Chapter 31: Mesh-current and nodal analysis

Example 1, page no. 546

In [1]:
from __future__ import division
import math
import numpy
import cmath
#initializing  the  variables:
V1  =  4;#  in  volts
V2  =  5;#  in  volts
R1  =  3;#  in  ohm
R2  =  5;#  in  ohm
R3  =  4;#  in  ohm
R4  =  1;#  in  ohm
R5  =  6;#  in  ohm
R6  =  8;#  in  ohm

#calculation:
 #The  mesh  currents  I1,  I2  and  I3  are  shown  in  Figure  31.2.  Using  Kirchhoff’s  voltage  law  in  3  loops
 #three  eqns  obtained
 #(R1  +  R2)*I1  -  R2*I2  =  V1
 #-1*R2*I1  +  (R2  +  R3  +  R4  +  R5)*I2  -  R4*I3  =  0
 #  -1*R4*I2  +  (R4  +  R6)*I3  =  -1*V2
 #using  determinants
d1  =  [[V1,  -1*R2,  0],[0,  (R2  +  R3  +  R4  +  R5),  -1*R4],[-1*V2,  -1*R4,  (R4  +  R6)]]
D1  =  numpy.linalg.det(d1)
d2  =  [[(R1  +  R2),  V1,  0],[-1*R2,  0,  -1*R4],[0,  -1*V2,  (R4  +  R6)]]
D2  =  numpy.linalg.det(d2)
d3  =  [[(R1  +  R2),  -1*R2,  V1],[-1*R2,  (R2  +  R3  +  R4  +  R5),  0],[0,  -1*R4,  -1*V2]]
D3  =  numpy.linalg.det(d3)
d  =  [[(R1  +  R2),  -1*R2,  0],[-1*R2,  (R2  +  R3  +  R4  +  R5),  -1*R4],[0,  -1*R4,  (R4  +  R6)]]
D  =  numpy.linalg.det(d)
I1  =  D1/D
I2  =  D2/D
I3  =  D3/D  
IR2  =  I1  -  I2
IR4  =  I2  -  I3


#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)current  in  the  5  ohm  resistance  is  ",round(IR2,2),"  A"
print  "\n  (b)current  in  the  1  ohm  resistance  is  ",round(IR4,2),"  A"


  Result  



  (a)current  in  the  5  ohm  resistance  is   0.44   A

  (b)current  in  the  1  ohm  resistance  is   0.69   A

Example 2, page no. 547

In [1]:
from __future__ import division
import math
import numpy
import cmath
#initializing  the  variables:
rv  =  100;#  in  volts
thetav  =  0;#  in  degrees
R1  =  5;#  in  ohm
R2  =  -1j*4;#  in  ohm
R3  =  4;#  in  ohm
R4  =  3j;#  in  ohm

 #calculation:
 #voltages
V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)
 #Currents  I1,  I2  with  their  directions  are  shown  in  Figure  31.03.
 #Two  loops  are  chosen.  The  choice  of  loop  directions  is  arbitrary.
 #using  kirchoff  rule  in  2  loops
 #two  eqns  obtained
 #(R1  +  R2)*I1  -  R2*I2  =  V
 #-1*R2*I1  +  (R3  +  R2  +  R4)*I2  =  0
 #using  determinants
d1  =  [[V,  -1*R2],[0,  (R3  +  R2  +  R4)]]
D1  =  numpy.linalg.det(d1)
d2  =  [[(R1  +  R2),  V],[-1*R2,  0]]
D2  =  numpy.linalg.det(d2)
d  =  [[(R1  +  R2),  -1*R2],[-1*R2,  (R3  +  R2  +  R4)]]
D  =  numpy.linalg.det(d)
I1  =  D1/D
I2  =  D2/D
I1mag  =  abs(I1)
 #Current  flowing  in  capacitor
Ic  =  I1  -  I2
 #Source  power  P
phi  =  cmath.phase(complex(I1.real,I1.imag))
P  =  V*I1mag*math.cos(phi)
Icmag  =  abs(Ic)


#Results
print  "\n\n  Result  \n\n"
print  "(a)current,I1 is  ",round(I1.real,2),"  +  (",round(  I1.imag,2),")i  A,  current,  I2  is",round(I2.real,2),"  +  (",round(I2.imag,2),")i  A"
print  "(b)current  in  the  capacitor  is  ",round(Icmag,2),"  A"
print  "(c)Source  power  P  is  ",round(abs(P),2),"  W"


