Finding equivalent resistance in complex circuit

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To find the equivalent resistance in a modified circuit with additional resistors, Kirchhoff's rules are necessary due to the complexity introduced by the Wheatstone bridge configuration. Traditional series and parallel methods are insufficient for this type of circuit. The discussion emphasizes the importance of understanding circuit topology when additional components are added. Participants highlight the need for systematic analysis rather than simple redrawing of the circuit. Ultimately, applying Kirchhoff's laws will provide the correct approach to solve for the equivalent resistance.
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In my previous question , situated here, I asked about finding the equivalent resistance of the following circuit :
5EMUd.png


I got some very good answers and some tips. Now what if the same circuit is modified by adding two more resistors on the free connecting wires like this :
attachment.php?attachmentid=62397&stc=1&d=1380724111.png


How should I find the equivalent resistance now ?
I probably can't redraw the circuit as mentioned in answers to my previous question since every wire has a resistor on it.
 

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That's a Wheatstone bridge. It cannot be solved by either parallel or series methods. You must use Kirchhoff's rules
 

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