Physics Courseware
Physics II
Resistivity
Problem 1.- A coaxial cable has an internal copper wire of 1.2mm diameter and an outer
cylindrical braided conductor of diameter 5.5mm. Calculate the resistance between the
two conductors, if the length of the wire is 20m and the resistivity of the dielectric in the
gap between the conductors is 250 Ωm.
dr
, where “dr” is the differential in the direction
A
of charge flow. In this case it is a radial flow. The area in the denominator of that
equation is the area that the charge flows through and in this case is the lateral area of a
cylinder whose length is 10m and whose radius is variable from 1.2mm to 5.5mm.
Solution: Consider the equation dR = ρ
So: dR = ρ
0.0055
R=
∫
0.0012
ρ
dr
and integrating:
2πr(20)
dr
ρ
 0.0055 
ln
=
 = 3.03 Ω
2πr(20) 2π (20)  0.0012 
Problem 2.- Find the resistance of a tapered copper wire that has a radius of 1mm on
one end and 1.5mm on the other. Consider the length of the wire to be 10m.
[ ρ Cu = 1.72 × 10 −8 Ωm ]
Solution: The radius of the wire is: r = 0.001 + 0.00005 x , where x is the distance from
one end.
2
The cross section area is: A = πr 2 = π (0.001 + 0.00005 x )
The differential of resistance is then:
dR =
ρdx
A
=
ρdx
π (0.001 + 0.00005 x )2
Integrating:
10
R=
ρdx
∫ π (0.001 + 0.00005 x )
2
0
R=−
ρ
+
=−
10
ρ
π (0.001 + 0.00005 x )0.00005 0
ρ
= 0.036 Ω
π (0.0015)0.00005 π (0.001)0.00005
Problem 2a.- A wire is uniformly tapered from a radius R1=1mm to a final radius R2=3.0
mm.
The length of the wire is 20 m, so the equation that describes the radius is:
r = 0.001 + 0.0001x
Calculate the resistance of the wire
[The resistivity of copper is 1.7x10-8 Ωm]
L
Solution: As can be shown by integrating R = ∫ ρ
0
R=ρ
L
,
πab
R = 1.7 × 10 −8
dx
b−a 

πa +
x
L 

2
where a and b are the radii at the two ends.
20
= 0.036Ω
3.1416(0.001)(0.003)
Problem 3.- Two wires, one of copper and one of aluminum have the same length and
same electric resistance. Which one is thicker? How do you know?
Solution: Aluminum has to be thicker to compensate for its higher resistivity.
Problem 4.- Which has more resistance a 100-W light bulb or a 75-W light bulb?
Explain.
Solution: Since Power=V2/R, the resistance of the 100-W light bulb is less than the one
of the 75-W light bulb.
Problem 5.- A coil is made with 10 meters of copper wire with diameter 0.65 mm.
Calculate the voltage drop in the wire with a current of I=2.5 Amps.
[The resistivity of copper is 1.7x10-8 Ωm]
Solution: The calculation is similar to the one we did in the lab:
L
10
= 1.7 × 10 −8
= 0.51 Ω
A
π (0.65 ×10 −3 ) 2 / 4
So the voltage drop is V = IR = 0.51× 2.5 = 1.3 volts
R=ρ
Problem 6.- What is the electric resistance along a 20m-long bar of copper that has a
thickness of 0.5cm and a width of 4cm.
[The resistivity of copper is 1.67x10-8 Ωm]
[Be careful with units, make any necessary conversions]
Solution: R = ρ
L
20
= 1.67 × 10 −8
= 0.00167 Ω
A
0.005 × 0.04
Problem 7.- What is the electric resistance along a pipe made of copper that has a length
of 15m, internal radius R1=10mm and external radius R2=12mm.
[The resistivity of copper is 1.67x10-8 Ωm]
[Be careful with units, make any necessary conversions]
Solution:
L
15
R = ρ = 1.67 × 10 −8
= 0.0018 Ω
2
A
π(0.012 − 0.010 2 )
Problem 8.- Explain the reason why metals have higher resistance at higher
temperatures?
Solution: In metals, when you increase the temperature you increase the frequency of
collisions between electrons and defects, other electrons and impurities, so the resistance
increases.
Problem 9.- Find the resistance of a 25m length of copper wire of diameter 2.85 mm and
compute the voltage drop when carrying a current of 12A.
ρ Copper = 1.68 x10 −8 Ωm
Resistivity of Copper
L
, where ρ is the resistivity, L
A
πd 2
is the length and A is the area of cross section of the wire, which is
, d being the
4
diameter. So:
L
4Lρ
R =ρ
= 2
2
 πd  πd


 4 
Solution: To find the resistance we use the formula R = ρ
With the values given in the problem:
R=
4(1.68x10 −8 Ωm)(25m)
= 0.066 Ω
(3.1416)(0.00285m) 2
To find the drop in voltage we use Ohm’s law:
V = RI =(0.066 Ω)(12 A)= 0.79 V