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Conference rusure::math

Title:Mathematics at DEC
Moderator:RUSURE::EDP
Created:Mon Feb 03 1986
Last Modified:Fri Jun 06 1997
Last Successful Update:Fri Jun 06 1997
Number of topics:2083
Total number of notes:14613

340.0. "Resistance across the cube" by CORVUS::THALLER () Tue Oct 08 1985 01:13

A 1" cube has its outline constructed out of wire whose resistance
is one ohm per inch.  What is the resistance between opposite corners?

T.RTitleUserPersonal
Name
DateLines
340.1ADVAX::J_ROTHTue Oct 08 1985 12:427
Opposite on the major diagonal? 5/6 ohm.

Interesting generalizations:  let the cubes faces be constructed of a
resistive surface with constant ohm/square resistivity.  Then construct
one of a solid of constant conductivity.

- Jim
340.2BEING::POSTPISCHILTue Oct 08 1985 13:0715
Since .1 did not explain how the result was achieved, let me point out that
the voltage at each of the three vertices adjacent to one of the vertices
is identical, by symmetry.  Hence, connecting these three vertices with wires
will not change anything.  Similarly, connecting the three vertices adjacent
to the diagonally opposite vertex will not change anything.  This leaves
us with a simple arrangement:  three one-ohm resistances in parallel, followed
serially by six in parallel, followed by three in parallel.

For parallel resistances, the combined resistance is the reciprocal of the sum
of the resistances (for three one-ohm resistances, 1/(1/1+1/1+1/1) = 1/3,
for six, 1/6).  For serial resistances, the combined resistance is the sum
of the resistances (1/3+1/6+1/3 = 5/6).


				-- edp