Physics 133: tutorial week 3,4
Capacitance, resistivity and resistance
20. How much charge flows from a 12.0 V battery when it is connected to a 2.0 µF capacitor?
(24 µC)
q = CV = 2.0 × 10−6 × 12 = 2.4 × 10−5 C .
21. The two plates of a capacitor carry +1500 µC and −1500 µC of charge respectively,
when the potential difference is 300 V. Calculate the capacitance.
(5 µF)
C =
1500 × 10−6
q
=
= 5 × 10−6 F = 5 µF .
V
300
22. Calculate the magnitude of the electric field between the plates of a 20 µF capacitor
if they are 2.0 mm apart and each has a charge of 300 µC.
(7500 V m−1 )
300 × 10−6
q
= 15 V .
=
C
20 × 10−6
15
V
= 7.5 × 103 V m−1 .
=
E =
d
2 × 10−3
V =
23. Three capacitors having capacitances of 0.16 µF, 0.22 µF and 0.47 µF are connected
in parallel and charged to a potential difference of 240 V.
(a) Determine the charge on each capacitor.
(b) What is the total capacitance of the combination?
(c) What is the total charge acquired?
(38.4 µC, 52.8 µC, 112.8 µC; 0.85 µF; 204 µC)
(a) The charge on the 0.16 µF capacitor q = CV
The charge on the 0.22 µF capacitor q = CV
The charge on the 0.47 µF capacitor q = CV
(b) For a parallel combination of capacitors,
Ceq = C1 + C2 + C3 = 0.16 + 0.22 + 0.47 =
(c) q = CV = 0.85 × 10−6 × 240 = 204 × 10−6
= 0.16 × 240 = 38.4 µC .
= 0.22 × 240 = 52.8 µC .
= 0.47 × 240 = 112.8 µC .
0.85 µC .
= 204 µC .
24. A 6.0 µF and a 4.0 µF capacitor are connected in series to a 60.0 V battery.
(a) Calculate the equivalent capacitance.
(b) What is the charge on each capacitor?
(c) Determine the voltage across each capacitor.
(2.4 µF; 144 µC; 24 V and 36 V)
(a) For a series combination of capacitors,
1
1
1
=
+
. Hence
Ceq
C1
C2
C1 × C2
6.0 × 4.0
24
C =
=
=
= 2.4 µF, .
C1 + C2
6.0 + 4.0
10
(b) The charge on the series combination is the same as the charge on each capacitor. Thus
q = CV = 2.4 × 60 = 144 µC .
q
144 × 10−6
(c) For the 6.0 µF capacitor V =
= 24 V .
=
C
6.0 × 10−6
−6
q
144 × 10
V =
= 36 V .
=
C
4.0 × 10−6
25. Calculate the energy stored in a 600 pF capacitor that is charged to 100 V.
(3 × 10−6 J)
Energy =
1
2 qV
=
2
1
2 CV
=
1
2
× 600 × 10−12 × 1002 = 3.0 × 10−6 J .
26. It takes 6.0 J of energy to move a 2000 µC charge from one plate of a 5.0 µF capacitor
to the other. Calculate the charge on each plate.
(0.015 C)
From the definition of potential difference we have
6.0
W
= 3.0 × 103 V .
=
V =
q
2000 × 10−6
The charge on each plate is therefore
q = CV = 5.0 × 10−6 × 3.0 × 103 = 15 × 10−3 C .
27. A 16.0 µF and a 4.0 µF capacitor are connected in parallel and charged by a 22.0 V
battery. What voltage is required to charge a series combination of the two capacitors
with the same total energy?
(55 V)
The capacitor combinations both have the same energy E = 12 CV 2 , where C is the equivalent
capacitance of the parallel or series combination. Hence
2
2
1
1
2 × Cparallel × Vparallel = 2 × Cseries × Vseries .
Therefore
16.0 × 4.0
×V2,
(16.0 + 4.0) × 222 =
16.0 + 4.0
which gives
2
1
20 × 222 2
V =
= 55 V .
64
28. 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? (5.4 × 104 C)
I =
q
,
t
∴
q = 15 × 3600 = 5.4 × 104 C .
29. A wire carries a current of 5 A. How many electrons are flowing past any point in
this wire per minute?
(1.9 × 1021 )
I = 5 A = 5 C s−1 = 5 × 60 C min−1 .
5 × 60
Hence in one minute
= 1.88 × 1021 electrons pass any point in the wire.
1.6 × 10−19
30. Calculate the resistance of a 2.0 m length of copper wire 0.15 mm in diameter. Take
(1.9 W)
ρCu = 1.7 × 10−8 W m.
R = ρ
ℓ
= 1.7 × 10−8 ×
A
π×
2.0
1
2
× 0.15 × 10−3
2 = 1.9 W .
31. A wire of length 0.24 m and diameter 3.0×10−5 m has a resistance of 160 W. Calculate
the resistivity of its material.
(4.7 × 10−7 W m)
π×
A
ρ = R = 160 ×
ℓ
1
2
× 3.0 × 10−5
0.24
2
= 4.71 × 10−7 W m .
32. Consider a cube 5 mm on a side, made of carbon. Estimate the resistance between a
(7 × 10−3 W)
pair of opposite faces given ρC = 3.5 × 10−5 W m.
R = ρ
5 × 10−3
ℓ
= 3.5 × 10−5 ×
= 7.0 × 10−3 W .
