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4 Electricity and Magnetism
6
Chapter 6 Transmission of Electrical Energy
Transmission of Electrical Energy
Practice 6.1 (p.291)
1
C
2
B
2
 4 

  0
2
equivalent stable current = 
2
LEDs allow current to flow in one direction
only. As a result, no current flows through the
=2A
7
circuit.
3
C
Vrms =
V1
4
5
Vrm2 s V12
=
2R
R
8
A
Ipeak =
(a)
Vpeak
R
(i)
=
(i)
d.c.
(b)
a.c.
(ii) Period = 0.02 s
1
Frequency =
= 50 Hz
0.02
(c)
(a)
V1
R
(ii) Period = 0.04 s
1
Frequency =
= 25 Hz
0.04
(b)
R
Vpeak = Pm axR = 2400 48 = 240 2 V
Root mean square voltage
Vpeak 240 2
=
=
= 240 V
2
2
2
Average power =
Pmax =
(c)
(i)
a.c.
3
(ii)
T = 0.0125
4
4 1
=
s
3 60
325
0
For Fig g,
2 2  ( 2 ) 2
=2A
2
0.02
0.04
time / s
–325
Equivalent stable current = mean of I 2
Practice 6.2 (p.310)
1
B
Thicker wires have smaller resistance.
For Fig h,
equivalent stable current =
Root mean square voltage
Vpeak 170
=
=
= 120 V
2
2
3
T = 0.025
2
2 1
 T = 0.025  =
s
3 60
1
Frequency = = 60 Hz
T
1
1
T= =
= 0.02 s
f 50
voltage / V
Frequency = 60 Hz
equivalent stable current =
Peak voltage Vpeak = 170 V
Vpeak = 2 Vrms = 2 × 230 = 325 V
 Period = 0.0125 
6
2
Vpeak
42  0
= 2.83 A
2
For Fig i,
New Senior Secondary Physics at Work (Second Edition)
 Oxford University Press 2015
 (1) is correct.
A laminated core reduces eddy currents.
 (2) is correct.
1
4 Electricity and Magnetism
2
Chapter 6 Transmission of Electrical Energy
If the secondary coil is winded on the right
After removing Y, the voltage across X does
arm, not all magnetic flux from the primary
not change.
coil passes through the secondary coil and
 (3) is incorrect.
more energy is lost.
The resistance of the secondary circuit
 (3) is incorrect.
increases.
B
 (2) is correct.
A steady current produces a steady magnetic
In the secondary circuit, by P =
field.
3
 (1) is incorrect.
R   P  (V constant)
A transformer may work with varying d.c.
Therefore, the input power of the transformer
 (3) is incorrect.
decreases.
 (1) is correct.
A
Since there is resistance in the transmission
cables, V3 < V2 and the system is not 100%
7
Output voltage = 16 
8
(a)
cannot be determined from the given
information.
Power lost
C
= I 2R = 2002 × 0.15 = 6000 W
Let IR be the current in the transmission cable.
P
Since T1 is ideal, we have IR =
.
V2
(b)
Potential difference across equipment
P2R
Power loss = I R R = 2
V2
On the other hand,
= 250 – 30 = 220 V
 The equipment can operate properly.
9
(a)
Voltage across it is VR = V2 – V3
V 2 (V  V3 ) 2
Power loss = R = 2
R
R
is produced in the soft-iron core. This
changing magnetic field induces an e.m.f.
in the secondary coil.
(1) and (3) reduces the resistance of the cables,
(b)
while (2) reduces the current flowing in the
6
B
Since the transformer is ideal, the secondary
10
(a)
the input power and the output power of the
Decrease the turns ratio.
N p Vp 220
Turns ratio =
=
=
= 22
N s Vs
10
(b)
Power = VI = 10 × 0.5 = 5 W
(c)
Since the transformer is ideal, the power
voltage depends only on the input voltage and
is not affected by the secondary circuit. Also,
When a changing current flows in the
primary coil, a changing magnetic field
D
cables.
Potential difference across cable
= IR = 200 × 0.15 = 30 V
2
5
4
= 64 V
1
Current in the cable
P 50 000
= =
= 200 A
V
250
efficient. The relationship between V1 and V4
4
V2
,
R
(d)
taken from the mains is 5 W.
P
5
Current = =
= 0.0227 A
V 220
transformer are the same.
2
New Senior Secondary Physics at Work (Second Edition)
 Oxford University Press 2015
4 Electricity and Magnetism
11
(a)
Chapter 6 Transmission of Electrical Energy
(c)
A
The magnetic field in the core will be
weakened and the secondary voltage
decreases.
B
C
Revision exercise 6
Concept traps (p.314)
The turns ratio of the transformer across
1
both A and B is 1.
An alternating current is one which reverses
VA + VB = 12
its direction periodically.
The number of turns connected to A is
the same as that connected to B.
 VA = VB = 6 V
2
T
3
F
A step-up transformer only increases the
The turns ratio for C is 2.
1
VC = 12  = 6 V
2
Therefore, all three bulbs work at their
transmitted voltage, but the transmitted power
is unchanged or decreased.
