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+// Exa 6.14

+format('v',7);clc;clear;close;

+// Given data

+C1 = 450;//capacitance in pF

+C1 = C1 * 10^-12;// in F

+C2 = 60;//capacitance in pF

+C2 = C2 * 10^-12;// in F

+// f1 = 1/(2*%pi*(sqrt(L*(C1+Cd))))     (i)

+// f2 = 1/(2*%pi*(sqrt(L*(C2+Cd))))     (ii)

+// and f2 = 2.5*f1      (iii)

+// from eq(i),(ii) and (iii)

+Cd = (C1 - (6.25*C2))/5.25;// value of self capacitance in F

+Cd = Cd * 10^12;// in pF

+disp(Cd,"The value of self capacitance in pF is");