 So far we have covered current, voltage, resistance and ohms law, so now it's time to start putting all this knowledge together and start analyzing some circuits. So if you have a look here, I've created a circuit here, it's a very simple circuit that contains a 9 volt battery, which is connected to using this breadboard here. It's just a component that I can use to connect other electronic components. I've got two resistors, so there's one resistor which is then connected to another resistor and then back to the negative terminal. So it's positive terminal of the battery, first resistor, second resistor to the negative terminal of the battery. So if I wanted to draw a circuit diagram of this circuit, it would look like this. So here we go, we have two one ohm resistors that go, so current would flow from the positive terminal straight to the first resistor, then through the second resistor and finally come round to the negative terminal. So how can I predict how much current is going to flow through this circuit here? Well the way that I can do that is I can use a very special, well there's two very special rules and they're called Kirchhoff's laws. But the key thing is they derive from two fundamental concepts in physics. The first one is conservation of energy. The second is conservation of charge. So these two principles lead to two Kirchhoff's laws. Conservation of energy states that if there's any loop in the circuit, so in this case our circuit forms a nice loop, then the total voltage, if you start at one point in the loop and you go all the way around, you must end with the same energy that you started with. So I'm going to start at this point here, I'll highlight it with yellow. If I start at this point in the circuit, if I go across the battery I gain nine volts of energy. So if I gain nine volts here then as I go around I must lose nine volts of energy across those two resistors. So that there is what we call the loop rule and that derives from conservation of energy. However there's also conservation of charge. And this is what we call the node rule. What that means is if I have a particular point in the circuit, maybe let's choose this point here. Then the amount of current that flows through this particular point in the circuit, the amount of current flying into it, must be equal to the amount of current flying out of it. So this is called the node rule. And all that means that there is going to be no charge building up at any point in the circuit. Therefore the charge going in must equal to the charge going out. So using these two rules I can make all sorts of predictions about circuits. And in this case here I'm going to use them to predict what the current is going to be through this circuit here. So let's see what do we know. I've got nine volts on this side. So that means that at this point here I must still have nine volts. And at this point here I'm at zero volts. So as the charges move across these resistors here they must lose nine volts of energy. And I also know that the amount of current that's going into this resistor must be equal to the amount of current going out of this resistor because of the node rule. So I know the current through each of these resistors must be the same. Let's put this into equations. I'm going to call this resistor, resistor one, and this resistor, resistor two. And I know that the voltage drop across both resistors, so the voltage across resistor one, plus the voltage drop across resistor two, must be equal to nine volts. And I also know from Ohm's law that the voltage across one of these resistors is going to be equal to the current times the resistance. So in this case I would have, then using conservation of charge or the node rule, I know that the current going across resistor one must be equal to the current going across resistor two. So I1 equals I2. So now I have a series of four equations. I can use all these equations together to find the current that's flowing through the circuit. So first of all, I'm going to substitute, first of all, I'm going to substitute equation two into equation three into equation one. So that will give me, and now I'm going to substitute equation four to get rid of I1. So now I can rearrange this to solve for I2. And in this case I know that both R1 and R2 are equal to one Ohm each, so that gives me, excuse me, nine volts over two Ohms, which is equal to 4.5 amps. So now I've worked out the amount of current that is flowing through this circuit.