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Electricity

Blazing goIITian

 Joined: 19 Mar 2012 Post: 421
24 Mar 2012 16:59:45 IST
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RESISTANCE
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# 12 WIRES , EACH OF RESISTANCE r ARE JOINED TOGETHER TO FORM A CUBE. FIND THE EQUIVALENT RESISTANCE IN THE EDGE OF CUBE

Blazing goIITian

Joined: 23 Dec 2011
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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
Posts: 421
24 Mar 2012 23:08:44 IST
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Forum Expert
Joined: 19 Oct 2006
Posts: 7804
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.

Let us consider each loop consisting of two resistors R in parallel connection as a single resistor whose resistance R1 is:

$\large \frac{1}{R_1}\,=\,\frac{1}{R}+\frac{1}{R}$$\large R_1\,=\,\frac{R}{2}$

Let us simplify the sketch of the circuit:

The expression for the resistance R2 between the junctions (ED) and (CF) reads:

$\large \frac{1}{R_2}\,=\,\frac{1}{R_1}+\frac{1}{\left(R_1+R+R_1\right)}$$\large \frac{1}{R_2}\,=\,\frac{1}{\frac{R}{2}}+\frac{1}{2\frac{R}{2}+R}\,=\,\frac{2}{R}+\frac{1}{2R}\,=\,\frac{5}{2R}$$\large R_2\,=\,\frac{2}{5}R$

The expression for resistance of the whole cube reads:

$\large \frac{1}{R_{AB}}\,=\,\frac{1}{R}+\frac{1}{R_1+R_2+R_1}$

Inserting $\large R_1\,=\,\frac{R}{2}$,  $\large R_2\,=\,\frac{2}{5}R$ we obtain RAB

$\large \frac{1}{R_{AB}}\,=\,\frac{1}{R}+\frac{1}{2\frac{R}{2}+\frac{2}{5}R}\,=\, \frac{1}{R}+\frac{1}{R+\frac{2}{5}R}\,=\,\frac{1}{R}+\frac{1}{\frac{7}{5}R}$$\large \frac{1}{R_{AB}}\,=\,\frac{1}{R}+\frac{5}{7R}\,=\,\frac{12}{7R}$

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

$\large R_{AB}\,=\,\frac{7}{12}R$

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