Re: Just out. Message #21 Posted by Gerson W. Barbosa on 1 May 2011, 9:56 p.m., in response to message #1 by Katie Wasserman
Very good stuff indeed, thanks!
I took the time to finally solve the cube resistor problem. I never did all the exercises in my textbooks (quite the contrary!). That one was in Resnick & Halliday's book, but I skipped in '82 :-)
7 o---------R---------o 6
|\ /|
| R R |
| \ 2 1 / |
| o-----R-----o-----------+
| | | | |
R R R R |
| | | | +--+--+
| o-----R-----o | | /|\ | 1A
| / 3 4 \ | | | |
| R R | +--+--+
|/ \| |
0 o---------R---------o 5 |
| |
| |
= =
For easy visualization the cube has been redrawn to 2-D. For convenience, let's use 1 ohm resistors. The circuit will be solved through nodal analysis.
Let's take node 0 (ground) as a reference and inject 1 amp into node 1. For convenience, let R = 1 ohm. The seven nodal equations can be readily obtained:
1) (V1 - V6) + (V1 - V2) + (V1 - V4) = 1
2) (V2 - V7) + (V2 - V1) + (V2 - V3) = 0
3) (V3 - V2) + (V3 - V4) + V3 = 0
4) (V4 - V5) + (V4 - V1) + (V4 - V3) = 0
5) V5 + (V5 - V6) + (V5 - V4) = 0
6) (V6 - V5) + (V6 - V7) + (V6 - V1) = 0
7) (V7 - V6) + (V7 - V2) + V7 = 0
Instead of putting the system of equations into matrix form before solving it, the equations may be used exactly as they are written and solved with help of
the hp-50g program described here. The program finds V1 = .833333333333, which implies the equivalent resistance across the nodes 0 and 1 is 5/6*R. Likewise,
when the 1 Amp current source is connected to the nodes 5 or 6, for instance, other equivalent resistances can be determined. They are 7/12*R and 3/4*R,
respectively.
Sets of equations to be used in the program:
%%HP: T(3)A(D)F(,);
{ 'V1-V6+V1-V2+V1-V4=1,' 'V2-V7+V2-V1+V2-V3=0,' 'V3-V2+V3-V4+V3=0,' 'V4-V5+V4-V1+V4-V3=0,' 'V5+V5-V6+V5-V4=0,' 'V6-V5+V6-V7+V6-V1=0,' 'V7-V6+V7-V2+V7=0,' }
%%HP: T(3)A(D)F(,);
{ 'V1-V6+V1-V2+V1-V4=0,' 'V2-V7+V2-V1+V2-V3=0,' 'V3-V2+V3-V4+V3=0,' 'V4-V5+V4-V1+V4-V3=0,' 'V5+V5-V6+V5-V4=1,' 'V6-V5+V6-V7+V6-V1=0,' 'V7-V6+V7-V2+V7=0,' }
%%HP: T(3)A(D)F(,);
{ 'V1-V6+V1-V2+V1-V4=0,' 'V2-V7+V2-V1+V2-V3=0,' 'V3-V2+V3-V4+V3=0,' 'V4-V5+V4-V1+V4-V3=0,' 'V5+V5-V6+V5-V4=0,' 'V6-V5+V6-V7+V6-V1=1,' 'V7-V6+V7-V2+V7=0,' }
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