Chapter 40: Field theory

Example 1, page no. 725

In [1]:
from __future__ import division
import math
#initializing  the  variables:
e0  =  8.85E-12;  
er  =  2.8;
l  =  1;#  in  m

#calculation:  
 #From  Figure  40.9
m  =  16;#  number  of  parallel  squares  measured  along  each  equipotential
n  =  6;#  the  number  of  series  squares  measured  along  each  line  of  force
C  =  e0*er*l*m/n


#Results
print  "\n\n  Result  \n\n"
print  "\n  capacitance  is  ",round(C*1E12,2),"pFarad."


  Result  



  capacitance  is   66.08 pFarad.

Example 2, page no. 725

In [2]:
from __future__ import division
import math
#initializing  the  variables:
e0  =  8.85E-12;  
er  =  3.4;
l  =  100;#  in  m

 #calculation:  
 #From  Figure  40.10
m  =  13;#  number  of  parallel  squares  measured  along  each  equipotential
n  =  4;#  the  number  of  series  squares  measured  along  each  line  of  force
C  =  e0*er*l*m/n


#Results
print  "\n\n  Result  \n\n"
print  "\n  capacitance  is  ",round(C*1E9,2),"nFarad."


  Result  



  capacitance  is   9.78 nFarad.

Example 3, page no. 726

In [3]:
from __future__ import division
import math
#initializing  the  variables:
e0  =  8.85E-12;  
er  =  2.7;
ri  =  0.0005;#  in  m
ro  =  0.006;#  in  m

 #calculation:  
 #capacitance  C
C  =  2*math.pi*e0*er/(math.log(ro/ri))


#Results
print  "\n\n  Result  \n\n"
print  "\n  capacitance  is  ",round(C*1E12,2),"pFarad."


  Result  



  capacitance  is   60.42 pFarad.

Example 4, page no. 727

In [4]:
from __future__ import division
import math
#initializing  the  variables:
C  =  80E-12;#  in  Farads
e0  =  8.85E-12;  
er  =  3.5;
d0  =  0.008;#  in  m

 #calculation:  
 #internal  diameter
di  =  d0*(math.e**(2*math.pi*e0*er/C))


#Results
print  "\n\n  Result  \n\n"
print  "\n  internal  diameter  is  ",round(di,2),"  m."


  Result  



  internal  diameter  is   0.09   m.

Example 5, page no. 728

In [5]:
from __future__ import division
import math
#initializing  the  variables:
e0  =  8.85E-12;  
er  =  3.5;
di  =  0.08;#  in  m
d0  =  0.032;#  in  m
r  =  0.03;#  in  m
V  =  40000;#  in  Volts

#calculation:  
 #capacitance  C
C  =  2*math.pi*e0*er/(math.log(di/d0))
 #dielectric  stress  at  radius  r,
E  =  V/(r*math.log(di/d0))
 #maximum  dielectric  stress,
Emax  =  V/((d0/2)*(math.log((di/d0))))
 #minimum  dielectric  stress,
Emin  =  V/((di/2)*(math.log((di/d0))))


#Results
print  "\n\n  Result  \n\n"
print  "\n  capacitance  is  ",round(C*1E12,2),"pF/km"
print  "\n  dielectric  stress  at  radius  r  is  ",round(E,2),"V/m"
print  "\n  maximum  dielectric  stress,  is  ",round(Emax,2),"V/m  minimum  dielectric  stress  ",round(  Emin,2),"V/m"


  Result  



  capacitance  is   212.4 pF/km

  dielectric  stress  at  radius  r  is   1455142.22 V/m

  maximum  dielectric  stress,  is   2728391.67 V/m  minimum  dielectric  stress   1091356.67 V/m

Example 6, page no. 729

In [1]:
from __future__ import division
import math
#initializing  the  variables:
e0  =  8.85E-12;  
er  =  3.5;
V  =  60000;#  in  Volts
f  =  50;#  in  Hz
Em  =  10E6;#  in  V/m


#calculation:  
 #core  radius,  a
a  =  V/Em
 #internal  sheath  radius,
b  =  a*math.e**1
 #capacitance
C  =  2*math.pi*e0*er/(math.log(b/a))
 #Charging  current
I  =  V*2*math.pi*f*C
 #charging  current  per  kilometre
Ipkm  =  I*1000


#Results
print  "\n\n  Result  \n\n"
print  "\n  core  radius  is  ",round(a*1000,2),"mm  and  internal  sheath  radius  ",round(b*1000,1),"mm"
print  "\n  capacitance  is  ",round(C*1E12,0),"pF/m"
print  "\n  the  charging  current  per  kilometre  ",round(Ipkm,2),"  A"


  Result  



  core  radius  is   6.0 mm  and  internal  sheath  radius   16.3 mm

  capacitance  is   195.0 pF/m

  the  charging  current  per  kilometre   3.67   A

Example 7, page no. 730

In [7]:
from __future__ import division
import math
#initializing  the  variables:
e0  =  8.85E-12;  
er  =  2.5;
di  =  0.08;#  in  m
d0  =  0.025;#  in  m
r  =  1000;#  in  m
V  =  132000;#  in  Volts
f  =  50;#  in  Hz
de  =  3.5E-3;#  rad.

