The topic of this problem is Kirchhoff's Current Law. We're going to use Kirchhoff's Current Law in this problem to determine I1. First of all, we want to restate Kirchhoff's Current Law. Kirchhoff's Current Law states that the algebraic sum of the currents entering any node at any instant in time is equal to 0. So we know that in this problem we have two nodes. We have a node at the top and we have a node at the bottom of the circuit. So we're going to take this algebraic sum of the currents flowing into the top node. And if we do that, we see that, first of all, on the left-hand side of the circuit we have 4 milliamps flowing down out of the top node through the resistor. So it's- 4 milliamps. And then we see that in the center leg we have a current source which is 2 milliamps and it's flowing into the top node. So it's a positive quantity, 10 milliamps. And then on the right-hand side we have I1 which is drawn in the circuit as flowing out of the top node. And so if we end up solving this, then we end up with an I1, for this problem, equal to 6 milliamps.