Engineering Electrical Circuit Question - Maximum Power Dispated in Resistor

AI Thread Summary
The discussion centers on finding the Thevenin equivalent resistance (R_Th) in a circuit with dependent sources. A participant initially questioned the validity of short-circuiting voltage sources and opening current sources while keeping the load resistor (R_L) in place. After several attempts, they calculated R_Th as 2 ohms using this method, which conflicted with results from other methods yielding 0.5 ohms. Ultimately, it was clarified that the correct approach involves using the open-circuit voltage and short-circuit current to find R_Th, confirming that R_L should be 0.5 ohms and emphasizing the importance of not manipulating dependent sources in this manner. The discussion concludes with a consensus on the correct methodology and results.
GreenPrint
Messages
1,186
Reaction score
0

Homework Statement



http://img855.imageshack.us/img855/6951/captureaww.png

Homework Equations





The Attempt at a Solution



Is one allowed to find R_{Th} by simply short circuiting all voltage source and opening all current sources and finding the equivalent resistance while keeping R_{L} in place and treating it's ends as stationary nodes? I know you can do this with independent sources but I'm not really sure about dependent sources. My professor said that he wasn't really sure.

When I use this method to find R_{Th} I get two ohms.

However I don't know if this is accurate as it's not what I get when I use other methods to find R_{Th}.

So is this a valid solution to this problem? If it is and R_{Th} is really two than I'm greatly concerned because I have tried solving this problem two other ways and don't get this for my answer. The two other methods I used to solve the problem give me the same answer as well. If it is than I'll post my other solutions if it is necessary.

Thanks for any help.
 

Attachments

  • Capture.PNG
    Capture.PNG
    12.3 KB · Views: 467
Last edited by a moderator:
Physics news on Phys.org
No, you have to divide the open-circuit voltage by the short-circuit current when there's a dependent source.
 
I get half a ohm for R th when I don't use this method, does this sound correct?
 
Last edited:
I got the same for the resistance with the highest power, which have to be the same as the Th resistance. What method did you use? ehild
 
You mean you got .5 ohm?

First Method
I found open circuit voltage of .25 V
short circuit current of 2 A
R th .5 ohm
Max power in Rl to be 1/32 W

Second Method
Considered Rl to be a variable. Found voltage across Rl as a function of Rl. Used power equation as a function of Rl, V/Rl. Took derivative set equal to zero and got
RL = .5
I plugged back into my equation for power in terms of Rl and got
1/32 W

I have spent several hours on this problem because I was under the assumption I could just short voltage sources and open the dependent current source and solve for Rth
I got 2 ohm when I did this
I then solved for the voltage across this resistor (I forget what I got I erased this)
and then plugged into the power equation
I was getting a different answer from the other two methods so I questioned if I could actually do this method or not.

Does the answers I got for the first two sound right?

Thanks for all your help!
 
Yes, RL=0.5 ohm. The second method is correct for sure.
I do not see yet how you got the open circuit voltage and short circuit current.
 
I think you meant you got 1/2 A for the short-circuit current, not 2 A. I got the same results you did then.
 
Oh you thanks a lot for all the help guys. The short circuit current should be .5 A and for the open circuit voltage I still get .25

.25/.5 = .5 ohms

I'm really glad I struggled through this and learned that I can't open circuit dependent current sources or short dependent voltage sources.

Thanks again guys.
 

Similar threads

Back
Top