import math
#variable declaration
Vrms=20; #in volts
Vm=20*1.41; #in volts
Rf=50.; #forward resistance in ohms
RL=1200; #load resistance in ohms
#Calculations&Results
Im=Vm/(Rf+RL); #peak load current
print'Im=%.4f A'%Im
Idc=Im/math.pi; #dc load current
print 'Idc=%.2e A'%Idc
Irms=Im/2;#rms load current
Irms1=math.sqrt((Irms**2)-(Idc**2))#rms ac load current
print 'rms ac load current is=%.3e A'%Irms1
Vdc=Idc*Rf; #Dc voltage across the diode
print 'Dc voltage across the diode=%.4f V'%Vdc
Pdc=Idc*Idc*RL; #Dc output power
print 'Dc output power=%.3f W'%Pdc
n=40.6/(1+(Rf/RL)); #conversion efficiency
print 'conversion efficiency=%.f %%'%n
s=Rf*100/RL; #Pertcentage regulation
print 'Percentage regulation=%.2f %%'%s
import math
#variable declaration
Rf=100; #forward resistance in ohms
Rl=1000.; #load resistance in ohms
n=10; #Primary to secondary turns ratio
Vp=240; #Primary input V(rms)
#Calculations&Results
Vm=24*(2**(1./2))/2; #secondary peak voltage from cenre tap
Vs=Vp/n; #Secondary input voltage
Im=Vm/(Rf+Rl); #peak current through the resistance in A
Idc=(2*Im)/math.pi; #DC Load current in A
print 'DC load current Idc=%.2e A'%Idc
I=Idc/2; #Direct current supplied by each diode in A
print 'Direct current supplied by each diode Idc=%.2e A'%I
Pdc=Idc*Idc*Rl; #DC power output
print 'Pdc=%.4f W'%Pdc
Irms=Im/(2**(1./2));
Vrp=math.sqrt((Irms*Irms)-(Idc*Idc))*Rl; #Ripple voltage in V
print 'Ripple voltage Vrp=%.2f V'%Vrp
M=(Rf*100)/Rl; #percentage regulation
print 'Percentage regulation=%.f %%'%M
n=81.2/(1+(Rf/Rl)); #Efficiency of rectification
print 'Efficiency of rectification=%.1f %%'%n
import math
#variable declaration
Rf=50.; #forward resistance in ohms
Rl=2500; #load resistance in ohms
Vp=30; #Primary input V(rms)
Vm=30*math.sqrt(2);
#Calculations&Results
Im=Vm/(2*Rf+Rl); #peak load current in A
Idc=2*Im/math.pi;
Vdc=Idc*Rl; #DC load voltage
print 'Vdc=%.f V'%Vdc
Irms=Im/math.sqrt(2);
Vrp=Rl*math.sqrt(((Irms*Irms)-(Idc*Idc))); #Ripple voltage in V
print 'Ripple voltage Vrp=%.1f V'%Vrp
M=(2*Rf/Rl)*100; #Percentage regulation
print 'Percentage regulation=%.f %%'%M
import math
#variable declaration
Vdc=20; #DC value in V
Vpp=1.; #Peak to peak ripple voltage in V
#Calculations&Results
Vp=Vpp/2; #Peak ripple voltage in V
Vrms=Vp/math.sqrt(2); #Vrms voltage in V
S=Vrms/Vdc; #Ripple Factor
print 'Ripple factor=%.4f'%S
T=S*100;
print 'Percentage Ripple=%.2f %%'%T
import math
#variable declaration
#For a full wave rectifier
#L-type LC filter
f=50#f=line frequency in Hz
w=2*math.pi*f
Vdc=10#Vdc=dc output voltage
Idc=100*10**-3#Idc=load current in Amperes
y=0.02#y=allowable ripple factor
#Calculations
#y=sqrt(2)/(12*(w^2)*L*C)
#Let L*C=a...............(1)
a=math.sqrt(2)/(y*12*(w**2))
RL=Vdc/Idc#RL=load resistance
#Lc=critical inductance
#Lc=RL/(3*w)
#For line frequency of 50Hz,Lc=RL/(300*%pi)
#Lc=RL/950
Lc=RL/950
L=0.1#Assumed inductance in henry
C=a/L#C=capacitance calculated from equation (1)
L1=1#Assumed inductance in henry
C1=a/L1#C1=capacitance calculated from equation (1)
Rb=950*L1#Rb=bleeder resistance for good voltage regulation
#Results
print "The designed values of the components for a full wave rectifier with L-type LC filter are:"
print "The load resistance RL is =%.f ohm"%RL
print "The critical inductance Lc is = %.1f H"%Lc
print "The inductance L is=%.1f H"%L
print "The capacitance C is %.f µF"%(C/10**-6)#C is converted in terms of microfarad
#In textbook 957µF is approximately taken as 600µF
print "\nBut if the inductance L designed is of the value = %.f H"%L1
print "the capacitance C will be of the value = %.f µF"%(C1/10**-6)#C1 is converted in terms of microfarad
print "So,a standard value of 50µF can be used in practice"
print "The bleeder resistance Rb for good voltage regulation is=%.f ohm"%Rb
print "\nAs Rb is much greater than RL,little power is wasted in Rb.This reflects the advantage of selecting L>Lc"