PHYSICS 120 : ELECTRICITY AND MAGNETISM
TUTORIAL QUESTIONS
CURRENT, RESISTANCE AND RESISTIVITY
Question 48
Most of the wiring in a typical house can safely handle about 15 A of current. At this current
level, how much charge flows through a wire in one hour?
Current is defined as the rate of change of charge, therefore
∆q
dq
=
I =
dt
∆t
∴ ∆Q = I∆t
= 15 × (60 × 60)
= 5.4 × 104 C
Question 49
A wire carries a current of 5.0 A. How many electrons are flowing past any point in this wire
per minute?
The amount of charge flowing past in one minute is
∆Q = I∆t = 5 × 60 = 300 C
The number of electrons that give this amount of charge is given by
Q = Ne
Q
∴N =
e
300
1.6 × 10−19
= 1.9 × 1021
=
Question 50
Calculate the resistance of a 2.0 m length of copper wire 0.15 mm in diameter. Take ρCu =
1.7 × 10−8 Ω m.
Resistance and resistivity are related by
l
R = ρ
A
∴ R = (1.7 × 10−8 ) ×
2
π(0.15/2 × 10−3 )2
= 1.9 Ω
page 1 of 7
assuming the wire is circular
PHYS120 ELECTRICITY AND MAGNETISM: TUTORIAL QUESTIONS
Question 51
A wire of length 0.24 m and diameter 3.0 × 10−5 m has a resistance of 160 Ω. Calculate the
resistivity of the material.
The resistivity is given by
A
l
160 × π × (3/2 × 10−5 )2
=
0.24
−7
= 4.7 × 10 Ω m
ρ = R
Question 52
Consider a cube 5.0 mm on a side, made of carbon. Estimate the resistance between a pair
of opposite faces given ρC = 3.5 × 10−5 Ω m.
The cross-sectional area through this cube is the area of one of the faces ((5.0 × 10−3 )2 m−2 )
and the length through it is simply the side 5.0 × 10−3 m. So the resistance through this
cube of carbon is
R=ρ
5.0 × 10−3
l
= 3.5 × 10−5 ×
= 0.70 × 10−2 Ω
A
(5.0 × 10−3 )2
Question 53
A 0.5 Ω wire is drawn out (stretched) to four times its original length. Assuming that the
density of the wire does not change, calculate its new resistance.
Since the density of the wire does not change, its volume V has to remain unchanged as
well, and it can be written as V = Al, where A is the cross-sectional area and l is its length.
The density ρ before and after stretching the wire is unchanged and it can be written as
ρ=
R1 A1
R2 A2
=
l1
l2
After stretching the length of the wire is four times the original length, therefore l2 = 4l1 .
For the volume to remain the same, the cross-sectional area A2 after stretching must be a
quarter of the original area A1 :
A2 l2 =
A1
4l1 = A1 l1 = V
4
page 2 of 7
PHYS120 ELECTRICITY AND MAGNETISM: TUTORIAL QUESTIONS
Solving for the new resistance gives
R2 A2
R1 A1
=
l1
l2
l2 A1
⇒ R2 = R1
l1 A2
4l1 A1
= R1
l1 A1 /4
= 16R1
= 16 × 0.5
= 8Ω
Question 55
A 100 W lightbulb has a resistance of about 12 Ω when cold and 140 Ω when ”on” (hot). Estimate the temperature of the filament when ”on” assuming a mean temperature-coefficient
of resistance of 6.0 × 10−3 ◦ C−1 .
The resistance as a function of temperature is given by
RT = R0 (1 + αT )
Let R0 = RCOLD be the resistance when ”cold” and RT = RHOT the resistance when ”hot”,
therefore
RHOT − RCOLD
T =
αRCOLD
140 − 12
=
6.0 × 10−3 × 12
= (1.8 × 103 )◦ C
Question 56
A coil of wire has resistance R0 at 0◦ C and a temperature coefficient of resistance α. If its
resistance is 20.0 Ω at 25.0◦ C and 25.0 Ω at 100◦ C, calculate α and R0 .
