Q1: Assume that in a certain 4-bit weighted ladder DAC, the input representing the most significant bit is applied to a 20 k resistor. What is the size of the resistor that represents the least significant bit?

ANS:D - 160 k

The opamp configuration of weighted resistor gives V=Rf(V2/R2+V1/R1+V0/R0) where conventionally R0=8k, R1=4k and R2=2k and Rf=1k.

Thus for Rf=20k, R0(concerned with LSB)=20*8k=160k.