Chapter 35: Maximum power transfer theorems and impedance matching

Example 1, page no. 620

In [1]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
rv  =  120;#  in  volts
thetav  =  0;#  in  degrees
Z  =  15  +  1j*20;#  in  ohm

 #calculation:  
 #voltage
V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)
 #maximum  power  transfer  occurs  when  R  =  mod(Z)
R  =  abs(Z)
 #the  total  circuit  impedance
ZT  =  Z  +  R
 #Current  I  flowing  in  the  load  is  given  by
I  =  V/ZT
Imag  =  abs(I)
 #maximum  power  delivered
P  =  R*I**2


#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)maximum  power  transfer  occurs  when  R  is  ",R,"  ohm"
print  "\n  (b)  maximum  power  delivered  is  ",abs(P),"  W"


  Result  



  (a)maximum  power  transfer  occurs  when  R  is   25.0   ohm

  (b)  maximum  power  delivered  is   180.0   W

Example 2, page no. 620

In [2]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
rv  =  120;#  in  volts
thetav  =  0;#  in  degrees
Z  =  15  +  1j*20;#  in  ohm

 #calculation:  
 #voltage
V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)
 #maximum  power  transfer  occurs  when  X  =  -1*imag(Z)  and  R  =  real(Z)
z  =  Z.real  -  1j*Z.imag
 #Total  circuit  impedance  at  maximum  power  transfer  condition,
ZT  =  Z  +  z
 #Current  I  flowing  in  the  load  is  given  by
I  =  V/ZT
Imag  =  abs(I)
 #maximum  power  delivered
P  =  Z.real*I**2


#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)maximum  power  transfer  occurs  when  Z  is  ",z.real,"  +  (",  z.imag,")i  ohm"
print  "\n  (b)  maximum  power  delivered  is  ",abs(P),"  W"


  Result  



  (a)maximum  power  transfer  occurs  when  Z  is   15.0   +  ( -20.0 )i  ohm

  (b)  maximum  power  delivered  is   240.0   W

Example 3, page no. 620

In [3]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
rv  =  200;#  in  volts
thetav  =  0;#  in  degrees
R1  =  100;#  in  ohm
C  =  1E-6;#  in  farad
f  =  1000;#  in  Hz

 #calculation:  
 #voltage
V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)
 #Capacitive  reactance,  Xc
Xc  =  1/(2*math.pi*f*C)
 #Hence  source  impedance,
z  =  R1*(1j*Xc)/(R1  +  1j*Xc)
 #maximum  power  transfer  is  achieved  when  R  =  mod(z)
R  =  abs(z)
 #Total  circuit  impedance  at  maximum  power  transfer  condition,
ZT  =  z  +  R
 #Current  I  flowing  in  the  load  is  given  by
I  =  V/ZT
Imag  =  abs(I)
 #maximum  power  transferred,
P  =  R*Imag**2


#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)maximum  power  transfer  occurs  when  R  is  ",round(R,2),"  ohm"
print  "\n  (b)  maximum  power  delivered  is  ",round(P,2),"  W"


  Result  



  (a)maximum  power  transfer  occurs  when  R  is   84.67   ohm

  (b)  maximum  power  delivered  is   127.9   W

Example 4, page no. 621

In [4]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
rv  =  60;#  in  volts
thetav  =  0;#  in  degrees
R1  =  4;#  in  ohm
XL  =  10;#  in  ohm
Xc  =  7;#  in  ohm
R2  =  XL*1j;#  in  ohm
R3  =  -1j*Xc;#  in  ohm

 #calculation:  
 #voltage
V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)
 #maximum  power  transfer  is  achieved  when
R  =  (R1**2  +  (XL  -  Xc)**2)**0.5
 #Hence  source  impedance,
ZT  =  R1  +  R2  +  R3  +  R
 #Current  I  flowing  in  the  load  is  given  by
I  =  V/ZT
Imag  =  abs(I)
 #maximum  power  transferred,
P  =  R*Imag**2


#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)maximum  power  transfer  occurs  when  R  is  ",R,"  ohm"
print  "\n  (b)  maximum  power  delivered  is  ",P,"  W"


  Result  



  (a)maximum  power  transfer  occurs  when  R  is   5.0   ohm

  (b)  maximum  power  delivered  is   200.0   W

Example 5, page no. 622

In [5]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
V  =  20;#  in  volts
R1  =  5;#  in  ohm
R2  =  15;#  in  ohm

#calculation:  
 #R  is  removed  from  the  network  as  shown  in  Figure  35.6
 #P.d.  across  AB,  E
E  =  (R2/(R1  +  R2))*V
 #Impedance  ‘looking-in’  at  terminals  AB  with  the  source  removed  is  given  by
r  =  R1*R2/(R1  +  R2)
 #The  equivalent  Th´evenin  circuit  supplying  terminals  AB  is  shown  in  Figure  35.7.  
    #From  condition  (2),  for  maximum  power  transfer
R  =  r
 #Current  I  flowing  in  the  load  is  given  by
I  =  E/(R  +  r)
 #maximum  power  transferred,
P  =  R*I**2


#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)maximum  power  transfer  occurs  when  R  is  ",R,"  ohm"
print  "\n  (b)  maximum  power  delivered  is  ",P,"  W"


