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// Grob's Basic Electronics 11e
// Chapter No. 25
// Example No. 25_4
clc; clear;
// What value of L resonates with a 106-pF C at 1000 kHz, equal to 1 MHz?
// Given data
C = 106*10^-12; // Capacitor=106 pFarad
fr = 1*10^6; // Resonant frequency=1 MHertz
A = %pi*%pi; // pi square
B = fr*fr; // Resonant frequency square
C = 1/(4*A*B*C);
disp (C,'The value of Inductor in Henry')
disp ('i.e 239 uF')
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