C
Chris Carlen
Guest
Hi:
Guy Macon wrote:
"Here is a classic that really shows the difference between algebraic
and RPN: calculate the resistance between points (a) and (b) using
only a calculator - no drawings or equations on paper.
(a)-- *----------- ONE OHM -------------* V1
|\ |\
| O O O
| N N N
| E E E
| |
| O O O
O H H H
N M M M
E \ V2 | \
*---- ONE OHM --------------------* V6
O | | |
H O | O
M N | N
| W | E
| | | |
| O | |
| H | O
| M | N
| V3 | | E
*----- ONE OHM -------------------* V4
\ | \ O
O | O H
N | N M
E | E |
| |
O | O |
H | H |
M | M |
\| V5 \|
*------------ ONE OHM ------------*--(b)"
I found this intriguing because it seemed silly that I could spend all
weekend studying Fourier transforms and "advanced" stuff like that, and
then be put off by a simple resistor problem. So I gave it a try...
I am thogroughly impressed by any folks who can solve this by eye, using
just the calculator's "arithmetic" capabilities whether it be RPN or
algebraic input.
I had to resort to pencil and paper and came up with the following
matrix solution for the node voltages for attaching a 1V source to
terminal A and ground to terminal B:
[ V1 ] [ 3 0 0 -1 0 -1 ]^-1 [ 1 ] [ 0.6 ]
[ V2 ] [ 0 3 0 0 -1 -1 ] [ 1 ] [ 0.6 ]
[ V3 ] [ 0 0 3 -1 -1 0 ] [ 1 ] [ 0.6 ]
[ V4 ] = [-1 0 -1 3 0 0 ] * [ 0 ] = [ 0.4 ]
[ V5 ] [ 0 -1 -1 0 3 0 ] [ 0 ] [ 0.4 ]
[ V6 ] [-1 -1 0 0 0 3 ] [ 0 ] [ 0.4 ]
Thus, R= V / Ib = 1 / (3*0.4/1) = 1/1.2 = 0.8333 ohms
Confirmed in LTSpice.
Fun problem. Wasn't what I had originally planned for my first hour of
the day.
Clues as to the "by eye" solution would be appreciated. I am no genius,
so usually must resort to methodical techniques such as writing out the
node voltage equations which is what I did after puzzling over it for 30
minutes. I thought a lot about superposition approaches, and re-wrote
the circuit in a planar form with one source applied to one of the three
branches attached to "a" with the other two grounded. There are some
symmetries that become more apparent after planarizing that could lead
to some further superposition approaches from that point. But is all
seems like more of a pain in the butt and more prone to error than
writing the node voltage eqns. So while a keen eye might come up with
an elegant solution, sometimes the brute force method delivers a result
long before that spark of insight hits, usually in the shower at the end
of the day (or week).
Good day!
--
____________________________________
Christopher R. Carlen
Principal Laser/Optical Technologist
Sandia National Laboratories CA USA
crcarle@sandia.gov
Guy Macon wrote:
"Here is a classic that really shows the difference between algebraic
and RPN: calculate the resistance between points (a) and (b) using
only a calculator - no drawings or equations on paper.
(a)-- *----------- ONE OHM -------------* V1
|\ |\
| O O O
| N N N
| E E E
| |
| O O O
O H H H
N M M M
E \ V2 | \
*---- ONE OHM --------------------* V6
O | | |
H O | O
M N | N
| W | E
| | | |
| O | |
| H | O
| M | N
| V3 | | E
*----- ONE OHM -------------------* V4
\ | \ O
O | O H
N | N M
E | E |
| |
O | O |
H | H |
M | M |
\| V5 \|
*------------ ONE OHM ------------*--(b)"
I found this intriguing because it seemed silly that I could spend all
weekend studying Fourier transforms and "advanced" stuff like that, and
then be put off by a simple resistor problem. So I gave it a try...
I am thogroughly impressed by any folks who can solve this by eye, using
just the calculator's "arithmetic" capabilities whether it be RPN or
algebraic input.
I had to resort to pencil and paper and came up with the following
matrix solution for the node voltages for attaching a 1V source to
terminal A and ground to terminal B:
[ V1 ] [ 3 0 0 -1 0 -1 ]^-1 [ 1 ] [ 0.6 ]
[ V2 ] [ 0 3 0 0 -1 -1 ] [ 1 ] [ 0.6 ]
[ V3 ] [ 0 0 3 -1 -1 0 ] [ 1 ] [ 0.6 ]
[ V4 ] = [-1 0 -1 3 0 0 ] * [ 0 ] = [ 0.4 ]
[ V5 ] [ 0 -1 -1 0 3 0 ] [ 0 ] [ 0.4 ]
[ V6 ] [-1 -1 0 0 0 3 ] [ 0 ] [ 0.4 ]
Thus, R= V / Ib = 1 / (3*0.4/1) = 1/1.2 = 0.8333 ohms
Confirmed in LTSpice.
Fun problem. Wasn't what I had originally planned for my first hour of
the day.
Clues as to the "by eye" solution would be appreciated. I am no genius,
so usually must resort to methodical techniques such as writing out the
node voltage equations which is what I did after puzzling over it for 30
minutes. I thought a lot about superposition approaches, and re-wrote
the circuit in a planar form with one source applied to one of the three
branches attached to "a" with the other two grounded. There are some
symmetries that become more apparent after planarizing that could lead
to some further superposition approaches from that point. But is all
seems like more of a pain in the butt and more prone to error than
writing the node voltage eqns. So while a keen eye might come up with
an elegant solution, sometimes the brute force method delivers a result
long before that spark of insight hits, usually in the shower at the end
of the day (or week).
Good day!
--
____________________________________
Christopher R. Carlen
Principal Laser/Optical Technologist
Sandia National Laboratories CA USA
crcarle@sandia.gov
