Chapter 5:Resistance and Wire Size

Problem no:5.1, Page no: 46

In [2]:
from __future__ import division

#Cal of Length of wire

#Initialization

L1=20                      #Length in ft

R1=13                      #Resistance in Ohm

R2=8                       #Resistance in Ohm


#Calculation

L2=(R2*L1)/R1

print "L2=",round(L2,1),"ft"

L2= 12.3 ft

Problem no.:5.2, Page no: 47

In [2]:
#Cal of Resistance

#Initialization

R1=2                       #Resistance in Ohm

D1=0.04                    #Diameter in inches

D2=0.10                    #Diameter in inches




#Cal of Resistance when diameter is 0.10 in

R2=(D1/D2)**2*R1

print "(a) The resistance when diameter is 0.10 in is",R2,"Ohm"




#Cal of Resistance when diameter is 0.01 in 

D2=0.01                    #Diameter in inches

R2=(D1/D2)**2*R1

print "(b) The resistance when diameter is 0.01 in is",int(R2),"Ohm"

(a) The resistance when diameter is 0.10 in is 0.32 Ohm
(b) The resistance when diameter is 0.01 in is 32 Ohm

Problem no:5.4, Page no:47

In [5]:
#Cal of Cross-sectional area

#Initialization

D=0.06408                    #Diameter in inches

#Calculation

D=0.06408/0.001              #Convert Diameter in mils

A=D**2

print "A_cmil=",int(A),"cmil"

A_cmil= 4106 cmil

Problem no:5.5, Page no:48

In [7]:
from __future__ import division

#Comparing the Resistances of wires

#Initialization

A10=10380                         #Area in cmil

A14=4106                          #Area in cmil

from decimal import *

#Calculation

R14_R10=A10/A14

print "The no. 14 wire has",round(R14_R10,2),"times as much resistance as the no. 10 wire"

The no. 14 wire has 2.53 times as much resistance as the no. 10 wire

Problem no:5.6, Page no:48

In [9]:
#Cal of Resistance

#Initialization

D_mil=0.080                       #Diameter in inches

l=1500                            #Length in ft

rho=10.4                          #Specific resistance in Ohm

#Calculation

D_mil=0.080/0.001                 #Diameter in mils

A_cmil=D_mil*D_mil                #Area in cmil

R=(rho*l)/A_cmil                  #Specific Resistance formula

print "R=",round(R,2),"ohm"

R= 2.44 ohm

Problem no:5.7, Page no:48

In [13]:
from __future__ import division

#Cal of Length of Nichrome wire

#Initialization

D_mil=20                         #Diameter in inches

rho=600                          #Specific resistance in Ohm

R=5                              #Resistance in ohm



#Calculation

A_cmil=D_mil*D_mil               #Area in cmil

l=(R*A_cmil)/rho #Specific Resistance formula

print "l=",round(l,2),"ft"

l= 3.33 ft

Problem no:5.8, Page no:48

In [14]:
#Cal of minimum diameter of wire

#initialization

l=400                     #Length in ft

R=1.5                     #Resistance in ohm

rho=10.4                  #Specific resistance for copper wire in cmil/ft

import math

#Calculation

A_cmil=(rho*l)/R

D_mil=math.sqrt(A_cmil)

print "D_mil=",round(D_mil,1),"mils"

D_mil= 52.7 mils

Problem no:5.14, Page no: 51

In [16]:
#Cal of Maximum length

#Initialization

A=6530                          #Area in cmil

V=5                             #Voltage in V

I=15                            #Current in A

#Calculation

l=(A*V)/(10.4*I)

print "l=",int(l),"ft"

l= 209 ft

Problem no:5.15, Page no: 52

In [1]:
#Cal of Resistance

#Initialization

l=80                            #length in m

A=2.5                           #Area in mm^2

rho=0.0175                      #Specific Resistance of copper wire in mm^2/m

#Calculation

R=(rho*l)/A

print "R=",R,"Ohm"

R= 0.56 Ohm

Problem no:5.16, Page no: 52

In [2]:
#Cal of Length of wire

#Iniatialization

A=0.1                        #Area in mm^2

R=3                          #Resistance in Ohm

rho=0.0175                   #Specific Resistance of copper wire in mm^2/m

#Calculation

l=(R*A)/rho

print "l= ",int(l),"m"

l=  17 m

Problem no:5.17, Page no: 52

In [3]:
#Cal of cross-sectional area

#Initialization

l=50                      #Length in m

R=0.2                     #Resistance in ohm

rho=0.0175                #Specific Resistance of copper wire in mm^2/m

#Calculation

A=(rho*l)/R

print "A=",round(A,1),"mm^2"

A= 4.4 mm^2