Chapter 2 - Conductivity of Metals

Exa2.1 page 50

In [41]:
#given data :
J=2.4  #in A/mm**2
J=2.4*10**6  #in A/m**2
n=5*10**28  #unitless
e=1.6*10**-19  # in coulomb
#Formula : J=e*n*v
v=J/(e*n) #in m/s
print "Drift velocity is : ",(v)," m/s or ",(v*10**3)," mm/s"

Drift velocity is :  0.0003  m/s or  0.3  mm/s

Exa2 page 50

In [42]:
#given data :
#Electron density
n=1*10**24 #unit less
#Electron charge
e=1.6*10**-19  # in coulomb
#Drift velocity
v=1.5*10**-2  # in meter per second
#cross-sectional area
A=1  # in centimeter square
A=1*10**-4  # in meter square
I=e*n*v*A # in ampere
print "Magnitude of current is :",(I)," A"

Magnitude of current is : 0.24  A

Exa2.3 page 50

In [43]:
#given data :
miu_e=7.04*10**-3  #in m**2/V-s
n=5.8*10**28   # in /m**3
e=1.6*10**-19  # in coulomb
m=9.1*10**-31 #in kg
#(i) Relaxation time,
tau=miu_e/e*m 
print "Relaxation time is : ",(tau)," second" 
sigma=(n*e*miu_e) 
#(ii) Resistivity of conductor,
rho=1/sigma 
print "Resistivity of conductor is : %0.3e"%rho," ohm-meter" 

Relaxation time is :  4.004e-14  second
Resistivity of conductor is : 1.531e-08  ohm-meter

Exa4 page 50

In [44]:
#given data :
rho=1.73*10**-8 #in ohm-meter
toh=2.42*10**-14   #in second
e=1.6*10**-19  #in C
m=9.1*10**-31 #in kg
sigma=1/rho 
#(i) Number of free electrons per m**3
print "Number of free electrons per cube meter is : ",(n)
n=(m*sigma)/(e**2*toh) 
#(ii) Mobility of electrons,
miu_e=(e*toh)/m 
print "Mobility of electrons is : %0.3e"%(miu_e)," m**2/V-s" 
#Note: Answer in the book is wrong

Number of free electrons per cube meter is :  5.8e+28
Mobility of electrons is : 4.255e-03  m**2/V-s

Exa5 page 51

In [45]:
#given data :
rho=1.54*10**-8  #in ohm-meter
#since sigma=1/roh
sigma=1/rho 
n=5.8*10**28   #unit less
e=1.6*10**-19  #in C (electron charge)
m=9.1*10**-31 #in kg (mass of electron)
#(i) Relaxation time
toh=(sigma*m)/(n*e**2) 
print "(i) Relaxation time of electrons is : %0.3e"%(toh)," seconds" 
#(ii) Mobility of electrons,
miu_e=(e*toh)/m 
print "(ii) Mobility of electrons is : %0.3e"%(miu_e)," m**2/V-s" 

(i) Relaxation time of electrons is : 3.980e-14  seconds
(ii) Mobility of electrons is : 6.997e-03  m**2/V-s

Exa2.6 page 51

In [46]:
#given data :
rho=1.7*10**-8  #in ohm-meter
#since sigma=1/roh
sigma=1/rho 
n=8.5*10**28   #unit less
e=1.6*10**-19  #in C (electron charge)
m=9.1*10**-31 #in kg
# Relaxation time
toh=(sigma*m)/(n*e**2) 
print " Relaxation time of electrons is : %0.3e"%(toh)," seconds" 

 Relaxation time of electrons is : 2.460e-14  seconds

Exa2.7 page 51

In [47]:
#given data :
E=100 #in V/m
rho=1.5*10**-8  #in ohm-meter
#since sigma=1/roh
sigma=1/rho 
n=6*10**28   #unit less
e=1.601*10**-19  #in C
m=9.107*10**-31 #in kg
# Relaxation time
toh=(sigma*m)/(n*e**2) 
print "(i) Relaxation time of electrons is : %0.3e"%(toh)," seconds" 
#Drift velocity
v=(e*E*toh)/m 
print "(ii) Drift velocity is : %0.3f"%(v)," m/s" 

