Kirt Blattenberger wrote:
I think you've answered
your own question. Altering any resistor value
so that it does not equal all the others disturbs
the symmetry of the cube, so currents in the
branch circuits at each node are not equal.
You might be tempted to think that by
doubling one resistor value it will cut the
current in that branch in half and therefore
the voltage across it will still be what it
was when at the original value, but the node
is not an ideal voltage (or current) source
that maintains the same value regardless of
the load; its value is dependent on the resistances
in all the other branches.
allows the assumptions that permit the simplified
solutions like the series/parallel equivalent
circuits, or my method that depends on Kirchhoff's
Does that answer your question,
or did I miss the point?
I'm still a bit confused. Mostly I think
it's the "equivalent circuits" that mess up the
math. Here is an example using 100 ohm resistors
and a 5 v supply. When the cube is modeled you get
82.7 ohms with 60 ma of total current. 20 ma through
each of the 3 branches and 10 ma distributing nicely
among the middle 6 resistors. When you draw it out
as an equivalent circuit of 3 parallel feeding 6
parallel outputting to 3 parallel it looks like
The circuit on the left behaves
like a normal series-parallel circuit and the math
and simulations come up with the same answer as
the cube simulation. However, if one of the middle
resistors is changed to 200 ohms, as is the case
on the right, then the math and a simulation of
circuit work as expected, yielding total resistance
of 84.85 ohms with 58.93 ma of current. 19.64 ma
in each input and output set of 3 resistors, 10.71
ma in the five 100 ohm parallel resistors and 5.36
ma through the 200 ohm resistor.
you model the cube and change one of these middle
resistors this is what you see:
resistance is now 85.3 ohms with 58.62 ma total
current. Each of the 3 branch input/output resistors
(for lack of a better description) has a different
total current. One branch has 19.83 ma, the second
20.69 ma and the last, the one with the 200 ohm
resistor, has 18.1 ma.
In the middle resistors,
the first branch divides to 10.34 and 9.48 ma respectively,
the second branch divides up to 9.48 and 11.21 ma,
and the third to 11.21 and 6.9 ma (the 6.9 ma flowing
through the 200 ohm resistor).
follows Kirchoff's law without a problem. The equivalent
circuits are just not equivalent unless the resistors
are all the same value.
My head hurts!