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4 Electricity and Magnetism
3
Chapter 3 Domestic Electricity
Domestic Electricity
By P 
V2
,
R
Practice 3.1 (p.133)
P2Ω = 0.0703ε2
1
C
P5Ω = 0.0781ε2
W is a unit of power, others are units of
P10Ω = 0.0391ε2
energy.
2
7
(a)
A
(b)
P = I2R  R (I constant)
 P1 : P2 : P3 = R1 : R2 : R3
3
C
1
V2

(V constant)
R
R
1 1 1
 P1 : P2 : P3 =
:
:
R1 R2 R3
(c)
P=
4
The answer obtained in (b)(ii) is smaller
than the rated current.
8
(a)
500 W
Let R be the resistance of one bulb and ε be
(b)
Current =
the e.m.f. of the battery.
V2
By P =
,
R
(c)
B
Pseries =
2
2R
Pparallel =
5
9
= 10 W
2
R
 
2
P 500
=
= 2.08 A
V 240
V 2 240 2
Resistance =
=
= 115 Ω
P
500
By P = I2R,
power of 5- resistor = 0.52  5 = 1.25 W
power of 1- resistor = 0.52  1 = 0.25 W
= 4Pseries = 4  10 = 40 W
Voltage across 2- resistor
= total voltage across 5- and 1- resistors
B
= 0.5  (5 + 1) = 3 V
Let R be the resistance of one bulb and ε be
Voltage across 1.5-Ω resistor = 6  3 = 3 V
V2
By P =
,
R
32
power of 2- resistor = = 4.5 W
2
32
power of 1.5- resistor =
=6W
1 .5
the e.m.f. of the battery.
P=
2
R
Total power dissipated by two bulbs
2 P
=
=
2R 2
6
V 2 220 2
=
= 24.2 Ω
P 2000
V 2 200 2
(i) Power =
=
= 1650 W
R 24 .2
V 200
(ii) Current = =
= 8.26 A
R 24 .2
V 220
Rated current = =
= 9.09 A
R 24 .2
Resistance =
B
16
1
Ω
 2=
1 1
3

5 10

I2Ω =
= 0.1875ε
Req
10
(a)
Req =
V2Ω = IR = 0.1875ε × 2 = 0.375ε
V5Ω = V10Ω = ε – V2Ω = 0.625ε
New Senior Secondary Physics at Work (Second Edition)
 Oxford University Press 2015
Appliance
Air
conditioner
Plasma TV
Power
Period
Energy
rating
of use /
consumed
/ kW
h
/ kW h
1.5
1.0
1.5
0.1
1.0
0.1
1
4 Electricity and Magnetism
Kettle
Water
heater
Lamp
Chapter 3 Domestic Electricity
2.0
0.1
0.2
3.5
0.2
0.7
0.5
1.0
0.5
12
Energy consumption difference per day
= (0.06  0.008) × 6 = 0.312 kW h
(b)
Air conditioner
(c)
Total energy consumed
= 1.5 + 0.1 + 0.2 + 0.7 + 0.5 = 3 kW h
11
(b)
Cost difference per day
14
(a)
= 0.312 × 1.1 = $0.3432
180  13.5
Days needed =
= 485
0.3432
V2
By P =
, for the same input voltage,
R
New reading
the circuit having the largest overall
= 14 212 + 3 = 14 215 kW h
resistance has the lowest power.
Total power = 100 × 2 = 200 W = 0.2 kW
The bulbs in circuit 1 are all connected
Total energy = 0.2 × 3.5 = 0.7 kW h
in series, so it has the largest resistance
Cost = 0.7 × 1.1 = $0.77
V2
(a) By P =
,
R
110 2
resistance of X =
= 201.7 
60
110 2
resistance of Y =
= 121 
100
and lowest power.
(b), (c)
The voltages across the bulbs are
calculated as follows.
VA = VB = VC = 2 V
VD = VE = 3 V
When connected in series to mains,
220 2
total power =
= 150 W
201 .7  121
(b)
VF = 6 V
If the resistance of a bulb is R, the
equivalent resistance of G and H is 0.5R.
0.5R
VG = VH =
×6=2V
0.5R  R
Current drawn when connected in series
220
=
= 0.682 A
201 .7  121
power of X
VI = 6  2 = 4 V
V2
P
 V2 (R constant)
R
= 0.6822  201.7 = 93.8 W > 60 W
VP
power of Y
 F is the brightest and A, B, C, G, H
= 0.6822  121 = 56.3 W < 100 W
are the dimmest.
By P = I2R,
X glows brighter than when operated at
13
2
(a)
rated value.
Practice 3.2 (p.148)
Y glows dimmer than when operated at
1
B
rated value.
2
B
Energy consumption difference per day
3
B
= (0.06  0.012) × 6 = 0.288 kW h
The current that can flow through the fuse is
Cost difference per day
smaller than 10 A.
= 0.288 × 1.1 = $0.3168
49.5  13.5
Days needed =
= 114
0.3168
By P = VI,
Power of cooker < 220 × 10 = 2200 W
New Senior Secondary Physics at Work (Second Edition)
 Oxford University Press 2015
4 Electricity and Magnetism
4
Chapter 3 Domestic Electricity
D
(b)
The appliance still works because
Fuse of the same fuse value should be used.
current only flows in the live wire and
 (1) is incorrect.
the neutral wire under normal condition.
Damaged cables and plugs should be replaced.
10
(a)
 (3) is incorrect.
5
L
N
E
blue
(b)
6
(a)
(b)
7
8
brown
The appliance has an insulating case so
P 1200
I1 = =
= 5.4545 A  5.45 A
V
220
I 5.4545
I2 = 1 =
= 2.73 A
2
2
Revision exercise 3
To prevent a current larger than normal
Concept traps (p.151)
from flowing through the circuit
P 500
I= =
= 2.27 A
V 220
no current can flow through the case.
1
F
This would keep the appliance live even when
the switch is open, but no current would flow
The fuse value should be slightly higher than
through the appliance unless a complete
the operating current, so the 3-A fuse should
circuit is formed (e.g. no one touches the
be used.
conducting part of the appliance).
(a)
2
T
Current changes its direction alternately in an
a.c., i.e. it can flow from the live wire through
the appliance to the neutral wire and vice
versa.
(b)
Any one of the following:
Thinner and cheaper cables can be used
in the ring main.
Multiple-choice questions (p.151)
3
Energy used = 3 × 0.06 kW × 5 h = 0.9 kW h
All sockets can be connected in parallel
to the ring main.
9
(a)
If the live wire gets loose and touches
B
Cost = 0.9 × 1.1 = $0.99
4
A
current will flow to the earth through the
Let R be the resistance of one bulb.
6 2 36
P=
=
R R
earth wire and blow the fuse in the live
Equivalent resistance of the circuit
wire. The appliance is prevented from
1 
 1

