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![[parent]](https://images.physicslibrary.org/images/uparrow.png) Viewing Message
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| ``Re: Question on resistance''
by rspuzio on 2009-05-28 03:53:29 |
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| The trick to this is to use symmetry to identify
equipotential points. The vertices of an n-dimensional
cube may be labeled by n-tuplets of + and - and two
vertices are the endpoints of an edge if and only if
their tuplets differ in exactly one slot. As the opposite
vertices between which we will consider the resistance
we may, without loss of generality, take as (+++++...) and
(-----...); put the former at a potential of 1 volt and
ground the latter. Then we note that any two vertices
whose tuplets have the same number of plus and minus
signs can be transformed into each other by a symmetry
operation which preserves the two vertices between which
we are measuring the resistance hence, by symmetry, they
will all be at the same potential.
Hence, the problem is equivalent to one in which we
collapse all the vertices at the same potential into a
single vertex. In this case, that is a problem in which
there are number of resistors between one node and the
next node in parallel. By replacing the parallel resistors
with their equivalents, all that is left is to sum series
resistors.
For instance, consider three dimensions. The nodes between
which we compute resistance are +++ and ---. ++-, +-+, and
-++ are at the same potential as are +--, -+-, and --+ (at
a different potential. There are 3 edges between +++ and
the nodes (++-, +-+, -++), 6 edges between (++-, +-+, -++)
and (+--, -+-, --+), and 3 nodes between (+--, -+-, --+) and
---. Thus the problem is equivalent to one with 4 nodes,
there being 3 parallel resistors between node 0 and node 1,
6 parallel resistors between node 1 and node 2, and 3
parallel resistors between nodes 2 and node 3, which is
equivalent to a 1/3 Ohm resistor between node 0 and node 1,
a 1/6 Ohm resistor between node 1 and node 2, and a 1/3 Ohm
resistor between node 2 and node 3, for a grand total of
5/6 Ohm as claimed.
For 4 dimensions, things go as follows:
++++
4 resistors
+++-, ++-+, +-++, -+++
12 resistors
++--, +-+-, -++-, +--+, -+--+, --++
12 resistors
+---, -+--, --+-, ---+
4 resistors
----
eqivalently:
1/4 + 1/12 + 1/12 + 1/4 = 2/3
In general, we will have the sum of 1/(m nCm).
Since it is late and I need to get some rest,
I won't fiddle with this sum now. |
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