# Electrical Circuit Question - Maximum Power Dispated in Resistor

1. Feb 21, 2013

### GreenPrint

1. The problem statement, all variables and given/known data

http://img855.imageshack.us/img855/6951/captureaww.png [Broken]

2. Relevant equations

3. 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.

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Last edited by a moderator: May 6, 2017
2. Feb 22, 2013

### vela

Staff Emeritus
No, you have to divide the open-circuit voltage by the short-circuit current when there's a dependent source.

3. Feb 22, 2013

### GreenPrint

I get half a ohm for R th when I don't use this method, does this sound correct?

Last edited: Feb 22, 2013
4. Feb 22, 2013

### ehild

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

5. Feb 22, 2013

### GreenPrint

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?

6. Feb 22, 2013

### ehild

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.

7. Feb 22, 2013

### vela

Staff Emeritus
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.

8. Feb 22, 2013

### GreenPrint

Oh ya 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.