The topic of this problem is Kirchhoff's Current Law and the problem is to determine I1, I4, and I6 in the circuit that's drawn below. In this circuit, we have a number of different components that are involved in a complex circuit and the cloud-like images represent a complex part of the circuit with certain currents flowing out of that complex part of the circuit at different points. So, at the top complex circuit, we have 60 milliamps flowing down through one leg. We have 20 milliamps flowing down through the right-most leg. And we have I1 which is flowing into that complex circuit from the left to the right. Similarly in the bottom part of the circuit there is another substance of the circuit, which is a complex circuit which has 40 miliamps flowing out, 30 miliamps flowing in, and I4 which we are going to determine also flowing out of that, part of the circuit. Again, we're using Kirchhoff's Current Law to determine I1, I4 and I6 so we treat each one of those complex circuits as a node. And so it's kind of a node that is encompassing part of the circuit. And flowing out of that node we have the three currents at the top, we have I1, I or 60 milliamps and 20 milliamps. So, we're going to start with the top and write Kirchhoff's current law for the top and again, before we do that, let's state what Kirchhoff's current law is. And that's, that the algebraic sum of all the currents entering any node at any instant in time is 0, so looking at the top node, the top node being the complex circuit element at the top, elements at the top, we have a current I1 flowing into it, or summing currents into that node. I1 is flowing in, it's a positive flowing in. We have 60 flowing out so it's a minus 60 milliamps and we also have 20 flowing out so it's a minus 20 milliamps and that's all the currents associated with the top complex circuit node, and we determine from that that I1 is equal to 80 milliamps. Similarly, we can look at the bottom complex circuit node. And the bottom complex circuit node has five currents associated with it. If we write the equation for that, then we have I4 flowing out, so it's a minus I4. Summing currents again, summing currents into that node. Plus 60 milliamps flowing into it at the top, right hand side, or left hand side. And then we also have 20 milliamps flowing in at the top right hand side. Now looking below that complex node that was flowing out of it, we have 40 milliamp flowing out so it's a minus 40 so it's flowing outward, and we have 30 flowing in so it's a 30 plus 30, That we add to our equation. So ultimately we end up with an equation for I4 that looks like this if we solve for I4, then we end up with an I4 equal to 70 milliamps. So now that we have those quantities, we can go back and we can find what I6 is. I6 can be solved by taking the algebraic sum of the currents flowing into the node at the left most side of the circuit. So in this case it's going to be minus I1 since it's flowing out, plus I4 since it's flowing in, minus I6 because it's flowing out of that left most node. And so if we plug in the values for these we have minus 80 milliamps plus 70 milliamps minus I6 is equal to 0. Solving for I6, we have an I6 equal to 80 minus 70 which gives us 10 miliamps and that's going to be a negative 10 miliamps because we have a negative sign in front of the I6
