# Chapter 33: Thevenin’s and Norton’s theorems

### Example 1, page no. 578

In [1]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
rv  =  200;#  in  volts
thetav  =  0;#  in  degrees
R1  =  5000;#  in  ohm
R2  =  20000;#  in  ohm
R3  =  -1j*120000;#  in  ohm
R4  =  150000;#  in  ohm

#calculation:
#voltage
V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)
#Initially  the  (150-i120)kohm  impedance  is  removed  from  the    circuit  as  shown  in  Figure  33.13.
#Note  that,  to  find  the  current  in  the  capacitor,
#only  the  capacitor  need  have  been  initially  removed  from  the  circuit.
#However,  removing  each  of  the  components  from  the  branch  through
#which  the  current  is  required  will  often  result  in  a  simpler  solution.
#From  Figure  33.13,
#current,  I1
I1  =  V/(R1  +  R2)
#The  open-circuit  e.m.f.  E  is  equal  to  the  p.d.  across  the  20  kohm  resistor,  i.e.
E  =  I1*R2
#Removing  the  V1  source  gives  the  network  shown  in  Figure  33.14.
#The  impedance,  z,  looking  in at  the  open-circuited  terminals  is  given  by
z  =  R1*R2/(R1  +  R2)
#The  Th´evenin  equivalent  circuit  is  shown  in  Figure  33.15,  where  current  iL  is  given  by
ZL  =  R3  +  R4
IL  =  E/(ZL  +  z)
ILmag  =  abs(IL)
#current  flowing  in  the  capacitor
Ic  =  ILmag
#P.d.  across  the  150  kohm  resistor,
Vr150  =  ILmag*R4

#Results
print  "\n\n  Result  \n\n"
print  "\n(a)the  current  flowing  in  the  capacitor  is  ",round(Ic*1000,2)," mA"
print  "\n(b)  the  p.d.  across  the  150  ohm  resistance  is  ",round(Vr150,2),"  V"


Result

(a)the  current  flowing  in  the  capacitor  is   0.82  mA

(b)  the  p.d.  across  the  150  ohm  resistance  is   122.93   V


### Example 2, page no. 579

In [2]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
V1  =  20;#  in  volts
V2  =  10;#  in  volts
R1  =  2;#  in  ohm
R2  =  1.5;#  in  ohm
L  =  235E-6;#  in  Henry
R4  =  3;#  in  ohm
f  =  2000;#  in  Hz

#calculation:
#The  impedance  through  which  current  I  is  flowing  is  initially  removed  from  the  network,  as  shown  in  Figure  33.17.
#From  Figure  33.17,
#current,  I1
I1  =  (V1  -  V2)/(R1  +  R4)
#the  open  circuit  e.m.f.  E
E  =  V1  -  I1*R1
#When  the  sources  of  e.m.f.  are  removed  from  the  circuit,
#the  impedance,  z,  ‘looking  in’  at  the  break  is  given  by
z  =  R1*R4/(R1  +  R4)
#The  Th´evenin  equivalent  circuit  is  shown  in  Figure  33.18,  where  inductive  reactance,
XL  =  2*math.pi*f*L
R3  =  1j*XL
#Hence  current
I  =  E/(R2  +  R3  +  z)

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


Result

the  current  I  is   2.7   +  ( -2.95 )i  A


### Example 3, page no. 580

In [3]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
rv  =  50;#  in  volts
thetav  =  0;#  in  degrees
R1  =  -1j*400;#  in  ohm
R2  =  300;#  in  ohm
R3  =  144j;#  in  ohm
R4  =  48;#  in  ohm

