Timeline for answer to How do you arrange six 6-ohm resistors to have a total resistance of 6-ohm? by tollin jose
Current License: CC BY-SA 3.0
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| May 20, 2014 at 6:13 | comment | added | candied_orange | He meant any square of identical resistors produce resistance identical to each resistor in the square. Thus by collapsing or expending squares you can avoid doing any calculations while you search for the resistor count you want. Doesn't really provide a rigorous algorithm for proving what resister counts would be impossible but it provides an elegant way to simplify the trial and error. It means needing to use 1 is the same as needing to use 4 or 9 or 16 ... | |
| Apr 5, 2014 at 3:01 | comment | added | tollin jose | ok you meant it is possible to make 6 ohm resistance using 9 six-ohm resistors. | |
| Apr 4, 2014 at 14:27 | comment | added | starblue | If you arrange resistors in a square you get the same resistance value again, because n times parallel divides the resistance by n and n times in series multiplies by n. It doesn't matter whether you first connect in series or parallel, i.e. you may choose to connect nodes of the same potential or not, without changing the resistance value. In your example R3 could be expanded to a 2x2 square, then you would get a 3x3 square overall. You could then make it regular by adding connections. | |
| Apr 4, 2014 at 11:59 | comment | added | tollin jose | @starblue can you make it more clear ? | |
| Apr 3, 2014 at 20:01 | comment | added | starblue | To get this circuit think of a square of 9 resistors and collapse the square in the lower left corner into a single resistor. | |
| Apr 3, 2014 at 18:05 | history | edited | tollin jose | CC BY-SA 3.0 |
added 111 characters in body
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| Apr 3, 2014 at 12:43 | history | answered | tollin jose | CC BY-SA 3.0 |