  Result  


(a)current,I1 is   10.17   +  ( 3.55 )i  A,  current,  I2  is 5.73   +  ( -8.74 )i  A
(b)current  in  the  capacitor  is   13.06   A
(c)Source  power  P  is   1017.06   W

Example 3, page no. 548

In [2]:
from __future__ import division
import math
import numpy
import cmath
#initializing  the  variables:
rv1  =  415;#  in  volts
rv2  =  415;#  in  volts
thetav1  =  120;#  in  degrees
thetav2  =  0;#  in  degrees
R  =  3  +  4j;#  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)
 #Two  mesh  currents  I1  and  I2  are  chosen  as  shown  in  Figure  31.4.
 #Two  loops  are  chosen.  The  choice  of  loop  directions  is  arbitrary.
 #using  kirchoff  rule  in  2  loops
 #two  eqns  obtained
 #2*R*I1  -  R*I2  =  V1
 #-1*R*I1  +  2*R*I2  =  V2
 #using  determinants
d1  =  [[V1,  -1*R],[V2, 2*R]]
D1  =  numpy.linalg.det(d1)
d2  =  [[2*R, V1],[-1*R, V2]]
D2  =  numpy.linalg.det(d2)
d  =  [[2*R, -1*R],[-1*R,  2*R]]
D  =  numpy.linalg.det(d)
I1  =  D1/D
I2  =  D2/D
I1mag  =  abs(I1)
 #line  current  IR
IR  =  I1
 #line  current  IB
IB  =  -1*I2
 #line  current  IY
IY  =  I2  -  I1


#Results
print  "\n\n  Result  \n\n"
print  "current, IR is",round(IR.real,2),"  +  (",round(  IR.imag,2),")i  A,  current,  IB  is",round(IB.real,2),"  +  (",round(  IB.imag,2),")i  A  and  current,  IY  is  ",round(IY.real,2),"  +  (",round(IY.imag,2),")i  A"


  Result  


current, IR is 38.34   +  ( 28.75 )i  A,  current,  IB  is -44.07   +  ( 18.82 )i  A  and  current,  IY  is   5.73   +  ( -47.58 )i  A

Example 4, page no. 551

In [4]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
ri  =  20;#  in  amperes
thetai  =  0;#  in  degrees
R1  =  10;#  in  ohm
R2  =  3j;#  in  ohm
R3  =  4;#  in  ohm
R4  =  16;#  in  ohm

#calculation:
 #current
I  =  ri*math.cos(thetai*math.pi/180)  +  1j*ri*math.sin(thetai*math.pi/180)
 #Figure  31.8  contains  two  principal  nodes  (at  1  and  B)  and  thus  only  one  nodal  equation  is  required. 
    #B  is  taken  as  the  reference  node  and  the  equation  for  node  1  is  obtained  as  follows. 
    #Applying  Kirchhoff’s  current  law  to  node  1  gives:
 #IX  +  IY  =  I
V1  =  I/((1/R4)  +(1/(R2    +R3)))
IY  =  V1/(R2  +  R3)
VAB  =  IY*R3


#Results
print  "\n\n  Result  \n\n"
print  "\n  voltage  VAB  is  ",round(VAB.real,2),"  +  (",round(  VAB.imag,2),")i  V"


  Result  



  voltage  VAB  is   62.59   +  ( -9.39 )i  V

Example 5, page no. 552

In [1]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
rv1  =  8;#  in  volts
rv2  =  8;#  in  volts
thetav1  =  0;#  in  degrees
thetav2  =  90;#  in  degrees
R1  =  5;#  in  ohm
R2  =  6j;#  in  ohm
R3  =  4;#  in  ohm
R4  =  3;#  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)
 #The  circuit  contains  no  principal  nodes.  
    #However,  if  point  Y  is  chosen  as  the  reference  node  then  an  equation 
    #may  be  written  for  node  X  assuming  that  current  leaves  point  X  by  both  branches
VX  =  ((V1/(R1  +  R3)  +  V2/(R2  +  R4))/(1/(R1  +  R3)  +  1/(R2  +  R4)))
VXY  =  VX