#calculation:
#voltage
V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)
#The  R3  and  R4  impedance  is  initially  removed  from  the  network  as  shown  in  Figure  33.20.
#From  Figure  33.20,
#current,  I
i  =  V/(R1  +  R2)
#the  open  circuit  e.m.f.  E
E  =  i*R2
#When  the  V  is  removed  from  the  circuit,  the  impedance,  z,  ‘looking  in’  at  the  break  is  given  by
z  =  R1*R2/(R1  +  R2)
#The  Th´evenin  equivalent  circuit  is  shown  in  Figure  33.21  connected  to  R#  and  R4,
#Hence  current
I  =  E/(R4  +  R3  +  z)
Imag  =  abs(I)
#the  power  dissipated  in  the  48  ohm  resistor
Pr48  =  R4*Imag**2

#Results
print  "\n\n  Result  \n\n"
print  "\n  the  power  dissipated  in  the  48  ohm  resistor  is  ",round(Pr48,2),"  W"


Result

the  power  dissipated  in  the  48  ohm  resistor  is   0.75   W

### Example 4, page no. 581

In [4]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
V  =  100;#  in  volts
R1  =  5;#  in  ohm
R2  =  20;#  in  ohm
R3  =  46;#  in  ohm
R4  =  50;#  in  ohm
R5  =  15;#  in  ohm
R6  =  60;#  in  ohm
R7  =  16;#  in  ohm
R8  =  80;#  in  ohm

#calculation:
#One  method  of  analysing  a  multi-branch  network  as  shown  in  Figure  33.22
#is  to  use  Th´evenin’s  theorem  on  one  part  of  the  network  at  a  time.
#For  example,  the  part  of  the  circuit  to  the  left  of  AA  may  be  reduced  to  a  Th´evenin  equivalent  circuit.
#From  Figure  33.23,
E1  =  (R2/(R1  +  R2))*V
z1  =  R1*R2/(R1  +  R2)
#Thus  the  network  of  Figure  33.22  reduces  to  that  of  Figure  33.24.
#The  part  of  the  network  shown  in  Figure  33.24  to  the  left  of  BB  may  be  reduced
#to  a  Th´evenin  equivalent  circuit,  where
E2  =  (R4/(R3  +  R4  +  z1))*E1
z2  =  R4*(z1  +  R3)/(R4  +  z1  +  R3)
#Thus  the  original  network  reduces  to  that  shown  in  Figure  33.25.
#The  part  of  the  network  shown  in  Figure  33.25  to  the  left  of  CC  may  be  reduced
#to  a  Th´evenin  equivalent  circuit,  where
E3  =  (R6/(R5  +  R6  +  z2))*E2
z3  =  R6*(z2  +  R5)/(R5  +  z2  +  R6)
#Thus  the  original  network  reduces  to  that  of  Figure  33.26,
#from  which  the  current  in  the  80  ohm  resistor  is  given  by
I  =  E3/(z3  +  R7  +  R8)

#Results
print  "\n\n  Result  \n\n"
print  "\n  the  current  flowing  in  the  80  ohm  resistor  is  ",round(I,2),"  A"


Result

the  current  flowing  in  the  80  ohm  resistor  is   0.2   A

### Example 5, page no. 582

In [5]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
rv  =  24;#  in  volts
thetav  =  0;#  in  degrees
R1  =  -1j*3;#  in  ohm
R2  =  4;#  in  ohm
R3  =  3j;#  in  ohm

#calculation:
#voltage
V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)
#Current  I1  shown  in  Figure  33.27  is  given  by
I1  =  V/(R1  +  R2  +  R3)
#The  Th´evenin  equivalent  voltage,  i.e.,  the  open-circuit  voltage  across  terminals  AB,  is  given  by
E  =  I1*(R2  +  R3)
#When  the  voltage  source  is  removed,  the  impedance  z  ‘looking  in’  at  AB  is  given  by
z  =  (R2  +  R3)*R1/(R1  +  R2  +  R3)
#Thus  the  Th´evenin  equivalent  circuit  is  as  shown  in  Figure  33.28.
#when  (3.75  +  i11)  ohm  impedance  connected  across  terminals  AB,
#the  current  I  flowing  in  the  impedance  is  given  by
R  =  3.75  +  11j;#  in  ohms
I  =  E/(R  +  z)
Imag  =  abs(I)
#the  p.d.  across  the(  3.75  +  i11)ohm  impedance.
VR  =  I*R
VRmag  =  abs(VR)