(i) Relaxation time of electrons is : 3.948e-14  seconds
(ii) Drift velocity is : 0.694  m/s

Exa2.8 page 52

In [48]:
from math import pi
#given data :
#Diameter of copper wire
d=2 #in milimeter
d=.002 #in meter
#conductivity of copper
nita=5.8*10**7 #in second per meter
#Electron mobility
miu_e=.0032 #in meter square per volt-second
#Applied electric field
E=20 #in mV/m
E=.02  #in V/m
e=1.6*10**-19 
#(i) From eq. (2.13)
#charge density
n=nita/(e*miu_e) #in per meter cube
print "(i) Charge density is : %0.3e"%(n)," /meter cube" 
#(ii) from eq. (2.9)
#current density
J=nita*E # in A/m**2
print "(ii) Current density is : ",(J)," A/m**2" 
#(iii) Current flowing in the wire I=J* Area of x-section of wire
# Area of x-section of wire= (pi*d**2)/4
I=(J*pi*d**2)/4 
print "(iii) Current flowing in the wire is : %0.2e"%(I)," A" 
#(iv) form eq.2.14
#Electron drift velocity
v=miu_e*E 
print "(iv) Electron drift velocity is : %0.1e"%(v)," m/s" 

(i) Charge density is : 1.133e+29  /meter cube
(ii) Current density is :  1160000.0  A/m**2
(iii) Current flowing in the wire is : 3.64e+00  A
(iv) Electron drift velocity is : 6.4e-05  m/s

Exa2.9 page 52

In [49]:
#given data
rho=0.5  # in ohm-meter
J=100  #in A/m**2
miu_e=0.4  #in m**2/V-s
E=J*rho  # since E=J/sigma
# Formula v=miu_e*E
v=miu_e*E 
print "Electron drift velocity is : ",(v)," m/s" 
print "Time taken by the electron to travel 10*10**-6 m in crystal =",
# let Time taken by the electron to travel 10*10**-6 m in crystal = t
t=(10*10**-6)/v 
print (t),"second" 

Electron drift velocity is :  20.0  m/s
Time taken by the electron to travel 10*10**-6 m in crystal = 5e-07 second

Exa10 page 52

In [50]:
#given data
miu_e=0.17 #in m**2/V-s
miu_h=0.035 #in m**2/V-s
nita_i=1.1*10**16  #in /m**3
e=1.6*10**-19 # in C (electron charge)
# Intrinsic conductivity,
sigma_i=(nita_i*e)*(miu_e+miu_h) 
IntrinsicResistivity=1/sigma_i 
print "Intrinsic resistivity is : %0.2e"%(IntrinsicResistivity)," ohm-meter" 

Intrinsic resistivity is : 2.77e+03  ohm-meter

Exa11 page 52

In [51]:
#given data
rho_i=2*10**-3  #in ohm-m  (there is miss printed in this line in the book)
sigma_i=1/rho_i 
miu_e=0.3 # in m**2/V-s
miu_h=0.1 #  in m**2/V-s
e=1.6*10**-19  # in C
# Formula sigma_i=nita_i*e*(miu_e+miu_h)
nita_i=sigma_i/(e*(miu_e+miu_h)) 
print "Carrier density is : %0.2e"%(nita_i)," /m**3" 

Carrier density is : 7.81e+21  /m**3

Exa2.13 page 56

In [52]:
from __future__ import division
#given data
R_15=250 # in ohm
R_t2=300  # in ohm
alpha=0.0039 # in degree C
t1=15 
#Formula R_t2 = R_15 * [1 + alpha1*(t2 - t1)]
t2=((R_t2/R_15)-1)/alpha+t1 
print "Temperature when its resistance is 300 ohms is : ",round(t2,1)," degree C" 

