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| author | priyanka | 2015-06-24 15:03:17 +0530 |
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| committer | priyanka | 2015-06-24 15:03:17 +0530 |
| commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
| tree | ab291cffc65280e58ac82470ba63fbcca7805165 /2642/CH1/EX1.3 | |
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Diffstat (limited to '2642/CH1/EX1.3')
| -rwxr-xr-x | 2642/CH1/EX1.3/Ex1_3.sce | 31 |
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diff --git a/2642/CH1/EX1.3/Ex1_3.sce b/2642/CH1/EX1.3/Ex1_3.sce new file mode 100755 index 000000000..efb217a2b --- /dev/null +++ b/2642/CH1/EX1.3/Ex1_3.sce @@ -0,0 +1,31 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 1 : REVIEW OF ELECRTIC CIRCUITS
+// Example : 1.3
+
+clc;clear; // clears the console and command history
+// Given data
+R1 = 5 // resistance in ohm
+R2 = 4 // resistance in ohm
+R3 = 9 // resistance in ohm
+R4 = 6 // resistance in ohm
+V1 = 10 // voltage in V
+V2 = 6 // voltage in V
+
+
+// caclulations
+// Remove R3 and find R_th by short circuiting V1 and V2 than R1 and R4 wiil be in parallel
+R = (R1*R4)/(R1+R4) // equivalent resistance in ohm
+// R is connected in series with R2
+R_th = R2+R // Thevenin's resistance in ohm
+I = 4/11 // current in the figure applying KVL in A
+V_th = (6*0.36)+6 // voltage in V
+I_9 = V_th/(R_th+R3) // current through R3 in A
+
+
+// display the result
+disp("Example 1.3 solution");
+printf(" \n Current through 9Ω \n I_9Ω = %.2f A ", I_9);
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