#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)  the  current  I  flowing  in  the  (3.75  +  i11)  impedance  is  given  by  is  ",round(Imag,2),"  A"
print  "\n  (b)  the  magnitude  of  the  p.d.  across  the  impedance  is  ",round(VRmag,2),"  V"


Result

(a)  the  current  I  flowing  in  the  (3.75  +  i11)  impedance  is  given  by  is   3.0   A

(b)  the  magnitude  of  the  p.d.  across  the  impedance  is   34.86   V

### Example 6, page no. 583

In [6]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
rv  =  16.55;#  in  volts
thetav  =  -22.62;#  in  degrees
R1  =  4;#  in  ohm
R2  =  1j*2;#  in  ohm
R3  =  1j*6;#  in  ohm
R4  =  3;#  in  ohm
R5  =  5;#  in  ohm
R6  =  -1*1j*8;#  in  ohm

#calculation:
#voltage
V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)
#The  capacitor  is  removed  from  branch  AB,  as  shown  in  Figure  33.30.
#Impedance,  Z
Z1  =  R3  +  R4  +  R5
Z  =  R1  +  (Z1*R2/(R2  +  Z1))
I1  =  V/Z
I2  =  (R2/(R2  +Z1))*I1
#The  open-circuit  voltage,  E
E  =  I2*R5
#If  the  voltage  source  is  removed  from  Figure  33.30,  the  impedance,  z,  ‘looking  in’  at  AB  is  given  by
z  =  R5*((R1*R2/(R1  +  R2))  +  R3  +  R4)/(R5  +  ((R1*R2/(R1  +  R2))  +  R3  +  R4))
#The  Th´evenin  equivalent  circuit  is  shown  in  Figure  33.31,
#where  the  current  flowing  in  the  capacitor,  I,  is  given  by
I  =  E/(z  +  R6)
Imag  =  abs(I)

#Results
print  "\n\n  Result  \n\n"
print  "\n  the  current  flowing  in  the  capacitor  of  the  network  is  ",round(Imag,2),"    A"


Result

the  current  flowing  in  the  capacitor  of  the  network  is   0.43     A

### Example 7, page no. 584

In [7]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
rv1  =  5;#  in  volts
rv2  =  10;#  in  volts
thetav1  =  45;#  in  degrees
thetav2  =  0;#  in  degrees
R1  =  8;#  in  ohm
R2  =  5;#  in  ohm
R3  =  3j;#  in  ohm
R4  =  4;#  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)
#Current  I1  shown  in  Figure  33.32  is  given  by
I1  =  V2/(R2  +  R3  +  R4)
#Hence  the  voltage  drop  across  the  5  ohm  resistor  is  given  by  VX  is  in  the  direction  shown  in  Figure  33.32,
Vx  =  I1*R2
#The  open-circuit  voltage  E  across  PQ  is  the  phasor  sum  of  V1,  Vx  and  V2,  as  shown  in  Figure  33.33.
E  =  V2  -  V1  -  Vx
#The  impedance,  z,  ‘looking  in’  at  terminals  PQ  with  the  voltage  sources  removed  is  given  by
z  =  R1  +  R2*(R3  +  R4)/(R2  +  R3  +  R4)
#The  Th´evenin  equivalent  circuit  is  shown  in  Figure  33.34  with  the  2  ohm  resistance  connected  across  terminals  PQ.
#The  current  flowing  in  the  2  ohm  resistance  is  given  by
R  =  2;#  in  ohms
I  =  E/(z  +  R)
Imag  =  abs(I)
#power  P  dissipated  in  the  2  ohm  resistor  is  given  by
Pr2  =  R*Imag**2

