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Twelve resistance wires of R ohm each are connected to form a skeleton cube . Calculate the equivalent resistance across an edge of the cube (without using Kirchhoff's Laws ) .
Suppose ABCDEFGH is the skeleton .Suppose a cell of emf E' is connected across edge DC , so that current i enters cube at D and leaves it at C .

2007-01-15 01:43:17 · 6 answers · asked by Anonymous in Science & Mathematics Physics

6 answers

We have to make something clear: There are two squares that use up all the vertexes: The lower one is ABCD and the upper one is EFGH so E is above A, F above B, G above C and H above D.

Due to symmetry, the potential of A and H is the same, as well as of B and G. That's why these points can be connected without changing the currents in any branch. Now, the problem is of mixed resistors. I will describe the net by assigning an ordered pair (ex. (1,2)) if there is one resistor r between the nodes. The nodes are:
1 = D, 2 = A = H, 3 = E, 4 = B = C, 5 = F, 6 = C
The net is:
(1,2), (1,2), (1,6), (2,3), (2,3), (2,5), (2,5), (3,4), (4,5), (4,5), (5,6), (5,6)
There are 12 ordered pairs, just as many as the edges in a cube. If an ordered pair appears more than once, you should draw as many resistors between the given nodes, as the times it appears.

Now, you can calculate it. I came up with

Re = 7*r/12

2007-01-15 02:09:20 · answer #1 · answered by Bushido The WaY of DA WaRRiOr 2 · 0 0

This is a tricky one. As far as I can see, the best way to approach it is to remove the wire between C and D, and calculate the equivalent resistance between C and D through the rest of the cube.

Then you have a simple problem: two resistors in parallel.

Good luck!

2007-01-15 02:13:05 · answer #2 · answered by Gnomon 6 · 0 0

I'm solving taking edges as follows:-
(AB, BC, CD, DA)=ABCDA,( EF, FG, GH, HE)=EFGHE, AF, DE, CH, BG
Symmetry suggests that the current entering DA has to come out from BC and entering DE has to come out from HC, for these conditions to be true at same time AF, BG, EF, HG have to be disconnected. After doing this we get tthree resistances of 3R, 3R and R in parallel
therefore, resultant is 3R/5 or 0.6R

2007-01-15 02:18:26 · answer #3 · answered by WhItE_HoLe 3 · 0 0

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2016-12-12 11:50:39 · answer #4 · answered by kluesner 4 · 0 0

MHT CET student ?

If yes, then this type of questions are out of your syllabus.
I appeared in 2005 and managed to clear it in 1st attempt.Now doing medicine!

Forget these questiona and practice from Bangui's numerical and textbook!

2007-01-15 03:57:27 · answer #5 · answered by Rohan 2 · 0 0

http://mathworld.wolfram.com/ResistanceDistance.html

7R/12

I didn't use Kirkhoff.

2007-01-15 02:35:34 · answer #6 · answered by Anonymous · 0 0

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