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```4
Electricity and Magnetism
6
Chapter 6 Transmission of Electricity
Transmission of Electricity
Practice 6.1 (p. 262)
1
D
2
D
For the sinusoidal current,
I 
Irms = 0
2
A square wave can have constant magnitude.
3
Therefore,
I 0
= 12.5
2
(a) This is an a.c.
Let T be the period.
= I2 =
I0 = 5 A
2  0.5T  2  0.5T
=2A
T
2
2
The peak value of the sinusoidal current is
5 A.
Alternative solution: Since the value of I2
(c) (i)
For sinusoidal current:
is always the same (= 22), the required
steady current = 2 2 = 2A.
(b) This is an a.c.
I
4
= 0 =
= 2.83 A
2
2
(c) This is a d.c.
Let T be the period.
For square current:
= I2 =
4
4 2  0.5T  2 2  0.5T
= 3.16 A
T
Billy is wrong. The rated value means that the
light bulb is working at an r.m.s. voltage of 220
V to give an average power of 60 W.
5
(a) The current shown in Figure a is an a.c.
while the current shown in Figure b is a
(ii) No. For the case of sinusoidal current,
d.c.
the current through the resistor
(b) Since the average power dissipated is the
becomes zero in some periods of time,
same for both currents, by P = Irms2R, the
and so the average power dissipated
r.m.s. value of the currents are the same.
decreases. Therefore, the average
Let I0 be the peak value of the sinusoidal
power dissipated is now smaller than
current.
that in the case of square current.
For the square current,
Irms = I 2 =
Practice 6.2 (p. 280)
5 2  0.5T
= 12.5
T
New Senior Secondary Physics at Work
1
1
C
 Oxford University Press 2010
4
Electricity and Magnetism
Chapter 6 Transmission of Electricity
2
D
(b) This is because power station needs a huge
3
A
amount of water for cooling down steam in
4
B
the condenser.
5
C
6
P 50  1000
(a) I = =
= 200 A
250
V
(c) The power loss in transmission cable will
be reduced if voltage is stepped up before
transmission.
Power lost in the cable
(d) This is because a.c. can be stepped up and
= I2R = 2002  0.15 = 6000 W
stepped down easily for transmission.
N p Vp 220
10 (a) Turns ratio =
=
=
= 22 : 1
N s Vs
10
(b) Voltage across cables
= IR = 200  0.15 = 30 V
Supply voltage at the hospital
(b) Power taken by the bulb
= 250  30 = 220 V
= VI = 10  0.5 = 5 W
Therefore, 220 V a.c. equipment rated can
7
(c) The efficiency of the transformer is 100%,
be used properly.
P 1000
I= =
= 4.545 A
220
V
i.e. no power loss. Hence,
power taken from the mains
= power taken by the bulb = 5 W
Voltage across cables
(d) Current drawn from the mains
P
5
= =
= 0.0227 A
V 220
= IR = 4.545  5 = 22.7 V
Voltage available for the appliance
8
= 220  22.7 = 197 V
l
(a) By R =  ,
A
11
At the transmission cable,
voltage = V2, power = 0.9P
By P = VI,
length of the copper cable
=
RA

=
0.5  3  10 6
16 .8  10 9
current flowing through the cable =
= 89.3 m
By P = I2R,
(b) Voltage across cables
2
 0.9 P 
  R
power loss in the cable = 
 V2 
= IR = 12  0.5 = 6 V
Supply voltage to house = 220 – 6 = 214 V
=
(c) Power supplied from the power pole
= VI = 220  12 = 2640 W
(d) Power loss in cables
12 By P =
= I2R = 122  0.5 = 72 W
0.81 P 2 R
V2 2
V2
,
R
resistance of each bulb =
Power loss percentage
72
 100 % = 2.73%
=
2640
9
0.9 P
V2
62
=18 
2
(a) By conservation of energy,
12Ip =
(a) Steam is needed in power station to drive
the steam turbine which in turn drives the
12 2
62
+
18  2 18
Ip = 0.5 A
alternator in the power station.