A
(5 × 10−3 )2
33. A 0.5 W 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.
(8 W)
Assume that the resistivity ρ (as well as the density) of the material is unchanged after stretching. Suppose R, ℓ and A and R′ , ℓ′ and A′ are the resistance, length and area of the sample
before and after stretching respectively. After stretching,
ℓ′ = 4 × ℓ and A′ = 14 × A , (since V = ℓ × A = ℓ′ × A′ = 4 × ℓ × A′ ) ,
hence
ℓ′
4×ℓ
ℓ
R′ = ρ ′ = ρ 1
= 16 × ρ = 16 × R = 8 W .
A
A
×
A
4
34. A 33 W resistor is made from a coil of copper wire whose total mass is 12 g. What
is the diameter of the wire and how long is it? Take dCu = 8.9 × 103 kg m−3 and
(0.18 mm, 51.2 m)
ρCu = 1.7 × 10−8 W m.
Elliminate A from R = ρℓ/A and V = Aℓ to get
1
RV 2
mass
12 × 10−3
ℓ =
where V =
= 1.348 × 10−6 m3 .
=
ρ
density
8.9 × 103
Hence
1
33 × 1.348 × 10−6 2
= 51.2 m .
ℓ =
1.7 × 10−8
2
Using V = Aℓ and A = πr we can determine the diameter. The result is d = 1.83 × 10−4 m .
35. A 100 W light bulb has a resistance of about 12 W when cold and 140 W when “on”
(hot). Estimate the temperature of the filament when “on”, assuming a mean
temperature-coefficient of resistance of 6 × 10−3 ‰−1 .
(∼ 1800‰)
Assume that Rcold ≈ R0 . Then
Rhot ≈ Rcold (1 + αThot )
which gives
140 − 12
T ≈
≈ 1800 ‰ .
12 × 6 × 10−3
36. A coil of wire has a resistance R0 at 0‰ and a temperature coefficient of resistance
α. If its resistance is 20 W at 25‰ and 25 W at 100‰, calculate α and R0 .
(3.64 × 10−3 ‰−1 ; 18.3 W)
We use RT = R0 (1 + αT ) . Thus
R25
R100
=
=
20 =
25 =
R0 (1 + α × 25) and
R0 (1 + α × 100) .
Solving the above relations for α and R0 we find α = 3.64 × 10−3 ‰−1 and R0 = 18.3 W .
37. An iron wire has a resistance of 5.90 W at 20‰ and a gold wire has a resistance of
6.70 W 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‰ to T‰ as 5.0 × 10−3 ‰−1 and 3.4 × 10−3 ‰−1 respectively.)
(166‰)
We denote iron and gold by the superscripts i and g respectively. Suppose the temperature at
which the iron and gold wires have the same resistance is T , then
R0i 1 + αi T = R0g (1 + αg T ) .
The R0 may be determined from R = R0 (1 + αT ) :
R0i =
5.90
W and R0g = 1 + 3.4 ×6.70
W.
1 + 5.0 × 10−3 × 20
10−3 × 20
Solving
6.70
5.90
1 + 5.0 × 10−3 T =
1 + 3.4 × 10−3 T
−3
−3
1 + 5.0 × 10 × 20
1 + 3.4 × 10 × 20
gives T = 166 ‰ .
38. Three 100 W 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?
(33.3 W; 66.7 W; 150 W; 300 W)
(a)
(b)
(c)
(d)
Each of the resistances in the diagrams (a) – (c) above R = 100 W . The equivalent resistance
Req for each of the above combinations is therefore:
(a) Req = R + R + R = 300 W ,
10000
R×R
= 100 +
= 150 W ,
(b) Req = R +
R+R
200
−1
1
1
= 66.7 W and
+
(c) Req =
100 100 + 100
−1
1
1
1
(d) Req =
= 33.3 W .
+
+
100 100 100
39. In each of the combinations below, calculate the equivalent resistance between points
A and B.
60 Ω
10 Ω
50 Ω
8Ω
5Ω
A
B
A
B
40 Ω
6Ω
(a)
(b)
8Ω
3Ω
1Ω
2Ω
4Ω
6Ω
A
A
10 Ω
B
24 Ω
8Ω
5Ω
B
30 Ω
(c)
3Ω
(d)
(74 W; 9.5 W; 7.5 W; 6.8 W)
60 × 40
= 74 W .
60 + 40
(10 + 8) × 6
= 9.5 W .
= 5+
(10 + 8) + 6
(a) Req = 50 +
(b) Req
(c) The resistance Rt of the top branch is Rt =
8 × 24
+ 3 + 1 = 10 W . Hence
8 + 24
10 × 30
= 7.5 W .
10 + 30
(d) The 6 W , 5 W and 3 W resistors on the right are in series and hence have an equivalent
resistance of 14 W . This equivalent resistance is in parallel with the 8 W resistor, which
gives a resistance of 14×8
14+8 = 5.09 W , which in turn is in series with the 4,W resistor yielding
4 + 5.09 = 9.09 W . This 9.09 W resistance is in parallel with the 10 W resistor giving an
equivalent resistance of 9.09×10
9.09+10 = 4.76 W in series with the 2 W resistor. The resistance
between A and B is therefore 2 + 4.76 = 6.76 W .
Req =
40. A standard resistor marked “5 ohms” is tested and found to have an actual resistance of 5.05 ohms. What length of nichrome wire of resistance 135 W m−1 must be
connected in parallel with the resistor to make the combined resistance of 5 ohms?
(3.74 m)
Let the resistance of the nichrome wire be R . R is in parallel with the 5.05 W resistor to give
an equivalent resistance of 5.00 W . Hence
1
1
1
+
=
, which gives R = 505 W .
R 5.05
5.00
The resistance of 1 m of nichrome wire is 135 W . Hence the length of nichrome wire required to
have a resistance of 505 W is
505
ℓ =
= 3.74 m .
135