The topic of this problem is Kirchhoff's Current Law. In this problem we have a circuit that has a current source IT in it, it has three resisters. The resisters have current flow associated with them as seen in the circuit. So we have 10 miliamps, 40 miliamps, and 20 miliamps going through each one of those individual resisters. So looking for i sub t. So we're going to use Kirchhoff's Current Law to be able to answer this problem. First of all, we want to restate what Kirchhoff's Current Law is. It is the algebraic sum of all the currents entering any node at any instant in time is zero. So if we look at the top node for instance, we have two nodes in this problem, right? We have one at the top and we have one at the bottom. So if we use Kirchhoff's Current Law at the top node and we take the algebraic sum of the currents entering the node at the top, then we can determine what the unknown current I sub t is. So looking at that, we see that I sub t is flowing into the nodes. So we're going to use Kirchhoff's Current Law and we're going to sum the currents into the node. So, it's I sub t flowing into the node at the top. And we have 10 milliamps flowing out of the nodes. So, it's going to be -10 milliamps for the current flowing out through the first resistor. Then, we have another 40 milliamps flowing through a second resistor. And we have 20 milliamps flowing out of the node and through the third and right most resister. And we know again, through Kirchhoff's Current Law, that the sum of those is going to be equal to zero. So if we take that and we solve for I sub T, then our I sub T is going to be equal to 10 plus 40 plus 20, or 70 milliamps.