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+//Chapter 2
+//Example 2.2
+//page 57
+//To find reactance of the conductor
+clear;clc;
+f=50; //frequency
+D=5.04; //diameter of the entire ACSR
+d=1.68; //diameter of each conductor
+Dsteel=D-2*d; //diameter of steel strand
+//As shown in fig
+D12=d;
+D13=(sqrt(3)*d);
+D14=2*d;
+D15=D13;
+D16=D12;
+//neglecting the central sttel conductor,we have the 6 possibilities
+D1=(0.7788*d)*D12*D13*D14*D15*D16;
+//we have total of 6 conductors,hence
+D2=D1;
+D3=D1;
+D4=D1;
+D5=D1;
+D6=D1;
+Ds=(D1*D2*D3*D4*D5*D6)^(1/(6*6));//GMR;
+//since the spacing between lines is 1m=100cm
+l=100;
+L=0.461*log10(l/Ds); //Inductance of each conductor
+Ll=2*L; // loop inductance
+Xl=2*%pi*f*Ll*10^(-3);//reactance of the line
+printf("\n\nInductance of each conductor=%0.4f mH/km\n\n",L);
+printf("Loop Inductance=%0.4f mH/km\n\n",Ll);
+printf("Loop Reactance=%f ohms/km\n\n",Xl);
+
+
+