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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 - 13
clear; clc; close;
//
// Given data
f1 = 250.0000;
f2 = 3000.0000;
K = 10.0000;
R = 20.0000 * 10^3;
Ri = 10.0000 * 10^3;
//
// Calculations Capacitance of Capacitor C2
C2 = 1/(2*%pi*f1*R)
// Calculations Capacitance of Capacitor C1
C1 = 1/(2*%pi*f2*R)
// Calculations Rf
Rf = (K *(f1 + f2) * Ri)/f2;
//
disp("Example 14-13 Solution : ");
printf(" \n C2 = Capacitance of Capacitor C2 = %.3f nanoFarad",C2*10^9)
printf(" \n C1 = Capacitance of Capacitor C1 = %.3f nanoFarad",C1*10^9)
printf(" \n R = Resistance of Resistor R = %.3f Kilo-ohm",R/1000)
printf(" \n Rf = Resistance of Resistor Rf = %.3f Kilo-ohm",Rf/1000)
printf(" \n Ri = Resistance of Resistor Ri = %.3f Kilo-ohm",Ri/1000)
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