# Chapter 6:Electrical Conducting and Insulating Materials¶

## Example 6.1,Page No:6.8¶

In [1]:
import math

#variable declaration
R75    = 57.2;     #resistance at 75 C in Ω
R25    = 55;       #resistance at 25 C in Ω
t1     = 25;       #temperature in C
t2     = 75        # temperature in C

#formula
#Rt = R0*(1+(alpha*t))
#calculation
alpha  = (R25-R75)/float((25*R75)-(75*R25));        #temperature cofficient

#result
print'temperature coefficient =%3.5f'%alpha,'K**-1';

temperature coefficient =0.00082 K**-1


## Example 6.2,Page No:6.9¶

In [2]:
import math

#variable declaration
R1     = 50;          #resistance in ohm at temperature  15°C
R2     = 60;          # resistance in ohm temperature  15°C
t1     = 15;          #temperature in °C
alpha  = 0.00425;     #temperature coefficient of resistance

#formula
#Rt = R0*(1+(alpha*t))
#Rt1/Rt2 = R0*(1+(alpha*t1))/R0*(1+(alpha*t2))
#calculation
R          = R2/float(R1);              #resistance in Ω
X          = 1+(alpha*t1);
t2         = ((R*X)-1)/float(alpha);    #temperature coefficient of resistance in °C

#result
print'temperature coefficient of resistance =%3.2f'%t2,'°C';

temperature coefficient of resistance =65.06 °C


## Example 6.3,Page No:6.9¶

In [4]:
import math

#variable declaration
t1         = 20;                #temperature in °C
alpha      = 5*10**-3;          #average  temperature coefficient at 20°C
R1         = 8;                 #resistance in Ω
R2         = 140;                #resistaance in Ω

#calculation
t2     = t1+((R2-R1)/float(R1*alpha));       #temperature in °C

#result
print'Hence temperature under normal condition is %3.2f'%t2,'°C';

Hence temperature under normal condition is 3320.00 °C


## Example 6.4,Page No:6.10¶

In [5]:
import math

#variable declaration
l     = 100;                      #length in cm
d     = 0.008;                    #diameter of wire in cm
R     = 95.5;                     #resistance in Ω
d    = 0.008;                      #diameter in cm

#formula
#R=p*l/A
#calculation
A    = (math.pi*d*d)/float(4);           #cross-sectional area
p  = (R*A)/float(l);                    #resistivity of wire in Ω-cm

#result
print'resistivity=%3.2e'%p,'Ω-m';

resistivity=4.80e-05 Ω-m


## Example 6.5,Page No:6.10¶

In [7]:
import math

#variable declaration
R0       =17.5;            #resistance at 0 degree c in Ω
alpha    =0.00428;         #temperature coefficient of copper in per °C
t        =16;              #temperature in °C

#calculations
Rt     = R0*(1+(alpha*t));               #resistance at 16 °C
P       = (R0/float(Rt))*100;            #percentage conductivity at 16 °C

#result
print'percentage conductivity=%3.2f'%P,'%';

percentage conductivity=93.59 %


## Example 6.10,Page No:6.30¶

In [8]:
import math

#variable declaration
l         = 60;                      #length in m
r2        = 38/float(2);             #radius of outer cylinder in m
r1        = 18/float(2);             #radius of inner cylinder in m
p         = 8000;                    #specific resistance in Ω-m

#calculation
R  = (p/float(2*math.pi*l))*math.log(r2/float(r1));     #insulation resistance of liquid resistor  in Ω

#result
print'insulation resistance=%3.0f '%R,'Ω';

insulation resistance= 16  Ω


## Example 6.11,Page No:6.30¶

In [10]:
import math

#variable declaration
d1   = 0.0018;                        #inner diameter  in m
d2   = 0.005;                         # outer diameter in m
R    = 1820*10**6;                     #insulation resistance in Ω
l    = 3000;                           #length in m

#calculations
r1  = d1/float(2);                   #inner radius  in m
r2  = d2/float(2);                   #outer radius in m
p   = (2*math.pi*l*R)/float(math.log(r2/float(r1)));      #resistivity of dielectric in Ω-m

#result
print'resistivity=%3.3e'%p,'Ω-m';

resistivity=3.358e+13 Ω-m


## Example 6.12,Page No:6.31¶

In [12]:
import math

#variable declaration
d1    = 0.05;          #inner diametr in m
d2    = 0.07;          #outer diameter in m
l     = 2000;          #length in m
p     = 6*10**12;      #specific resistance in Ω-m

#calculations
r1  = d1/float(2);                   #inner radius  in m
r2  = d2/float(2);                   #outer radius in m
R = (p/float(2*math.pi*l))*(math.log(r2/float(r1)));       #insulation resistance

#result
print'insulation resistance =%1e'%R,'Ω';
print' Note: calculation mistake in textbook in calculating insulating resistance';

insulation resistance =1.606537e+08 Ω
Note: calculation mistake in textbook in calculating insulating resistance


