 Let's take a look at a numerical example of Kirchhoff's current rule. And I'm doing a simple one with just three things. And in this particular case, I've got it set up so that current 1, which has a value of 5.2 amps, and current 2, which has a value of 1.65 amps, are both going into the junction. So they're on the left hand side of the equation. And current 3, which I don't know, is going out, and it's on the right hand side of the equation. Plugging in the numbers I get this, which is just a simple addition, and that means that my actual value here is just 6.85 amps. Now I can change this around a little bit to show you some different examples. So I could actually change the direction of current 2 and make it going out instead of going in. Well, that would change my base equation because now the i2 has to go over to the right hand side of the equation in both the symbolic and in the numeric one, which will change my answer. Now I have to actually subtract the 1.65 over to the other side, leaving me with a value of 3.55 amps. You can change things around here in different configurations until you figure out what the answer should be. As long as you know all but one of the currents, Kirchhoff's current rule will allow you to find the unknown one. I'm going to do a few more examples in one more video.