# Chapter 2 : Conducting Materials¶

## Example 2.1 Page No : 21¶

In [1]:
# Variables
l = 300     #in meters
a = 25*(10**-6)     #in meter square
d15 = 2.7     #density at 15 degree C in ohm-meter
k0 = 0.004     #temp coefficient in ohm/degree C at 0 degree C
t = 15.;T = 50.     #in degree C

# Calculations
R15 = d15*(l/a)
k15 = k0/(1+(k0*t))
R50 = R15*(1+k15*(T-t))

# Results
print 'The value of Resistance at 15 degree C is %3.2f.ohms '%(R15)
print 'The value of Resistance at 50 degree C is %3.2f.ohms'%(R50)

The value of Resistance at 15 degree C is 32400000.00.ohms
The value of Resistance at 50 degree C is 36679245.28.ohms


## Example 2.2 Page No : 21¶

In [3]:
# Variables
R20 = 400.     # in ohms
k0 = 0.0038
t = 20.;T = 80.     #degree C

# Calculations
k1 = k0/(1+(k0*t))
R80 = R20*(1+k1*(T-t))

# Results
print 'The value of Resistance at 80 degree C is %3.4f ohms'%(R80)

The value of Resistance at 80 degree C is 484.7584 ohms


## Example 2.3 Page No : 22¶

In [4]:
# Variables
t = 15.     #degree C
R15 = 250.;RT = 300.     #ohms
k0 = 0.0038     #ohm/degree C

# Calculations
k1 = k0/(1+(k0*t))
T = (((RT/R15)-1)/k1)+t     #math.since RT = R15{1+k1*(T-t)}

# Results
print 'The value of Temperature at 300 ohm resistance is %3.1f degree C'%(T)

The value of Temperature at 300 ohm resistance is 70.6 degree C


## Example 2.4 Page No : 22¶

In [6]:
import math

# Variables
#Part (a)
d = 0.4*(10**-3)     #diameter in meter
a = math.pi*(d**2)/4     #area in meter square
p1 = 100*(10**-8)     #resistivity of nichrome in ohm-meter
R = 40     #resistance in ohms

#Part(b)
d = 0.4*(10**-3)     #diameter in meter
a = 12.6*(10**-8)     #area in meter square
p2 = 1.72*(10**-8)     #resistance of copper wire in ohm-meter
R = 40     #resistance in ohms

# Calculations
l1 = R*a/p1
l2 = R*a/p2

# Results
print 'Thus the length of heater element with nichrome wire is %2.1f meter '%(l1)
print 'Thus the length of heater element with copper wire is %2.1f meter'%(l2)

Thus the length of heater element with nichrome wire is 5.0 meter
Thus the length of heater element with copper wire is 293.0 meter


## Example 2.5 Page No : 23¶

In [7]:
# Variables
R0 = 80     #in ohms
t = 40     # in degree C
k0 = 0.0043

# Calculations
R40 = R0*(1+(k0*t))

# Results
print 'The value of Resistance at 40 degree C is %3.2f ohms'%(R40)

The value of Resistance at 40 degree C is 93.76 ohms


## Example 2.6 Page No : 23¶

In [9]:
# Variables
R80 = 50.;R28 = 40.     # resistance in ohms
t = 28.;T = 80.     # temp in degrees

# Calculations
k28 = ((R80/R28)-1)/(T-t)     #math.since RT = Rt{1+k*(T-t)}
k0 = k28/(1-k28*t)     # math.since k28 = k0/(1+k0*t)

# Results
print 'The value of Temperature coefficient  at 28 degree C is %3.4f ohms per degree C '%(k28)
print 'The value of Temperature coefficient  at 0 degree C is %3.4f ohms per degree C'%(k0)

The value of Temperature coefficient  at 28 degree C is 0.0048 ohms per degree C
The value of Temperature coefficient  at 0 degree C is 0.0056 ohms per degree C


## Example 2.7 Page No : 24¶

In [10]:
import math

# Variables
l = 1000     # length in meters
d = 0.09/100     # diameter in meters
p = 1.724*(10**-8)     # specific resistance in ohm meter

# Calculations
a = math.pi*(d**2)/4     # area in meter square
R = p*l/a     #resistance in ohms

# Results
print 'The value of Resistance is %3.2f ohms'%(R)

The value of Resistance is 27.10 ohms


## Example 2.8 Page No : 24¶

In [11]:
# Variables
R20 = 50.     # resistance in ohms
T = 60.;t = 20.     # temp in degree C
k0 = 0.00427     #temp coefficient at zero degreeC

