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+//CHAPTER 2- STEADY-STATE ANALYSIS OF SINGLE-PHASE A.C. CIRCUIT
+//Example 8
+
+disp("CHAPTER 2");
+disp("EXAMPLE 8");
+
+//Given
+//V=300.cos(314.t+20) volts
+//i=15.cos(314.t-10) Amp
+//
+//VARIABLE INITIALIZATION
+//V=300.cos(314.t+20) volts
+//V=300.sin(314.t+110) volts as cos(theta)=sin(theta+90)
+//i=15.cos(314.t-10) Amp
+//i=15.sin(314.t+80) Amp as cos(theta)=sin(theta+90)
+//Now
+V=300/sqrt(2);                   //in Volts
+angle_V=110;                     //in degrees
+I=15/sqrt(2);                    //in Amperes
+angle_I=80;                      //in degrees
+
+//SOLUTION
+Z=V/I;				//circuit impedance
+angle_Z=angle_V-angle_I;	//angle between current and voltage
+disp(sprintf("The circuit impedance is %d Ω",Z));
+disp(sprintf("The phase angle is %d degrees",angle_Z));
+//Pav=Vm*Im.cos (phi) in RL circuit
+Pav=V*I*cos(angle_Z*(%pi/180));  //to convert angle_z from degrees to radians  
+disp(sprintf("The average power drawn is %7.2f W",Pav));// textboo answer is 1949.85 w
+
+//END