## Example 6.13,Page No:6.31¶

In [13]:
import math

#variable declaration
a     = 110*10**-3;                  #area in m**2
d     = 2;                           #thickness in mm
er    = 5;                           #relative permitivity
E     = 12.5*10**3;                  #electric field strength in V/mm
e0    = 8.854*10**-12;               #charge of electron in coulombs

#calculations
A     = a*a;                                 #area in m**2
C     = e0*((er*A)/float(d*10**-3))          #capacitance in F
V     = E*(d);
Q     = (C)*(V)                             #charge on capacitor in C

#result
print'capacitance =%3.2e'%C,'F';
print' charge=%3.3e'%Q,'C';

capacitance =2.68e-10 F
charge=6.696e-06 C


## Example 6.14,Page No:6.31¶

In [10]:
import math

#variable declaration
I       = 15*10**-3;                #current in A
t       = 5;                        #time in s
V       = 1000;                    #voltage in volts
d       = 10**-3;                   #thickness in m
a       = 120*10**-3;

#calculation
A       = a**2                            #area in m**2
Q      = I*t;                             #charge on capacitor in C
#since charge and electric field are equal
phi     = Q;                              #electric flux in mc
D      = Q/float(A);                      #electric flux density in c/m**2
E      = V/float(d);                      #electric field strength in dielectric

#result
print'charge=%3.2e'%Q,'C';
print' electric flux=%4.3f'%(phi*10**3),'mc';
print' electric flux density=%3.2f'%D,'c/m**2';
print' electric field strength=%2.3e'%E,'V/m';

charge=7.50e-02 C
electric flux=75.000 mc
electric flux density=5.21 c/m**2
electric field strength=1.000e+06 V/m


## Example 6.15,Page No:6.32¶

In [11]:
import math

#variable declaration
n        = 12;                   #number of plates
er       = 4;                     #relative permitivty
d        = 1.0*10**-3;            #distance between plates in m
A       = 120*150*10**-6;         #area in m**2
e0      = 8.854*10**-12;          # in F/m

#calculation
c      = (n-1)*e0*er*A/float(d);         #capacitance in F

#result
print'capacitance=%3.4e'%c,'F';

capacitance=7.0124e-09 F


## Example 6.16,Page No:6.32¶

In [12]:
import math

#variable declaration
e0     = 40000;        #dielectric strength  in volts/m
d      = 33000;        #thickness in kV

#calculations
t      = d/float(e0);         #required thickness of insulation in mm

#result
print'thickness=%3.2f'%t,'mm';

thickness=0.82 mm


## Example 6.17,Page No:6.32¶

In [13]:
import math

#variable declaration
C      = 0.03*10**-6;           #capacitance in F
d      = 0.001;                 #thickness in m
er     = 2.6;                   #dielectric constant
e0     = 8.85*10**-12;          #dielectric strength
E0     = 1.8*10**7

#formula
#C=e0*er*A/d
#e0=v/d
#calculation
A      = (C*d)/float(e0*er);        #area of dielectric needed in m**2
Vb      = E0*d;                     #breakdown voltage in m

#result
print'area = %3.2f'%A,'m**2';
print' breakdown voltage=%3.1e'%Vb,'V';

area = 1.30 m**2
breakdown voltage=1.8e+04 V


## Example 6.18,Page No:6.33¶

In [14]:
import math

#variable declaration
C          = 0.035*10**-6;                      #capacitance in F
tangent    = 5*10**-4;                          #power factor
f          = 25*10**3;                          #frequency in Hz
I          = 250;                              #current in A

#calculation
V      = I/float(2*math.pi*f*C)                  #voltage across capacitor in volts
P      = V*I*tangent;                            #dielectric loss in watts

#result
print'dielectric loss=%3.1f'%P,'watts';

dielectric loss=5684.1 watts


## Example 6.19,Page No:6.33¶

In [15]:
import math

#variable declaration
Q        = 20*10**-6;                #charge of electron in coulomb
V        = 10*10**3;                 #potential in V
e0       = 8.854*10**-12;            #absolute permitivity
d        = 5*10**-4;                 #separation between plates in m
er       = 10;                       #dielectric constant

#formula
#Q=CV
#C=er*e0*A/d
C       = Q/float(V);
A       = (C*d)/float(er*e0);         #area in m**2

#result
print'area=%1e'%A,'m**2';

area=1.129433e-02 m**2


## Example 6.20,Page No:6.35¶

In [16]:
import math

#variable declaration
n  = 3.0*10**28;              #number of electrons per m**3
t   = 3*10**-14;              #time in s
m   = 9.1*10**-31;            #mass of electron in kg
L   = 2.44*10**-8;            #lorentz number in ohm W/K**2
T  = 300;                    #temperature in kelvin
e   = 1.6*10**-19;             #charge of electron in coulomb

#calculation
sigma  = (n*(e**2)*t)/float(m);    #electrical conductivity in (ohm-m)**-1
K      = sigma*T*L;

#result
print'electrial conductivity=%3.2e'%sigma,'(Ω-m)**-1';
print'lorentz number = %3.2f'%K,'W/mK';

electrial conductivity=2.53e+07 (Ω-m)**-1
lorentz number = 185.33 W/mK
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