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D
Electricity and Magnetism
Chapter 16 Electricity in the Home
16 Electricity in the Home
Practice 16.1 (p. 82)
1
C
2
D
6
Voltage across 3- resistor = 0.5  3
= 1.5 V
Voltage across 1- resistor = 0.5  1
2
By P =
V
,
R
= 0.5 V
220 2
Power of the bulb =
= 100 W
484
3
B
Voltage across 2- resistor = 1.5 + 0.5 = 2 V
V 2
Current passing 2- resistor = = = 1 A
R 2
Energy consumed = 2  0.1  3.5
Total current in the circuit = 1.5 A
Voltage across resistor R = 6  2 = 4 V
4
V
Resistance of resistor R = =
= 2.67 
I 1 .5
= 0.7 kW h
Cost = 0.7  0.9 = $0.63
4
V2
,
R
V 2 220 2
R=
=
= 24.2 
P 2000
By P =
By P =
V2
,
R
0 .5 2
= 0.25 W
1
22
power of 2- resistor =
=2W
2
1.52
power of 3- resistor =
= 0.75 W
3
42
power of resistor R =
=6W
2.67
V
6
By I = , current in the circuit =
= 0.5 A
R
12
power of 1- resistor =
By P = IV,
P 2000
I= =
= 9.09 A
V
220
V 220
(Or by I = =
= 9.09 A)
R 24 .2
The resistance of the heating element is
24.2  and the current through the heating
5
By V = IR,
element is 9.09 A.
E
(a) By P = ,
t
500
P=
= 500 W
1
7
By P  I 2 R ,
Power of the 2- light bulb = 0.52  2
= 0.5 W
The power of the heater is 500 W.
P
(b) By I = ,
V
500
I=
= 2.08 A
240
Power of the 10- light bulb = 0.52  10
= 2.5 W
Larger power gives a brighter bulb. Thus 10-
light bulb is brighter when the switch is open.
8
The current through the heater is 2.08 A.
2
V
,
P
240 2
R=
= 115.2 
500
From the calculations in Practice 15.4 Q11:
Current flowing through 10- resistor = 0.3 A
(c) By R =
Current flowing through 5- resistor = 0.6 A
Current flowing through 2- resistor = 0.9 A
The resistance of the heater is 115.2 .
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D
Electricity and Magnetism
Chapter 16 Electricity in the Home
By P  I 2 R ,
12
Power dissipated by 10- resistor = 0.3  10
2
Let R be the resistance of a bulb,
and V be the voltage of the battery.
= 0.9 W
When the bulbs are in series, the equivalent
Power dissipated by 5- resistor = 0.6  5
2
resistance is 2R. When they are in parallel, the
= 1.8 W
equivalent resistance = R/2.
Power dissipated by 2- resistor = 0.9  2
2
When the bulbs are in series, by P =
= 1.62 W
V2
2R
V2
R
20
10 
The 5- resistor dissipates the most power.
9
From the calculations in Practice 15.4 Q12,
current passing 3-, 4- and 12- resistors
are 1 A, 0.75 A and 0.25 A respectively.
When the bulbs are in parallel, by P =
By P  I R ,
2
Power of 3- resistor = 12  3
new power 
=3W
Power of 4- resistor = 0.752  4
= 2.25 W
Power of 12- resistor = 0.25  12
= 0.75 W
and V be the voltage of the battery.
V
,
R
When the resistors are in series,
Ry
voltage across Y =
V
Rx  R y
110 2
= 302.5 
40
110 2
resistance of hair dryer Y =
= 151.25 
80
resistance of hair dryer X =
2
 Ry