Temperature when its resistance is 300 ohms is :  66.3  degree C

Exa2.15 page 57

In [53]:
#given data
alpha0=0.0038 # in ohm/ohm/degree C
t1=20  #in degree C
alpha20=1/(1/alpha0+t1) 
R1=400 #in ohm
#Formula R2=R1*[1+alpha20*(t2-t1)]
R2=R1*(1+alpha20*(80-20)) 
print "Resistance of wire at 80 degree C si : ",round(R2,1)," ohm"

Resistance of wire at 80 degree C si :  484.8  ohm

Exa2.16 page 57

In [54]:
# given data
R1 = 50 # ohm
R2 = 57.2 # ohm
t1 = 25 # degree C
t2 = 70 # degree C


from sympy import symbols, solve, N
alfa0, R0 = symbols('alfa0 R0') # temperature coefficient at 0 degree C

# Accrding to formula :
#r1 = R0(1+t1*alfa0)
#r2 = R0(1+t2*alfa0)
r1byr2 = R1/R2
alfa0 = solve((1+t1*alfa0)/(1+t2*alfa0)-r1byr2)[0]
alfa0 = N(alfa0, 3)

print "alpha0 = ",alfa0," ohm/ohm/degree C" 

alpha0 =  0.00348  ohm/ohm/degree C

Exa2.17 page 57

In [55]:
# given data
R1 = 45 # ohm
R2 = 59 # ohm
t1 = 25 # degree C
t2 = 75 # degree C


from sympy import symbols, solve, N
alfa0, R0 = symbols('alfa0 R0') # temperature coefficient at 0 degree C

# Accrding to formula :
#r1 = R0(1+t1*alfa0)
#r2 = R0(1+t2*alfa0)
r1byr2 = R1/R2
alfa0 = solve((1+t1*alfa0)/(1+t2*alfa0)-r1byr2)[0]
alfa0 = N(alfa0, 5)

print "alpha0 = %0.2e"%alfa0," ohm/ohm/degree C" 

alpha0 = 7.37e-03  ohm/ohm/degree C

Exa2.18 page 58

In [56]:
# given data
R1 = 3.146 # ohm
R2 = 3.767 # ohm
t1 = 40 # degree C
t2 = 100 # degree C


from sympy import symbols, solve, N
alfa0, R0 = symbols('alfa0 R0') # temperature coefficient at 0 degree C

# Accrding to formula :
#r1 = R0(1+t1*alfa0)
#r2 = R0(1+t2*alfa0)
r1byr2 = R1/R2
alfa0 = solve((1+t1*alfa0)/(1+t2*alfa0)-r1byr2)[0]
alfa0 = N(alfa0, 3)

print "Temperature coefficient of resistance at 40 degree C = ",
alpha40=1/(1/alpha0+40) 
print round(alpha40,5)
#Formula R1 = R0 * (1+40*alpha0)
R0=R1/(1+40*alpha0) 
print "Resistance of platinum coil at 0 degree C is : ",round(R0,3)," ohm " 

Temperature coefficient of resistance at 40 degree C =  0.0033
Resistance of platinum coil at 0 degree C is :  2.731  ohm

Exa2.19 page 58

In [57]:
# given data
R1 = 18 # ohm
R2 = 20 # ohm
R3 = 21 # ohm
t1 = 20 # degree C
t2 = 50 # degree C
ts = 15 # degree C # surrounding temperature

from sympy import symbols, solve, N
alfa0, R0, t = symbols('alfa0 R0 t') # temperature coefficient at 0 degree C

