Given circuit. with v_(s)=50mvAccording to the concept of virtual ground ie. resistame between node a and b will be considered to be negligiblehence quadV_(a)=V_(b)=V_(s):. current through resistor of 20k. will be.{:[i=(V_(s))/(R)=(50mV)/(20kOmega)],[i=2.5 xx10^(-6)Amp]:}using KVL.{:[V_(0) ... See the full answer