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+//Book - Power System: Analysis & Design 5th Edition
+//Authors - J. Duncan Glover, Mulukutla S. Sarma, and Thomas J. Overbye
+//Chapter - 10 ; Example 10.2
+//Scilab Version - 6.0.0 ; OS - Windows
+clc;
+clear;
+Irelay=200                  //Current through the relay in Amperes
+CTratio=100/5;              //CT ratio
+Zs=0.082;                   //Secondary resistance of a 100:5 CT in Ohm
+IZB=[8 0.8; 8 3];           //Secondary output current in Amperes and burden resistance in Ohm
+E=(Zs+IZB(1,2))*IZB(1,1);   //Secondary Excitation voltage in Volts
+Ie=0.40                     //Secondary Excitation current for the secondary voltage from figure !0.8 in Amperes
+I=CTratio*(IZB(1,1)+Ie);    //Primary current of the CT in Amperes
+printf('\nCase: a');
+if (Irelay>I) then
+    printf('\nWith the burden resistance of %0.2f Ohm, the minimum primary current required is %d Amperes.\nTherefore the relay will operate because of the 200 Amperes fault current',IZB(1,2),I)
+else
+    printf('\nWith the burden resistance of %0.2f Ohm, the minimum primary current required is %d Amperes.\nTherefore the relay will not operate because of the 200 Amperes fault current',IZB(1,2),I);
+end
+E=(Zs+IZB(2,2))*IZB(2,1);   //Secondary Excitation voltage in Volts
+Ie=30                       //Secondary Excitation current for the secondary voltage from figure !0.8 in Amperes
+I=CTratio*(IZB(2,1)+Ie);    //Primary current of the CT in Amperes
+printf('\n\nCase: b');
+if (Irelay>I) then
+    printf('\nWith the burden resistance of %0.2f Ohm, the minimum primary current required is %d Amperes.\nTherefore the relay will operate because of the 200 Amperes fault current',IZB(2,2),I)
+else
+    printf('\nWith the burden resistance of %0.2f Ohm, the minimum primary current required is %d Amperes.\nTherefore the relay will not operate because of the 200 Amperes fault current',IZB(2,2),I);
+end