Download IV Ch 3

4
Electricity and Magnetism
3
Chapter 3 Domestic Electricity
Domestic Electricity
Practice 3.1 (p. 119)
1
6
voltage across them is the same. By
C
W is a unit of power while the others are units
P=
of energy.
2
3
(a) Power ratio = 6 : 3 : 2
1 unit refers to 1 kW h.
100
 3.5  0.9 = $0.63
Cost = 2 
1000
(b) Power ratio = 4 : 3 : 1
1
1
1
(c) Power ratio =
:
:
R X RY R Z
V2
,
R
V 2 220 2
R=
=
= 24.2 
P 2000
7
(a) By P =
power of 1- resistor = 0.52  1 = 0.25 W
Voltage across 2- resistor
= total voltage across 3- and 1- resistors
= 0.5  (3 + 1) = 2 V
V 2 200 2
=
= 1650 W
R
24 .2
V 200
(ii) I = =
= 8.26 A
R 24 .2
E 500
(a) P = =
= 500 W
t
1
(b) (i)
By P = I2R,
power of 3- resistor = 0.52  3 = 0.75 W
24.2 .
Voltage across R = 6  2 = 4 V
2
Current through R = 0.5 + = 1.5 A
2
V2
By P =
,
R
22
power of 2- resistor =
=2W
2
P=
(b) By P = VI,
P 500
I= =
= 2.08 A
V 240
By P = VI,
power of R = 1.5  4 = 6 W
V2
,
R
V 2 240 2
R=
=
= 115.2 
500
P
(c) By P =
5
1
1
1
V2
, their power ratio is
.
:
:
R X RY RZ
R
B
The resistance of the heating element is
4
Since the bulbs are connected in parallel, the
8
When the switch is open, the two bulbs are in
series. Therefore the current passing through
the bulbs is the same.
Since the bulbs are connected in series, current
By P = I2R, for the same I, a larger resistance
passing them is the same. And, by P = I2R,
gives a larger power. Therefore, the 10- light
their power ratio is RX : RY : RZ.
bulb dissipates a larger power, and so it is
(a) Power ratio = 1 : 2 : 3
brighter.
(b) Power ratio = 3 : 4 : 12
9
(c) Power ratio = RX : RY : RZ
From the calculations in Practice 2.4 Q10:
Current flowing through 10- resistor = 0.3 A
Current flowing through 5- resistor = 0.6 A
Current flowing through 2- resistor = 0.9 A
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4
Electricity and Magnetism
Chapter 3 Domestic Electricity
V2
,
R
V2
10 =
2R bulb
By P = I2R,
By P =
power dissipated by 10- resistor = 0.32  10
= 0.9 W
power dissipated by 5- resistor = 0.62  5
V2
= 20
R bulb
= 1.8 W
power dissipated by 2- resistor = 0.92  2
When they are in parallel, the equivalent
R
resistance = bulb .
2
V2
New power =
R
V2
=
R bulb
2
 V2 

= 2
R

bulb


= 1.62 W
The 5- resistor dissipates the most power.
10 From the calculations in Practice 2.4 Q11,
current passing 3-, 4- and 12- resistors
are 1 A, 0.75 A and 0.25 A respectively.
By P = I2R,
power of 3- resistor = 12  3 = 3 W
power of 4- resistor = 0.752  4 = 2.25 W
power of 12- resistor = 0.252  12 = 0.75 W
11
By P =
V2
,
R
= 2  20
= 40 W
110 2
resistance of hairdryer X =
= 15.1 
800
110 2
resistance of hairdryer Y =
= 10.1 
1200
14 (a) Let RX be the resistance of X,
RY be the resistance of Y,
and V be the voltage of the battery.
When the resistors are in series,
RY
V
voltage across Y =
R X  RY
When connected in series to 220-V mains,
220
current drawn =
= 8.73 A
15 .1  10 .1
 RY