=
 =R
RR RR
the metal body of the appliance, a large
becoming live.
New Senior Secondary Physics at Work (Second Edition)
 Oxford University Press 2015
1
3
4 Electricity and Magnetism
Total power =
5
Chapter 3 Domestic Electricity
6 2 36
=
=P
R
R
 (1) is incorrect.
If there are too many sockets, the main ring
A
may be overloaded.
For the microwave oven,
V 2 110 2
R=
=
= 12.347 Ω
980
P
220
V
I= =
= 17.8 A > 15 A
R 12 .347
 (3) is incorrect.
For the bulbs in (2),
220 2
R=
 6 = 4840 Ω
60
V
220
I= =
= 0.0455 A < 15 A
R 4840
P
For the bulbs in (3), by I = ,
V
100
I=
 4 = 1.82 A < 15 A
220
6
9
Equivalent resistance of the three resistors
1
1 
1
= 
 = 2.25 Ω
 3 4  5
Current through the bulb =
= IR = 2 × 2.25 = 4.5 V
E.m.f. of the battery = 4.5 + 3 = 7.5 V
10
minimized. To do so, the heating elements
should be connected in parallel.
11
from the live wire to the neutral and earth
A fault does not always result in a current
wires through the heating element. The size of
flowing in the earth wire. For example, a fault
the current in the live wire is equal to the
develops when the live wire touches the
operating current so the fuse would not break.
neutral wire. In this case, no current flows in
 (2) is incorrect.
the earth wire.
If the heating element is broken, the circuit is
 (3) is incorrect.
not complete and no current would flow
A
through the live wire, so the fuse would not
If both sockets draw a current of 15 A, it is not
break.
enough to use cables rated less than 15 A in
the ring main, as shown below.
15 A
L
A
If the insulation at Y is worn out, current flows
 (2) is incorrect.
8
V2
, to boil water the fastest, P
R
should be maximized and hence R should be
A
wires under normal conditions.
A
By P 
A
Current flows through the live and neutral
P 6
= =2A
V 3
Voltage across the resistor network
The switch should be installed at the live wire.
7
C
12
A
When connected in parallel, the voltage across
V2
the resistor r is constant. By P 
, the
R
15 A
30 A
 (3) is incorrect.
15 A
power dissipated by r is constant.
15 A
4
New Senior Secondary Physics at Work (Second Edition)
 Oxford University Press 2015
4 Electricity and Magnetism
Chapter 3 Domestic Electricity
When connected in series, the voltage across r
r
is given by Vr 
V and the power
Rr
V2
r
V2.
dissipated is Pr  r 
r
( R  r) 2
13
(HKCEE 2009 Paper 2 Q41)
14
(HKCEE 2009 Paper 2 Q20)
15
(HKDSE 2012 Paper 1A Q33)
16
(HKDSE 2013 Paper 1A Q31)
19
(a)
(a)
= 9.09 + 4.55 + 0.682
V
Apply I = .
R
(i)
= 14.3 A
1M
Current through the body
220
=
= 0.0022 A
100  10 3
(ii) Current through the body
220
=
= 0.22 A
1  10 3
(b)
(b)
power
supply
1M
It is safe to do so.
1A
It disconnects the circuit when the
Live wire.
neutral wire, the appliances are still at
high potential after the circuit is opened.
1A
20
(a)
X: live wire; Y: neutral wire
(b)
As the electric potential at wire Y is
always zero,
bottom of
staircase
1A
(c)
1A
1A
Wire X has a large changing potential.
1A
(Lamp and power supply correctly
(b)
1A
nothing will happen when it touches the
metal case.
(Two-way switches)
1A
If the circuit breaker is connected to the
1A
1A
top of
staircase
1A
to avoid danger due to overloading. 1A
(c)
lamp
(a)
< 15 A
current is too large
1A
Do not touch any electrical appliances
with wet hands.
18
1M
Total current
Conventional questions (p.153)
17
P
.
V
2000
Ikettle =
= 9.09 A
220
1000
Iiron =
= 4.55 A
220
150
ITV =
= 0.682 A
220
Apply I =
drawn)
1A
A large current will flow through wires
(Correct labels)
V2
By P =
,
R
V 2 240 2
R

