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Blazing goIITian

 Joined: 19 Mar 2012 Post: 421
25 Mar 2012 10:46:07 IST
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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 . 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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Blazing goIITian

Joined: 8 Sep 2011
Posts: 437
26 Mar 2012 00:33:29 IST
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like it if its useful.....

Blazing goIITian

Joined: 8 Sep 2011
Posts: 437
26 Mar 2012 00:36:41 IST
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give a like if its useful...

Blazing goIITian

Joined: 19 Mar 2012
Posts: 421
26 Mar 2012 10:30:33 IST
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IN THE NCERT CLASS 12 PHYSICS, IT IS GIVEN THAT ALL 3 BRANCHES HAVE THE SAME CURRENT.

Blazing goIITian

Joined: 19 Mar 2012
Posts: 421
26 Mar 2012 10:35:20 IST
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WHY ALL 3 BRANCHES SHOULD NOT HAVE SAME CURRENT. 3I ENTERS AND SPLITS INTO 3 BRANCHES,SO EACH BRANCH SHOULD HAVE I CURRENT.GUYS PLEASE GIVE ME THE WHOLE SOLUTION TO THIS PROBLEM. I HAVE SPEND A LOT OF TIME ON THIS.......

Blazing goIITian

Joined: 19 Mar 2012
Posts: 421
26 Mar 2012 10:42:33 IST
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I HAVE SOLVED THE PROBLEM OF FINDING OUT THE RESISTANCE BETWEEN DIAGONAL FACES WHICH IS GIVEN IN NCERT.I AM NOT ABLE TO SOLVE THAT HOW TO FIND IT BETWEEN THE EDGE OF CUBE. PLEASE POST A DETAILED SOLUTION OF THIS. PLEASE........

Blazing goIITian

Joined: 19 Mar 2012
Posts: 421
26 Mar 2012 11:51:57 IST
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Blazing goIITian

Joined: 19 Mar 2012
Posts: 421
26 Mar 2012 11:53:25 IST
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Blazing goIITian

Joined: 19 Mar 2012
Posts: 421
26 Mar 2012 12:06:29 IST
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Blazing goIITian

Joined: 19 Mar 2012
Posts: 421
26 Mar 2012 12:09:28 IST
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Blazing goIITian

Joined: 19 Mar 2012
Posts: 421
26 Mar 2012 12:14:34 IST
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Blazing goIITian

Joined: 19 Mar 2012
Posts: 421
26 Mar 2012 12:39:04 IST
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hey, guyzzz c'mon you all are genius , juz give me shortcut.

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Joined: 19 Oct 2006
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26 Mar 2012 13:33:43 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

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

Blazing goIITian

Joined: 19 Mar 2012
Posts: 421
26 Mar 2012 16:57:50 IST
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guyz..... please tell me a short method of solving this using Kirchoff law & symmetry.

Blazing goIITian

Joined: 19 Mar 2012
Posts: 421
26 Mar 2012 18:35:33 IST
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HI, ISN'T THERE ANYBODY TO PROVIDE ME A COMPLETE SOLUTION.

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