Chapter 32: The superposition theorem

Example 1, page no. 564

In [1]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
rv1  =  100;#  in  volts
rv2  =  50;#  in  volts
thetav1  =  0;#  in  degrees
thetav2  =  90;#  in  degrees
r1  =  25;#  in  ohm
R  =  20;#  in  ohm
r2  =  10;#  in  ohm

 #calculation:
 #voltage
V1  =  rv1*math.cos(thetav1*math.pi/180)  +  1j*rv1*math.sin(thetav1*math.pi/180)
V2  =  rv2*math.cos(thetav2*math.pi/180)  +  1j*rv2*math.sin(thetav2*math.pi/180)
 #The  circuit  diagram  is  shown  in  Figure  32.7.  Following  the  above  procedure:
 #The  network  is  redrawn  with  the  50/_90°  V  source  removed  as  shown  in  Figure  32.8
 #Currents  I1,  I2  and  I3  are  labelled  as  shown  in  Figure  32.8.
I1  =  V1/(r1  +  r2*R/(R  +  r2))
I2  =  (r2/(r2  +  R))*I1
I3  =  (R/(r2  +  R))*I1
 #The  network  is  redrawn  with  the  100/_0°  V  source  removed  as  shown  in  Figure  32.9
 #Currents  I4,  I5  and  I6  are  labelled  as  shown  in  Figure  32.9.
I4  =  V2/(r2  +  r1*R/(r1  +  R))
I5  =  (r1/(r1  +  R))*I4
I6  =  (R/(r1  +  R))*I4
 #Figure  32.10  shows  Figure  32.9  superimposed  on  Figure  32.8,  giving  the  currents  shown.
 #Current  in  the  20  ohm  load,
I20  =  I2  +  I5
 #Current  in  the  100/_0°  V  source
IV1  =  I1  -  I6
 #Current  in  the  50/_90°  V  source
IV2  =  I4  -  I3


#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)current  in  the  20  ohm  load  is  ",round(I20.real,2),"  +  (",round(I20.imag,2),")i  A"
print  "\n  (b)Current  in  the  100/_0deg  V  source  is  ",round(IV1.real,2),"  +  (",round(IV1.imag,2),")i  A"
print  "\n  (b)Current  in  the  50/_90deg  V  source  is  ",round(IV2.real,2),"  +  (",round(IV2.imag,2),")i  A"


  Result  



  (a)current  in  the  20  ohm  load  is   1.05   +  ( 1.32 )i  A

  (b)Current  in  the  100/_0deg  V  source  is   3.16   +  ( -1.05 )i  A

  (b)Current  in  the  50/_90deg  V  source  is   -2.11   +  ( 2.37 )i  A

Example 2, page no. 566

In [2]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
V1  =  12;#  in  volts
V2  =  20;#  in  volts
R1  =  5;#  in  ohm
R2  =  4;#  in  ohm
R3  =  2.5;#  in  ohm
R4  =  6;#  in  ohm
R5  =  2;#  in  ohm

#calculation:
 #Removing  the  20  V  source  gives  the  network  shown  in  Figure  32.12.
 #Currents  I1  and  I2  are  shown  labelled  in  Figure  32.12
Re1  =  (R4*R5/(R4  +  R5))  +  R3
Re2  =  Re1*R2/(Re1    +  R2)  +  R1
I1  =  V1/Re2
I2  =  (R2/(Re1  +  R2))*I1
 #Removing  the  12  V  source  from  the  original  network  gives  the  network  shown  in  Figure  32.14.
 #Currents  I3,  I4  and  I5  are  shown  labelled  in  Figure  32.14.
Re3  =  (R1*R2/(R1  +  R2))  +  R3
Re4  =  Re3*R4/(Re3  +  R4)  +  R5
I3  =  V2/Re4
I4  =  (R4/(Re3  +  R4))*I3
I5  =  (R1/(R1  +  R2))*I4
 #Superimposing  Figure  32.14  on  Figure  32.12  shows  that  the  current  flowing  in  the  4  ohm  resistor  is  given  by
Ir4  =  I5  -  I2


#Results
print  "\n\n  Result  \n\n"
print  "\ncurrent  in  the  4  ohm  resistor  of  the  network  is  ",round(Ir4,2),"  A"


  Result  



current  in  the  4  ohm  resistor  of  the  network  is   0.48   A

Example 3, page no. 567

In [1]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
rv1  =  30;#  in  volts
rv2  =  30;#  in  volts
thetav1  =  45;#  in  degrees
thetav2  =  -45;#  in  degrees
R1  =  4;#  in  ohm
R2  =  4;#  in  ohm
R3  =  1j*3;#  in  ohm
R4  =  -1j*10;#  in  ohm

