diff options
| author | priyanka | 2015-06-24 15:03:17 +0530 |
|---|---|---|
| committer | priyanka | 2015-06-24 15:03:17 +0530 |
| commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
| tree | ab291cffc65280e58ac82470ba63fbcca7805165 /608/CH28/EX28.02 | |
| download | Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.tar.gz Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.tar.bz2 Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.zip | |
initial commit / add all books
Diffstat (limited to '608/CH28/EX28.02')
| -rwxr-xr-x | 608/CH28/EX28.02/28_02.sce | 14 |
1 files changed, 14 insertions, 0 deletions
diff --git a/608/CH28/EX28.02/28_02.sce b/608/CH28/EX28.02/28_02.sce new file mode 100755 index 000000000..e4422ccdf --- /dev/null +++ b/608/CH28/EX28.02/28_02.sce @@ -0,0 +1,14 @@ +//Problem 28.02: An R–L–C series circuit is comprised of a coil of inductance 10 mH and resistance 8 ohm and a variable capacitor C. The supply frequency is 1 kHz. Determine the value of capacitor C for series resonance.
+
+//initializing the variables:
+R = 8; // in ohms
+L = 0.010; // IN Henry
+f = 1000; // in Hz
+
+//calculation:
+//At resonance
+//capacitance C
+C = 1/(L*(2*%pi*f)^2)
+
+printf("\n\n Result \n\n")
+printf("\n capacitance, C is %.2E F\n",C)
\ No newline at end of file |