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| author | priyanka | 2015-06-24 15:03:17 +0530 |
|---|---|---|
| committer | priyanka | 2015-06-24 15:03:17 +0530 |
| commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
| tree | ab291cffc65280e58ac82470ba63fbcca7805165 /2240/CH26/EX25.6 | |
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initial commit / add all books
Diffstat (limited to '2240/CH26/EX25.6')
| -rwxr-xr-x | 2240/CH26/EX25.6/EX25_6.sce | 19 |
1 files changed, 19 insertions, 0 deletions
diff --git a/2240/CH26/EX25.6/EX25_6.sce b/2240/CH26/EX25.6/EX25_6.sce new file mode 100755 index 000000000..55b57d698 --- /dev/null +++ b/2240/CH26/EX25.6/EX25_6.sce @@ -0,0 +1,19 @@ +// Grob's Basic Electronics 11e
+// Chapter No. 25
+// Example No. 25_6
+clc; clear;
+// What is the ac resistance of the coil in A series circuit resonant at 0.4 MHz develops 100 mV across a 250-uH L with a 2-mV input.
+
+// Given data
+
+Vo = 100*10^-3; // Output voltage=100 mVolts
+Vi = 2*10^-3; // Input voltage=2 mVolts
+L = 250*10^-6; // Inductor=250 uHenry
+f = 0.4*10^6; // Frequency=0.4 MHertz
+pi = 3.14;
+
+Q = Vo/Vi;
+Xl = 2*pi*f*L;
+
+rs = Xl/Q;
+disp (rs,'The Ac Resistance of Coil in Ohms')
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