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+//CHAPTER 2- STEADY-STATE ANALYSIS OF SINGLE-PHASE A.C. CIRCUIT
+//Example 13
+
+disp("CHAPTER 2");
+disp("EXAMPLE 13");
+
+//given
+//load of impedance 1+j.1 ohm connected AC Voltage
+//AC Voltage represented by V=20.sqrt(2).cos(wt+10) volt
+
+//to find
+//current in form of i=Im.sin(wt+phi) A
+// real power 
+
+//Equations to be used
+//real Power pr=Vrms.Irms.cos (phi)
+//             =(Vm/sqrt(2)).(Im/sqrt(2)).cos(phi)
+// apparent power pa=Vrms.Irms
+//		    =(Vm/sqrt(2)).(Im/sqrt(2))
+//
+//VARIABLE INITIALIZATION
+z1=1+(%i*1);                    //impedance in rectangular form in Ohms
+v=20*sqrt(2);                   //amplitude of rms value of voltage in Volts
+
+//SOLUTION
+function [z,angle]=rect2pol(x,y);
+z=sqrt((x^2)+(y^2));            //z is impedance & the resultant of x and y
+angle=atan(y/x)*(180/%pi);      //to convert the angle from radians to degrees
+endfunction;
+
+//solution (i)
+[z,angle]=rect2pol(1,1); 
+v=v/sqrt(2);
+angle_v=100;                    //v=(20/sqrt(2))*sin(ωt+100)
+I=v/z;                          //RMS value of current
+angle_I=angle_v-angle;
+Im=I*sqrt(2);
+disp(sprintf("(i) The current in load is i = %d sin(ωt+%d) A",Im,angle_I));
+
+//solution (ii)
+pr=(v/sqrt(2))*(I*sqrt(2))*cos(angle*(%pi/180));
+disp(sprintf("(ii) The real power is %4.0f W",pr));
+
+//solution (iii)
+pa=(v/sqrt(2))*(I*sqrt(2));
+disp(sprintf("(ii) The apparent power is %6.2f VAR",pa));
+
+//END
+
+
+