**Analysis for Challenge 2**

R_{IN} = 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 i_{100}, as follows:

i_{n } = i/(2^{n})

Thus,

i_{100} = i/(2^{100})

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: http://en.wikipedia.org/wiki/Resistor_ladder