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+//Chapter-12, Example 12.5, Page 344
+//=============================================================================
+clc
+clear
+//INPUT DATA
+Imax=400*10^-3;//maximum value of current in mA
+Iav=150*10^-3;//average value of current in mA
+Vs=100;//maximum value of secondary voltage in V
+//CALCULATIONS
+//we know that maximum value of current does not exceed 80 percentage
+Imax1=0.8*Imax;//maximum value of current in mA
+Vm=sqrt(2)*(Vs);//maximum value of secondary voltage in V
+Rl=(Vm)/(Imax1);//value of load resistor in ohms
+Vdc=(2*Vm)/(%pi);//DC(load) voltage
+Idc=Vdc/(Rl);//DC load current in A
+PIV=2*Vm;//PIV of each diode
+mprintf("Thus value of load resistor,voltage,current and PIV of each diode are %1.0f ohms,%d V,%1.3f A and %3.1f V respectively",Rl,Vdc,Idc,PIV);
+//=================================END OF PROGRAM======================================================================================================