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diff --git a/3556/CH14/EX14.13/Ex14_13.sce b/3556/CH14/EX14.13/Ex14_13.sce
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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)
+