Equivalent resistance in this network

AI Thread Summary
To find the equivalent resistance in the given circuit, it's recommended to redraw the circuit into a more familiar configuration while maintaining the series-parallel relationships. This approach simplifies the analysis and helps clarify the connections. The user initially confused the circuit with a Wheatstone bridge but confirmed it is not one. Additional resources, like forum stickies, can provide further guidance. Overall, visualizing the circuit differently can significantly aid in solving the problem.
Smash55
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http://img571.imageshack.us/img571/3774/circuitb.png

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For my a level homework i have to find the equivalent resistance in this circuit, i don't want anyone to do my homework i would just like to be pointed in the right direction.

We have been doing a bit about wheatstone bridges, so i think its related to that but I am having trouble working out how to start, i think its because of how its laid out.

Thanks for any help
Sam

Ps sorry for the crudely drawn image
 
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It's not a Wheatstone Bridge.

The first step, if you are confused, is to try and redraw the circuit into something more familiar, while keeping all the series-parallel relationships the same. This is what it'll look like:
http://img37.imageshack.us/img37/4938/circuitx.th.jpg

That enough to solve it yourself?

I'll put my answer in spoiler tags. No fair peeking until you've done it yourself!
RT = 14.888... ohms
 
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Smash55 said:
http://img571.imageshack.us/img571/3774/circuitb.png

Uploaded with ImageShack.us

For my a level homework i have to find the equivalent resistance in this circuit, i don't want anyone to do my homework i would just like to be pointed in the right direction.

We have been doing a bit about wheatstone bridges, so i think its related to that but I am having trouble working out how to start, i think its because of how its laid out.

Thanks for any help
Sam

Ps sorry for the crudely drawn image

Jiggy-Ninja said:
It's not a Wheatstone Bridge.

The first step, if you are confused, is to try and redraw the circuit into something more familiar, while keeping all the series-parallel relationships the same. This is what it'll look like:
http://img37.imageshack.us/img37/4938/circuitx.th.jpg

That enough to solve it yourself?

I'll put my answer in spoiler tags. No fair peeking until you've done it yourself!
RT = 14.888... ohms
I thought I should point out this Sticky at the top of this EE forum:

https://www.physicsforums.com/showthread.php?t=224442"

Regards
 
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Jiggy-Ninja said:
It's not a Wheatstone Bridge.

The first step, if you are confused, is to try and redraw the circuit into something more familiar, while keeping all the series-parallel relationships the same. This is what it'll look like:
http://img37.imageshack.us/img37/4938/circuitx.th.jpg

That enough to solve it yourself?

I'll put my answer in spoiler tags. No fair peeking until you've done it yourself!
RT = 14.888... ohms

Thanks for the help, that does make it a lot easier, i did try something similar before but i the diagonal kept confusing me.
 
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dlgoff said:
I thought I should point out this Sticky at the top of this EE forum:

https://www.physicsforums.com/showthread.php?t=224442"

Regards

Sorry should have read that first
 
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