#Results
print  "\n\n  Result  \n\n"
print  "\n  voltage  VXY  = ",round(abs(VXY),2),"/_",round(cmath.phase(complex(VXY.real, VXY.imag))*180/math.pi,2),"deg V"


  Result  



  voltage  VXY  =  9.12 /_ 52.13 deg V

Example 6, page no. 553

In [6]:
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:
 #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)
 #There  are  only  two  principal  nodes  in  Figure  31.10  so  only  one  nodal  equation  is  required.  
    #Node  2  is  taken  as  the  reference  node.
 #The  equation  at  node  1  is  I1  +  I2  +  I3  =  0
Vn1  =  ((V1/R1  +  V2/R3)/(1/R1  +  1/R2  +  1/R3))
I1  =  (Vn1  -  V1)/R1
I2  =  Vn1/R2
I3  =  (Vn1  -  V2)/R3


#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 \n and  current,  I3  is  ",round(I3.real,2),"  +  (",round(I3.imag,2),")i  A"


  Result  



  current,  I1  is   -3.16   +  ( 1.05 )i  A, 
 current,  I2  is     1.05   +  ( 1.32 )i  A 
 and  current,  I3  is   2.11   +  ( -2.37 )i  A

Example 7, page no. 554

In [3]:
from __future__ import division
import math
import numpy
import cmath
#initializing  the  variables:
rv1  =  25;#  in  volts
rv2  =  25;#  in  volts
thetav1  =  0;#  in  degrees
thetav2  =  90;#  in  degrees
R1  =  2;#  in  ohm
R2  =  -1j*4;#  in  ohm
R3  =  5;#  in  ohm
R4  =  4j;#  in  ohm
R5  =  2.5;#  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)
 #The  equation  at  node  1
 #Vn1*(1/R1  +  1/R2  +  1/R3)  -  Vn2/R3  =  V1/R1
 #The  equation  at  node  2
 #Vn1*(-1/R3)  +  Vn2*(1/R4  +  1/R5  +  1/R3)  =  V2/R5
 #using  determinants
d1  =  [[V1/R1,  -1/R3],[V2/R5,  (1/R4  +  1/R5  +  1/R3)]]
D1  =  numpy.linalg.det(d1)
d2  =  [[(1/R1  +  1/R2  +  1/R3),  V1/R1],[-1/R3,  V2/R5]]
D2  =  numpy.linalg.det(d2)
d  =  [[(1/R1  +  1/R2  +  1/R3),  -1/R3],[-1/R3,  (1/R4  +  1/R5  +  1/R3)]]
D  =  numpy.linalg.det(d)
Vn1  =  D1/D
Vn2  =  D2/D
 #current  in  the  j4  ohm  inductance  is  given  by:
I4  =  Vn2/R4
 #current  in  the  5  ohm  resistance  is  given  by:
I3  =  (Vn1  -  Vn2)/R3
 #active  power  dissipated  in  the  2.5  ohm  resistor  is  given  by
P5  =  R5*((Vn2  -  V2)/R5)**2
 #magnitude  of  the  active  power  dissipated
P5mag  =  abs(P5)

#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)  the  voltage  at  nodes  1  and  2  are ",round(abs(Vn1),1),"/_",round(cmath.phase(complex(Vn1.real, Vn1.imag))*180/math.pi,2),"deg  V  and ",round(abs(Vn2),1),"/_",round(cmath.phase(complex(Vn2.real, Vn2.imag))*180/math.pi,1),"deg  V"
print  "\n  (b)the  current  in  the  j4  ohm inductance = ",round(abs(I4),2),"/_",round(cmath.phase(complex(I4.real, I4.imag))*180/math.pi,2),"deg  A"
print  "\n  (c)the  current  in  the  5  ohm resistance  = ",round(abs(I3),2),"/_",round(cmath.phase(complex(I3.real, I3.imag))*180/math.pi,2),"deg  A"
print  "\n  (d)  magnitude  of  the  active  power  dissipated  in  the  2.5  ohm  resistance  is  ",round(P5mag,2),"  W"


  Result  



  (a)  the  voltage  at  nodes  1  and  2  are  17.1 /_ -5.3 deg  V  and  15.8 /_ 93.2 deg  V

  (b)the  current  in  the  j4  ohm inductance =  3.95 /_ 3.23 deg  A

  (c)the  current  in  the  5  ohm resistance  =  4.99 /_ -44.06 deg  A

  (d)  magnitude  of  the  active  power  dissipated  in  the  2.5  ohm  resistance  is   34.4   W