# Calculations
R0 = R20/(1+(k0*t))
R60 = R0*(1+(k0*T))

# Results
print 'The value of Resistance at 0 degree C is %3.2f ohms '%(R0)
print 'The value of Resistance at 60 degree C is %3.2f ohms'%(R60)

The value of Resistance at 0 degree C is 46.07 ohms
The value of Resistance at 60 degree C is 57.87 ohms


## Example 2.9 Page No : 24¶

In [12]:
# Variables
k20 = 1/254.5     # temperature coefficient at 20 degreeC
p0 = 1.6*(10**-6)     # resistivity at 0 degree C in ohm-cm
t = 20
T = 50     #temp in degree C

# Calculations
k0 = k20/(1-(t*k20))     #temperature coefficient at 0 degreeC
p50 = p0*(1+(T*k0))     # resistivity at 50 degree C in ohm-cm
k50 = 1/(T+(1/k0))     #temperature coefficient at 50 degreeC

# Results
print 'Thus the temperature coefficient at 50 degree C is %3.4f '%(k0)
print 'Thus the resistivity at 50 degree C is %e in ohm-cm'%(p50)

Thus the temperature coefficient at 50 degree C is 0.0043
Thus the resistivity at 50 degree C is 1.941151e-06 in ohm-cm


## Example 2.10 Page No : 25¶

In [15]:
# Variables
R15 = 50
RT = 58     # resistance in ohms
t = 15     # te mp in degree C
k0 = 0.00425     # temp coefficient at 0 degree C

# Calculations
R0 = R15/(1+(k0*t))     # resistance at 0 degree C in ohms
T = ((RT/R0)-1)/k0     # temp in degree C

# Results
print 'The value of Resistance at 0 degree C is %3.1f ohms '%(R0)
print 'The value of Temperature at 58 ohm resistance is %3.4f degree C'%(T)

The value of Resistance at 0 degree C is 47.0 ohms
The value of Temperature at 58 ohm resistance is 55.0471 degree C


## Example 2.11 Page No : 25¶

In [16]:
# Variables
R25 = 50.
R70 = 57.2     # resistance in ohms
t = 25.
T = 70     # temp in degree C

# Calculations
#math.since Rt = R0(1+(k0*t))
k0 = (R70-R25)/((R25*T)-(R70*t))

# Results
print 'The temp coefficient at 0 degree C is %3.3f'%(k0 )

The temp coefficient at 0 degree C is 0.003


## Example 2.12 Page No : 26¶

In [17]:
# Variables
R0 = 15.5     # resistance in ohms
t = 16     #in degree C
k0 = 0.00428     #temp coefficient

# Calculations
R16 = R0*(1+(k0*t))
G = (R0/R16)*100     # math.since conducmath.tance = reciprocal of esismath.tance

# Results
print 'The value of Resistance at 16 degree C is %3.4f ohms '%(R16)
print 'The value of percentage conductivity at 16 degree C is %3.2f percent'%(G)

The value of Resistance at 16 degree C is 16.5614 ohms
The value of percentage conductivity at 16 degree C is 93.59 percent


## Example 2.13 Page No : 26¶

In [18]:
# Variables
RT = 144.
R20 = 10.     # in ohms
t = 20.     # in degree C

# Calculations
k20 = 5*(10**-3)     #temp coefficient at 20 degree C
T = (((RT/R20)-1)/k20)+t

# Results
print 'The value of temp required for tungsten bulb is %4.2f degree C'%(T)

The value of temp required for tungsten bulb is 2700.00 degree C


## Example 2.14 Page No : 27¶

In [19]:
# Variables
V15 = 250
Vt = 250     #voltage in volts
I15 = 5
It = 4     #current in amperes
T = 15     #temp in degree C

# Calculations
R15 = V15/I15     #resistance in ohms at 15 degreeC
Rt = Vt/It     #resistance at t degreeC
k0 = 0.0038
R0 = R15/(1+(k0*T))
t = ((Rt/R0)-1)/k0

# Results
print 'Resistance at 15 degree C is %3.1f ohms '%(R15)
print 'Resistance at t degree C is %3.1f ohms '%(Rt)
print 'Resistance at 0 degree C is %3.2f ohms '%(R0)
print 'Temperature t is %3.2f degree C'%(t)

Resistance at 15 degree C is 50.0 ohms
Resistance at t degree C is 62.0 ohms
Resistance at 0 degree C is 47.30 ohms
Temperature t is 81.76 degree C


## Example 2.15 Page No : 28¶

In [20]:
# Variables
n = 100     #no of slots
c = 12     #conductors per slot
Lm = 300     # mean length of turn in cm
a = 1.5*0.2     #cross section of each conductor in cm**2
s = 1.72*(10**-6)     #specific resistance of copper at 20 degreeC
p = 4     # poles
t = 20
T = 75     #temp in degreeC
k0 = 0.00427     #temp coefficient of resistivity for copper