# Accrding to formula :
#r1 = R0(1+t1*alfa0)
#r2 = R0(1+t2*alfa0)
#r3 = R0(1+t*alfa0)
r1byr2 = R1/R2
alfa0 = solve((1+t1*alfa0)/(1+t2*alfa0)-r1byr2)[0]
alfa0 = N(alfa0, 3)

r3byr2 = R3/R2
t = solve(r3byr2 - (1+alfa0*t)/(1+alfa0*t2), t)[0]
tr = t-ts # temp. rise
print "Temperature rise = %0.f degree C" %tr

Temperature rise = 50 degree C

Exa2.20 page 59

In [58]:
from __future__ import division
from fractions import Fraction
#given data
alpha20=1/254.5 # in ohm/ohm/degree C
t2=60 #degree C
t1=20 #degree C
rho0=1.6*10**-6 
alpha60=1/(1/alpha20+(t2-t1)) 
print "Temperature coefficient of resistance at 60 degree C is : ",Fraction(alpha60).limit_denominator(1000),"or",round(alpha60,5)," ohm/(ohm/degree C)" 
#from alpha20=1/(1/alpha0+20)
alpha0=1/(1/alpha20-20) 
#Formula rho60=rho0*(1+alpha0*t)
rho60=rho0*(1+alpha0*t2) 
print "Specific resistance at 60 degree C is : %0.4e"%(rho60)," ohm-cm"

Temperature coefficient of resistance at 60 degree C is :  2/589 or 0.0034  ohm/(ohm/degree C)
Specific resistance at 60 degree C is : 2.0094e-06  ohm-cm

Exa2.21 page 59

In [59]:
#given data
R=95.5 #in ohm
l=1 #in meter
d=0.08 #in mm
d=d*10**-3 #in meter
a=(pi*d**2)/4 
#Formula R=rho*l/a
rho=R*a/l 
print "Resistance of the wire material is : %0.3e"%(rho)," ohm-meter"

Resistance of the wire material is : 4.800e-07  ohm-meter

Exa2.22 page 59

In [60]:
#given data
R=4 #in ohm
d=0.0274 #in cm
d=0.000274 #in meter
rho=10.3 #in miu ohm-cm
rho=10.3*10**-8 #in ohm-m
a=(pi*d**2)/4 

#Formula R=rho*l/a
l=R*a/rho 
print "Lenght of wire is : %0.2f"%(l)," meters"

Lenght of wire is : 2.29  meters

Exa2.23 page 60

In [61]:
#given data
V=220 # in V
W=100 #in watt
R100=V**2/W #in ohm
alpha20=0.005 
t1=20 
t2=2000 
# since R100=R20*[1+alpha20*(t2-t1)]
R20=R100/(1+alpha20 * (t2-t1)) 
I20=V/R20 
print "Current flowing at the instant of switching on a 100 W metal filament lamp is : ",round(I20,2)," A"

Current flowing at the instant of switching on a 100 W metal filament lamp is :  4.95  A

Exa2.24 page 60

In [62]:
from fractions import Fraction
#given data
t2=50 # in degree C
t1=20  # in degree C
R1=600 # in ohm
R2=300 # in ohm

# Let resistance of 600 ohm resistance at 50 degree C = R_600
R_600=R1*(1+(t2-t1)*.001) # in ohm
# Let resistance of 300 ohm resistance at 50 degree C = R_300
R_300=R2*(1+(t2-t1)*.004) # in ohm
R_50=R_600+R_300 # in ohm
print "Resistance of combination at 50degree C is : ",(R_50)," ohm"
R_20=R1+R2 # in ohm
alpha_20=(R_50/R_20-1)/(t2-t1) 
alpha_50=1/(1/(alpha_20)+(t2-t1)) 
print "Effective temperature coefficient of combination at 50 degree C is : ",Fraction(alpha_50).limit_denominator(1000),"per degree C"

Resistance of combination at 50degree C is :  954.0  ohm
Effective temperature coefficient of combination at 50 degree C is :  1/530 per degree C

