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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/EX14.17/Ex14_17.sce | |
| parent | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (diff) | |
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initial commit / add all books
Diffstat (limited to '3556/CH14/EX14.17/Ex14_17.sce')
| -rw-r--r-- | 3556/CH14/EX14.17/Ex14_17.sce | 28 |
1 files changed, 28 insertions, 0 deletions
diff --git a/3556/CH14/EX14.17/Ex14_17.sce b/3556/CH14/EX14.17/Ex14_17.sce new file mode 100644 index 000000000..1b6d76732 --- /dev/null +++ b/3556/CH14/EX14.17/Ex14_17.sce @@ -0,0 +1,28 @@ +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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