 #calculation:
 #voltage
V1  =  rv1*math.cos(thetav1*math.pi/180)  +  1j*rv1*math.sin(thetav1*math.pi/180)
V2  =  rv2*math.cos(thetav2*math.pi/180)  +  1j*rv2*math.sin(thetav2*math.pi/180)
 #The  network  is  redrawn  with  V2  removed,  as  shown  in  Figure  32.17.
 #Current  I1  and  I2  are  shown  in  Figure  32.17.  From  Figure  32.17,
Re1  =  R4*(R2  +  R3)/(R4  +  R3  +  R2)
Re2  =  Re1  +  R1
 #current
I1  =  V1/Re2
I2  =  (R4/(R2  +  R3  +  R4))*I1
 #The  original  network  is  redrawn  with  V1  removed,  as  shown  in  Figure  32.18
 #Currents  I3  and  I4  are  shown  in  Figure  32.18.  From  Figure  32.18,
Re3  =  R1*(R2  +  R3)/(R1  +  R3  +  R2)
Re4  =  Re3  +  R4
I3  =  V2/Re4
I4  =  (R1/(R2  +  R3  +  R1))*I3
 #If  the  network  of  Figure  32.18  is  superimposed  on  the  network  of  Figure  32.17,  
    #it  can  be  seen  that  the  current  in  the  (4+i3)  ohm  impedance  is  given  by
Ir4i3  =  I2  -  I4


#Results
print  "\n\n  Result  \n\n"
print  "current  in  (4  +  i3)  ohm  impedance  of  the  network  is  ",round(Ir4i3.real,2),"  +  (",round(  Ir4i3.imag,2),")i  A"


  Result  


current  in  (4  +  i3)  ohm  impedance  of  the  network  is   2.15   +  ( 0.42 )i  A

Example 4, page no. 568

In [5]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
E1  =  5  +  0j;#  in  volts
E2  =  2  +  4j;#  in  volts
Z1  =  3  +  4j;#  in  ohm
Z2  =  2  -  5j;#  in  ohm
Z3  =  6  +  8j;#  in  ohm

#calculation:
 #The  original  network  is  redrawn  with  E2  removed,  as  shown  in  Figure  32.20.
 #Currents  I1,  I2  and  I3  are  labelled  as  shown  in  Figure  32.20.
Ze1  =  Z3*Z2/(Z3  +  Z2)
Ze2  =  Ze1  +  Z1
 #current
I1  =  E1/Ze2
I2  =  (Z2/(Z3  +  Z2))*I1
I3  =  (Z3/(Z3  +  Z2))*I1
 #The  original  network  is  redrawn  with  E1  removed,  as  shown  in  Figure  32.22
 #Currents  I4,  I5  and  I6  are  shown  labelled  in  Figure  32.22  
    #with  I4  flowing  away  from  the  positive  terminal  of  the  E2  source.
Ze3  =  Z3*Z1/(Z3  +  Z1)
Ze4  =  Ze3  +  Z2
I4  =  E2/Ze4
I5  =  (Z1/(Z3  +  Z1))*I4
I6  =  (Z3/(Z3  +  Z1))*I4
 #If  the  network  of  Figure  32.18  is  superimposed  on  the  network  of  Figure  32.17,  
    #it  can  be  seen  that  the  current  in  the  (4+i3)  ohm  impedance  is  given  by
i1  =  I1  +  I6
i2  =  I3  +  I4
i3  =  I2  -  I5
 #magnitude
i1mag  =  abs(i1)
i2mag  =  abs(i2)
E1mag  =  abs(E1)
E2mag  =  abs(E2)
 #phase
phi1  =  cmath.phase(complex(i1.real,i1.imag))
phi2  =  cmath.phase(complex(i2.real,i2.imag))
 #voltage  across  the(6  +  i8)  ohm  impedance
V6i8  =  i3*Z3
V6i8m  =  abs(V6i8)
 #power
P  =  (E1mag*i1mag*math.cos(phi1))  +  (E2mag*i2mag*math.cos(phi2  -  cmath.phase(complex(E2.real,E2.imag))))


#Results
print  "\n\n  Result  \n\n"
print  "\n(a)currents  are: \n ",round(i1.real,2),"  +  (",round(  i1.imag,2),")i  A, \n ",round(i2.real,2),"  +  (",round(i2.imag,2),")i  A \n and  ",round(i3.real,2),"  +  (",round(i3.imag,2),")i  A"
print  "\n(b)current  in  the  (6  +  i8)  ohm  resistor  of  the  network  is  ",round(V6i8m,2),"  V"
print  "\n(c)the  total  active  power  delivered  to  the  network  is  ",round(P,2),"  W"