= 115.2 Ω
P
500
1A
X and Z via the metal case due to a short
At 220 V,
220 2
Power =
115 .2
= 420 W
New Senior Secondary Physics at Work (Second Edition)
 Oxford University Press 2015
1M
(d)
circuit
1A
and the fuse would blow.
1A
No current would flow through the metal
case as the circuit is incomplete, and the
fuse would not blow.
1M
1A
1A
However, if someone touches the metal
case, an electric shock would result. 1A
5
4 Electricity and Magnetism
21
(a)
Chapter 3 Domestic Electricity
The switch and the fuse should be placed
at the live wire.
1A
so a 10-A fuse should be used.
23
(a)
Otherwise, the lamp is still live even if
the switch is off or the fuse blows.
If one bulb is broken, the circuit is not
complete and all the bulbs go out.
1A
(b)
1A
touches the conducting part of the lamp.
The new bulb shares less voltage when
1A
compared to the original one.
live
larger share of voltage.
Or
1A
The equivalent resistance of all the bulbs
(Correct positions of fuse and switch)
decreases.
1A
increases.
1M
The 0.5-A fuse should be used.
24
(a)
(i)
(ii) No.
one of W, X, and Y will exceed the
will burn out.
the fuse will blow.
1A
Afterwards, the circuit is not
1M
complete and no current flows
through Z.
(b)
RA
l
 4  10 3 