#Results
print  "\n\n  Result  \n\n"
print  "\n  power  P  dissipated  in  the  2  ohm  resistor  is  ",round(Pr2,2),"  W"


Result

power  P  dissipated  in  the  2  ohm  resistor  is   0.07   W

### Example 8, page no. 585

In [8]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
rv  =  30;#  in  volts
thetav  =  0;#  in  degrees
R1  =  15;#  in  ohm
R2  =  40;#  in  ohm
R3  =  20j;#  in  ohm
R4  =  20;#  in  ohm
R5  =  5j;#  in  ohm
R6  =  5;#  in  ohm
R7  =  -1j*25;#  in  ohm

#calculation:
#voltage
V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)
#The  R7  is  initially  removed  from  the  network,  as  shown  in  Figure  33.36
Z1  =  R1
Z2  =  R2
Z3  =  R3  +  R4
Z4  =  R5  +  R6
#P.d.  between  A  and  C,
Vac  =  (Z1/(Z1  +  Z4))*V
#P.d.  between  B  and  C,
Vbc  =  (Z2/(Z2  +  Z3))*V
#Assuming  that  point  A  is  at  a  higher  potential  than  point  B,  then  the  p.d.  between  A  and  B  is
Vab  =  Vac  -  Vbc
#the  open-circuit  voltage  across  AB  is  given  by
E  =  Vab
#Point  C  is  at  a  potential  of  V  .  Between  C  and  A  is  a  volt  drop  of  Vac.  Hence  the  voltage  at  point  A  is
Va  =  V  -  Vac
#Between  points  C  and  B  is  a  voltage  drop  of  Vbc.  Hence  the  voltage  at  point  B
Vb  =  V  -  Vbc
#Replacing  the  V  source  with  a  short-circuit  (i.e.,  zero  internal  impedance)
#gives  the  network  shown  in  Figure  33.37(a).  The  network  is  shown  redrawn  in  Figure  33.37(b)
#and  simplified  in  Figure  33.37(c).  Hence  the  impedance,  z,  ‘looking  in’  at  terminals  AB  is  given  by
z  =  Z1*Z4/(Z1  +  Z4)  +  Z2*Z3/(Z2  +  Z3)
#The  Th´evenin  equivalent  circuit  is  shown  in  Figure  33.38,  where  current  I  is  given  by
I  =  E/(z  +  R7)
Imag  =  abs(I)

#Results
print  "\n\n  Result  \n\n"
print  "\n    the  current  flowing  in  the  capacitor  is  ",round(Imag,2),"  A  in  direction  from  B  to  A."


Result

the  current  flowing  in  the  capacitor  is   0.13   A  in  direction  from  B  to  A.

### Example 9, page no. 589

In [9]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
V  =  5;#  in  volts
R1  =  2;#  in  ohm
R2  =  3;#  in  ohm
R3  =  -1j*3;#  in  ohm
R4  =  2.8;#  in  ohm

#calculation:
#The  branch  containing  the  R4  is  short-circuited,  as  shown  in  Figure  33.48.
#The  R2  in  parallel  with  a  short-circuit  is  the  same  as  R2  in  parallel  with  0  ohm
#giving  an  equivalent  impedance  of
Z1  =  R2*0/(R3  +  0)
#Hence  the  network  reduces  to  that  shown  in  Figure  33.49,  where
Isc  =  V/R1
#If  the  Voltage  source  is  removed  from  the  network  the  input  impedance,  z,  ‘looking-in’
#at  a  break  made  in  AB  of  Figure  33.48  gives
z  =  R1*R2/(R1  +  R2)
#The  Norton  equivalent  network  is  shown  in  Figure  33.51,  where  current  I  is  given  by
I  =  (z/(z  +  R4  +  R3))*Isc