Power of Y = 
 V  / Ry
 Rx  R y



If resistance of X is doubled, the power of
When working in 220 V,
2
 Ry

Y becomes 
 V  / Ry .
 2 Rx  Ry



2
220
= 160 W
302 .5
220 2
power of hair dryer Y =
= 320 W
151 .25
power of hair dryer X =
Thus the power dissipated by Y will be
smaller.
Thus both hair dryers will not work at their
(b) When the resistors are in parallel,
rated values.
11
By P =
V2
V2
V2
 2
 2  40 W
R/ 2 V
V
/2
20
40
Ry be the resistance of Y,
2
By P =
V2
,
R
13 (a) Let Rx be the resistance of X,
2
10
V2
,
R
voltage across Y = voltage across X
V2
, the power P dissipated is halved
R
=V
2
if the resistance R is doubled.
Power of Y =
V
Ry
If resistance of X is doubled, the power
remains unchanged.
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D
Electricity and Magnetism
Chapter 16 Electricity in the Home
Practice 16.2 (p. 96)
1
B
2
B
3
B
8
By I =
P
,
V
current passing the cooker =
500
= 2.27 A
220
Since the current limit of the fuse for the
If the 10-A fuse can properly protect a rice
cooker should be slightly higher than the
cooker, the current passing the rice cooker
current flowing in the cooker, a 3-A fuse
should be slightly smaller than 10 A.
P
By I = ,
V
P
I=
< 10
220
should be used.
9
The socket is still ‘live’ when the bulb is taken
away. One will get an electric shock if he/she
touches the metal part of the socket.
 P < 2200 W
Correct wiring:
The power of the rice cooker is slightly less
than 2200 W and the closest value is 2000 W.
4
A
5
10 (a)
6
7
P
,
V
1200
I1 =
= 5.45 A
220
By I =
In a ring main, current flows by two paths.
I
I2 = 1
2
5.45
=
= 2.73 A
2
(b) In a ring main circuit, the current in a
It is because water vapour in the bathroom
are connected in parallel, so that
may short-circuit the mains socket and people
appliances connected to the socket can
will get an electric shock when they touch the
operate at 220 V and work independently.
cable is half of the total current of the
circuit. Hence, thinner and cheaper cables
can be used. Also, all sockets in the circuit
socket.
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D
Electricity and Magnetism
Chapter 16 Electricity in the Home
The equivalent resistance of the circuit
11 (a)

= (3  3) 1  (3  3) 1
By P =

1
=3
V2
,
R
total power dissipated by the circuit
=
(b) The hair-dryer has an insulating plastic
3
A
4
D
case. Even if a fault develops, the case is
cable overheats and causes a fire.
The current passing an appliance can be found
P
by I = .
V
P 980
Microwave oven: I = =
= 4.45 A
V 220
P 1600
Washer-dryer: I = =
= 7.27 A
220
V
P 4  100
Lamp: I = =
= 1.82 A
220
V
The earth wire is a safety wire. It connects the
None of them would overload the circuit.
safe to touch since a current will not flow
in it.
12
The fuse is a short length of thin wire which
overheats and melts when too much current
flows through it. If an appliance overloads, the
fuse blows and breaks the circuit before the
13
62
= 12 W
3
metal body of an electrical appliance to the
5
(HKCEE 2005 Paper II Q21)
earth. If the live wire gets loose and touches
the metal body of the appliance, a large
Section B
current flows to the earth and blows the fuse in
6
D
the live wire. This breaks the circuit and
7
A
prevents people from getting an electric shock.
Connection in A:
Equivalent resistance

Revision exercise 16
= 1001  1001  1001
Multiple-choice (p. 100)
Connection in B:
Section A
Equivalent resistance
1
D
= 100 + 1001  1001
2
A

By R =

1
= 33.3 
= 150 
Equivalent resistance
= 100 + 100 + 100 = 300 
Connection in D:
62
= 3  = resistance of the resistor R
12
Equivalent resistance

= 50 + 1001  1001
New Physics at Work (Second Edition)
1
Connection in C:
V2
,
P
resistance of each bulb
=

