please help! this is one problem that ive been stuck on
(1)Hooln: Given that !a) When v_(b)=0,v_(x)0v_(a)^(')==> from virtual grand concept, the potential at (2) is eguat to potential at (0., Now, Appy, Kcl(2)(2)=>quadnu_(2)=(v_(a)R_(4))/(R_(3)+R_(y))Now, Apply, iscl 20{:[=>(V_(1)-0)/(R_(1))+(V_(1)-V_(x))/(R_(2))=0],[=>quad(v_(1))/(R_(1))+(v_(1)-v_(x))/(R_(2))=0],[=>quad(V_(1)R_(2)+V_(1)R_(2)-V_(2)R_(1))/(R_(1)R_(2))=0],[=>quadV_(1)(R_(1)+R_(2))=V_(x)R_(1)],[=>nu_(1)=((nu_(x)R_(1))/(R_(1)+R_(2)))],[" But "nu_(1)=v_(2)quad(:'" virtual Ground ")" concept. "],[=>(V_(x)R_(1))/(R_(1)+R_(2))=(V_(a)R_(y))/(R_(3)+R_(y))],[=>quadnu_(x)=[(R_(1)+R_(2))/(R_(3)+R_(4))]((R_(4))/(R_(1)))*V_(a)quadv_(01 r)]:}(b) When V_(a)=0,quadV_(x) due to V_(a)!={:[=>(V_(2)-0)/(R_(3))+(V_(2)=0)/(R_(4))=0],[=>(V_(2)R_(3)+V_(2)R_(3))/(R_(3)R_(4))=0],[=>V_(2)(R_(3)+R_(4))=0],[=>quadV_(2)=0]:}lly Apply kcl@ (1),{:[quadquad(v_(1)-v_(b))/(R_(1))+(v_(1)-v_(x))/(R_(2))=0], ... See the full answer