Chapter 30: Introduction to network analysis

Example 1, page no. 536

In [4]:
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
R2  =  20;#  in  ohm
R3  =  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  branch  currents  and  their  directions  are  labelled  as  shown  in  Figure  30.4
 #Two  loops  are  chosen.  loop  ABEF,  and  loop  BCDE
 #using  kirchoff  rule  in  3  loops
 #two  eqns  obtained
 #(R1  +  R2)*I1  +  R2*I2  =  V1
 #R2*I1  +  (R2  +  R3)*I2  =  V2
I1  =  (3*V1  -  2*V2)/(3*(R1  +  R2)  -  2*(R2))
I2  =  (V2  -  R2*I1)/(R2  +  R3)
I  =  I1  +  I2


#Results
print  "\n\n  Result  \n\n"
print  "\n  current,  I1  is  ",round(I1.real,2),"  +  (",round( I1.imag,2),")i  A,  \n current,  I2  is    ",round(I2.real,2),"  +  (",round(  I2.imag,2),")i  A and "
print   " total  current,  I  is  ",round(I.real,2),"  +  (",round( I.imag,2),")i  A"


  Result  



  current,  I1  is   3.16   +  ( -1.05 )i  A,  
 current,  I2  is     -2.11   +  ( 2.37 )i  A and 
 total  current,  I  is   1.05   +  ( 1.32 )i  A

Example 2, page no. 537

In [4]:
from __future__ import division
import math
import numpy
#initializing  the  variables:
V  =  8;#  in  volts
R1  =  1;#  in  ohm
R2  =  2;#  in  ohm
R3  =  3;#  in  ohm
R4  =  4;#  in  ohm
R5  =  5;#  in  ohm
R6  =  6;#  in  ohm

#calculation:
 #Currents  and  their  directions  are  assigned  as  shown  in  Figure  30.6.
 #Three  loops  are  chosen  since  three  unknown  currents  are  required.  The  choice  of  loop  directions  is  arbitrary.
    #loop  ABCDE,  and  loop  EDGF  and  loop  DCHG
 #using  kirchoff  rule  in  3  loops
 #three  eqns  obtained
 #R5*I1  +  (R6  +  R4)*I2  -  R4*I3  =  V
 #-1*R1*I1  +  (R6  +  R1)*I2  +  R2*I3  =  0
 #  R3*I1  -  (R3  +  R4)*I2  +  (R2  +  R3  +  R4)*I3  =  0
#using  determinants
d1  =  [[V, (R6  +  R4), -1*R4],[0, (R6  +  R1),  R2], [0,  (-1*(R3  +  R4)), (R2  +  R3  +  R4)]]
D1  =  numpy.linalg.det(d1)
d2  =  [[R5, V,  -1*R4],[-1*R1,  0,  R2],[ R3,  0,  (R2  +  R3  +  R4)]]
D2  =  numpy.linalg.det(d2)
d3  =  [[R5,  (R6  +  R4),  V],[-1*R1,  (R6  +  R1),  0],[ R3,  (-1*(R3  +  R4)),  0]]
D3  =  numpy.linalg.det(d3)
d  =  [[R5,  (R6  +  R4),  -1*R4],[-1*R1,  (R6  +  R1),  R2],[ R3,  (-1*(R3  +  R4)),  (R2  +  R3  +  R4)]]
D  =  numpy.linalg.det(d)
I1  =  D1/D
I2  =  D2/D
I3  =  D3/D  
#Current  in  the  2  ohm  resistance
I  =  I1  -  I2  +  I3
#power  dissipated  in  the  3  ohm  resistance
P3  =  R3*I**2


#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)current  through  2  ohm  resistor  is  ",round(I2,3),"  A"
print  "\n  (b)power  dissipated  in  the  3  ohm  resistor  is  ",round(P3,3),"W"


