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+//Example6.16  // for the a current to voltage converter show a) Rin = (Rf/1+Aop)  b) Rf = 10 K ohm Aop = 1000 
+clc;
+clear;
+close;
+
+//a) The input resistance given as
+//Rin = (Rf)/(1+Aop) ;
+
+// The input resistance of the circuit can be written as
+//Rin = (V1/i!);
+
+//  the feedback current of the given circuit is defined as
+//i1 =(V1-Vo)/RF ;
+
+// the feedback resistance RF is 
+//RF =(V1-Vo)/i1 ;
+
+// The output voltage Vo is
+//Vo = -Aop*V1 ;
+
+//by using this output feedback currenty i1 can be reformed as
+//i1 = (V1-(-Aop*V1))/RF ;
+
+//i1 = V1*(1+Aop)/RF ;
+
+// Then Rin Becomes 
+//Rin =Rf/(1+Aop);
+
+Rf =10*10^3 ;
+Aop = 1000 ;
+
+// the input current and output voltage of the circuit are defined as
+//i1 =(Rs)/(Rs+Rin) ;
+// Vo = -(Aop*(RF/1+Aop))*i1 ;
+
+//the input resistance Rin is 
+Rin =(Rf/(1+Aop)) ;
+
+// subsituting the value of RF Aop Rin and Vo we get 
+RF = 10 ;
+Rin = RF/(1+Aop)
+disp('The input resistance Rin is = '+string(Rin)+ ' ohm ');
+
+Aop = 1000 ;
+//(1000/1001)*(Rs/(Rs*0.00999))> 0.99 ;
+// by solving above equation we get 
+Rs = 1.099  ;  // K ohm 
+disp(' The value of Resistance Rs is = '+string(Rs)+ ' K ohm ');