I decided to create this circuit and solve it with mesh analysis. Then, I thought I could find the Thevenin Equivalent, but I have no idea where to start! I can't get to understand how to get the thevenin voltage when I have the open circuit right in the middle. If you could help me out with the way I should do it, that would be great. Thanks! (I'm trying to solve it myself, so I'm looking more for an explanation than an answer itself)
@Mylane, are you trying to solve for Vin by only doing Thevenin or will you like to solve for Vin in multiple ways?
Here is one way without doing Thevenin Equivalent:
The voltage across the 6 ohm and 4 ohm resistors together is the same voltage as 3Vx. Vx is the same voltage across 4 ohms resistor.
Going anticlockwise in the last loop with Ohm's Law (V = I*R):
-3Vx + 6ip + Vx = 0
6ip = 3Vx - Vx
ip = 2Vx/6
Since the 6 ohm and 4 ohm resistors are in series, they have the same current flowing through them
Thus Vx = 4 x ip (Ohm's Law)
Let's call the current going DOWN (not UP) thru 2 ohm and 12 ohm resistors since they are in series i2
So when adding the current at the 12 ohm and 1 Amp Current Source node, (assuming all positive currents are leaving that node)
-1/2ip - i2 + 1A = 0
i2 = 1 - 1/2ip
Thus the voltage across the 2 ohm resistor, let's call that V2
Using Ohm's Law V2 = 2 * i2
Looking at the middle upper loop,
Going clockwise with Ohm's Law (V = I*R):
-Vin - V2 - 8v + 6ip = 0
Vin = 6ip - V2 - 8v
Vin = 6ip - (-3ip) - 8v
Vin = 9ip - 8v
For some reason everything seems to cancel each other out. Did you make up this circuit yourself or was it given to you?