On Monday 20 August 2007 20:42, M H Stein wrote: Yeah, if I wanted to take the time and find my protoboard or something similar to hold it, and... Oh my. :-) -- Member of the toughest, meanest, deadliest, most unrelenting -- and ablest -- form of life in this section of space, ?a critter that can be killed but can't be tamed. ?--Robert A. Heinlein, "The Puppet Masters" - Information is more dangerous than cannon to a society ruled by lies. --James M Dakin

It's slightly harder if the resistors are not either indidual pin-pin, or n resistors with one common pin, or a resistor chain with pins at all the tappings. If you assume the resistors link pins (although there may be 'deltas' and other parallel paths in the network), you can actually solve it by taking 2 sets of measurments : 1) Resistance between each pin an all the other pins shorted together. 2) Measuring the output voltage, using the network as a portential divider. Link all but 2 pins together (and call that ground), apply a known voltage to one of the 'free' pins and measure the voltage on the other 'free' pin. Repeat for all combinations. Now, if you consider 2 pins a and b with a resistor Rab between them and an effective restor of Rbg between pin b and all the other pins (not including a and b) shorted togther (this will be all the 'other' resistors connected to pin b in parallel), then: Fro mthe first set of measurements, you get the resistance from pin b to all other pins shorted. THis is Rab // Rbg (since you've conencted pin a to the common point too) get Vout = Vin * (Rbg / (Rab + Rbg)). Solve those for Rab, and repeat. I did this once for a resistor array in an HP printer.... -tony