# Calculations
L = n*c*Lm     #total length of conductors
Ls = L/p     #length of conductors in each parallel path
s0 = s*(1-(k0*t))
RT = (s0*Ls)/a

# Results
print 'Thus specific resistance at 0 degree C is %e ohm-cm '%(s0)
print 'Thus resistance at working temp of 75 degree C is %3.4f ohm'%(RT)

Thus specific resistance at 0 degree C is 1.573112e-06 ohm-cm
Thus resistance at working temp of 75 degree C is 0.4719 ohm


## Example 2.16 Page No : 28¶

In [22]:
# Variables
a = 15.     #cross section area in cm**2
l = 100000     #length in cm
p0 = 7.6*(10**-6)     #specific resistance at 0 degree C in ohm-cm
k0 = 0.005     #temp coefficient at 0 degree C
t = 50     #temp in degree C

# Calculations
p50 = p0*(1+(t*k0))     #resistivity at 50 degree C
R50 = p50*(l/a)

# Results
print 'Thus resistance at 50 degree C is %3.5f ohms '%(R50)

Thus resistance at 50 degree C is 0.06333 ohms


## Example 2.17 Page No : 29¶

In [23]:
# Variables
I2 = 27.5     #current of No.25 wire in Amperes
d = 1./2     #math.since I1/I2 = 1/2

# Calculations
I1 = I2*(d**(3./2))

# Results
print 'Thus fusing current of No.33 wire is %3.3f amperes '%(I1)

Thus fusing current of No.33 wire is 9.723 amperes


## Example 2.18 Page No : 30¶

In [26]:
import math

# Variables
sAl = 2.85*(10**-6)
sCu = 1.7*(10**-6)     #specific resistance in ohm-cm
gAl = 2.71
gCu = 8.89     #specific gravity
cAl = 5000.
cCu = 10000.     #math.cost per tonne

# Calculations
#P = V**2/R, power is same for both so resistance must also be same
#so R = (p*l)/(pi*d**2) = (p*l)/(pi*d'**2)
Kd = math.sqrt(sAl/sCu)     #Kd = d/d'
Km = (Kd**2)*(gAl/gCu)
Kc = Km*(cAl/cCu)

# Results
print 'Thus the ratio of diameters is %3.3f '%(Kd)
print 'Thus the ratio of weights is %3.4f '%(Km)
print 'Thus the ratio of math.costs is %3.4f'%(Kc)

Thus the ratio of diameters is 1.295
Thus the ratio of weights is 0.5111
Thus the ratio of math.costs is 0.2555


## Example 2.19 Page No : 33¶

In [27]:
# Variables
R1 = 18.6     #resistacne in ohms
Kl = 5.     #math.since l2 = 5*l1
Ka = 3     # math.since a2 = 3*a1

# Calculations
R2 = R1*Kl/Ka
# resistivity is same because wires are of same material

# Results
print 'Thus the resistance of another conductor is %3.1f ohms'%(R2)

Thus the resistance of another conductor is 31.0 ohms


## Example 2.20 Page No : 57¶

In [28]:
# Variables
m = 1.     #mass in kg
S = 4200.     #specific heat of water
T2 = 100.;T1 = 15.     # temp in degree C
W = 500     #wattage rating of kettle in volts
t = 15*60     # time in sec

# Calculations
H = m*S*(T2-T1)     #heat utilised in J
Hd = W*t     #heat developed in J
He = (H/Hd)*100     #Heat efficiency

# Results
print 'Heat utilised is %6.2f Joules '%(H)
print 'Heat developed is %6.2f Joules '%(Hd)
print 'Thus heat efficiency is %3.2f percent'%(He)

Heat utilised is 357000.00 Joules
Heat developed is 450000.00 Joules
Thus heat efficiency is 79.33 percent


## Example 2.21 Page No : 58¶

In [29]:
# Variables
m = 3.6     #mass in kg
S = 4200.     #specific heat of water
T2 = 95.;T1 = 15.     # temp in degree C
e = 0.84     #efficiency of kettle
Ei = H/e     #Energy input in J
W = 1000     #rating of kettle in watts

# Calculations
H = m*S*(T2-T1)     #heat utilised in J
t = (Ei/W)/60     #time taken in min

# Results
print 'Heat utilised is %7.2f Joules '%(H)
print 'Energy input is %8.2f Joules '%(Ei)
print 'Thus time taken is %2.1f min '%(t)

Heat utilised is 1209600.00 Joules
Energy input is 425000.00 Joules
Thus time taken is 7.1 min