 V 
R X  RY

power of Y = 
RY
power of hairdryer X
= I2R = 8.732  15.1 = 1150 W
power of hairdryer Y
= I2R = 8.732  10.1 = 770 W
=
Therefore, they will not work at their rated
values when they are connected in series to a
By P =
V2
, the power P dissipated is halved
R
R X
 RY 2
Therefore the power dissipated by Y will
if the resistance R is doubled.
13
V 2 RY
If resistance of X is doubled, the power of
V 2 RY
Y becomes
.
2 R X  RY 2
220-V mains supply.
12
2
be smaller.
Let Rbulb be the resistance of one bulb, and V
be the voltage of the battery.
When the bulbs are in series, the equivalent
resistance is 2Rbulb.
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4
Electricity and Magnetism
Chapter 3 Domestic Electricity
(b) When the resistors are in parallel,
7
voltage across Y = voltage across X = V
may short-circuit the mains socket and people
2
Power of Y =
This is because water vapour in the bathroom
V
RY
will get an electric shock when they touch the
socket.
If resistance of X is doubled, the power
8
dissipated by Y remains unchanged.
By P = VI,
Practice 3.2 (p. 134)
current passing the cooker
P 500
= =
= 2.27 A
V 220
1
B
Since the current limit of the fuse for the
2
B
cooker should be slightly higher than the
If the 10-A fuse can properly protect a rice
current flowing in the cooker, a 3-A fuse
cooker, the current passing the rice cooker
should be used.
9
should be slightly smaller than 10 A.
By P = VI,
overheats and melts when too much current
I < 10
flows through it. If an appliance overloads, the
 P = VI = 220I < 2200 W
fuse blows and breaks the circuit before the
The power of the rice cooker is slightly less
cable overheats and causes a fire.
10
than 2200 W and the closest value is 2000 W.
3
The fuse is a short length of thin wire which
The socket is still ‘live’ when the bulb is taken
away. One will get an electric shock if he/she
A
touches the metal part of the socket.
4
Correct wiring:
5
6
By P = VI,
P 1200
I1 = =
= 5.45 A
V
220
11 (a)
In a ring main, current flows by two paths.
I 5.45
I2 = 1 =
= 2.73 A
2
2
220
V
(a) (i) I = =
= 4.4  104 A
R 500  1000
220
V
(ii) I = =
= 2.2 A
R 0.1  1000
(b) Do not touch any electrical appliances
with wet hands.
New Senior Secondary Physics at Work
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4
Electricity and Magnetism
Chapter 3 Domestic Electricity
(b) In a ring main circuit, the current in a
Revision exercise 3
cable is half of the total current of the
Multiple-choice (p. 139)
circuit. Therefore, thinner and cheaper
1
D
cables can be used. Also, all sockets in the
2
B
circuit are connected in parallel, so that
3
D
appliances connected to the socket can
4
A
operate at 220 V and work independently.
By P =
12 (a)
V2
,
R
resistance of each bulb
=
V 2 62
=
= 3  = resistance of the resistor R
P 12
Equivalent resistance of the circuit

= 3  31  3  31

1
=3
Total power dissipated by the circuit
=
(b) The hairdryer has an insulating plastic
5
A
6
D
case. Even if a fault develops, the case is
V 2 62
=
= 12 W
3
R
The current passing an appliance can be found
P
by I = .
V
P 980
Microwave oven: I = =
= 4.45 A
V 220
safe to touch since a current does not flow
in it.
13 (a) If the live wire touches the metal case of
the appliance, a large current flows to the
not get an electric shock in case that you
 15 A
P 1600
Washer-dryer: I = =
= 7.27 A  15 A
220
V
P 4  100
Lamp: I = =
= 1.82 A  15 A
220
V
accidentally touch the metal case.
None of them would overload the circuit.
earth and blows the fuse in the live wire.
The neutral wire is earthed. If it touches
the metal case of the appliance, you would
(b) The earth wire connects the metal case of
7
D
an electrical appliance to the earth. If a
8
D
fault develops and the metal body of the
9
A
appliance becomes ‘live’, a large current
Connection in A:
flows to the earth and blows the fuse in the
Equivalent resistance
live wire. This breaks the circuit and
= 1001  1001  1001

prevents people from getting an electric
1
= 33.3 
Connection in B:
shock.
Equivalent resistance

