0:20

The unknown currents are I1 and I2.

Â So to find these currents, we're going to use Kirchhoff's Current Law,

Â and we're going to sum the currents into the nodes.

Â So the first thing we have to do when we solve problems using

Â Kirchhoff's Current Law is we have to identify the nodes in the problem.

Â So we have a common node at the bottom.

Â We have a node on the left hand side of the circuit, and

Â we have a node on the right hand side of the circuit.

Â So there's three nodes associated with this circuit.

Â So if we want to use Kirchhoff's Current Law, we can sum the current into any one

Â of those nodes and see if we can solve for quantities that we have of interest.

Â So let's start with the node on the left hand side of the circuit.

Â So we want to sum the currents into the left hand node at the top.

Â And so if we do that, we see first of all, that we have I1 flowing into that node.

Â So it's I1 and

Â then we have three amps flowing out through the left most resistor.

Â So it's a -3 mA, and we also have I2,

Â which is flowing out of that part of the circuit as well.

Â And so that node has the equation as shown above,

Â so that is node 1 and we'll assume that is equation 1.

Â If we go to the other side of the circuit and

Â look at node 2, and we do the same thing.

Â We sum the currents into that node, then we see that I1 is flowing out, so

Â it's -I1, and we have 12mA that's flowing up into node 2 through the voltage source.

Â So it's +12 mA.

Â 2:25

So, we can see that we can solve for I1, and then alpha mean for I2.

Â So, I1 in this case is going to be 8 mA,

Â and if I1 is 8 mA, then I2 using

Â equation 1 turns out to be 5 mA.

Â