Exa2.25 page 61

In [63]:
#given data
toh=1.73#in micro-ohm-cm
tohDesh=1.74 #in micro-ohm-cm
sigma=1/toh # conductivities of pure metal
sigmaDesh=1/tohDesh #conductivities metal with impurity
PercentImpurity=((sigma-sigmaDesh)/sigma)*100 
print " Percent impurity in the rod is : %0.4f"%(PercentImpurity)," %"

 Percent impurity in the rod is : 0.5747  %

Exa2.26 page 64

In [64]:
#given data
ElectricalResistivity=2.86*10**-6 #in ohm-cm
sigma=1/ElectricalResistivity 
T=273+20 # in Kelvin (Temperature)
#Formula K/(sigma*T)=2.44*10**-8
K=(2.44*10**-8*T*sigma) 
print "Thermal conductivity of Al = %0.2f"%K

Thermal conductivity of Al = 2.50

Exa2.27 page 69

In [65]:
#given data
E_AC=16*10**-6 #in V per degree C
E_BC=-34*10**-6 #in V per degree C
#By law of successive contact (or intermediate metals)
E_AB=E_AC-E_BC #in V/degree C
E_AB=E_AB*10**6 # in miu V/degree C
print "EMF of iron with respect to constantan is : ",(E_AB)," micro V/degree C"

EMF of iron with respect to constantan is :  50.0  micro V/degree C

Exa2.28 page 69

In [66]:
#given data
E_AC=7.4 #in miu V per degree C
E_BC=-34.4 #in miu V per degree C
#By law of successive contact (or intermediate metals)
E_AB=E_AC-E_BC #in miu V/degree C
E_AB=E_AB*10**-6 # in  V/degree C
# Let Thermo-emf for a temperature difference of 250 degree C = EMF_250
EMF_250=E_AB*250 # in V
EMF_250=EMF_250*10**3 #in mV
print "Termo-emf for a temperature difference of 250 degree C is ",(EMF_250)," mV" 

Termo-emf for a temperature difference of 250 degree C is  10.45  mV

Exa2.29 page 70

In [67]:
#given data
#Take iron as metal A and copper as metal B with respect to lead
#For metal A:
p_A=16.2 
q_A=-0.02 
#For metal B:
p_B=2.78 
q_B=+0.009 
p_AB=p_A-p_B 
q_AB=q_A-q_B 
T2=210 #in degree C
T1=10 # in degree C
E=p_AB*(T2-T1)+q_AB/2*(T2**2-T1**2) 
print "Thermo-electric emf is : ",(E)," micro V" 
Tn=-p_AB/q_AB 
print "Neutral temperature is : ",round(Tn,0)," degree C" 

Thermo-electric emf is :  2046.0  micro V
Neutral temperature is :  463.0  degree C

Exa2.30 page 70

In [68]:
from math import ceil
#given data
p_A=17.34 
q_A=-0.0487 
p_B=1.36 
q_B=+0.0095 
p_AB=p_A-p_B 
q_AB=q_A-q_B 
T2=210 #in degree C
T1=10 # in degree C
E=p_AB*(T2-T1)+q_AB/2*(T2**2-T1**2) #in miu V
E=E*10**-3 #in m V
print "Thermo-electric emf is : ",(ceil(E))," m V" 
Tn=-p_AB/q_AB 
print "Neutral temperature is : ",(ceil(Tn))," degree C" 
Tc=10 # in degree C
Ti=Tn+(Tn-Tc) 
print "Temperature of inversion is : ",(ceil(Ti))," degree C" 
E_max=15.98*(275-10)-1/2*0.0582*(275**2-10**2) #in miu V
E_max=E_max*10**-3 # in mV
print "Maximum possible thermo-electric emf at neutral temperature that is at 275 degree C is : %0.3f"%(E_max)," mV" 