 #calculation:
 #core  radius,  a
a  =  d0/2
 #internal  sheath  radius,
b  =  di/2
 #capacitance
C  =  2*math.pi*e0*er*1E3/(math.log(b/a))
 #Charging  current
I  =  V*2*math.pi*f*C
 #power  loss
P  =  (2*math.pi*f*C*math.tan(de))*V**2


#Results
print  "\n\n  Result  \n\n"
print  "\n  (a)capacitance  for  a  1  km  length  is  ",round(C*1E6,2),"uF"
print  "\n  (b)the  charging  current  ",round(I,2),"A/km"
print  "\n  (c)power  loss  ",round(P,2),"  W"


  Result  



  (a)capacitance  for  a  1  km  length  is   0.12 uF

  (b)the  charging  current   4.96 A/km

  (c)power  loss   2289.78   W

Example 8, page no. 732

In [8]:
from __future__ import division
import math
#initializing  the  variables:
e0  =  8.85E-12;  
er  =  3.2;
di  =  0.06;#  in  m
d0  =  0.020;#  in  m

 #calculation:
 #core  radius,  a
a  =  d0/2
 #internal  sheath  radius,
b  =  di/2
 #capacitance
C  =  2*math.pi*e0*er/(math.log(b/a))


#Results
print  "\n\n  Result  \n\n"
print  "\n  capacitance  per  m  of  length  is  ",round(C*1E9,2),"nF"


  Result  



  capacitance  per  m  of  length  is   0.16 nF

Example 9, page no. 736

In [9]:
from __future__ import division
import math
#initializing  the  variables:
e0  =  8.85E-12;  
er  =  1;
D  =  0.05;#  in  m
d  =  0.005;#  in  m
l  =  200;#  in  m

 #calculation:
 #capacitance
C  =  math.pi*e0*er/(math.log(D/(d/2)))
 #capacitance  of  a  200  m  length
C200  =  C*l


#Results
print  "\n\n  Result  \n\n"
print  "\n  capacitance  of  a  200  m  length  is  ",round(C200*1E6,5),"uF"


  Result  



  capacitance  of  a  200  m  length  is   0.00186 uF

Example 10, page no. 736

In [10]:
from __future__ import division
import math
#initializing  the  variables:
e0  =  8.85E-12;  
er  =  1;
D  =  1.2;#  in  m
r  =  0.004;#  in  m
f  =  50;#  in  Hz
V  =  15000;#  in  Volts
l  =  1000;#  in  m

 #calculation:
 #capacitance
C  =  math.pi*e0*er/(math.log(D/r))
 #capacitance  of  a  1  km  length
Cpkm  =  C*l
 #Charge  Q
Q  =  Cpkm*V
 #Charging  current
I  =  V*2*math.pi*f*Cpkm


#Results
print  "\n\n  Result  \n\n"
print  "\n  capacitance  per  1km  length  is  ",round(Cpkm*1E9,2),"nF"
print  "\n  Charge  Q  is  ",round(Q*1E6,2),"uC"
print  "\n  Charging  current  is  ",round(I,2),"  A"


  Result  



  capacitance  per  1km  length  is   4.87 nF

  Charge  Q  is   73.12 uC

  Charging  current  is   0.02   A

Example 11, page no. 737

In [11]:
from __future__ import division
import math
#initializing  the  variables:
e0  =  8.85E-12;  
er  =  1;
I  =  0.015;#  in  Amperes
d  =  1.25;#  in  m
r  =  800;#  in  m
f  =  50;#  in  Hz
V  =  10000;#  in  Volts

 #calculation:
 #capacitance
C  =  I/(2*math.pi*f*V)
 #required  maximum  value  of  capacitance
Cmax  =  C/r
 #maximum  diameter  of  each  conductor
D  =  2*d/(math.e**(math.pi*e0*er/Cmax))


#Results
print  "\n\n  Result  \n\n"
print  "\n  required  maximum  value  of  capacitance  is  ",round(Cmax*1E12,2),"pF/m"
print  "\nthe  maximum  diameter  of  each  conductor  is  ",round(D,2),"  m"


