So in our case, our load is 6K, and it would represent the right hand side

of our circuit in our Norton's equivalent circuit.

What we have to come up with is the two R,

the two elements on the left hand side of the circuit.

We have to come up with the current source I short circuit, and

we have to come up with R Thevenin's.

These two are equivalent representations.

This is an equivalent representation for the circuit that we have originally.

We take the original circuit, and we reduce everything except for the load.

We've taken the load out.

We reduce everything except for the load to this circuit.

If we do that, then we can solve the circuit very easily using current division

to find what the current is through the load resistance, or

ultimately what the voltage is across the load resistance.

So, we need to come up with I short circuit and R thevenins.

In order to do that, we redraw our circuit for the short circuit condition.