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+//CHAPTER 1- D.C. CIRCUIT ANALYSIS AND NETWORK THEOREMS 
+//Example 42
+
+disp("CHAPTER 1");
+disp("EXAMPLE 42");
+
+//VARIABLE INITIALIZATION
+v=10;                               //in Volts
+I=0.5;                              //in Amperes
+r1=4;                               //top LHS resistance in Ohms
+r2=2;                               //top RHS resistance in Ohms
+r3=2;                               //first resistance in Ohms
+r4=2;                               //second resistance in Ohms
+
+//SOLUTION
+
+//using Thevenin's theorem
+rth=(r1*r3)/(r1+r3);
+vth=v*(r3/(r1+r3));                 //Thevenin voltage
+R=(40-(56*I))/(24*I);               //solving for R directly
+disp(sprintf("(i) By Thevenin Theorem, the value of R is %d Ω",R));
+
+//v1=(10R+4)/(3R+4)........eq(1)    //using nodal analysis at node 1
+//v1=1+R...................eq(2)    //using nodal analysis at node 2
+//the following the quadratic equation is formed when both the equations are compared
+//(3)R^2+(-3)R+(0)=0
+//solving the quadratic equation 
+a=3;
+b=-3;
+c=0;
+D=(b^2)-(4*a*c);                    //discriminant  
+R1=(-b+sqrt(D))/(2*a);
+R2=(-b-sqrt(D))/(2*a);
+if(R1==1) then
+disp(sprintf("(ii) By Nodal analysis, the value of R is %d Ω",R1));
+else 
+disp(sprintf("(ii) By Nodal analysis, the value of R is %d Ω",R1));
+end;
+
+//END