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| author | prashantsinalkar | 2017-10-10 12:27:19 +0530 |
|---|---|---|
| committer | prashantsinalkar | 2017-10-10 12:27:19 +0530 |
| commit | 7f60ea012dd2524dae921a2a35adbf7ef21f2bb6 (patch) | |
| tree | dbb9e3ddb5fc829e7c5c7e6be99b2c4ba356132c /3556/CH14/EX14.13/Ex14_13.sce | |
| parent | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (diff) | |
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initial commit / add all books
Diffstat (limited to '3556/CH14/EX14.13/Ex14_13.sce')
| -rw-r--r-- | 3556/CH14/EX14.13/Ex14_13.sce | 33 |
1 files changed, 33 insertions, 0 deletions
diff --git a/3556/CH14/EX14.13/Ex14_13.sce b/3556/CH14/EX14.13/Ex14_13.sce new file mode 100644 index 000000000..072fe0805 --- /dev/null +++ b/3556/CH14/EX14.13/Ex14_13.sce @@ -0,0 +1,33 @@ +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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