Thevenin Equivalent Voltage: why ignore the 3-kΩ resistor?
Computing a Thevenin Equivalent requires two steps:


Obtain the Thevenin Impedance \$Z_{TH}\$
Obtain the Thevenin Voltage \$V_{TH}\$


In order to compute \$Z_{TH}\$ you have to "turn off" the independent current source. As it is a current generator you will need to place an open circuit there. This will "disconnect" the 4k resistor from the circuit, giving you \$Z_{TH} = 7k\Omega + 3k\Omega = 10k\Omega\$, as the resistors are in series.

In order to obtain \$V_{TH}\$ you need an open circuit between the terminals A and B. Having this open circuit will cause no current to flow through the 3k resistor, as it would violate Kirchhoff's First Law. That current would have no way to get "back" to the circuit, so you would be diminishing the electron count!

Another way to look at it is as if you had an infinite resistor between A and B. The air acts as an insulator, so given a real valued voltage drop in A and V \$V_{AB} = \alpha\$ you would get:
$$I = \lim_{R \to \infty}\frac{V_{AB}}{R} = \lim_{R \to \infty}\frac{\alpha}{R} = 0$$
No matter how you want to look at it, you have no current flowing there. This leaves the current flowing only in the left loop. Having no current in the 3k resistor means we will have no voltage drop there. Remember that we want to find \$V_{TH} = V_A - V_B\$. Following the circuit we see that the only drop involved is the one in the 7k resistor. By Ohm's Law, that drop will be \$V_{DROP} = I·R\$, so:
$$V_A = V_B - I·R \to V_A - V_B = V_{TH} = - I·R$$
Being I = 8mA and R = 7k\$\Omega\$.

This is the Thevenin Equivalent and why the 3k resistor doesn't play a role in the Thevenin Voltage.







simulate this circuit &ndash; Schematic created using CircuitLab