12 WIRES , EACH OF RESISTANCE r ARE JOINED TOGETHER TO FORM A CUBE. FIND THE EQUIVALENT RESISTANCE IN THE EDGE OF CUBE . I DID IT USING KIRCHOFF LAW N FOUND IT TO BE VERY LONG.PLEASE GIVE ME A SHORTCUT TO SOLVE IT. I WILL BE REALLY THANKFUL!
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In this circuit the vertices E and D (F and C) are equivalent, they are on the same potential. The resistance of the whole cube is not changed by merging these vertices into one. Let us merge the vertices E and D (F and C) into one junction, redraw the circuit into the plane and supplement each cube edge with a resistor. The resistance of each edge is R.
Let us consider each loop consisting of two resistors R in parallel connection as a single resistor whose resistance R1 is:
Let us simplify the sketch of the circuit:
The expression for the resistance R2 between the junctions (ED) and (CF) reads:
The expression for resistance of the whole cube reads:
Inserting , we obtain RAB
Thus the total resistance of the cube between the vertices A and B is
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