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+//Calculate the most economical cross sectional area
+clear;
+clc;
+//soltion
+//given
+id=0.1;//interest & depreciation charges
+pf=.8;//lagging
+P=10^6;//Watts//load
+V=11000;//volts//supply voltage
+cst=.15;//rs// cost of energy per unit
+d=1.75*10^-6;//ohm cm//specific resistance
+l=1000//m//length of the cable
+t=3000;//hours
+printf("Annual cost of 2 core feeder cable is Rs(30 + 500a)per meter\n");
+R=(d*1000*100);//ohm//resistance of conductor
+printf("Resistance of each conductor= %f/a \n", R);
+i=P/(V*pf);//ampere
+printf("Current in each conductor= %f A\n", i);
+P2=500*10^3*id;//energy lost per annum
+printf("Capital cost= P2*a= %d*a \n", P2);
+P3=(2*i^2*R*t*cst)/1000;//kWh//annual cost of energy loss
+printf("Energy loss per annum= P3/a= %f/a \n",P3);
+a=sqrt(P3/P2);
+printf("Economic cross section of conductor is= √(P3/P2)= %f square cm \n",a);
+printf("Diameter of conductor= %f cm \n", sqrt(4*a/%pi));