voltage that we find in this example, hypothetical example, would that be equal to XF? The answer is yes. Correct.

So Vth that we find from here would be equal to the final value. The final value is basically the source voltage that is being applied to the capacitor. And in this case, it would be the Thevenin voltage. Yes.

So the question being asked is, in general terms, I think what the question that's being asked is, how can we use a capacitor, right? Applications of a capacitor in real life. Am I understanding the question correctly, right?

Applications. So one of the most common examples I like to give was in a camera flash. You know, the big ones that are attached on top of a professional camera. If you've ever, have you ever heard like a whining noise?

In that flash, what is happening is there is a capacitor that's being charged from the batteries. Usually there are like four AA batteries, so six volts of energy, six volts of charge that's supplied to the capacitor. And when you click the camera shutter button, all of that energy stored in the capacitor is released all at once. And so there's a flashlight, there's a light bulb that goes off like a flash, hence the name flash.

Does that make sense? So there's a capacitor. And when you click on the shutter button, that energy is released all at once.

Even closer, everybody uses a capacitor every time you touch your touchscreens, because these screens are now capacitive touchscreens. When you touch it with a bare skin, it causes a capacitance. I mean, you know, the human skin along with the screen causes a capacitance. That change in the capacitance is detected and is registered as a touch, converted into an electrical signal. Previously, it used to be resistive touchscreens.

Resistive touchscreens, you had to apply pressure. So when you applied pressure on it, it would change the resistance value. And depending on the spot where it changes the resistance, that is transferred into an electrical signal.

Resistive touchscreens do not work with skin. It requires like a sharp point or a sharp object. Hence, we used to have those styluses to use those, you know, to use those screens. So as I was saying, capacitors, one of the common example is where you store, you know, store energy that is released all at once.

And then of course, your capacitive touchscreens. Also, there is there are multiple applications we look at another application in chapter six which is frequency response which is when we use it for electronic filtering because the response of a capacitor to an ac circuit is different well not different but that's where we'll apply okay let's look at the next slide here where i show a universal time constant chart Remember, I stated all the examples are exactly this. All the equations are exactly the same, right?

You have e raised to minus t over tau, right? e raised to minus t over tau. Whether it's charging or discharging, this is the only equation you need to understand.

Because it's e raised to minus t over tau. It's exponential relationship. This here shows a charging waveform. And this here shows a discharging waveform. Doesn't matter what kind of circuit it is.

Doesn't matter what kind of value the capacitor inductor has. Doesn't matter the value of the resistors. It will always have this exact waveform. That does not change.

That's one of the, for me, it's like one of the mysteries of nature. When we are going into more like philosophy in that sense. How is it that everything behaves exactly the same way? That's just the property.

It has an exponential relationship with time, whether it's charging, whether it's discharging, it remains exactly the same. In terms of pure numbers, we can plug in values of time to any value we want, 0.01, 0.005, 0.0003, and we can find the value of the world's current at any given point in time. But if we plug in the value of time constant as multiples of tau, the time constant. we obtain that universal time constant chart. But look at the first time constant.

When it's charging, e raised to minus t, e raised to minus 1, e raised to 0, of course, is 1. e raised to minus 1 would be 0.368. So when it's charging, what happens is it acquires more than 50%, 63.8%. or 63.2% of its charge after the first time constant.

So if you look at this chart, you can see that when it's charging within a very short period of time, meaning the first time constant, it acquires more than half of its charge, exponential, right? There's an initial burst, and then it kind of slowly flattens out. Discharging, exactly the same thing happens.

it loses more than 60% of its charge after the first time constant. So it's a drastic drop, and then it flattens out towards zero. And that's what that equation signifies.

So whether it's charging or whether it's discharging, it acquires or it loses more than 60% of the charge or the stored energy. But it's a mathematical expression. So if you plug in a value of T, as like 100 seconds, 1,000 seconds, a million seconds, or a million days, mathematically, you will still obtain some small value of that voltage or current. right? But realistically speaking, after a five-time constant, it acquires more than 99% of a charge, or it loses more than 99% of its energy.

So if it's charging, it's more than 99%. If it's discharging, it's less than 1%. So after five-time constants, we put an end to it by standard accepted practice, by standard accepted convention. So after a five-time constant, We say it is either fully charged or fully discharged. Okay, fully charged or fully discharged.

All right, so Monday, we will look at the example problems. Quiz on Monday is based on chapter three. Again, I will post a announcement later this afternoon. It will be on piecewise linear analysis.

I'm letting you know right now. Piecewise linear analysis. So you know what you have to study for.

And there are a couple of extra solved examples on the website. It's been there since the beginning of the quarter. There is a PDF with solved examples.

Go through those examples. Go through the examples that are solved in class. And then prepare accordingly. So piecewise linear analysis quiz on Monday.

All right. Have a good weekend. I will see you all on Monday.

Yes.

Transcript for:Transient Analysis Lecture

Transcript for:
Transient Analysis Lecture