=100 + 1001  1001
New Senior Secondary Physics at Work

4

1
= 150 
 Oxford University Press 2010
4
Electricity and Magnetism
Chapter 3 Domestic Electricity
Connection in C:
New meter reading at 8pm
Equivalent resistance
= 14 212 + 3.2
= 100 + 100 + 100 = 300 
= 14 215.2 kW h
(1A)
(Complete circuit.)
(1A)
Connection in D:
3
(a)
Equivalent resistance
= 50 + 50 + 50 = 150 
Connection A has the lowest resistance. By P
=
V2
, A should give the largest power and
R
boil water fastest.
10
(HKCEE 2003 Paper II Q36)
11
(HKCEE 2006 Paper II Q39)
12
(HKCEE 2006 Paper II Q41)
13
(HKCEE 2007 Paper II Q18)
(Correct connection of the two-way
switches.)
(Correct labels.)
Conventional (p. 140)
1
(b) Apply P =
(1M)
P 27
=
= 3 A (1A)
9
V
V
.
R
V 2 240 2
=
= 115.2 
500
P
Current through bulb Y
R=
= current in the main circuit
For the bulb rated at ‘120 V, 75 W’:
=5–3=2A
R=
(1A)
Power dissipated by bulb Y
V 2 120 2
=
= 192 
P
75
(1A)
(1A)
The filament in the bulb rated at ‘240 V,
= VI = 9  2 = 18 W
(1A)
500 W’ has a lower resistance than that in
(a)
Power Time
Energy
rating / switched consumed /
kW
on / h
kW h
Air conditioner 1.5
1.0
1.5
Plasma TV
0.3
1.0
0.3
Kettle
2.0
0.1
0.2
Water heater
3.5
0.2
0.7
Lights
0.5
1.0
0.5
4
Appliance
(b) Air conditioner has cost the most.
the bulb rated at ‘120 V, 75 W’.
P
(a) By I = ,
V
(1A)
(1M)
current through the kettle I1
2000
=
= 9.09 A,
220
current through the iron I2
1000
=
= 4.55 A,
220
(4  0.5A)
current through the TV set I3
150
=
= 0.682 A
220
(1A)
(c) Energy consumed between 7 pm and 8 pm
= 1.5 + 0.3 + 0.2 + 0.7 + 0.5
= 3.2 kW h
(1M)
For the bulb rated at ‘240 V, 500 W’:
– current through bulb X
2
(1A)
2
By P = VI,
Current through bulb X =
(1A)
(1A)
Total current = I1 + I2 + I3 = 14.3 A < 15 A
(1A)
(1A)
New Senior Secondary Physics at Work
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4
Electricity and Magnetism
Chapter 3 Domestic Electricity
Therefore, it is safe to operate these
(b) As the new bulb has a lower resistance
appliances at the same time from the
socket.
than the original one, the equivalent
(1A)
resistance in the circuit decreases. (1A)
V
By I = , a larger current flows in the
R
(b) The circuit breaker cuts off the current if it
exceeds a certain value.
(1A)
circuit.
This prevents the circuit from overheating,
(1A)
2
which may result in a fire.
5
By P = I R, the power of each working
(1A)
bulb increases with the current, and is
(a) Average energy consumption per month
1106
=
(1M)
12
larger than that it should be.
8
Energy required to boil away the water
= 0.5  4200  (100 – 20) + 0.5  2.26  106
= 92.2 kW h
= 92.2  3.6  106 J
= 3.32  10 J
8
(1M)
(1A)
= 1 298 000 J
(1M)
E 1 298 000
Time required = =
= 649 s (1A)
2000
P
1 298 000
Cost = 0.9 
= $0.3245
(1A)
60  60  1000
(b) Bills for using the air conditioner
= 0.9  92.2
(1M)
= $83.0
(1A)
(c) RC-XC98J is the most efficient.
(1A)
9
2
6
(a) By P =
V
,
R
(1M)
V 2 220 2
=
= 26.9 
P 1800
wires in this circuit branches into two
(1A)
paths
around the room.
(ii) Live: brown
(1A)
slightly larger than the normal current,
7
(1A)
(1A)
Earth: yellow/green
(1A)
(iii) The circuit breaker should be
connected to the live wire.
(1A)
This ensures that no part of the
(1A)
hairdryer and cable is ‘live’ when the
there is no alternative pathway for the
circuit breaker breaks the circuit.
