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authorSiddharth Agarwal2019-09-03 18:27:40 +0530
committerSiddharth Agarwal2019-09-03 18:27:40 +0530
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treee1bc17aae137922b1ee990f17aae4a6cb15b7d87 /Working_Examples/83/CH11/EX11.2/example_11_2.sce
parent52a477ec613900885e29c4a0b02806a415b4f83a (diff)
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+//Chapter 11
+//Example 11.2
+//page 408
+//To find fault current and voltage across the grounding resistor
+clear;clc;
+
+X1eq=(%i*0.18)/2;
+X2eq=(%i*0.15)/2;
+Z0eq=(%i*0.10)+3*(2*20/(11^2));
+
+Ea=1;
+
+//calculation of fault current
+printf('\nFault current is given by ');
+If=(3*Ea)/(X1eq+X2eq+Z0eq)
+
+//current in grounding resistor
+Ifg=abs(If)*(20/(11*sqrt(3)));
+printf('\n\nCurrent through grounding resistor Ifg=%0.2fkA',Ifg);
+
+//voltage across grounding resistor
+Vgr=abs(If*(2*20/(11^2))*(11/sqrt(3)));
+printf('\n\nVoltage across grounding resistor Vgr=%0.2fkV\n\n',Vgr); \ No newline at end of file