#Results
print  "\n\n  Result  \n\n"
print  "\n    the  current  flowing  in  the  capacitor  is  ",round(I.real,2),"  +  (", round(I.imag,2),")i  A"


Result

the  current  flowing  in  the  capacitor  is   0.48   +  ( 0.36 )i  A

### Example 10, page no. 589

In [10]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
V1  =  20;#  in  volts
V2  =  10;#  in  volts
R1  =  2;#  in  ohm
R2  =  1.5;#  in  ohm
R3  =  2.95j;#  in  ohm
R4  =  3;#  in  ohm

#calculation:
#The  inductive  branch  is  initially  short-circuited,  as  shown  in  Figure  33.53.
#From  Figure  33.53,
I1  =  V1/R1
I2  =  V2/R4
Isc  =  I1  +  I2
#If  the  voltage  sources  are  removed,  the  impedance,  z,  ‘looking  in’  at  a  break  made  in  AB  is  given  by
z  =  R1*R4/(R1  +  R4)
#The  Norton  equivalent  network  is  shown  in  Figure  33.54,  where  current  I  is  given  by
I  =  (z/(z  +  R2  +  R3))*Isc

#Results
print  "\n\n  Result  \n\n"
print  "\n    the  current  flowing  in  the  inductive  branch  is  ",round(I.real,2),"  +  (",round(  I.imag,2),")i  A"


Result

the  current  flowing  in  the  inductive  branch  is   2.7   +  ( -2.95 )i  A

### Example 11, page no. 590

In [1]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
V  =  10;#  in  volts
R1  =  1;#  in  ohm
R2  =  4;#  in  ohm
R3  =  4;#  in  ohm
R4  =  -2j;#  in  ohm

#calculation:
Isc = V/R3
z = 1/(1/R2 + 1/R3 + 1/R4)
I = (z/(R1 + z))*Isc
pd1 = abs(I)*R1

#Results
print  "\n\n  Result  \n\n"
print  "\n the magnitude of the p.d. across the 1 ohm resistor is ",round(pd1, 2),"V"


Result

the magnitude of the p.d. across the 1 ohm resistor is  1.58 V

### Example 12, page no. 591

In [12]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
rv  =  20;#  in  volts
thetav  =  0;#  in  degrees
R1  =  2;#  in  ohm
R2  =  4;#  in  ohm
R3  =  3j;#  in  ohm
R4  =  -1j*3;#  in  ohm

#calculation:
#voltage
V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)
#Terminals  AB  are  initially  short-circuited,  as  shown  in  Figure  33.61.
#The  circuit  impedance  Z  presented  to  the  voltage  source  is  given  by
Z  =  R1  +  R4*(R2  +  R3)/(R2  +  R3  +  R4)
#Thus  current  I  in  Figure  33.61  is  given  by
I  =  V/Z
Isc  =  ((R2  +  R3)/(R2  +  R3  +  R4))*I
#Removing  the  voltage  source  of  Figure  33.60  gives  the  network  Figure  33.62  of  Figure  33.62.
#Impedance,  z,  ‘looking  in’  at  terminals  AB  is  given  by
z  =  R4  +  R1*(R2  +  R3)/(R2  +  R3  +  R1)
#The  Norton  equivalent  network  is  shown  in  Figure  33.63.
R  =  5;#  in  ohms
#Current  IL
IL  =  (z/(z  +  R))*Isc
ILmag  =  abs(IL)
#the  power  dissipated  in  the  5  ohm  resistor  is
Pr5  =  R*ILmag**2

#Results
print  "\n\n  Result  \n\n"
print  "\n  the  power  dissipated  in  the  5  ohm  resistor  is  ",round(Pr5,2),"  W"


