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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 /174/CH12 | |
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initial commit / add all books
Diffstat (limited to '174/CH12')
| -rwxr-xr-x | 174/CH12/EX12.1/example12_1.sce | 29 | ||||
| -rwxr-xr-x | 174/CH12/EX12.1/example12_1.txt | 2 |
2 files changed, 31 insertions, 0 deletions
diff --git a/174/CH12/EX12.1/example12_1.sce b/174/CH12/EX12.1/example12_1.sce new file mode 100755 index 000000000..53a5dee3a --- /dev/null +++ b/174/CH12/EX12.1/example12_1.sce @@ -0,0 +1,29 @@ +// To find percentage error
+// Modern Electronic Instrumentation And Measurement Techniques
+// By Albert D. Helfrick, William D. Cooper
+// First Edition Second Impression, 2009
+// Dorling Kindersly Pvt. Ltd. India
+// Example 12-1 in Page 360
+
+
+clear; clc; close;
+
+// Given data
+R = 1; //Resistance of the wire in ohm
+R_L = 10*10^3; //Load resistance in ohm
+I_supply = 50*10^-3; //power supply current in A
+V_out = 1; //output of the amplifier in V
+
+//Calculations
+V_L = (V_out+(I_supply*R))*R_L/(2*R+R_L);
+printf("The load voltage calculated = %0.2f\n",V_L);
+
+%error = ceil((V_L -V_out)/V_L*100);
+printf("The percentage error is about %d %%, which is unacceptable in most systems",%error);
+
+//Result
+// The load voltage calculated = 1.05
+// The percentage error is about 5 %, which is unacceptable in most systems
+
+
+
diff --git a/174/CH12/EX12.1/example12_1.txt b/174/CH12/EX12.1/example12_1.txt new file mode 100755 index 000000000..00c3fe9b5 --- /dev/null +++ b/174/CH12/EX12.1/example12_1.txt @@ -0,0 +1,2 @@ +The load voltage calculated = 1.05
+The percentage error is about 5 %, which is unacceptable in most systems
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