4
F
rated value, so PA = PB = PC = 2 W.
The function of the soft-iron core is to ensure
Power input of transformer
that all field lines from the primary coil pass
= power output = 2 + 2 + 2 = 6 W
through the secondary coil.
Current in primary coil
P 6
= =
= 0.5 A
V 12
(b)
F
5
T
The voltage of a dry cell is steady and cannot
induce an e.m.f. in the secondary coil of a
The turns ratio of the transformer for
transformer.
each of the bulb is 2.
VA = VB = VC = 12 
1
=6V
2
Multiple-choice questions (p.314)
6
B
All bulbs work at their rated value.
Vrms = V 2 =
Similar to (a), the current in the primary
coil is 0.5 A.
(c)
12
(a)
(b)
7
The turns ratio of the transformer for
1
each of the bulb is
. Similar to (b),
2
the current in the primary coil is 0.5 A.
2
1
Voltage ratio =
=
3.5 7
7
Number of turns = 100  = 350
2
V02  ( V0 ) 2
= V0
2
C
Since there are more turns in the secondary
coil than in the primary coil, this is a step-up
transformer.
 (1) is correct.
The input can be a varying d.c.
 (3) is incorrect.
8
D
The mains voltage 220 V is the r.m.s. value.
 (1) is incorrect.
New Senior Secondary Physics at Work (Second Edition)
 Oxford University Press 2015
3
4 Electricity and Magnetism
Chapter 6 Transmission of Electrical Energy
The power dissipated in the bulb is maximum
Therefore
when the voltage across it is at a peak value.
13
withstand an r.m.s. current of 5 A, which has a
peak current larger than 5 A.
 (3) is correct.
The output power is less than the input power,
so the efficiency is less than 100% and the
transformer is not ideal.
 (1) and (3) are correct.
The turns ratio is equal to the voltage ratio as
long as there is no flux leakage.
D
Voltage across PS =
For sinusoidal a.c., a 5-A a.c. fuse can
C
14
(HKALE 2007 Paper 2A Q37)
15
(HKALE 2008 Paper 2A Q19)
16
(HKCEE 2011 Paper 2 Q42)
17
(HKDSE 2014 Paper 1A Q30)
Conventional questions (p.316)
18
(a)
1A
C
Vrms =
Irms =
11
12
4
A.c. can be easily stepped-up and
stepped-down while steady d.c. cannot.
 (2) is incorrect.
10
=6V
240
= 120 V
2
1
Voltage across PQ = 120 
=6V
1  3  16
1 3
Voltage across PR = 120 
= 24 V
1  3  16
 (2) is correct.
9
40 2
 V0 = 339 V
When a 220-V voltage is applied, the bulb
dissipates a power of 60 W.
V0
17
Stepping up the voltage before
= 12.02 V
transmitting power though long cables
2
reduces the power loss.
50
P
=
= 4.16 A
Vrms 12 .02
1A
As a result, power transmission over
B
long distances has a smaller power loss
Energy E dissipated in one period
2
 2
T I 
T
= I0 R    0  R    2
4 3
4 

5 2
= I 0 RT
9
E 5 2
Average power = = I 0 R
T 9
if a.c. is used.
(b)
1A
Factors affecting the efficiency (any one
of the following):
1A
Resistance of wires
Magnetization and demagnetization of
the core
Induced currents in the core
C
Ways to improve efficiency
Let V0 be the input peak voltage.
V
Peak voltage across PS = 0
2
V0
V
1
Peak voltage across PQ = 
= 0
2 1  3  16 40
V
R.m.s. voltage across PQ = 0
40 2
(corresponding to the above):
1A
Use thicker wires.
Use a soft iron core.
Use a laminated core.
(c)
19
(a)
Use thicker cables.
l
By R =  ,
A
1A
1M
New Senior Secondary Physics at Work (Second Edition)
 Oxford University Press 2015
4 Electricity and Magnetism
l=
(b)
(c)
RA

=
Chapter 6 Transmission of Electrical Energy
0.5  3  10 6
= 89.3 m
1.68  10 8
=
1A
=2A
Voltage across cable
= IR = 12 × 0.5 = 6 V
1M
Supply voltage = 220 – 6 = 214 V
1A
Power supply = VI
1M
(b)
(d)
Power loss
= I 2R = 122 × 0.5 = 72 W
20
(a)
Ip =
1A
Percentage power loss
72
=
100 % = 2.73 %
2640
Vp
Turns ratio =
Vs
(b)
= 0.1 W
1A
(c)
(c)
Operating power = VI
(d)
Consider the step-down transformer.
Np
20
Vs =
VBD =
 12 = 240 V
1M
Ns
1
1A
Voltage drop across cables
1M
= IR = 0.1 × 10 = 1 V
VAC = 240 + 1 = 241 V
Voltage of the input a.c.
Np
1
Vs =
=
 241 = 12.1 V
Ns
20
Or
1A
21
(a)
1A
P
Current through bulb =
V
1M
New Senior Secondary Physics at Work (Second Edition)
 Oxford University Press 2015
Since the step-up transformer is 100%
Input power = 24 + 0.1 = 24.1 W
input power
Input voltage =
input current
24 .1
=
= 12.1 V
2
flowing in the printer will be twice its
printer.
