how to ca;culate wheatstone bridge by giving some examples

see, basically it will consist of a arrangement of 5 resistance, in which they shall be arranged in a way like of two triangelesjoined by their hypotnus. when they need to be solved the part of hypotenuse is then not considered, so basically you are just left with the 4 corners of square. but remember the ratio of the side from where hyoptenuse have been removed should be same, now when you have imagined it then you can solve it as any other question of resistance. hope you are able to understand, because i can`t attach images.

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