Determine the Norton equivalent with respect to terminals a and b.

However, closer consideration of the circuit reveals a problem with this approach: the 30 V source is not in series with a resistor (thus we cannot apply a source transform thereto), nor will any source transforms involving other components put a resistor in series with the 30 V source, thus we cannot reduce the circuit down to the Norton equivalent by source transforms alone. (One might think that we could treat the wire as a 0 ohm resistor in series with the source, but what would this yield for the value of the current source?)

SO, our next decision is what approach to take.

We could partially reduce the circuit. There are two source transforms immediately available, but in neither case will a transform allow us to combine any components thus reducing component count.

If we eventually decide to use nodal analysis, the circuit would be easier to solve (for V_oc and I_sc) in its current form.
If we decide to use mesh analysis. transforming the two current sources would reduce 4 meshes to 2 meshes, so that might be warranted.

We also need to consider the possibility of determining R_Th directly by killing the independent sources. This looks rather simple to me, so let's find R_Th first.

This may seem a bit odd due to the wire in parallel with the 10 K resistor, but we can just treat it exactly that way. Combining any non-zero resistance in parallel with an ideal wire (0 ohms) will always yield an equivalent resistance of 0 ohms (thus an ideal wire). Another way of saying this is that the 10 K resistor is shorted - ANY current entering the parallel combination will ALL go through the wire. This of course means that the voltage across the 10 K resistor is zero, as expected.

Replacing the 10 K and the wire with a wire as described above yields

Hopefully it is now obvious that the 15 K and 5 K are in parallel, yielding

R_Th = 3.75K

I will use nodal analysis.

The 30 volt source directly connects the reference node to node A, thus

V_A = 30 V

Now writing KCL at node B (summing current out of the node) gives

(V_B - 30) / 15K + 3m + V_B / 5K = 0

Multiply through by 15K and solve

V_B - 30 + 45 + 3 V_B = 0
4 V_B = - 15
V_B = - 3.75 V

Thus the Thevenin equivalent is

or equivalently

and the Norton equivalent (by source transform) is

or equivalently

Although we could do a standard nodal analysis, I will use the "seat of the pants" method.

An interesting idiom, that. As I understand, this originated in the early days of aviation. The pilot would feel the forces applied to his or her backside via the chair in the aeroplane, thus detecting what the craft was doing and "flying by the seat of the pants". Now, of course, this phrase is applied to any situation where one "feels" their way through it without a formal structured approach (such as nodal analysis).

We know that V_A = 30 V (relative to the reference node).

We also know that V_B = 0 V (node B is shorted to the reference node)

Thus I₁₅ = (30 - 0) / 15K = 2 mA

Consideration of KCL indicates that I_x must equal 1 mA. (I₁₅ + I_x = 3m)

Since the current through the 5K resistor must be 0 (it is shorted out), KCL implies that

I_sc = - I_x = - 1 mA

Which verifies that the Norton equivalent obtained above is correct.