Result

the  power  dissipated  in  the  5  ohm  resistor  is   22.54   W

### Example 13, page no. 592

In [13]:
from __future__ import division
import math
import numpy
import cmath
#initializing  the  variables:
rv1  =  5;#  in  volts
rv2  =  10;#  in  volts
thetav1  =  45;#  in  degrees
thetav2  =  0;#  in  degrees
R1  =  8;#  in  ohm
R2  =  5;#  in  ohm
R3  =  3j;#  in  ohm
R4  =  4;#  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)
#Terminals  PQ  are  initially  short-circuited,  as  shown  in  Figure  33.65.
#Currents  I1  and  I2  are  shown  labelled.  Kirchhoff’s  laws  are  used.
#For  loop  ABCD,  and  moving  anticlockwise,
#I1*(R2  +  R3  +  R4)  +  I2*(R3  +  R4)  =  V2
#For  loop  DPQC,  and  moving  clockwise,
#R2*I1  -  R1*I2  =  V2  -  V1
#Solving  Equations  by  using  determinants  gives
d1  =  [[V2,  (R3  +  R4)],[(V2  -  V1),  -1*R1]]
D1  =  numpy.linalg.det(d1)
d2  =  [[(R2  +  R3  +  R4),  V2],[R2,  (V2  -  V1)]]
D2  =  numpy.linalg.det(d2)
d  =  [[(R2  +  R3  +  R4),  (R3  +  R4)],[R2,  -1*R1]]
D  =  numpy.linalg.det(d)
I1  =  D1/D
I2  =  D2/D
#the  short-circuit  current  Isc
Isc  =  I2
#The  impedance,  z,  ‘looking  in’  at  a  break  made  between  P  and  Q  is  given  by
z  =  R1  +  R2*(R3  +  R4)/(R2  +  R3  +  R4)
#The  Norton  equivalent  circuit  is  shown  in  Figure  33.66,  where  current  I  is  given  by
R  =  2;#in  ohm
I  =  (z/(z  +  R))*Isc
Imag  =  abs(I)

#Results
print  "\n\n  Result  \n\n"
print  "\n  the  magnitude  of  the  current  flowing  5  ohm  resistor  is  ",round(Imag,2),"  A"


Result

the  magnitude  of  the  current  flowing  5  ohm  resistor  is   0.19   A

### Example 15, page no. 595

In [14]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
E1  =  12;#  in  volts
E2  =  24;#  in  volts
Z1  =  3;#  in  ohm
Z2  =  2;#  in  ohm
R1  =  4j;#  in  ohm
R2  =  1.8;#  in  ohm

#calculation:
Z3  =  R1  +  R2
#For  the  branch  containing  the  E1  source,  conversion  to  a  Norton  equivalent  network  gives
Isc1  =  E1/Z1
#For  the  branch  containing  the  E2  source,  conversion  to  a  Norton  equivalent  circuit  gives
Isc2  =  E2/Z2
#Thus  Figure  33.73  shows  a  network  equivalent  to  Figure  33.72.  From  Figure  33.73,
#the  total  short-circuit  current
Isc  =  Isc1  +  Isc2
#the  total  impedance  is  given  by
z  =  Z1*Z2/(Z1  +  Z2)
#Thus  Figure  33.73  simplifies  to  Figure  33.74.
#The  open-circuit  voltage  across  AB  of  Figure  33.74,  E
E  =  Isc*z
#the  impedance  ‘looking  in’  at  AB,is  z
#the  Th´evenin  equivalent  circuit  is  as  shown  in  Figure  33.75.
R  =  1.8  +  4j;#  in  ohm
#when  R  impedance  is  connected  to  terminals  AB  of  Figure  33.75,  the  current  I  flowing  is  given  by
I  =  E/(z  +  R)
Imag  =  abs(I)

#Results
print  "\n\n  Result  \n\n"
print  "\n  the  magnitude  of  the  current  flowing  (1.8  +  i4)  ohm  resistor  is  ",round(Imag,2),"  A"


Result

the  magnitude  of  the  current  flowing  (1.8  +  i4)  ohm  resistor  is   3.84   A

