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authorprashantsinalkar2017-10-10 12:27:19 +0530
committerprashantsinalkar2017-10-10 12:27:19 +0530
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+//Chapter 9 Example 8//
+clc
+clear
+// from the diagram dab,dbc,dac are distances from each conductor to other in the transmission linr//
+dab=5;
+dbc=4;
+dac=6;
+// diameter of conductor=d,radius of each conductor=r,capacitance of phase a to neutral plane=can,equivalent distance=deq//
+d=0.025;// in m//
+deq=(dab*dbc*dac)^(1/3);
+printf("\n Equivalent distance deq = %.2f m\n",deq);
+r=d/2;
+cab=(2*%pi*(10^-9/(36*%pi))*10^12)/(log(deq/r));// 10^12 is conversion factor//
+printf("\n The capacitance of phase a to neutral plane = %.2f pF/m\n",cab);