Thermo-electric emf is :  2.0  m V
Neutral temperature is :  275.0  degree C
Temperature of inversion is :  540.0  degree C
Maximum possible thermo-electric emf at neutral temperature that is at 275 degree C is : 2.037  mV

Exa2.31 page 72

In [69]:
#given data
rho=146*10**-6# in ohm-cm
a=1 #in cm**2
l=1 #in cm
# let current = i
i=0.06 #in amp 
R=rho*l/a #in ohm
# Let potential difference per degree centigrade = P
P=i*R # By Ohm's law
print "Potential difference per degree centigrade is : ",(P)," volt" 

Potential difference per degree centigrade is :  8.76e-06  volt

Exa2.32 page 73

In [70]:
from sympy.mpmath import quad
#given data
T_lower=10 # in degree C
T_upper=150 # in degree C

# T for iron at any temperature T degree C w.r.t. lead is given by (17.34-0.0487 T)*10**-6 and that for copper by (1.36-.0095 T)*10**-6
T_i = lambda T: (17.34-0.0487*T)*10**-6 #Thermo-electric power
T_c = lambda T: (1.36-0.0095*T)*10**-6 #Thermo-electric power

# Thermo-electric power, P=dE/dT
# or dE=P*dT
# Thermo-emf for copper between temperature 10 degree C and 150 degree C,


E_c= quad(T_c,[T_lower,T_upper]) 
# Thermo-emf for iron between temperature 10 degree C and 150 degree C,
E_i= quad(T_i,[T_lower,T_upper]) 
# Thermo-emp for copper-iron thermo-couple
E=E_i-E_c 
print "Thermo-emf for iron between temperature 10 degree C and 150 degree C is : ",(E*10**6)," micro V" 

Thermo-emf for iron between temperature 10 degree C and 150 degree C is :  1798.16  micro V

Exa2.34 page 79

In [71]:
#given data
Hc_0=8*10**5 #in A/m
Tc=7.26 #in K
T=4 #in K
Hc_T=Hc_0*(1-(T/Tc)**2)
print "The critical value of magnetic field at T=4 K is : %0.4e"%(Hc_T)," A/m" 

The critical value of magnetic field at T=4 K is : 5.5715e+05  A/m

Exa2.35 page 79

In [72]:
from __future__ import division
#given data
Hc=7900 #in A/m
d=1 #in mm
r=d/2 #in mm
r=r*10**-3 #in m
Ic=2*pi*r*Hc 
print "Critical current is : %0.3f"%(Ic)," A" 

Critical current is : 24.819  A

Exa2.36 page 79

In [73]:
#given data
Hc_0=8*10**4 #in A/m
Tc=7.2 #in K
T=4.5 #in K
d=1 #in mm
r=d/2 #in mm
r=r*10**-3 #in m
Hc=Hc_0*(1-(T/Tc)**2)
print "The critical field at T=4.5 K is : %0.3e"%(Hc)," A/m" 
Ic=2*pi*r*Hc 
print "Critical current is : %0.2f"%(Ic)," A" 

The critical field at T=4.5 K is : 4.875e+04  A/m
Critical current is : 153.15  A

Exa2.37 page 86

In [74]:
from math import sqrt
# Formula R=rho*l/a
#putting value for copper wire
R=2 # in ohm
l=100 #in meter
rho=1.7*10**-8 # (for copper)
a=rho*l/R #in meter
a=a*10**6 # in mm
# Formula a=pi/4*d**2
d_copper=sqrt(a*4/pi)  #  (d_copper is diameter for copper)

# Formula R=rho*l/a
#putting value for Aluminium wire
R=2 # in ohm
l=100 #in meter
rho=2.8*10**-8 # (for aluminium)
a=rho*l/R #in meter
a=a*10**6 # in mm
# Formula a=pi/4*d**2
d_aluminium=sqrt(a*4/pi)  #  (d_aluminium is diameter for aluminium)
DiaRatio=d_aluminium/d_copper  #  (DiaRatio is ratio of diameter of aluminium and copper)
print "The diameter of the aluminium wire is ",round(DiaRatio,2)," times that of copper wire" 

