Blazing goIITian

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24 Mar 2012 16:59:45 IST
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pooja garg's Avatar

Blazing goIITian

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24 Mar 2012 17:35:07 IST
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(7/5)R cud b solved by taking symmetry about the edge 

Blazing goIITian

Joined: 19 Mar 2012
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24 Mar 2012 23:08:44 IST
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no. the answer is 7R/12
edison's Avatar

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28 Mar 2012 11:53:59 IST
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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.

The plotting of the circuit in the plane

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 simplyfied 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 R_1\,=\,\frac{R}{2},  R_2\,=\,\frac{2}{5}R we obtain RAB


Thus the total resistance of the cube between the vertices A and B is


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