New Senior Secondary Physics at Work
2
 Oxford University Press 2010
4
Electricity and Magnetism
Chapter 6 Transmission of Electricity
Conventional (p. 284)
(b) By conservation of energy,
2
12Ip =
6
3
18
1
Ip = 0.5 A
(c) By conservation of energy,
62
62
12Ip =
2+
18
18
Ip = 0.5 A
Switch S
Switch S is closed
is open
case (i)
case(ii)
Power of bulb X
P
P
4P
V
V
V
I
2I
2I
(6  1A)
13 (a) Vp = 1  5 = 5 V
2
Vs = 3.5  5 = 17.5 V
Vp
5
Voltage ratio =
=
= 0.286 : 1
V s 17 .5
(b) By
Np
(1M)
current passing through the bulb
P 24
= =
=2A
V 12
Vp
,
Vs
V
17 .5
Ns = Np  s = 100 
= 350
Vp
5
Ns
=
(a) By P = VI,
(b) By P = VI,
current in the cables
24
P
= =
= 0.1 A
V 12  20
The number of turns in the secondary coil
Power loss in the cables
is 350.
= I2R = 0.12  10 = 0.1 W
(c) When one of the C-cores is removed, just a
(c) By P = VI,
small fraction of the magnetic field lines
through the secondary coil. This reduces
the induced voltage across the secondary
coil.
Therefore, the voltage ratio,
Vs
much larger than the turns ratio,
3
, would be
Np
Ns
.
2
A
3
B
4
(HKALE 2004 Paper II Q27)
5
(HKALE 2007 Paper II Q37)
= I 2R
(1A)
(1M)
(1A)
(1M)
= 1000  0.1
2
= 100 000 W = 100 kW
(1A)
(iii) By P = VI,
voltage available at the village
P
=
I
220 000  100 000
=
1000
= 120 V
New Senior Secondary Physics at Work
(1A)
(ii) Power lost in the cables
Multiple-choice (p. 284)
B
By P = VI,
current in the cables
P 220 000
= =
= 1000 A
220
V
Revision exercise 6
1
(a) (i)
(1M)
(1M)
current in the 12-V a.c. supply
P
=
V
24  0.1
=
= 2.01 A
12
coming out of the primary coil would ‘cut’
Vp
(1A)
3
(1A)
 Oxford University Press 2010
4
Electricity and Magnetism
(Or
Physics in articles (p. 287)
By V = IR,
(a) An alternating current varies periodically with
voltage available at the village
time in direction
= 220  IR
while a direct current always travels in one
= 220  1000  0.1
direction,
= 120 V
(b) (i)
Chapter 6 Transmission of Electricity
(1A))
A step-down transformer
(1A)
with turns ratio 300 : 1
(1A)
(1A)
(1A)
as shown in the following figures.
should be connected to the
transmission cable.
(Correct circuit diagram.)
(1A)
(Correct labels.)
P 220 000
(ii) I = =
= 3.33 A
66 000
V
(1A)
(1M)
Power loss in cables
= I 2R
(1M)
= 3.33  0.1
2
= 1.11 W
(1A)
The power loss in cables is much
(Correct example of a.c.)
(1A)
(Correct example of d.c.)
(1A)
(b) Transformers can step up/down voltages.(1A)
reduced, so power should be
transmitted at 66 kV.
(1A)
(c) Any one of the following:
(1A)
(c) The power loss in transmission can be reduced
if electricity is transmitted at a high voltage.
(1A)
Shorten the length of transmission cables.
The voltage of a.c. can be stepped up and down
Reduce the resistance of cables.
4
(HKCEE 2000 Paper I Q10)
easily by transformers while the voltage of d.c.
5
(HKCEE 2005 Paper I Q12)
cannot.
6
(HKALE 2005 Paper II Q4)
Since the power loss in d.c. transmission is
(1A)
large, a lot of generating units are needed along
the path of transmission.
New Senior Secondary Physics at Work
4
(1A)
 Oxford University Press 2010
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