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+//Example 6// Ch 5
+clc;
+clear;
+close;
+// given data
+d=0.001;
+a = 0.1*10^-2;//radii of concentric circle in meters
+b = 2.1*10^-2;//radii of concentric circle in meters
+p = 101.3;//gas pressure in kPa
+p1=3*p;
+p2=5*p;
+C = -2400.4;//constant value
+A = 0.027;//constant value
+As = 10^8;//avalanche size
+ri = 0.0772;//in m
+V01 = [log(10^8)-{(C*p1)*(ri-a)}]*(b-a)/[A*{(1/a)-(1/ri)}];
+V02 = [log(10^8)-{(C*p2)*(ri-a)}]*(b-a)/[A*{(1/a)-(1/ri)}];
+printf("corona onset voltage at 3atmp is %f kV \n",V01)
+printf("corona onset voltage at 5atmp is %f kV \n",V02)
+V01pos = 41.9;//in kV at 3 atmp
+V01neg = 42.2;//in kV at 3 atmp
+V02pos = 69.2;//in kV at 5 atmp
+V02neg = 69.8;//in kV at 5 atmp
+printf("positive corona onset voltage %f kV \n",V01pos)
+printf("negative corona onset voltage %f kV \n",V01neg)
+printf("positive corona onset voltage %f kV \n",V02pos)
+printf("negative corona onset voltage %f kV \n",V02neg)
+//answer given in the book is wrong
+
+//acc. to eq N2>=N1 and Neph>=1 with increase of gas pressure improves the dielectric strength of the gas since breakdown voltage increses with gas pressure