Cant remember how to solve this question


Joined Jan 28, 2005
The star-delta approach can help you get to the answer. There are several other techniques that can be used to solve the problem as well. KVL, KCL, Thevenin's, Millman's, and superposition are but a few of the tools that can be used to solve this problem.


Thread Starter


Joined Oct 17, 2005
Do you know the easiest way? I remember solving this before without the use of a star-delta conversion... I think! I think i did it using the thevenin equivalent resistance. I found that and said that Rl has to equal this for maximum power transfer. Although everytime i try it now i get the wrong answer to what is written beside it.

I was under the impression that the thevenin resistance here is R2&R4 in parallel with R3&R5. Although i am not to sure on what to do with R1 in this case.

Penny for your thoughts!

Thread Starter


Joined Oct 17, 2005
I guess what i am trying to ask here is how does the current see the 6 ohm resistor? Does the current want to flow through [R3 + (R5 // R1 +R4) ] // [R2 + (R4 // R1 + R5) ] ?? I am sure that if i do a delta star convertion it will be alot easier but anyone know of an easy way to do this?

The Electrician

Joined Oct 9, 2007
They want you to find the resistance seen looking into the two nodes where RL is connected. The current source is replaced with an open circuit. Then use any method that works. I get 4.34151 ohms.


Joined Apr 2, 2007
to me converting the delta to star seems much easier than any other method.
[R3 + (R5 // R1 +R4) ] // [R2 + (R4 // R1 + R5) ]
at a glance this equation seems a bit confusing and incorrect as i dont think u can add
R1 +R4 as they don't seem to be in series...there is a node between them.
did u put in the value to get the answer...maybe we will have another look at it if the answer fortunately/unfortunately comes out to be the same.


Joined Jan 7, 2008
i am also interested in this topic. if anyone can show the detail solution for this question, i think it will be nice for people to learn.


Joined May 19, 2004
to thevinize this circuit: a wheatstone bridge in parallel with a single resistor R1, as said remove the current source (leave an open), remove RL and look back between the terminals left by the missing RL. after 2 sketches of the new resistance circuit, i realized the new circuit is but another wheatstone bridge. R2 and R7 on the top legs, R1 on the center leg, and R4 and R5 on the bottom legs. now just solve for Rth.


Joined Apr 27, 2007
Nice, I was looking for an answer to this kind of problem, but I was expecting solutions involving matrices. I didn't know that this could be that simple. I was not aware of the DY transformation.


Check the attachment for the details.
Can the transformation be applied to R1, R3, R5 (converting a wye to a delta) or any other three resistors that share a common point?