1A
efficient,
1A
This overlarge current can damage the
1M
Consider the step-up transformer.
If the transformer is not used, the current
rated value.
1A
1M
= 66 W
1A
output power
(d) Efficiency =
100 % 1M
input power
output power
Input power =
efficiency
66
=
98 %
(e)
Since the step-up transformer is 100%
Current drawn from input
N
20
= s Is =
 0.1 = 2 A
Np
1
= 110 × 0.6
= 67.3 W
1A
efficient,
1M
= 0.6  2
= 0.849 A
1M
= 0.1 × 10
1A
Peak current = I rms  2
Ns
1
Is =
 2 = 0.1 A
Np
20
2
220
=
110
=2
1M
Power loss = I 2R
1M
1A
Since the step-down transformer is
100% efficient,
Ip Ns
By
=
,
Is N p
= 220 × 12
= 2640 W
24
12
22
(a)
1M
1M
1A
The one in Figure m is an a.c. and the
one in Figure n is a d.c.
1A
5
4 Electricity and Magnetism
(b)
Chapter 6 Transmission of Electrical Energy
Since the resistance and average power
23
are the same in the two cases, the r.m.s.
current must be the same.
Power of X
Consider Figure n.
Voltmeter
5 0
5
=
2
2
2
Irms =
reading
1M
Ammeter
Consider Figure m.
Peak current = I rms  2
=
5
2
reading
(i)
open
case (i)
case (ii)
P
P
4P
V
V
V
I
2I
2I
1M
6 × 1A
24
 2
=5A
(c)
S is closed
S is
(a)
1A
current
I0
Magnetic field
 NI
= 0
l
4π  10 7  2500  6
=
0.2
1M
= 0.0942 T
(b)
time
(I is zero for more than half a
period)
1A
(Current passes through only when
it is greater than a certain positive
value)
I0
25
1M
1M
= 0.113 V
1A
(c)
Reduce the turns ratio.
1A
(d)
Zero
1A
(a)
A sudden change in the primary current
1A
current
Induced e.m.f.
N
=
t
NAB
=
t
500  12  10 4  0.0942
=
0.5
1A
causes a sudden change in the magnetic
time
field in the transformer.
(Correct graph)
1A
This induces a large e.m.f. in the
1A
secondary coil.
(ii) The average power decreases in the
1A
case of Figure m and unchanged in
The large number of turns in the
the case of Figure n.
secondary coil also contributes to the
1A
high secondary voltage.
As seen from the above graphs,
energy is sometimes not dissipated
in the former case
1A
(b)
1A
Increase the rate of change of the
primary current.
1A
but the energy dissipation is not
This will increase the rate of change of
affected in the later case.
the magnetic flux through the secondary
1A
coil, thus increase the induced e.m.f.
1A
6
New Senior Secondary Physics at Work (Second Edition)
 Oxford University Press 2015
4 Electricity and Magnetism
26
(a)
(i)
Chapter 6 Transmission of Electrical Energy
Maximum current
P
=
V
440 000
=
220
(c)
25 kV
a.c. input
= 2000 A
25-turn coil
= I 2R
1M
= 20002 × 0.1
= 4 × 105 W
1A
(iii) Maximum voltage drop
(d)
= IR
1M
= 2000 × 0.1
= 200 V
(i)
1A
Turns ratio =
=
Vp
1M
Vs
(Correct drawing of structure)
1A
(Correct labels)
1A
(Correct turns ratio)
Ip
30 N s
= slope of graph ≈
=
Is
1 Np
1A
The transformer is ideal.
1A
28
(HKCEE 2009 Paper 1 Q12)
29
(HKCEE 2011 Paper 1 Q10)
30
(HKDSE Sample Paper 2008 Paper 1B Q13)
1A
(ii) If the voltage is higher, the current
in the cables is smaller.
(a)
400-turn coil
Experiment questions (p.319)
66 000
220
= 300
27
400 kV
a.c. output
1A
(ii) Maximum power loss
(b)
soft-iron core
1M
1A
Physics in article (p.320)
31
(a)
The magnetic field produced by the two
By P = I 2R, the smaller the current,
currents cancel each other
1A
the smaller the power loss.
so the meter’s reading is zero.
1A
1A
An a.c. reverses its direction periodically
(b)
1A
inductance
while a d.c. always travels in one
direction.
Both of them make use of the mutual
1A
and works only with a current that is
1A
current
(c)
changing.
1A
No,
1A
The coil is parallel to the magnetic field
d.c.
of the a.c.
1A
The change in magnetic field through the
time
coil is always zero.
a.c.
1A
(d) It cannot be used to measure a steady d.c.
(b)
(Correct example of a.c.)
1A
(Correct example of d.c.)
1A
To transmit power at a high voltage. 1A
New Senior Secondary Physics at Work (Second Edition)
 Oxford University Press 2015
1A
(Or other reasonable answers)
32
(HKDSE 2012 Paper 1B Q9)
7