The temperature dependent resistance is given by
RT = R0 (1 + αT )
therefore
R25 = R0 (1 + 25α)
R100 = R0 (1 + 100α)
Dividing (I) by (II) gives
R25
20
1 + 25α
=
=
R100
25
1 + 100α
1
= 3.64 × 10−3 ◦ C−1
∴α =
275
page 3 of 7
(I)
(II)
PHYS120 ELECTRICITY AND MAGNETISM: TUTORIAL QUESTIONS
From (I), the resistance R0 at 0◦ C is
20
1 + 20α
20
=
1 + 25/275
= 18.3 Ω
R0 =
Question 57
An iron wire has a resistance of 5.90 Ω at 20◦ C and a gold wire has a resistance of 6.70 Ω
at the same temperature. At what temperature T do the wires have the same resistance?
(Take the mean temperature-coefficients of resistance of iron and gold over the range from
20◦ C to T )◦ C as 5.00 × 10−3 ◦ C−2 and 3.40 × 10−3 ◦ C−1 respectively.
For iron, the resistance at 20◦ C is given by
(Fe)
R20 = R0 (1 + 20α)
therefore the resistance of the iron wire at 0◦ C is
5.90
(Fe)
R0
=
1 + 20 × 5.00 × 10−3
= 5.36 Ω
For the gold wire, its 0◦ C resistance is
(Au)
R0
6.70
1 + 20 × 3.40 × 10−3
= 6.27 Ω
=
The resistances will be equal at a temperature T , therefore
(Fe)
RT
(Au)
= RT
(Fe)
(Au)
R0 (1 + α(Fe) T ) = R0
(1 + α(Au) T )
(Au)
∴T =
R0
(Fe)
(Fe)
− R0
(Au)
R0 α(Fe) − R0
α(Au)
6.27 − 5.36
=
(5.36 × 5.00 × 10−3 ) − (6.27 × 3.40 × 10−3 )
= 166◦ C
page 4 of 7
PHYS120 ELECTRICITY AND MAGNETISM: TUTORIAL QUESTIONS
Question 58
Three 100 Ω resistors can be connected together in four different ways, making series and/or
parallel combinations. What are these four ways and what is the net resistance in each case?
Two combinations are for the three resistors to all be in series (i) or all in parallel (ii) with
one another. The other two cases involve one resistor in series with a parallel combination
of the other two, and finally one resistor in parallel with a series combination of the other two.
(i) The net or equivalent resistnace REQ is
REQ = R1 + R2 + R3 = 3R = 300 Ω
since the three resistors are the same.
(ii)
1
3
3
=
⇒ REQ = 33.3 Ω
=
REQ
R
100
(iii)
REQ = R +
(R2 )
3
= R = 150 Ω
R+R
2
(iv)
REQ =
2 R2
(R + R) × R
=
= 66.7 Ω
(R + R) + R
3R
Question 59
In each of the combinations below, calculate the equivalent resistance between points A and
B.
(a)
REQ = 50 +
60 × 40
= 74 Ω
60 + 40
(b)
page 5 of 7
PHYS120 ELECTRICITY AND MAGNETISM: TUTORIAL QUESTIONS
REQ = 5 +
(10 + 8) × 6
= 9.5 Ω
(10 + 8) + 6
(c)
REQ =
8×24
8+24
8×24
8+24
+ 3 + 1 × 30
= 7.5 Ω
+ 3 + 1 + 30
(d)
The equivalent circuit becomes
The equivalent resistance between points A and B is then
REQ
10 ×
=2+
10 +
100
11
100
11
=2+
page 6 of 7
100
= 6.8 Ω
21
PHYS120 ELECTRICITY AND MAGNETISM: TUTORIAL QUESTIONS
Question 60
A standard resistor marked ”5 Ohms” is tested and found to have an actual resistance of
5.05 Ω. What length of nichrome wire of resistance 135 Ω m−1 must be connected in parallel
with the resistor to make the combined resistance 5.00 Ω?
A length of wire of resistance R is connected in parallel to the 5.05 Ω resistor so that the
equivalent resistor is 5.00 Ω.
5.00 =
R × 5.05
R + 5.05
∴R=
5.00 × 5.05
= 505 Ω
5.05 − 5.00
Since the resistance per unit length of the wire is 135 Ω m−1 , the length of wire l needed to
obtain a resistance of 505 Ω is
505
= 3.74 m
l=
135
page 7 of 7