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+//chapter23
+//example23.5
+//page515
+
+Vm=400 // V
+v=150 // V
+R_L=200 // ohm
+
+// since v=Vm*sin(theta), we get
+
+theta=asin(v/Vm)*180/%pi // in terms of degrees
+
+V_av=Vm*(1+cos(theta*%pi/180))/(2*%pi)
+I_av=V_av/R_L
+P=V_av*I_av
+
+printf("firing angle = %.2f degrees \n",theta)
+printf("average output voltage = %.3f V \n",V_av)
+printf("average current for load of 200 ohm = %.3f A \n",I_av)
+printf("power output = %.3f W \n",P)
+
+// the accurate answer for power output is 75.250 W but in book it is given as 75.15 W