# Homework / test. Thevenin equivalent

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) • Mar 28 '16 at 6:22
• Prakash Darji, "Link only answers" are discouraged. That being said, it's sorta warranted, and I do highly recommend the second link. As for your specific hangup, Mylane, the open in the center is good, even necessary for getting the Thevenin voltage. It's all the other jazz that's not doing you any favors. You should try the process on some simpler circuits first if you don't get what to do. This way may feel too trivial, but you'll come to understand the merits of the process just from going through the motions. Again, that second link is a really good reference. Mar 28 '16 at 6:37
• There's a schematic editor in the editor toolbar, it will allow you to make a much more readable, less fuzzy circuit. Also, you should post your attempt at solving the circuit. Mar 28 '16 at 14:35

@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

= Vx/3

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?

• I did it myself. Does it really cancel out? :0 And I'm just practicing thevenin equivalent. That's why I'm asking where should I start. Mar 28 '16 at 13:30
• So this is not the original circuit? Did you take out the load at Vin? At nodes a-b? Mar 28 '16 at 13:34
• Yes, an 8 Ohm resistor used to be there. I just thought that was the best position to take out the load and calculate the thevenin equivalent. Mar 28 '16 at 13:36
• @Mylane, Okay now I understand. You created the circuit with an 8 ohm resistor and try to get the thevenin equivalent looking at it from the 8 ohm resistor load. There is nothing wrong with that. Yes it cancels out. Can you get another problem from a book/online with answers, so that we can check if we get it right? Also doing thevenin equivalent with dependent sources is not the easiest thing to do, but nontheless we'll get it done. Do u know how to get the Thevenin resistance? Mar 28 '16 at 13:40
• Well, kinda. I'm lost with this circuit. I know how to get it in a circuit where the a-b part is on a side of the circuit. When it is right in the middle, I get lost. I do it with the tension source and getting the current with mesh analysis using superposition. And I can't get to find a circuit that looks similar to this one in a book with answers, most of them look alike. Mar 28 '16 at 13:48