Can a Dependent source be dependent on itself?

Question

I am looking for the Thevenin Resistance here. However, I have no clue what to do with \$i_b\$ and the source dependent on it. This is the entire circuit. \$i_b\$ is the current through the 5\$k\Omega\$ resistor. In this case, There is an open circuit across terminals a and b and so there is no current flowing through the 10\$k\Omega\$ resistor. Meaning that all of the current flowing in the circuit is \$i_b\$. That presents this problem, because there's a dependent source that is dependent on the current \$i_b\$ and they are in series. Hence, the confusion and the question.

Equations: KVL along outer loop CCW:

Sidenote: \$I_{sc}\$, short circuit current across terminals a & b, is easily obtainable by a KVL on the outer loop, the relationship of the current through the 10\$k\Omega\$ resistor and \$i_b\$, (Which is \$4i_b = i_b + 3i_b\$ :: KCL in the middle top node), and it is -1.8mA.

Answer 1

So, with the equations, you end up with ib = 4*ib. Therefore, ib == 0.

Because the right side is open, the only loop is the left-hand one. While mathematically, ib==0 solves the problem, in practice, the circuit is unstable (it has positive feedback), and ib would tend to go to +infinity or -infinity if you actually built it.

Answer 2

There are several ways of determining the Thevenin equivalent resistance. One way is to open the output and measure measure the open circuit voltage and resistance. However, that doesn't work here because you end up with a nonsense term 4 ib = ib. So you have to use a different method. I think they penalty want you to use the "test source" technique here, where you attach a test voltage source to the output.