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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 /2093/CH1/EX1.13/exa_1_13.sce | |
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initial commit / add all books
Diffstat (limited to '2093/CH1/EX1.13/exa_1_13.sce')
| -rwxr-xr-x | 2093/CH1/EX1.13/exa_1_13.sce | 18 |
1 files changed, 18 insertions, 0 deletions
diff --git a/2093/CH1/EX1.13/exa_1_13.sce b/2093/CH1/EX1.13/exa_1_13.sce new file mode 100755 index 000000000..05990a6a5 --- /dev/null +++ b/2093/CH1/EX1.13/exa_1_13.sce @@ -0,0 +1,18 @@ +// Exa 1.13
+clc;
+clear;
+close;
+// Given data
+Rf= 250;// in kohm
+// Output voltage expression, Vo= -5*Va+3*Vb
+// and we know that for a difference amplifier circuit,
+// Vo= -Rf/R1*Va + [R2/(R1+R2)]*[1+Rf/R1]*Vb
+// Comparing both the expression, we get
+// -Rf/R1*Va= -5*Va, or
+R1= Rf/5;// in kohm
+disp(R1,"The value of R1 in kohm")
+// and
+R2= 3*R1^2/(R1+Rf-3*R1)
+disp(R2,"The value of R2 in kohm")
+
+// Note : Answer in the book is wrong
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