26 .9  π  
2 

=
2
(c)
1A
Z has a lower resistance than Y. By
P = I2R, the power dissipated in Z
1M
is smaller,
1A
so Z glows dimmer than Y.
1A
2
(ii) Y is the brightest among them. W
and X have the same brightness.
= 1.69  10–4  m
1A
P 1800
Rated current = =
= 8.18 A 1A
V
220
than the rated current,
(i)
1A
1A
The fuse value should be slightly larger
6
voltage limit before Z and the bulb
As the total current is larger than 0.5 A,
Resistivity =
1A
circuit increases, the voltage across
Total current = 0.2975  2 = 0.595 A
= 26.9 Ω
1A
By V = IR, as the current in the
1A
V2
Resistance of coil =
P
220 2
=
1800
1A
Z has a lower resistance than the
others.
1M
(b)
1A
Thus, the current through the bulbs
Resistance of the lamp
V 2 220 2
=
=
= 806.7 Ω
P
60
Current through the circuit
V
240
= =
= 0.2975 A  0.298 A 1A
R 806 .7
(a)
1A
The power of the other bulbs increases.
neutral
22
1A
All the other bulbs in the chain have a
(b)
(d)
1A
The power of the other bulbs increases.
An electric shock may result if someone
(c)
1A
1A
(iii) When Z is removed, the voltage
across Y increases and therefore Y
glows brighter.
1A
1A
New Senior Secondary Physics at Work (Second Edition)
 Oxford University Press 2015
4 Electricity and Magnetism
25
(a)
Chapter 3 Domestic Electricity
The voltage across W and X stays
Also, some of the energy is absorbed by
the same, so their brightness does
the water as latent heat of vaporization
not change.
to turn the water to steam.
1A
The shutters of the socket holes have to
(c)
be opened by the earth pin which the
two-pin plug does not have.
(b)
1A
A longer earth pin ensures that the earth
supply and S is connected to P, a
wire is connected before the live wire.
large current will flow through the
kettle and the 20-A fuse will break.
Method 1:
metal case to the earth pin.
1A
If the kettle is connected to a 110-V
1A
supply and S is connected to Q, a
If a fault develops, a large current flows
small current will flow through the
between the live and earth wires and
kettle. The kettle will still work,
break the fuse. This prevents a large
but with a smaller power.
1A
2
V
(iii) By P =
,
R
V 2 110 2
R1 =
=
= 6.05 Ω
1A
P 2000
V 2 220 2
R1 + R2 =
=
= 24.2 Ω
P 2000
current from flowing through the user’s
body.
1A
Method 2:
Use double insulation.
1A
Current cannot flow through the
R2 = 24.2 – 6.05 = 18.2 Ω
insulating case even if a fault develops.
1A
Energy consumed
15
= 1 kW 
h  30 = 7.5 kW h
60
Cost = 7.5  1.1 = $8.25
26
(a)
(i)
27
(a)
= 1.298 × 106 J
1A
Time needed
E 1.298 10 6
= =
= 649 s
P
2000
1M
(b)
1A
can no longer fully immerse in the water.
1A
1M
= 2000 × 6 × 60
= 720 000 J
1A
(c)
Some of the energy provided by the
kettle is lost to the surroundings and the
kettle itself.
Much energy is lost in the process. 1A
V2
By P 
, if the voltage is halved, the
R
1
power becomes
of the original. 1A
4
The time needed is 4 times longer. 1A
1A
(d)
New Senior Secondary Physics at Work (Second Edition)
 Oxford University Press 2015
1A
As the water is boiled away, the heater
(ii) Energy provided by kettle
= Pt
1M + 1M
= 0.5(4200)(100 – 20) + 0.5(2.26 × 106)
1M
= 2 × 4200 × (100 – 25)
= 630 000 J
1A
Energy needed
= mcΔT + mlf
Energy absorbed by water
= mcΔT
(b)
1A
(ii) If the kettle is connected to a 220-V
Use a three-pin plug and connect the
(d)
(1) P
(2) Q
1A
(c)
(i)
1A
His suggestion is incorrect.
1A
7
4 Electricity and Magnetism
The total resistance is doubled when the
by R  
two heaters are in series.
1A
V2
By P 
, the total power of the two
R
increases.
heaters is halved to that of a single
increases.
heater.
(HKCEE 2010 Paper 1 Q9)
29
(HKDSE 2014 Paper 1B Q8)
V 12
(a) (i) rp = =
=6Ω
I 2
1 1
(ii)
= 8
Rp 6
6
Rp =
8
= 0.75 Ω
(b)
(c)
1A
1A
and are inexpensive.
(c)
1A
The equivalent resistance of the speaker
decreases if they are connected in
1M
1A
parallel.
1M
1A
Therefore, the cables have a larger share
of voltage,
1A
and the voltage across the speakers
decreases.
For the heater,
(b)
1A
Copper wires have a lower resistance,
1A
1M
1A
Therefore, the power output of the
speakers decreases.
(iii) For one strip,
P = VI = 12 × 2 = 24 W
l
, the resistance of the wire
A
Hence the voltage drop due to the cable
1A
28
30
Chapter 3 Domestic Electricity
P = 24 × 8 = 192 W ≈ 200 W 1A
RA
ρ=
1M
l
6  2.4 10 4  2.0 10 3
=
1M
0.9
= 3.2  106 Ω m
1A + 1A
2
V
(i) By P 
, if the applied voltage
R
is the same and an equal power is
expected, the equivalent resistance
must be the same.
0.75
(ii) rs =
= 0.0938 Ω
8
1A
1A
(iii) The parallel circuit would be
chosen. If one of the strips is
damaged, others still work if a
parallel circuit is used.
1A
Physics in article (p.157)
31
(a)
If the length is increased or the
cross-sectional area is decreased,
8
New Senior Secondary Physics at Work (Second Edition)
 Oxford University Press 2015