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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 /608/CH10/EX10.20/10_20.sce | |
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initial commit / add all books
Diffstat (limited to '608/CH10/EX10.20/10_20.sce')
| -rwxr-xr-x | 608/CH10/EX10.20/10_20.sce | 20 |
1 files changed, 20 insertions, 0 deletions
diff --git a/608/CH10/EX10.20/10_20.sce b/608/CH10/EX10.20/10_20.sce new file mode 100755 index 000000000..962796c86 --- /dev/null +++ b/608/CH10/EX10.20/10_20.sce @@ -0,0 +1,20 @@ +//Problem 10.20: The current I flowing in a resistor R is measured by a 0–10 A ammeter which gives an indication of 6.25 A. The voltage V across the resistor is measured by a 0–50 V voltmeter, which gives an indication of 36.5 V. Determine the resistance of the resistor, and its accuracy of measurement if both instruments have a limit of error of 2% of f.s.d. Neglect any loading effects of the instruments.
+
+//initializing the variables:
+I = 6.25; // in Amperes
+Im = 10; // max in Amperes
+V = 36.5; // in volts
+Vm = 50; // max in volts
+e = 2; // in %
+
+//calculation:
+R = V/I
+Ve = e*Vm/100
+Ve1 = Ve*100/V // in %
+Ie = e*Im/100
+Ie1 = Ie*100/I // in %
+em = Ve1 + Ie1
+Re = em*R/100
+
+printf("\n\n Result \n\n")
+printf("\n Resistance R = %.2f ohms(+-)%.2f ohms\n",R,Re)
\ No newline at end of file |