  Result  



(a)currents  are: 
  0.57   +  ( 0.62 )i  A, 
  0.56   +  ( 1.33 )i  A 
 and   0.01   +  ( -0.71 )i  A

(b)current  in  the  (6  +  i8)  ohm  resistor  of  the  network  is   7.09   V

(c)the  total  active  power  delivered  to  the  network  is   9.29   W

Example 5, page no. 571

In [1]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
rv1  =  50;#  in  volts
rv2  =  30;#  in  volts
thetav1  =  0;#  in  degrees
thetav2  =  90;#  in  degrees
R1  =  20;#  in  ohm
R2  =  5;#  in  ohm
R3  =  -1j*3;#  in  ohm
R4  =  8;#  in  ohm
R5  =  8;#  in  ohm

#calculation:
 #voltage
V1  =  rv1*math.cos(thetav1*math.pi/180)  +  1j*rv1*math.sin(thetav1*math.pi/180)
V2  =  rv2*math.cos(thetav2*math.pi/180)  +  1j*rv2*math.sin(thetav2*math.pi/180)
 #The  network  is  redrawn  with  the  V2  source  removed,  as  shown  in  Figure  32.26.
 #Currents  I1  to  I5  are  shown  labelled  in  Figure  32.26.  
 #current
Re1  =  R4*R5/(R5  +  R4)  +  R3
Re2  =  Re1*R2/(R2  +  Re1)
I1  =  V1/(Re2  +  R1)
I2  =  (Re1/(R2  +  Re1))*I1
I3  =  (R2/(Re1  +  R2))*I1
I4  =  (R4/(R4  +  R5))*I3
I5  =  I3  -  I4
 #The  original  network  is  redrawn  with  the  V1  source  removed,  as  shown  in  Figure  32.27.
 #Currents  I6  to  I10  are  shown  labelled  in  Figure  32.27
Re3  =  R1*R2/(R1  +  R2)
Re4  =  Re3  +  R3
Re5  =  Re4*R4/(Re4  +  R4)
Re6  =  Re5    +  R5
I6  =  V2/Re6
I7  =  (Re4/(Re4  +  R4))*I6
I8  =  (R4/(Re4  +  R4))*I6
I9  =  (R1/(R1  +  R2))*I8
I10  =  (R2/(R1  +  R2))*I8
 #current  flowing  in  the  capacitor  is  given  by
Ic  =  I3  -  I8
 #magnitude  of  the  current  in  the  capacitor
Icmag  =  abs(Ic)

i1  =  I2  +  I9
i1mag  =  abs(i1)
 #magnitude  of  the  p.d.  across  the  5  ohm  resistance  is  given  by
Vr5m  =  i1mag*R2
 #Active  power  dissipated  in  the  20  ohm  resistance  is  given  by
i2  =  I1  -  I10
i2mag  =  abs(i2)
phii2  =  cmath.phase(complex(i2.real,i2.imag))
Pr20  =  R1*(i2mag)**2
 #Active  power  developed  by  the  V1
P1  =  rv1*i2mag*math.cos(phii2)
 #Active  power  developed  by  V2  source
i3  =  I6  -  I5
i3mag  =  abs(i3)
phii3  =  cmath.phase(complex(i3.real,i3.imag))
P2  =  rv2*i3mag*math.cos(phii3  -  (thetav2*math.pi/180))
 #Total  power  developed
P  =  P1  +  P2


#Results
print  "\n\n  Result  \n\n"
print  "\n(a)the  magnitude  of  the  current  flowing  in  the  capacitor  is  ",round(Icmag,2),"  A"
print  "\n(b)  the  p.d.  across  the  5  ohm  resistance  is  ",round(Vr5m,2),"  V"
print  "\n(c)the  active  power  dissipated  in  the  20  ohm  resistance  is  ",round(Pr20,0),"  W"
print  "\n(d)the  total  active  power  taken  from  the  supply  is  ",round(P,1),"  W"


  Result  



(a)the  magnitude  of  the  current  flowing  in  the  capacitor  is   2.11   A

(b)  the  p.d.  across  the  5  ohm  resistance  is   5.85   V

(c)the  active  power  dissipated  in  the  20  ohm  resistance  is   111.0   W

(d)the  total  active  power  taken  from  the  supply  is   191.9   W