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+//Chapter 3
+//Example 3.3
+//page 88
+//To claculate the capacitance to neutral and charging current of a double circuit three phase transmission line
+clear;clc;
+
+//After deriving the equation for Cn from the textbook and starting calculation from Eq 3.36 onwards
+
+r=0.865*10^(-2); frequency=50; v=110e3;
+h=6; d=8; j=8; //Referring to fig given in the textbook
+
+i=((j/2)^2+((d-h)/2)^2)^(1/2);
+f=(j^2+h^2)^(1/2);
+g=(7^2+4^2)^(1/2);
+
+
+Cn=4*%pi*8.85*10^(-12)/(log((((i^2)*(g^2)*j*h)/((r^3)*(f^2*d)))^(1/3)));
+
+Cn=Cn*1000 ; //Cn is in per m.to convert it to per km,we multiply by 1000
+WCn=2*%pi*frequency*Cn;
+
+Icp=(v/sqrt(3))*WCn;
+
+Icc=Icp/2;
+
+printf("\nTotal capacitance to neutral for two conductors in parallel = %0.6f uF/km \n\n",Cn*10^(6));
+printf("Charging current/phase = %0.3f A/km \n\n",Icp);
+printf("Charging current/conductor = %0.4f A/km \n\n",Icc);
+
+
+