Xcos examples from textbooks and for blocksHEADmaster

1 files changed, 40 insertions, 0 deletions

diff --git a/Working_Examples/83/CH6/EX6.1/example_6_1.sce b/Working_Examples/83/CH6/EX6.1/example_6_1.sce
new file mode 100755
index 0000000..cb0ae70
--- /dev/null
+++ b/Working_Examples/83/CH6/EX6.1/example_6_1.sce
@@ -0,0 +1,40 @@
+//Chapter 6
+//Example 6.1
+//page 195
+//To Ybus using singular transformation
+
+clear;clc;
+printf('Let us solve this problem by giving values given in the table 6.1 instead of keeping it in variables');
+
+y10=1;y20=1;y30=1;y40=1;
+y34=2-%i*6;y23=0.666-%i*2;
+y12=2-%i*6;y24=1-%i*3;
+y13=1-%i*3;
+
+Y=[y10 0 0 0 0 0 0 0 0;
+   0 y20 0 0 0 0 0 0 0;
+   0 0 y30 0 0 0 0 0 0;
+   0 0 0 y40 0 0 0 0 0;
+   0 0 0 0 y34 0 0 0 0;
+   0 0 0 0 0 y23 0 0 0;
+   0 0 0 0 0 0 y12 0 0;
+   0 0 0 0 0 0 0 y24 0;
+   0 0 0 0 0 0 0 0 y13];
+A=[1 0 0 0;
+   0 1 0 0;
+   0 0 1 0;
+   0 0 0 1;
+   0 0 1 -1;
+   0 -1 1 0;
+   1 -1 0 0;
+   0 -1 0 1;
+   -1 0 1 0];
+printf('\n\n Ybus matrix using singular transformation for the system of fig.6.2 is \n Ybus= ');
+Y=A'*Y*A;
+disp(Y);
+// for verification let us calculate as given in the text book
+printf('\n\n For verification,calculating Ybus substituting as given in the text book\n Ybus(verifiaction)=');
+Yveri=[(y10+y12+y13) -y12 -y13 0;-y12 (y20+y12+y23+y24) -y23 -y24;-y13 -y23 (y30+y13+y23+y34) -y34;0 -y24 -y34 (y40+y24+y34)];
+disp(Yveri);
+   
+