Okay, hey, before we go on to the next course, like I wanted to stop and explain the resistor and an inductor in a DC circuit, like I show you here in this, this slide, and L and n R has a time constant. We spoke about RC time constants. Previously, you know in a previous course, well, L and R we also have a time constant. All right, the time constant for that is L divided by r. So, what I'm showing you here is I've got a 10 ohm resistor, and a one Henry inductor. I just do the math in my time, one time constant is zero dot one second and just like on an RC time constant, okay for it to go to a steady state, either steady current here or steady voltage level here. It will take five plus time constants and we're going to talk about that right now.

Let's clear the slide and when I close the switch in this particular circuit, all right, what happens is I get again I get a back emf. So the minute I close this switch, I get an influx of current. This current produces a back end On my inductor, all right, and that's what we're showing you here. All right at that instant right there, we have a back emf, that actually at that instant is equal to the source voltage, in this case 10 volts. So therefore, if I've got two voltage sources opposing each other, for an instant, we have no current flow, and I spoke about that earlier in a couple of courses back. So as when I close the switch, and I don't get a rapid change of current current, the change in current stays the same.

My back emf decreases to zero. And then if you look at this curve here, That's my current and voltage across the resistor here. That goes from zero up to its maximum value, right and flattens out. So it's just like the RC time constant it the first time constant is 63%. Well, if you if you look here, when I'm looking at the voltage across my my coil, this this this one here, at the first time constant, it discharges 63% and I'm left with 37% across the coil. All right, the second time constant, it discharges 86% I'm left with 14% across the coil.

On the third time constant, it discharges 990 6%, I'm left with 4%. Then the fourth 98, left with 2%. And then the fifth 99, left with 1%. And then we just, we just keep getting on ideally, it never truly discharges. And ideally, it never really truly charges but it gets so minuscule at that point. It's not worth mentioning.

All right. So again, with RC time constant, after five time constants, I was either fully charged or fully discharged for all intents and purposes. And the same thing here with a with a coil and a resistor in series. Again, this is my time constant T. For one time constant equals L divided by R and again, I've done the math. And in this particular case, it's point one seconds. All right, so let's stop here and go to the next slide.

Okay, we put a problem here and let's look at this. It says an applied DC voltage of 10 volts will produce a steady state current of 100 milliamp hours in 100 Ohm coil. Alright, so here's my hundred ohm resistor is my coil. How much is the current after point two seconds and after one second? Well, we did our, our math here and we know that the time constant is L over R i do my math and I get one time constant is point two seconds right there. All right.

So if you look at this 100 milliamp hours of steady state current well as You know that once we close the switch at some point, this is my current and voltage curve through my resistive element. And we're going to get 100% here and that were would represent 100 milliamp hours how do we get that? Well it would be 10 volts divided by 100 ohms is my 10 volts is my hundred ohms. And if I do the math, I get 100 milliamp hours All right, so the first time constant, which would be point two seconds, what am I going to get? I'm going to get 63% of that. So 63% of 100 is 63 milliamp hours and then at Five plus time constants, we're going to get 100%, which B would be 100 milliamp hours.

So that's, that's the answer to the problem. If you look at the next slide, I'll, I'll clear this off. We'll go to the next one. Well, it's explained there. All right, basically, here's the answer I kind of explained it on the last slide, but since point two seconds is one time constant, I then is 60 p 63% of 100 milliamp hours which equals 63 milliamp hours after five time constants. Okay.

Oh one second. The current will reach steady state value of 100 milliamp hours and remain at this value as long as the applied voltage stays at 100. So again, we at the first time constant, my my current value will go up to 63%. Have the calculated value again 63 milliamp ORS and after five plus time constants, we hit my steady state. And that will be 100 milliamp because he added as long as the bat as long as we're connected to 10 volts and that's, that's a given. All right, let's go to the next slide.

All right, and this slide it's, it's basically what we've done previously, the only twist is we've added a large value resistor, which is one megohm or 1 million ohms. And one of the things that I just want to mention, I probably didn't stress this on the previous slides, but that 100 DOM and I think at the very first slide, it was 10 ohms that could be the actual coil resistance in this, this coil. You probably heard me stated that everything has resistance, conductors have resistance and so forth. Alright, so this is rubber Presenting the coil resistance and I make a little note here. So basically, if you look without taking into consideration, the coil resistance because the total up are in that circuit is 1,000,100 ohms. And quite honestly that that 100 ohms is very, very, very small compared to the million ohms.

