Timer said go. Here is the charge. If you make a move, we have a current. I. If we have more charges, we have more current. Moving charges in a straight line at this gives us a constant current. Direct current or DC. Moving them back and forth creates an alternating current, AC, although we just look at DC for now. Why are charges moved? Because they want to move down their potential. What does that mean? When you have a battery, they will have a positive side and a negative side. We say that a positive side has a high potential, the negative side has a low potential. As the rule in certain analysis, charges want to move from high potential to low potential. How badly they want to move is called voltage. So, if you have a low voltage, something like three volts, you might have a 40 milliampere. If you have a high voltage, something like nine volts, you might have a current of 41 milliampers. The ratio between voltage and current is called the resistance. So, if the voltage is 1000 times as high as current, we say that the resistance is 1000 ohms. Let's look at the sky's action. This is the symbol for voltage source. It can be something like a battery or anything else. This is the symbol for resistor. It can be anything that you select. It does think of it as a lamp. This example, a 9 volt battery with a 90 ohm resistor, gives us a current of 100 milliampers. The tackle heart prompt will need a new loop. We'll care just current loop. It's pretty easy actually. It does just have the current that goes in, must also come out. So, if this wire has five amperes, this one is four amperes, they know that this one has a one ampere, another related to the school of the current of voltage load. It does just that if you go around the loop, you must end up with zero voltage. This makes sense, because if it didn't, we could just go around twice, get a higher voltage. Then again, again, and so on, they eventually have an influence voltage. Instead, it doesn't make any sense, you'll just get zero to go around the loop. Let's apply this load. Let's try to find a voltage in these resistors. While a current load, they both have the same current, so wires don't break up. Let's say the current goes clockwise, and you use the voltage load in this loop. So, we have three I, this two I, plus nine equals zero. Keep in mind that the sign of voltage source matters. Had we turned to voltage source around, we would have minus nine in this equation. Softing this gives us I equals minus two amperes. The minus sign tells us that we're wrong about guessing the current with clockwise, so it goes counterclockwise. The voltage is in this resistors, and four voltage and six voltage. Now, there's a trick. There are two algorithms for solving any voltage and current in a cell. I'll show you one here called match analysis. It goes as follows. Make up curves in all small loops in a cell. To make a simple for you, let's just say they all go clockwise. Then just apply a curve just voltage load to each loop. You might be confused when two loops meet, but don't worry. You just need to realize that, since you go in opposite directions, there's nothing current, or have difference between them. Where we switch the signs depending on which loop we're looking. Now, you have some equations. All you have to do is just solve them. Have you solved them? Well... Solved. Which brings us to the next here. Death and death. This here will almost seem too good to be true. It doesn't have no matter what circuit you may have. But truly big or complicated or whatever. As long as it has two terminals, we can simplify it as a voltage source and resistor. So how do we do it? Well, the voltage is the voltage, so that's sort of the same way as we did before. Resistance is a bit hot. First, you need to know how to add to resistors. Remember in a brief example, when the two are, just the three are, just add it. That was the good thing. In fact, whenever two resistors are right behind each other, we say they're in series. Whenever two resistors are in series, we can just add them, and then again. And then again, as many times as we look. The second way to add resistors when they are in parallel. That is, when the start and ends are attached to the same point. Then we can add one over to resistance and finally flip that. So here we would have one third plus one sixth equals three six. Flip that so that the resistance will be two ohms. And let's go back to the definite theorem. To find resistance, we just have to add all the resistors in the overshook. Since we don't need the voltage sources, we can just ignore so let's do it. What 12th plus 1th, 4th, 12th, flip it to get three ohms. They are in series and also three ohms with one ohm. It's four ohms. So resistance is four ohms. Need. Debring. Okay, so till you can both. Let's first talk about the capacitor. The capacitor consists of two plates. But it's a bit weird since there aren't even a connector. When you connect a battery to the capacitor, those two are charged. We say that the capacitor is charging up. It will continue charging up. The field has the same voltage as the battery. Now if we were to increase the voltage on the capacitor, there would increase the current. We would lower the voltage on the capacitor. And it would also lower the current. So the higher the current, the higher the change the voltage. Remember, not the voltage because it's possible for that to be a voltage, no current. It's a change in voltage. The ratio of current to changing in voltage is called the capacitance C. The inductor is very similar. But here instead of the current increasing in changing voltage, another voltage increase in changing current. In this case, the ratio is called the inductance L. Let's apply these components now. Because we're looking at the change in voltage with current, we have to add the time component. The way this is done is by creating a circuit that has a battery added at a specific time, which you should call D or Z. Now to solve for the voltage in the capacitor, we basically thought of the exact same procedure we did before. The only thing that has changed now is that we're inductor, right? Three times the IDD, the SAT is the voltage. And for the capacitor, well, that's what we're trying to solve. So let's go to V. Then by the voltage law, we get this differential equation. Now we're done. What if it is just solved? How do we solve it? Well... So there you have it, that's our solution. How do you find these two constants C1 and C2? We apply the initial conditions that we have. That is, we check what the voltage was on this capacitor. If we were we used to do the battery and we check what the current was before we looked on the battery. Without a battery, there's no current. So DVTT equals 0. And if we assume like a capacitor was in charge, then V equals 0. Combine all these facts gives us a full solution. So now let's take a quick look into the world of maths before we take a look at the scary AC curve. In AC cellulants, the curve is moving back and forth. The maths and the processes do go, but they're nice tricks to make it simpler. Instead of thinking of it going back and forth, we imagine it moving in a circle. It's going up the imaginary dimension. We can look at real components to see the real current at our top. Remember that the circle does actually exist. We just made it up to make it maths-y. So now, from your maths process, you remember that complex numbers moving in a circle as this equation. We're going to write it in phase 4, which looks like this. Comparing these two equations, we have one in the phaser domain and one in the time domain. Now let's relate this to circuit elements. Current is simply this in phase 4. Voltage is current times resistance, so it says, relationship between voltage and current in a inductor contains a derivative in a time domain. If take this derivative in time domain, we see that all you have to do is multiply the j romega out like what is for a capacitor. It's 1 over j romega C. So you have it or relations. As you can see, inductance and capacitance, can we see the sort of imaginary resistance in the phaser domain? For that reason, we have to group all things together and name it, impedance. So now, let's go back to this example and try to find the current. Instead of adding the battery at t equals 0, the battery will have a voltage of cosine of 60t. The only difference in solving this now is that we simply write everything in the phaser domain to first. This can be done with the noise released to the hand before. But all we have to do is add the impedance us sensor in a series. So, add Zil. Then always to do so via the voltage by impedance and we got the current. But now we have to address one final question. Why did we calculate all these voltages and currents? Well, that's, calculate the power. That's what you eventually want. Power is a product of voltage and current. But as you know, easy current can make this a bit tricky to evolve. There's so much to say that power and more advancements for anus circuit, like Fourier transform, but blood transform is so that I can make several videos for those topics. So, I will say it to you for those videos and in the meantime, leave a comment on what your opinion is on this video.