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)
```

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)
```

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)
```

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)
```

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)
```

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)
```

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)
```

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)
```

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)
```

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)
```

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 )
```

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)
```

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)
```

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)
```

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)
```

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)
```

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)
```

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)
```

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)
```

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)
```

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)
```