20

1
= 100 
 Oxford University Press 2006
D
Electricity and Magnetism
By P =
Chapter 16 Electricity in the Home
V2
, the connection in A should give
R
For the bulb rated at ‘120 V, 75 W’:
V2
,
P
120 2
R=
= 192 
75
By R =
the largest power and boil water fastest.
8
D
9
(HKCEE 2003 Paper II Q36)
(1A)
The filament in the bulb rated at ‘240 V,
Conventional (p. 101)
500 W’ has a lower resistance than that in
Section A
1
By I =
P
,
V
Current through bulb X =
3
(1M)
27
=3A
9
= current in the main circuit
(1A)
current through the TV set I3
150
=
= 0.682 A
220
Power dissipated by bulb Y
2
(1M)
current through the iron I2
1000
=
= 4.55 A,
220
– current through bulb X
= 9  2 = 18 W
(1A)
current through the kettle I1
2000
=
= 9.09 A,
220
(1A)
Current through bulb Y
=5–3=2A
the bulb rated at ‘120 V, 75 W’.
P
(a) By I = ,
V
(1A)
(a)
(1A)
Total current = I1 + I2 + I3 = 14.3 A < 15 A
(1A)
Hence, it is safe to operate these
appliances at the same time from the
socket.
(1A)
(b) The circuit breaker cuts off the current
flow if it exceeds a certain value.
(Complete circuit.)
This prevents the circuit from overheating,
(1A)
which may result in a fire.
(Correct connection of the two-way
switch.)
(1A)
(Correct labels.)
(1A)
4
(1A)
In ring main, current flows by two paths.(1A)
Thinner and cheaper cables can be used
because only half the current in ring main
(b) For the bulb rated at ‘240 V, 500 W’:
V2
By R =
,
(1M)
P
240 2
R=
= 115.2 
(1A)
500
flows in the ring of cable.
(1A)
Sockets in the ring main are connected in
parallel,
(1A)
so that a fault in any socket will not affect the
others.
New Physics at Work (Second Edition)
(1A)
21
(1A)
 Oxford University Press 2006
D
5
Electricity and Magnetism
Chapter 16 Electricity in the Home
If the live wire touches the metal body of the
8
(a) By R =
appliance, a large current flows to the earth
and blows the fuse in the live wire.
(1A)
V2
,
P
Resistance of the coils =
Since the neutral wire is connected to the earth
(1A)
P
(b) By I = ,
V
If the neutral wire touches the metal body of
the appliance, you would not get an electric
current through the wire
1800
=
= 8.18 A
220
shock in case that you accidentally touch the
6
220 2
1800
= 26.9 
at the local substation, its voltage is zero.(1A)
neutral wire.
(1M)
(1A)
(a)
(1A)
Since the correct fuse has a fuse value
Appliance
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
slightly larger than the normal current,
(1A)
it is suitable to use a 10-A fuse.
9
(b) Air conditioner has cost the most.
(HKCEE 2003 Paper I Q8)
10 (a) (i)
(1A)
(c) Energy consumed between 7 pm and 8 pm
(1A)
(1M)
(1M + 1A)
The current in a filament is 0.6 A.
= 1.5 + 0.3 + 0.2 + 0.7 + 0.5
= 3.2 kW h
Voltage of the mains supply
= 40  6 = 240 V
V
(ii) By I = ,
R
6
I=
= 0.6 A
10
(2A)
(1A)
(iii) The current in the circuit is 0.6 A.
(1A)
(1A)
The new meter reading
(b) (i)
= 14 212 + 3.2
= 14 215.2 kW h
7
changes to heat in the filament of the
(1A)
light bulb.
(a) Average energy consumption per month
1224
=
= 102 kW h
(1M + 1A)
12
(1A)
When the filament is hot enough, the
heat changes to light energy.
(b) The bills for using the air conditioner
= 0.9  102 = $91.8
Electrical energy in the circuit
(1A)
(ii) Period that the lights are left on
= 12  12  60  60
(1M + 1A)
(c) AH-AP09CV is the most efficient. (1A)
= 518 400 s
(1A)
Total energy used
= Pt = 144  518 400
(1M)
= 74 649 600 J
= 74.6496 MJ
74 649 600
 0 .9
(iii) Cost =
60  60  1000
= $18.7
New Physics at Work (Second Edition)
22
(1A)
(1M)
(1A)
 Oxford University Press 2006
