Hey guys. So in this video, we're going to talk about a very simple but very helpful rule that you can use to make solving circuits easier. Now I should note that a lot of professors and textbooks don't cover this until a little bit later on, but we're going to do it here because it's going to help you immediately in problem-solving. So let's check it out. Alright, so the rule is called Kirchhoff's or Kirchhoff's, however you prefer, his junction rule. Okay, Now remember that resistors in series always have to have the same currents. And that's because if you have 2 resistors in series like this, the wire doesn't break. So all the current going through this first resistor here, I one, let's call it the current through the first resistor, let's call it \( I_{1} \). And then this current \( I_{2} \) have to be the same because the charges have nowhere to go. So they just keep flowing. Okay? Now currents will change only if the wire splits into 2 or more parts. So for example, let's say if you have, this here and then it splits into 2 wires. Right? \( R_{2} \), \( R_{3} \). So these currents will not be the same. \( I_{1} \) will not be the same as \( I_{2} \) and \( I_{1} \) will not be the same as \( I_{3} \). It splits. Now the point where it splits, this point right here is called a junction or a node. K? So that's the special point right there. And Kirchhoff's or Kirchhoff's junction rule says that the current into a junction is always equal to the current out of the junction. Current in equals current out. And this flows from the conservation of charge. Right? So if this was not the case, then what would happen is that charge would accumulate over here. And you can't have that because charges are just always flowing. So if something like 3 amps goes in and I know that 2 amps goes out here, it means that 1 amp has to be here because the charge is conserved and it has to keep flowing. That's it. This is one of the simplest ideas in physics. This is called his junction rule, but it's also sometimes called currents law. K. So it's a law, it's a law, junction rule, current law. Cool? Alright.
So let's do a quick example here. Super simple. What is the voltage across the \( 2 \Omega \) resistor? So what is the voltage across this guy here? And I want to remind you of Ohm's law, which is an equation that ties the voltage, the currents, and the resistance of a resistor. And the idea of Ohm's law, if you remember, is that if you know 2 of these, you can find the 3rd. Okay? And you'll see how we're going to use it. So let's see. I want to find the voltage of the \( 2 \Omega \). So I know the resistance is \( 2 \Omega \). I am looking for the voltage. Is there an equation that ties these guys together? Yes. It's this one here. So I can write \( V = IR \). And the problem is I know \( R \), I'm looking for \( V \) but I don't have \( I \). So is there a way for me to figure out \( I \) from the diagram? What is the current going through here? Let's call this \( I_{2} \) since it's going through the 2, which by the way is the same as \( I_{2} \) over here. Notice that I have 4 amps going into this node or junction. I have 3 coming out and I have \( I_{2} \) coming out. If 4 goes in this way and then 4 have to come out, 3 is already coming out so one has to be going this way. That's it. That current is 1. So I can do 1 amp times the resistance. The resistance is \( 2 \Omega \). 1 amp times \( 2 \Omega \) is 2 and the units of voltage is volts. So the answer is simply 2 volts. Cool. That's it. Super simple but super helpful. Let's get going.