charge to flow through when the circuit is
New Senior Secondary Physics at Work
(1A)
Neutral: blue
(1A)
(a) Since the bulbs are connected in series,
open at that broken bulb.
(1A)
and forms a large ‘ring’ which loops
Since the correct fuse has a fuse value
it is suitable to use a 10-A fuse.
The circuit is called ring main. (1A)
of the live, the neutral and the earth
(b) By P = VI,
current through the wire
P 1800
= =
= 8.18 A
V
220
(a) (i)
This circuit is so called because each
resistance of the coils
=
(1A)
(1A)
(1A)
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4
Electricity and Magnetism
Chapter 3 Domestic Electricity
(b) The earth pin is longer than other pins.
(b) (i)
(1A)
It is designed to open ‘shutters’ on the live
L stands for live.
(1A)
N stands for neutral.
(1A)
(ii) & (iv)
and the neutral holes on the socket (1A)
and to ensure that the earth wire is
connected before the live wire.
(1A)
(c) Method 1:
Change the metal case to a plastic case.
(1A)
Plastic is an insulator. Even if the live wire
touches the case and the cable has no earth
wire installed, users will not get an electric
shock.
(1A)
Method 2:
(1A)
(Switch S on the live side.)
(1A)
(iii) If the fuse blows, the appliance is not
Change the two-pin plug to a three-pin
live to touch.
plug and, in the cable of the hairdryer, add
(1A)
(v) When switch is at the off position,
an earth wire connecting the case of the
hairdryer to the earth.
(Fuse F on the live side.)
the element is not live.
(1A)
(1A)
(vi)
When the earth wire is used, even if the
live wire touches the case, current will
flow to the earth through the earth wire.
This prevents users from getting an
electric shock.
(1A)
(d) Total energy consumed by the hairdryer
= Pt
(1M)
15
=1
 30 = 7.5 kW h
60
(1M)
Cost = 7.5  0.9 = $6.75
(1A)
(New element in parallel with the
original one.)
10
(HKCEE 2003 Paper I Q8)
(Switch allowing independent
11
(HKCEE 2004 Paper I Q10)
operation.)
12
(HKCEE 2005 Paper I Q11)
13 (a) (i)
All the bulbs light up.
(ii) Two-way switching.
(vii) Left: d.c.
(1A)
(1A)
(1A)
(1A)
Middle: d.c.
(1A)
(1A)
Right: a.c.
(1A)
(iii) To switch on from one place in the
home, move switch 1 to B.
(1A)
Then move switch 2 to C to switch
off from a different place.
New Senior Secondary Physics at Work
(1A)
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4
Electricity and Magnetism
Chapter 3 Domestic Electricity
14 (a) D
(b) (i)
(1A)
Total resistance
1 1 1
=   
 4 4 8
1
(1M)
= 1.6 
(ii) By V = IR,
V 12
I= =
= 7.5 A
R 1 .6
(1A)
(1M)
(1A)
(c) Brighter lamps mean that the power
dissipated by each lamp is larger.
(1A)
2
By P =
V
, the resistance of each lamp is
R
smaller, so the resistance of the circuit is
smaller.
(1A)
Physics in articles (p. 145)
(a) The resistance R of the cable is inversely
proportional to its cross-sectional area A and
directly proportional to its length l, i.e.
l
(1A)
R .
A
If the resistance R of the cables is larger, the
voltage drop across the wire is larger and the
power output is smaller.
(1A)
(b) Copper cables are used because the resistance
of copper is very small
(1A)
and copper cables are cheap.
(1A)
(c) This is because gold is not easily oxidized and
oxidized wire has a poor contact with other
metal parts.
New Senior Secondary Physics at Work
(1A)
8
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