D
Electricity and Magnetism
Chapter 16 Electricity in the Home
Section B
(b) The earth pin is longer than other pins.
11 (a) Since the bulbs are connected in series,
(1A)
there is no alternative pathway for the
It is designed to open ‘shutters’ on the live
charge to flow through when the circuit is
and the neutral holes on the socket (1A)
broken at that broken bulb.
and to ensure that the earth wire is
(1A)
(b) When the broken bulb short-circuits the
connected before the live wire.
contacts in the socket, the equivalent
resistance in the circuit decreases.
(c) Method 1:
(1A)
Change the metal case to a plastic case.
A larger current flows in the circuit. (1A)
(1A)
By P  I R , the power of each working
Since plastic is an insulator, even if the
bulb increases with the current.
live wire touches the case and the cable
2
12
(1A)
(1A)
Energy required to boil away the water
has no earth wire installed, users will not
= 0.5  4200  (100 – 20) + 0.5  2.26  10
= 1 298 000 J
6
get an electric shock.
(1M + 1A)
Method 2:
1 298 000
Time required =
2000
= 649 s
1 298 000
Cost = 0.9 
= $0.3245
60  60  1000
13 (a) (i)
(1A)
Change the two-pin plug to three-pin plug
and, in the cable of the hair-dryer, add an
(1A)
earth wire, which connects the case of the
(1A)
hair-dryer to the Earth.
(1A)
The circuit is called ring main. (1A)
When the earth wire is used, even if the
This circuit is called as such because,
live wire touches the case, current will
in this circuit, each of the live, the
flow to the Earth through the earth wire.
neutral and the earth wires
This prevents users from getting electric
(1A)
shocks.
branches into two paths and forms a
(d) Total energy consumed by the hair-dryer
large ‘ring’ which loops around the
room.
(1A)
(ii) Live: brown
(1A)
Neutral: blue
(1A)
Earth: yellow/green
(1A)
(1A)
= Pt
(1M)
= 1  0.25  30 = 7.5 kW h
(1A)
Cost = 7.5  0.9 = $6.75
(1A)
(iii) The circuit breaker should be
connected to the live wire.
(1A)
This is because it ensures that no part
of the hair-dryer and cable is ‘live’
when the circuit breaker breaks the
circuit.
New Physics at Work (Second Edition)
(1A)
23
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D
Electricity and Magnetism
14 (a) (i)
Chapter 16 Electricity in the Home
Physics in articles (p. 106)
Negative charge is on the toner.(1A)
(ii) Objects of unlike charges attract and
(a) The resistance R of the cable is inversely
toner particles (negatively charged)
proportional to its cross-sectional area A (1A)
are attracted to the drum (positively
and directly proportional to its length l, (1A)
l
i.e. R  .
A
charged).
(1A)
In order to pull toner particles onto
(b) Copper cables are used because the resistance
paper from the drum, the paper has to
be more positively charged than the
drum, so that the attractive force on
(1A)
metal parts.
(1A)
Q
,
t
0.15
I=
= 0.06 A
2 .5
and copper cables are cheap.
oxidized wire has a poor contact with other
greater than that towards the drum.
By I =
(1A)
(c) It is because gold is not easily oxidized and
toner particles towards the paper is
(b) (i)
of copper is very small
(1A)
(1M)
(1A)
The current in the wire is 0.06 A.
(ii) By E = QV,
(1M)
energy = 0.15  200
= 30 unit joule (J) (1A + 1A)
(c) (i)
laser
printer
(Correct connection.)
P
(ii) By I = ,
V
500
I=
= 2.17 A
230
(2A)
(1M)
(1A)
A fuse of 3 A should be used. (1A)
15
(HKCEE 2003 Paper I Q10)
16
(HKCEE 2005 Paper I Q11)
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24
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