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clc
// Fundamental of Electric Circuit
// Charles K. Alexander and Matthew N.O Sadiku
// Mc Graw Hill of New York
// 5th Edition
// Part 2 : AC Circuits
// Chapter 14 : Frequency Response
// Example 14 - 17
clear; clc; close;
//
// Given data
f0_upper = 1600.0000 * 10^3;
f0_lower = 540.0000 * 10^3;
L = 10^-6
//
// Calculations C1 The High End of The AM Band
C1 = 1/(4*(%pi)^2*(f0_upper)^2*L);
// Calculations C1 The Low End of The AM Band
C2 = 1/(4*(%pi)^2*(f0_lower)^2*L);
//
disp("Example 14-18 Solution : ");
printf(" \n C1 = Capacitance of Capacitor The High End of The AM Band = %.3f nanoFarad",C1*10^9)
printf(" \n C2 = Capacitance of Capacitor The Low End of The AM Band = %.3f nanoFarad",C2*10^9)
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