Analysis for Challenge 2

RIN = R

The circuit is known as an R-2R ladder.  Starting with the right-most circuitry we
have R+R in parallel with 2R, which is R.  This R is in series with an R, making 2R.
This new 2R is in parallel with 2R, which is R.  We then proceed from right-to-left,
until there is a final 2R in parallel with 2R, which is R.

Assume a voltage is applied at the terminals on the left.  Assume the input current
is i.  This i “sees” 2R in parallel with 2R, so it divides into i/2 and i/2.  One i/2 goes
vertically down to the lower line, and then returns to the voltage source.  The
other i/2 proceeds to the right through the first horizontal R.  It then “sees” 2R in
parallel with 2R, so it divides into i/4 and i/4.  This divide-by-2 (binary) pattern
continues all the way to the nth loop.  If, as in this challenge, there are 100 2R
resistors, then the final current division would yield i100, as follows:

in    = i/(2n)
Thus,
i100 = i/(2100)

Theoretically this is an attractive observation, but where, as here, n is a very large
number, it cannot (by reason of resistor tolerances) be sustained beyond a single
digit (perhaps 6, or 7, or 8).  Why we might care about how large n can be is a
discussion left to the world of A-to-D and D-to-A converters.

Want to Know More?

Here’s one reference: 