The diameter of the aluminium wire is  1.28  times that of copper wire

Exa2.38 page 99

In [75]:
from math import log
#given data
l=60 # in cm
l=l*10**-2 #in meter
d=20 # in cm 
d=d*10**-2 #in meter
D=35 # in cm 
D=D*10**-2 #in meter
r1=d/2 
r2=D/2 
rho=8000 # in ohm-cm
rho=80 # in ohm-m
# Let Insulation resistance of the liquid resistor = Ir
Ir=(rho/(2*pi*l))*log(r2/r1) 
print " Insulation resistance of the liquid resistor is : ",round(Ir,2)," ohm"

 Insulation resistance of the liquid resistor is :  11.88  ohm

Exa2.39 page 100

In [76]:
#given data
R_desh=1820 # in M ohm
R_desh=R_desh*10**6 # in ohm
d1=1.5 # in cm
d1=d1*10**-2 # in meter
d2=5 # in cm
d2=d2*10**-2 # in meter
l=3000 # in meter
r1=d1/2 
r2=d2/2 

rho= (2*pi*l*R_desh)/log(r2/r1) 
print "Resistivity of dielectric is : %0.2e"%(rho)," ohm meter"

Resistivity of dielectric is : 2.85e+13  ohm meter

Exa2.40 page 100

In [77]:
# given data
# First Case:
r1=1.5/2 # in cm
# let radius thickness of insulation = r1_t
r1_t=1.5 # in cm
r2=r1+r1_t 
R_desh=500 # in M ohm
R_desh=R_desh*10**6 # in ohm
# Second case:
r1_desh=r1 # in cm (as before)
# let radius thickness of insulation = r2_t
r2_t=2.5 # in cm
r2_desh=r1+r2_t 
# since Insulation resistance , R_desh= sigma/(2*pi*l)*log(r2/r1) and
#                               R1_desh= sigma/(2*pi*l)*log(r2_desh/r1_desh)
# Dividing R1_desh by R1, We get
# R1_desh/R_desh = log(r2_desh/r1_desh)/log(r2/r1)
# Let R = R1_desh/R_desh, Now
R= log(r2_desh/r1_desh)/log(r2/r1) 
R1_desh=R*R_desh 
print "New insulation resistance is : ",round(R1_desh*10**-6,2)," M ohm" 

New insulation resistance is :  667.36  M ohm

Exa2.41 page 101

In [78]:
from math import exp

# given data
t1=20 # in degree C
t2=36 # in degree C
alpha_20=0.0043 # in per degree C  (Temperature Coefficient)
InsulationResistance=480*10**6 # in ohm
copper_cond_res=0.7 # in ohm  (copper conductor resistance)
l=500*10**-3 # in kilo meter (length)
R1_desh=InsulationResistance * l # in ohm

# From Formula log(R2_desh)= log(R1_desh-K*(t2-t1))
# K= 1/(t2-t1)*log(R1_desh/R2_desh)
# since when t2-t1=10 degree C and R1_desh/R2_desh= 2

K=1/10*log(2) 

# (i) Insulation resistance at any temperature t2, R2_desh is given by
logR2_desh= log(R1_desh)-(t2-t1)/10* log(2) 
R2_desh= exp(logR2_desh)
  
print "(i) Insulation resistance at any temperature : ",round(R2_desh*10**-6,2)," Mega ohm" 
  
# (ii) 
R_20= copper_cond_res/l # in ohm
R_36=R_20*(1+alpha_20*(t2-t1))
    
print "(ii) Resistance at 36 degree C is : ",(R_36)," ohm"
    

(i) Insulation resistance at any temperature :  79.17  Mega ohm
(ii) Resistance at 36 degree C is :  1.49632  ohm