Simple parallel circuit problem

[Gerson W. Barbosa]

(08-25-2015 11:07 PM)Garth Wilson Wrote:  Often overlooked, and I don't know why since this is what's happening: Add the conductances, then invert again to get resistance:

a // b = 1/ (1/a + 1/b)

I think the alternative form a // b = a*b/(a + b) was an easier option long ago, when these calculations were done by hand. Dividing 1 by a large number (more than two significant digits) was a cumbersome calculation to do manually, I remember. For this particular problem, i = 1/2048 + 512/163 might be better.

Gerson.