  Result  



  required  maximum  value  of  capacitance  is   5.97 pF/m

the  maximum  diameter  of  each  conductor  is   0.02   m

Example 12, page no. 739

In [12]:
from __future__ import division
import math
#initializing  the  variables:
e0  =  8.85E-12;  
er  =  1;
C  =  10E-9;#  in  Farad
V  =  1000;#  in  Volts
t  =  10E-6;#  in  sec

 #calculation:
 #energy  stored,Wf
Wf  =  C*V*V/2
 #average  power  developed
Pav  =  Wf/t


#Results
print  "\n\n  Result  \n\n"
print  "\n  the  energy  stored  is  ",Wf,"J"
print  "\nthe  average  power  developed  is  ",Pav,"  W"


  Result  



  the  energy  stored  is   0.005 J

the  average  power  developed  is   500.0   W

Example 13, page no. 739

In [13]:
from __future__ import division
import math
#initializing  the  variables:
e0  =  8.85E-12;  
er  =  1;
Q  =  5E-3;#  in  Coulomb
W  =  0.625;#  in  Joules

 #calculation:
 #voltage  across  the  plates
V  =  2*W/Q
 #Capacitance  C
C  =  Q/V


#Results
print  "\n\n  Result  \n\n"
print  "\n  voltage  across  the  plates  is  ",V,"  V"
print  "\n  Capacitance  C  is  ",C*1E6,"uF"


  Result  



  voltage  across  the  plates  is   250.0   V

  Capacitance  C  is   20.0 uF

Example 14, page no. 740

In [14]:
from __future__ import division
import math
#initializing  the  variables:
e0  =  8.85E-12;  
er  =  10;
C  =  0.01E-6;#  in  Farad
E  =  10E6;#  in  V/m
V  =  2500;#  in  Volts

 #calculation:
 #thickness  of  ceramic  dielectric,
d  =  V/E
 #cross-sectional  area  of  plate
A  =  C*d/(e0*er)
 #Maximum  energy  stored,
W  =  C*V*V/2


#Results
print  "\n\n  Result  \n\n"
print  "\n  thickness  of  ceramic  dielectric  is  ",d*1000,"mm"
print  "\n  cross-sectional  area  of  plate,  is  ",round(A,2),"m2"
print  "\n  Maximum  energy  stored  is  ",round(W,3),"  J"


  Result  



  thickness  of  ceramic  dielectric  is   0.25 mm

  cross-sectional  area  of  plate,  is   0.03 m2

  Maximum  energy  stored  is   0.031   J

Example 15, page no. 740

In [15]:
from __future__ import division
import math
#initializing  the  variables:
e0  =  8.85E-12;  
er  =  2.3;
A  =  0.02;#  in  m2
C  =  400E-12;#  in  Farad
V  =  100;#  in  Volts

 #calculation:
 #energy  stored  per  unit  volume  of  dielectric,
W  =  ((C*V)**2)/(2*e0*er*A**2)


#Results
print  "\n\n  Result  \n\n"
print  "\n  energy  stored  per  unit  volume  of  dielectric  is  ",round(W,2),"  J/m3"


  Result  



  energy  stored  per  unit  volume  of  dielectric  is   0.1   J/m3

Example 16, page no. 744

In [16]:
from __future__ import division
import math
#initializing  the  variables:
u0  =  4*math.pi*1E-7;  
ur  =  1;
a  =  0.001;#  in  m
b  =  0.004;#  in  m

 #calculation:
 #inductance  L
L  =  (u0*ur/(2*math.pi))*(0.25  +  math.log(b/a))


#Results
print  "\n\n  Result  \n\n"
print  "\n  inductance  L  is  ",round(L*1E6,2),"uH/m"


  Result  



  inductance  L  is   0.33 uH/m

Example 17, page no. 744

In [17]:
from __future__ import division
import math
#initializing  the  variables:
u0  =  4*math.pi*1E-7;  
ur  =  1;
da  =  0.010;#  in  m
L  =  4E-7;#  in  H/m

 #calculation:
 #diameter  of  the  sheath
db  =  da*(math.e**(L/(u0*ur/(2*math.pi))))


#Results
print  "\n\n  Result  \n\n"
print  "\n  diameter  of  the  sheath  is  ",round(db,2),"  m"


  Result  



  diameter  of  the  sheath  is   0.07   m

Example 18, page no. 745

In [18]:
from __future__ import division
import math
#initializing  the  variables:
u0  =  4*math.pi*1E-7;  
ur  =  1;
e0  =  8.85E-12;
er  =  3;
da  =  0.010;#  in  m
db  =  0.025;#  in  m
l  =  7500;#  in  m

