Hello,
I'm having a lot of trouble solving for the Thevenin equivalent for this circuit. The dependent source is really throwing me off.
My attempt:
The independent current source has 2A, since the resistors are both 1 ohm, I assumed the current will split evenly.
Using V=IR, Vx=1 V
To find Vth, I solve for the Voc voltage using KVL, and I called V over the 1Ω resistor on top V1 and V over the dependent source current V2.
Vx=V1+V2
1=1+V2
V2=0

I'm pretty sure that's wrong, because I'm asked for available power for the network as well. If anyone can tell me whats wrong and some tips for finding the Thevenin resistance, Norton current and available power, that would be appreciated.

http://i1255.photobucket.com/albums/hh622/theaznintds/47dec865.jpg

 

When you have a dependent source in a circuit, you O/C independent current sources and S/C voltage sources.

There is a specific technique to get a Thevenin equivalent when dependent sources are involved, have you read up on it?

To get Rth for this specific circuit you can place a test voltage at the load after having O/C the independent current source, and solving Rth = Vth/Iin with a test voltage value of your choice.

 

Hello,
I'm having a lot of trouble solving for the Thevenin equivalent for this circuit. The dependent source is really throwing me off.
My attempt:
The independent current source has 2A, since the resistors are both 1 ohm, I assumed the current will split evenly.
Using V=IR, Vx=1 V
To find Vth, I solve for the Voc voltage using KVL, and I called V over the 1Ω resistor on top V1 and V over the dependent source current V2.
Vx=V1+V2
1=1+V2
V2=0

I'm pretty sure that's wrong, because I'm asked for available power for the network as well. If anyone can tell me whats wrong and some tips for finding the Thevenin resistance, Norton current and available power, that would be appreciated.

http://i1255.photobucket.com/albums/hh622/theaznintds/47dec865.jpg

On what basis do you claim that the current from the 2A source is going to split evenly between the two 1Ω resistors? Would you make such a claim if the dependent current source were removed? Would you make such a claim if the two resistor's were in series? In general, under what conditions would such a claim be valid? Do those conditions apply in this circuit?

Now, that's not to say that it might not work out such that it does split evenly, but if it does, it will by by coincidence and not for the reason that I suspect you are thinking of.

 

When you have a dependent source in a circuit, you O/C independent current sources and S/C voltage sources.

There is a specific technique to get a Thevenin equivalent when dependent sources are involved, have you read up on it?

To get Rth for this specific circuit you can place a test voltage at the load after having O/C the independent current source, and solving Rth = Vth/Iin with a test voltage value of your choice.

A simpler way is just to analyze the circuit under two conditions: As drawn (open circuit output) and calculate the open-circuit voltage (this IS the Thevenin voltage). The short circuit the output and calculate the output current (this IS the Norton current). The equivalent resistance is then just the ratio of the two.