  Result  



  (a)maximum  power  transfer  occurs  when  R  is   3.75   ohm

  (b)  maximum  power  delivered  is   15.0   W

Example 6, page no. 622

In [6]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
rv  =  100;#  in  volts
thetav  =  30;#  in  degrees
R1  =  5;#  in  ohm
R2  =  5;#  in  ohm
R3  =  10j;#  in  ohm

#calculation:  
 #voltage
V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)
 #Resistance  R  and  reactance  X  are  removed  from  the  network  as  shown  in  Figure  35.9
 #P.d.  across  AB,
E  =  ((R2  +  R3)/(R1  +  R2  +  R3))*V
 #With  the  source  removed  the  impedance,  z,  ‘looking  in’  at  terminals  AB  is  given  by:
z  =  (R2  +  R3)*R1/(R1  +  R2  +  R3)
 #The  equivalent  Th´evenin  circuit  is  shown  in  Figure  35.10.  From  condition  3, 
    #maximum  power  transfer  is  achieved  when  X  =  -1*imag(z)  and  R  =  real(z)
X  =  -1*z.imag
R  =  z.real
Z  =  R  +  1j*X
 #Current  I  flowing  in  the  load  is  given  by
I  =  E/(z  +  Z)
Imag  =  abs(I)
 #maximum  power  transferred,
P  =  R*Imag**2


#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)maximum  power  transfer  occurs  when  R  is  ",R,"  ohm  and  X  is  ",  X,"  ohm"
print  "\n  (b)  maximum  power  delivered  is  ",round(P,2),"  W"


  Result  



  (a)maximum  power  transfer  occurs  when  R  is   3.75   ohm  and  X  is   -1.25   ohm

  (b)  maximum  power  delivered  is   416.67   W

Example 7, page no. 624

In [7]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
Ro  =  448;#  in  ohm
tr  =  8;#  turn  ratio  N1/N2

 #calculation:  
 #The  equivalent  input  resistance  r  of  the  transformer  must  be  Ro  for  maximum  power  transfer.
r  =  Ro
RL  =  r*(1/tr)**2


#Results
print  "\n\n  Result  \n\n"
print  "\n  the  optimum  value  of  load  resistance  is  ",RL,"  ohm"


  Result  



  the  optimum  value  of  load  resistance  is   7.0   ohm

Example 8, page no. 624

In [8]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
Zo  =  450  +  1j*60;#  in  ohm
ZL  =  40  +  1j*19;#  in  ohm

 #calculation:  
 #transformer  turns  ratio  tr  =  (N1/N2)
Zomag  =  abs(Zo)
ZLmag  =  abs(ZL)
tr  =  (Zomag/ZLmag)**0.5


#Results
print  "\n\n  Result  \n\n"
print  "\n  the  transformer  turns  ratio  is  ",round(tr,2),""


  Result  



  the  transformer  turns  ratio  is   3.2

Example 9, page no. 625

In [9]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
V1  =  240;#  in  volts
V2  =  1920;#  in  volts
R1  =  5;#  in  ohms
R2  =  1600;#  in  ohms

#calculation:  
 #The  network  is  shown  in  Figure  35.12.
 #turn  ratio  N1/N2  =  V1/V2
tr  =  V1/V2
 #Equivalent  input  resistance  of  the  transformer,
RL  =  R2
r  =  RL*tr**2
 #Total  input  resistance,
Rin  =  R1  +  r
 #primary  current,  I1
I1  =  V1/Rin
 #For  an  ideal  transformer  V1/V2  =  I2/I1
I2  =  I1*(V1/V2)
 #Power  dissipated  in  the  load  resistance
P  =  RL*I2**2


#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)  primary  current  flowing  is  ",I1,"  A"
print  "\n  (b)  Power  dissipated  in  the  load  resistance  is  ",P,"W"


  Result  



  (a)  primary  current  flowing  is   8.0   A

  (b)  Power  dissipated  in  the  load  resistance  is   1600.0 W

Example 10, page no. 625

In [10]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
rv  =  30;#  in  volts
thetav  =  0;#  in  degrees
r  =  20000;#  in  ohms
tr  =  20;#  turn  ratio

#calculation:
 #voltage
V  =  rv*math.cos(thetav*math.pi/180)  + 1j*rv*math.sin(thetav*math.pi/180)  
 #The  network  diagram  is  shown  in  Figure  35.13.
 #For  maximum  power  transfer,  r1  must  be  equal  to
r1  =  r
 #load  resistance  RL
RL  =  r1/tr**2
 #The  total  input  resistance  when  the  source  is  connected  to  the  matching  transformer  is
RT  =  r  +  r1
 #Primary  current
I1  =  V/RT
 #N1/N2  =  I2/I1
I2  =  I1*tr
 #Power  dissipated  in  load  resistance  RL  is  given  by
P  =  RL*I2**2


#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)the  value  of  the  load  resistance  is  ",RL,"  ohm"
print  "\n  (b)  Power  dissipated  in  the  load  resistance  is  ",abs(P*1000),"mW"


  Result  



  (a)the  value  of  the  load  resistance  is   50.0   ohm

  (b)  Power  dissipated  in  the  load  resistance  is   11.25 mW