So it has very, very little effect it's, it's going to affect it like point 01 percent. All right, so, we're going to do our math and we've got a new time constant here of of 20 microseconds and basically everything is the same except for the time. So right here, this is my first time constant. So what will happen here is the current or the voltage across my resistance is going to be 63% and my voltage or background voltage across my inductor is going to be 37% of 10 volts. All right, and then the inductor is going to keep decaying this way. So on the second second time constant, we're only going to have 14% across the coil, the third 4%, the fourth 2% and the fifth 1%.

And ideally, after five time constants, I should be I should be have discharged 100%. So that should be zero percent across the coil. Ideally, we're not, it's going to be very, very small. And again, my current and voltage across my resistive elements, all right, is going to be in phase and they're going to increase. So again after the first time, constant 63% Under the supply voltage, and 60% of the calculated current after five time constants, the second 86% the third 96%, the fourth 98%, and the fifth 99%. And then after five plus time constants is going to be 100%.

So, that's pretty much it. And just remember the time constant formula, t equals L over R. And we're good on this one. We've got one more slide in this section. And we'll do that and where we will end this course at that point. Okay, looking here on on this I just want to present a concept here. And these this this curve should be familiar.

Right there and right there. This is My what? Again, that's my voltage and my current across my resistor. And this is the back EMF across my coil. I've taken the switch out and what I'm showing you right here, as I'm applying a voltage, what we call a step voltage, or a square wave. And what this square wave will do, it will go from zero to some voltage, I'll call it plus v right now.

And it will stay steady for a certain amount of time. And then it will decay rapidly down to zero. All right. So how do we get these waveforms and that's what I'm going to explain right now. How would get intuitively I'm not going to put numbers in there. I just want you to intuitively understand why we getting these waveform shapes and that's what I'm going to explain.

At this point, all right, so let's clear the slide and we'll do that. Alright, so if I look at my my square wave on my step voltage here, we notice that we get a very sharp rise. That goes up very, very quickly. And if you remember, what did we have we said E sub l equals L times dI dt. All right? Well, di stands for a change of current, over a change of time.

Well, if you look at this, because it's such such a very sharp rise, we're going to have a very quick rise time, change in time and we're going to have a quick change in current D dt. So we're going to create a magnetic field across that coil for an instance. And that's my back emf. And here it is here. So if you look at this curve right there, that sector curve there. So we hit if I call this plus V, then this is my plus v here.

All right, I'm going to decay, right? Gonna decay. Let's say that this the width here is five plus time constant. So from here to here, is five plus time constants. Alright, so I'm going to decay, and we're going to decay down to zero, but what happens right here Well, I'm going to get another fast edge. And in this in this instance, instead of going from zero to plus, as I've done over here, I'm going to go to plus the negative, so my current flow is going to be in the opposite direction.

So what is that doing that setting up a back emf but in the opposite direction, and that's this peak here. So we could actually say that that's a minus v with respect to my amplitude and my coil, let's say appear was five volts, well, then this is going to be five volts. And this is going to be five volts. And again, we come over here and then what happens here we get another sharp rise which causes this one to rise and Then we got this one here, and I didn't show it. But that's we're gonna get something like there. And it just, it just keeps going on as long as I got a square wave.

I keep generating that, that waveform across the coil. Right? That's, and that's a voltage waveform. All right. So let's clear the slide and see how we get VR. Okay, here we go.

I just want to point out that I've made an adjustment here that should be zero. And if you look at VR, All right, here's my my inductor curve right there. If you look at this here is the opposite of it right there. All right. form on my voltage source, so my square wave goes to zero. So now my current goes to zero.

And then what happens here we get an increase in voltage, which will give an increase in current. And so I get this right there. And then it keeps on going as long as I have a continuous square wave these waveforms just repeat and repeat, repeat until either right turn it off or I run out of power, whatever, but it just goes on. So that that's pretty much it here. And we're going to end this course. And the next one we're going to do is RLC.

We did an RC, we just did an RL and now we're going to do an RLC. And we'll see how by putting a capacitor inductor and a resistor in a circuit and we use a sine wave a step voltage. We'll look at, look how that reacts and and how we can use that to our advantage. All right with that said that this is all from Al's electronic classroom saying, take care. We'll see in the next course.