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+// To find the value of the unknown resitance in the series of resistances in a circuit.
+
+clc;
+clear;
+
+R1=20;
+
+V=220;
+
+P=50;
+
+R=poly([0 1],'R','c');
+Rt=R1+R;
+
+I=V/Rt;
+
+A=(I^2)*R;// To get the characteristic eqaution to find R. 
+B=A-50;
+C=B(2);
+
+rts=roots(C);// To find the two resistances
+
+R=round(10000.*rts)./10000;// Rounding off to four decimal points.
+
+Rt=R1+R;// Total resistance
+
+I=V./Rt;// Currents
+
+pow=(I.^2)*(R)';
+
+power=diag(pow);
+
+disp(B(2),'The Characteristic polynomial to find resistance R equated to zero is')
+
+disp('ohms',R,'The solution of the above equation yields two resistances')
+
+disp('Now to check which resistance is suitable by finding out the power dissipated by each of them')
+
+disp('watts',power,'The Power dissipated by both the resistors are')
+
+disp('ohms',R(1),'From comparison with the given value (50 watts), We find that the suitable resistance is')
+
+// The higher resistance is preferred because it limits the amount of current, ( Please see the current ratings of the resistors (Heating effect)) 
+
+
+