#calculation:
 #inductance  per  metre  length
L  =  (u0*ur/(2*math.pi))*(0.25  +  math.log(db/da))
 #Since  the  cable  is  7500  m  long,
L7500  =  L*7500
 #capacitance  C
C  =  2*math.pi*e0*er/(math.log(db/da))
 #//Since  the  cable  is  7500  m  long,
C7500  =  C*7500


#Results
print  "\n\n  Result  \n\n"
print  "\ninductance  is  ",round(L7500*1000,2)," mH"
print  "\ncapCItance  is  ",round(C7500*1E6,2),"uF"


  Result  



inductance  is   1.75  mH

capCItance  is   1.37 uF

Example 19, page no. 748

In [19]:
from __future__ import division
import math
#initializing  the  variables:
u0  =  4*math.pi*1E-7;  
ur  =  1;
e0  =  8.85E-12;
er  =  3;
D  =  1.2;#  in  m
a  =  0.008;#  in  m

 #calculation:
 #inductance  per  metre  length
L  =  (u0*ur/(math.pi))*(math.log(D/a))


#Results
print  "\n\n  Result  \n\n"
print  "\ninductance  is  ",round(L*1E6,2),"uH/m"


  Result  



inductance  is   2.0 uH/m

Example 20, page no. 748

In [1]:
from __future__ import division
import math
#initializing  the  variables:
u0  =  4*math.pi*1E-7;  
ur  =  1;
e0  =  8.85E-12;
er  =  1;
l  =  1000;#  in  m
D  =  0.8;#  in  m
a  =  0.01/2;#  in  m

 #calculation:
 #inductance  per  metre  length
L  =  (u0*ur/(math.pi))*(0.25  +  math.log(D/a))
 #Since  the  cable  is  1000  m  long,
L1k  =  L*l
 #capacitance  C
C  =  math.pi*e0*er/(math.log(D/a))
 #//Since  the  cable  is  1000  m  long,
C1k  =  C*l


#Results
print  "\n\n  Result  \n\n"
print  "\ninductance  is  ",round(L1k*1000,2)," mH"
print  "\ncapcitance  is  ",round(C1k*1E9,2),"nF"


  Result  



inductance  is   2.13  mH

capcitance  is   5.48 nF

Example 21, page no. 749

In [21]:
from __future__ import division
import math
#initializing  the  variables:
L  =  2.185E-6;#  in  H/m
u0  =  4*math.pi*1E-7;  
ur  =  1;
a  =  0.012/2;#  in  m

 #calculation:
 #distance  D
D  =  a*math.e**((L*math.pi)/(u0*ur)  -  0.25)


#Results
print  "\n\n  Result  \n\n"
print  "\ndistance  D  is  ",round(D,2),"  m"


  Result  



distance  D  is   1.1   m

Example 22, page no. 752

In [22]:
from __future__ import division
import math
#initializing  the  variables:
L  =  0.2;#  in  H
I  =  0.05;#  in  Amperes
u0  =  4*math.pi*1E-7;  
ur  =  1;

#calculation:
 #energy  stored  in  inductor
W  =  L*I*I/2
 #current  I
I  =  (2*2*W/L)**0.5


#Results
print  "\n\n  Result  \n\n"
print  "\nenergy  stored  in  inductor  is  ",round(W*1000,2),"mJ"
print  "\ncurrent  I  is  ",round(I,2),"A"


  Result  



energy  stored  in  inductor  is   0.25 mJ

current  I  is   0.07 A

Example 23, page no. 752

In [23]:
from __future__ import division
import math
#initializing  the  variables:
B  =  0.05;#  in  Tesla
A  =  500E-6;#  in  m2
l  =  0.002;#  in  m
u0  =  4*math.pi*1E-7;  

#calculation:
 #energy  stored
W  =  (B**2)/(2*u0)
 #Volume  of  airgap
v  =  A*l
 #energy  stored  in  airgap
W  =  W*v


#Results
print  "\n\n  Result  \n\n"
print  "\nenergy  stored  in  the  airgap  is  ",round(W*1E6,2),"uJ"


  Result  



energy  stored  in  the  airgap  is   994.72 uJ

Example 24, page no. 752

In [24]:
from __future__ import division
import math
#initializing  the  variables:
B  =  0.8;#  in  Tesla
A  =  500E-6;#  in  m2
l  =  0.002;#  in  m
u0  =  4*math.pi*1E-7;  
ur  =  1;
e0  =  8.85E-12;
er  =  1;

#calculation:
 #energy  stored  in  mag.  field
W  =  (B**2)/(2*u0)
 #electric  field
E  =  (2*W/(e0*er))**0.5


#Results
print  "\n\n  Result  \n\n"
print  "\nelectric  field  strength  is  ",round(E/1E6,2),"MV/m"


  Result  



electric  field  strength  is   239.89 MV/m