  Result  



  (a)current  through  2  ohm  resistor  is   0.203   A

  (b)power  dissipated  in  the  3  ohm  resistor  is   1.267 W

Example 3, page no. 539

In [3]:
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:
 #Currents  I1  and  I2  with  their  directions  are  shown  in  Figure  30.8.
 #Two  loops  are  chosen  with  their  directions  both  clockwise.loop  ABEF  and  loop  BCDE,
 #using  kirchoff  rule  in  3  loops
 #two  eqns  obtained
 #(Z1  +  Z3)*I1  -  Z3*I2  =  E1
 #-1*Z3*I1  +  (Z2  +  Z3)*I2  =  E2
I1  =  ((Z2  +  Z3)*E1  +  Z3*E2)/((Z2  +  Z3)*(Z1  +  Z3)  -  Z3*Z3)
I2  =  -1*(E1  -  (Z1  +  Z3)*I1)/Z3
I3  =  I1  -  I2


#Results
print  "\n\n  Result  \n\n"
print  "current,  I1  is  ",round(I1.real,2),"  +  (",round(  I1.imag,2),")i  A,\n current,  I2  is    ",round(I2.real,2),"  +  (",round( I2.imag,2),")i  A and "
print   " current,  I3  is  ",round(I3.real,2),"  +  (",round(  I3.imag,2),")i  A"


  Result  


current,  I1  is   0.57   +  ( 0.62 )i  A,
 current,  I2  is     0.56   +  ( 1.33 )i  A and 
 current,  I3  is   0.01   +  ( -0.71 )i  A

Example 4, page no. 541

In [8]:
from __future__ import division
import math
import numpy
import cmath
#initializing  the  variables:
rv1  =  10;#  in  volts
rv2  =  12;#  in  volts
rv3  =  15;#  in  volts
thetav1  =  0;#  in  degrees
thetav2  =  0;#  in  degrees
thetav3  =  0;#  in  degrees
R1  =  4;#  in  ohm
R2  =  -5j;#  in  ohm
R3  =  8;#  in  ohm
R4  =  4;#  in  ohm
R5  =  3j;#  in  ohm

#calculation:
 #voltages
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)
V3  =  rv3*math.cos(thetav3*math.pi/180)  +  1j*rv3*math.sin(thetav3*math.pi/180)
 #Currents  I1,  I2  and  I3  with  their  directions  are  shown  in  Figure  30.10.
 #Three  loops  are  chosen.  The  choice  of  loop  directions  is  arbitrary.  loop  ABGH,  and  loopBCFG  and  loop  CDEF
Z4  =  R4  +  R5
 #using  kirchoff  rule  in  3  loops
 #three  eqns  obtained
 #R1*I1  +  R2*I2  =  V1  +  V2
 #-1*R3*I1  +  (R3  +  R2)*I2  +  R3*I3  =  V2  +  V3
 #  -1*R3*I1  +  R3*I2  +  (R3  +  Z4)*I3  =  V3
 #using  determinants
d1  =  [[(V1  +  V2),  R2,  0],[(V2  +  V3),  (R3  +  R2),  R3],[V3,  R3,  (R3  +  Z4)]]
D1  =  numpy.linalg.det(d1)
d2  =  [[R1,  (V1  +  V2),  0],[-1*R3,  (V2  +  V3),  R3],[-1*R3,  V3,  (R3  +  Z4)]]
D2  =  numpy.linalg.det(d2)
d3  =  [[R1,  R2,  (V1  +  V2)],[-1*R3,  (R3  +  R2),  (V2  +  V3)],[-1*R3,  R3,  V3]]
D3  =  numpy.linalg.det(d3)
d  =  [[R1,  R2,  0],[-1*R3,  (R3  +  R2),  R3],[-1*R3,  R3,  (R3  +  Z4)]]
D  =  numpy.linalg.det(d)
I1  =  D1/D
I2  =  D2/D
I3  =  D3/D  
I3mag  =  abs(I3)


#Results
print  "\n\n  Result  \n\n"
print  "\n  magnitude  of  the  current  through  (4  +  i3)ohm  impedance  is  ",round(I3mag,2),"  A"


  Result  



  magnitude  of  the  current  through  (4  +  i3)ohm  impedance  is   1.84   A