Step 1 :

Calculation of $V_{Th}$

From the given figure , in order to find  $V_{Th}$ let us remove the  resistor of $10 \Omega$ from the circuit.

$\therefore V_{x}=-25(5I_1)$

$\therefore V_x=-125I_1$........(1)

Applying KVL to the mesh 1,

$12-1000I_1-2(-125I_1)=0$

$\therefore I_1=0.016A$

Putting the value of $I_1$in equation (1),

$\therefore V_x=-125I_1=-125(0.016)=-2V$

Writing …