Example 8, page no. 556

In [8]:
from __future__ import division
import math
import numpy
import cmath
#initializing  the  variables:
ri  =  25;#  in  amperes
thetai  =  0;#  in  degrees
R1  =  4;#  in  ohm
R2  =  3j;#  in  ohm
R3  =  5;#  in  ohm
R4  =  10j;#  in  ohm
R5  =  20j;#  in  ohm

 #calculation:
 #current
I  =  ri*math.cos(thetai*math.pi/180)  +  1j*ri*math.sin(thetai*math.pi/180)
 #Node  3  is  taken  as  the  reference  node.
 #At  node  1,
 #V1*(1/(R1  +  R2)  +  1/R3)  -  V2/R3  =  I
 #The  equation  at  node  2
 #V1*(-1/R3)  +  V2*(1/R4  +  1/R5  +  1/R3)  =  0
 #using  determinants
d1  =  [[I,  -1/R3],[0 , (1/R4  +  1/R5  +  1/R3)]]
D1  =  numpy.linalg.det(d1)
d2  =  [[(1/(R1  +  R2)  +  1/R3),  I],[-1/R3,  0]]
D2  =  numpy.linalg.det(d2)
d  =  [[(1/(R1  +  R2)  +  1/R3),  -1/R3],[-1/R3,  (1/R4  +  1/R5  +  1/R3)]]
D  =  numpy.linalg.det(d)
V1  =  D1/D
V2  =  D2/D
 #the  voltage  between  point  X  and  node  3  is
VX  =  V1*R2/(R1  +  R2)
 #Thus  the  voltage
VY  =  V2
VXY  =  VX  -  VY


#Results
print  "\n\n  Result  \n\n"
print  "\n  voltage  VXY  is  ",round(VXY.real,2),"  +  (",round(  VXY.imag,2),")i  V"


  Result  



  voltage  VXY  is   -16.16   +  ( -15.05 )i  V

Example 9, page no. 557

In [3]:
from __future__ import division
import math
import numpy
import cmath
#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:
 #In  Figure  31.13,  the  reference  node  is  shown  at  point  A.
 #At  node  1,
 #V1*(1/R1  +  1/R6  +  1/R5)  -  V2/R1  -  V3/R5  =  V/R5
 #The  equation  at  node  2
 #V1*(-1/R1)  +  V2*(1/R2  +  1/R1  +  1/R3)  -  V3/R3  =  0
 #At  node  3
 #  -  V1/R5  -  V2/R3  +  V3*(1/R4  +  1/R3  +  1/R5)  =  -1*V/R5
#using  determinants
d1  =  [[V/R5,  -1/R1,  -1/R5],[0, (1/R2  +  1/R1  +  1/R3),  -1/R3],[-1*V/R5, -1/R3,  (1/R4  +  1/R3  +  1/R5)]]
D1  =  numpy.linalg.det(d1)
d2  =  [[(1/R1  +  1/R6  +  1/R5),  V/R5,  -1/R5],[-1/R1,  0,  -1/R3],[-1/R5,  -1*V/R5,  (1/R4  +  1/R3  +  1/R5)]]
D2  =  numpy.linalg.det(d2)
d3  =  [[(1/R1  +  1/R6  +  1/R5),  -1/R1,  V/R5],[-1/R1,  (1/R2  +  1/R1  +  1/R3),  0],[-1/R5,  -1/R3,  -1*V/R5]]
D3  =  numpy.linalg.det(d3)
d  =[[(1/R1  +  1/R6  +  1/R5),  -1/R1,  -1/R5],[-1/R1,  (1/R2  +  1/R1  +  1/R3),  -1/R3],[-1/R5,  -1/R3,  (1/R4  +  1/R3  +  1/R5)]]
D  =  numpy.linalg.det(d)
Vn1  =  D1/D
Vn2  =  D2/D
Vn3  =  D3/D  
 #the  current  in  the  2  ohm  resistor
I2  =  Vn2/R2
 #power  dissipated  in  the  3  ohm  resistance
P3  =  R3*((Vn2  -  Vn3)/R3)**2


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


  Result  



  (a)current  through  2  ohm  resistor  is   0.19   A

  (b)power  dissipated  in  the  3  ohm  resistor  is   1.27   W