### Example 16, page no. 596

In [15]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
V1  =  5;#  in  volts
V2  =  10;#  in  volts
i  =  0.001;#  in  Amperes
R1  =  1000;#  in  ohm
R2  =  4000;#  in  ohm
R3  =  2000;#  in  ohm
R4  =  200;#  in  ohm
R5  =  -1j*4000;#  in  ohm

#calculation:
#For  the  branch  containing  the  V1  source,  conversion  to  a  Norton  equivalent  network  gives
Isc1  =  V1/R1
z1  =  R1
#For  the  branch  containing  the  V2  source,  conversion  to  a  Norton  equivalent  circuit  gives
Isc2  =  V2/R2
z2  =  R2
#Thus  the  circuit  of  Figure  33.76  converts  to  that  of  Figure  33.77.
#The  above  two  Norton  equivalent  networks  shown  in  Figure  33.77  may  be  combined,
#since  the  total  short-circuit  current  is
Isc  =  Isc1  +  Isc2
#the  total  impedance  is  given  by
Z1  =  z1*z2/(z1  +  z2)
#Both  of  the  Norton  equivalent  networks  shown  in  Figure  33.78  may  be  converted  to  Th´evenin  equivalent  circuits.
#Open-circuit  voltage  across  CD  is
Ecd  =  Isc*Z1
#the  impedance  ‘looking  in’  at  CD  is  Z1
#Open-circuit  voltage  across  EF
Eef  =  i*R3
#the  impedance  ‘looking  in’  Figure  33.79  at  EF
Z2  =  R3
#Thus  Figure  33.78  converts  to  Figure  33.79.
#Combining  the  two  Th´evenin  circuits  gives  e.m.f.
E  =  Ecd  -  Eef
#impedance  z
z  =  Z1  +  Z2
#the  Th´evenin  equivalent  circuit  for  terminals  AB  of  Figure  33.76  is  as  shown  in  Figure  33.80.
Z3  =  R4  +  R5
#If  an  impedance  Z3  is  connected  across  terminals  AB,  then  the  current  I  flowing  is  given  by
I  =  E/(z  +  Z3)
Imag  =  abs(I)

#Results
print  "\n\n  Result  \n\n"
print  "\n  the  current  in  the  capacitive  branch  is  ",  Imag*1000,"mA"


Result

the  current  in  the  capacitive  branch  is   0.8 mA

### Example 17, page no. 597

In [16]:
from __future__ import division
import math
import cmath
#initializing  the  variables:
V  =  5;#  in  volts
i  =  0.004;#  in  Amperes
R1  =  2000;#  in  ohm
R2  =  1000j;#  in  ohm

#calculation:
#Converting  the  Th´evenin  circuit  to  a  Norton  network  gives
Isc1  =  V/R2
#Thus  Figure  33.81  converts  to  that  shown  in  Figure  33.82.
#The  two  Norton  equivalent  networks  may  be  combined,  giving
Isc  =  Isc1  +  i
z  =  R1*R2/(R1  +  R2)
#This  results  in  the  equivalent  network  shown  in  Figure  33.83.
#Converting  to  an  equivalent  Th´evenin  circuit  gives  open  circuit  e.m.f.  across  AB,
E  =  Isc*z
#Thus  the  The´venin  equivalent  circuit  is  as  shown  in  Figure  33.84.
R  =  600  -  800j;#  in  ohms
#When  a  R  impedance  is  connected  across  AB,  the  current  I  flowing  is  given  by
I  =  E/(z  +  R)
Imag  =  abs(I)
#the  power  dissipated  in  the  R  resistor  is
PR  =  R.real*Imag**2

#Results
print  "\n\n  Result  \n\n"
print  "\n  the  power  dissipated  in  the  (600  -  i800)  ohm  resistor  is  ",round(PR*1000,2),"mW"


Result

the  power  dissipated  in  the  (600  -  i800)  ohm  resistor  is   19.68 mW