Xcos examples from textbooks and for blocksHEADmaster

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diff --git a/Working_Examples/83/CH5/EX5.1/example_5_1.sce b/Working_Examples/83/CH5/EX5.1/example_5_1.sce
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+//Chapter 5
+//Example 5.1
+//page 132
+//To find sending-end voltage and voltage regulation
+clc;clear;
+
+load1=5000; //kW
+pf=0.707;
+Vr=10000; //receiving end voltage
+R=0.0195*20;
+X=2*%pi*50*0.63*10^-3*20;
+
+//to find sending end voltage and voltage regulation
+I=load1*1000/(Vr*pf);
+Vs=Vr+I*(R*pf+X*sin(acos(pf)));
+voltage_regulation=(Vs-Vr)*100/Vr;
+printf('\n\nReceiving current =I=%d A\n',I);
+printf('Sending end voltage =Vs=%d V\n',Vs);
+printf('Voltage Regulation=%0.2f %%',voltage_regulation);
+
+//to find the value of the capacitor to be connected in parallel to the load
+voltage_regulation_desi=voltage_regulation/2;
+Vs=(voltage_regulation_desi/100)*Vr+Vr;
+//by solving the equations (i) and (ii)
+pf=0.911;
+Ir=549;
+Ic=(Ir*(pf-%i*sin(acos(pf))))-(707*(0.707-%i*0.707));
+Xc=(Vr/imag(Ic));
+c=(2*%pi*50*Xc)^-1;
+printf('\n\nCapacitance to be connected across the load so as to reduce voltage regulation by half of the above voltage regulation is given by :\n C = %d uF\n',c*10^6);
+
+//to find efficiency in both the cases
+//case(i)
+losses=I*I*R*10^-3;
+n=(load1/(load1+losses))*100;
+printf('\n Efficiency in : \nCase(i) \t n=%0.1f%%',n);
+//caase(ii)
+losses=Ir*Ir*R*10^-3;
+n=(load1/(load1+losses))*100;
+printf('\nCase(ii) \t n=%0.1f%%',n);
+
+
+
diff --git a/Working_Examples/83/CH5/EX5.1/result_example_5_1.txt b/Working_Examples/83/CH5/EX5.1/result_example_5_1.txt
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+
+Receiving current =I=707 A
+Sending end voltage =Vs=12174 V
+Voltage Regulation=21.75 %
+
+Capacitance to be connected across the load so as to reduce voltage regulation by half of the above voltage regulation is given by :
+ C = 87 uF
+
+ Efficiency in : 
+Case(i) 	 n=96.2%
+Case(ii) 	 n=97.7%