Now we're ready to talk about resistors and Ohm's law. So a resistor is common circuit element. And we use this zig zag line to represent the tortuous path electrons have to take to pass through this resistor. And these are for the value of resistors. There's this symbol here, the node of resistor. And lets say that there's a voltage potential difference of v applied across the resistor. Then there's going to be a current flowing through the resistor from high potential to low potential and Ohm's law tells you the relationship between the voltage, the current And the resistance. So, it's just v equals i times r. Or, if you want to rewrite it, the current that's going to flow through this resistor is the applied potential difference divided by the resistance. So, for a fixed voltage level A high resistance is going to allow a very small current to flow. A lower resistance of course, would let more current flow. So you can kind of think of resistance as sort of the analogy of fluid flowing through a pipe or a straw. So let's say I have two vessels filled with water and a pipe connecting them, and let's say I come along and I put a higher pressure in this one, say put a piston and press on it? water is going to flow through the pipe and if the pipe is small and has a very large flow resistance And a very small flow. If I have the same situation with a much larger pipe. Then the same pressure difference or the same tensile difference is going to give you a much larger current or a much larger flow. Okay, now that you've been introduced to Ohm's law, to start building circuits we need one other thing. We need a battery. So a battery is just a source of voltage. And the representation of a battery is is this, showing the, the The cathode and anode plates of the battery. The positive side of the battery is the higher potential, this is lower potential side of the battery. So a typical AA battery has a potential difference of about 1.5 volts between the plus terminal and the minus terminal. So we can take that battery and we can connect it with a resistor across it and a very simple circuit here. So by this battery is going to establish a potential difference across the resistor and a current's going to flow. And so the voltage across the resistor is v. So using ohm's law, the current flowing through the resistor, and that's also the same current has to flow through the battery so that current is coming from the battery. So the current coming from the battery flowing through the resistor is just v over r Now this suggests a a method for solving a circuit like this, and what this really is is a very simple illustration of Kirchoff's voltage law. Kirchoff's voltage law says that the sum of the voltages going around a closed path in a circuit Has to sum up to 0. So it's the sum of the voltage across every element. If I add them all up and follow a closed path, it's 0. So in this case, if I start at point A and I go through the battery this direction, so I'm traversing the circuit in the direction that I've defined. The current flow to be positive. So I start at point a and I go through the battery from the low side to the high side, and I picked up a voltage v and then I dropped a voltage r. There's a voltage drop across the resistor. I'm losing energy. And the sum of the VI pickup and the V lost across the resistor has to be zero. So, you can think of this, really, as kind of a roller coaster. So, if I start at some point in the roller coaster and I follow the closed loop with ups and downs, I come back to the same point And the change in my potential, I go up to a higher potential energy and then and where I he, hit the top of the hill I have maximum potential energy and I'm, I'm barely moving in the car and then the car goes down the, down the hill and the potential energy's converted into kinetic energy. But if I take and I keep track of how much potential energy I have at every point going around this closed loop, when I come back, all of those gains when I go up hills, and loses of potential energy when I go downhill, they all have to add up to zero because after all I ended up back at the same point that I started at. So, let's take another look at this same circuit. And apply Kirchhoff's Voltage Law to come to the same conclusion. So this is, this is trivial. But, it really illustrates the methodology, which is the important thing. So, Kirchhoff's voltage law says the sum of the voltages is zero. So, we start at a, and we're going to go clockwise around the circuit. So, the voltage increases from minus to plus by an amount v. You want to go up through the batter and then there is a voltage drop of v when I go through the resistor. So using Kirchhoff's voltage law I go up V, so I pick up a plus V, then I drop a voltage which is equal to in size, equal to I times R. So the sum of the voltage picked up plus the voltage dropped, or the voltage lost. Which is minus i r, has to be zero. And so this is trivial to solve, it says v equals i r or i is v over r, which is Ohm's law. So this is a lot of formality to retrieve Ohm's law. But it's now put us in a position using Kirchhoff's voltage law. To start to solve more complicated circuits. So this circuit voltage divider is actually useful in audio applications. Like for volume control. So let's say I have a voltage, I have a battery, and I connect two resistors in series across that battery. Now I'm going to have, I have 1 loop and there's a current that's going to flow round this loop. And now I have what I want to find is I want to calculate the current and I want to find the voltage at B. Now the voltage at A I'm going to define to be 0. If you go back. We were talking about, we're free to set the zero of potential to, the zero of electric potential to anything we want. And so I'm going to just arbitrarily say, this point of the circuit is grounded. So this is zero potential. Then if I take the negative side of a battery and connect it to that ground the battery is going to give me a boost of plus 1.5 volts if it's a double A battery when I go from the negative to the positive terminal. And then this is a high potential up here. It's a one, one and a half volts. And I'm going to lose some of that voltage by going through this resistor and then the rest of it by going through that resistor.