4 Electricity and Magnetism Chapter 6 Transmission of Electrical Energy
6 Transmission of Electrical Energy
Practice 6.1 (p.291)
1 C equivalent stable current =
2 B
LEDs allow current to flow in one direction =2A
only. As a result, no current flows through the
7 Pmax =
circuit.
3 C Vpeak = = = 240 V
Vrms = Root mean square voltage
= = = 240 V
Average power = =
8 (a) Peak voltage Vpeak = 170 V
4 A
Root mean square voltage
Ipeak = =
= = = 120 V
5 (a) (i) d.c.
(b) T = 0.025
(ii) Period = 0.04 s
Frequency = = 25 Hz T = 0.025 = s
(b) (i) a.c.
Frequency = = 60 Hz
(ii) Period = 0.02 s
Frequency = = 50 Hz (c) T= = = 0.02 s
(c) (i) a.c. Vpeak = Vrms = × 230 = 325 V
(ii) T = 0.0125
Period = 0.0125 = s
Frequency = 60 Hz
6 Equivalent stable current =
For Fig g, Practice 6.2 (p.310)
equivalent stable current = =2A 1 B
Thicker wires have smaller resistance.
For Fig h, (1) is correct.
equivalent stable current = = 2.83 A A laminated core reduces eddy currents.
(2) is correct.
For Fig i,
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If the secondary coil is winded on the right the input power and the output power of the
arm, not all magnetic flux from the primary transformer are the same.
coil passes through the secondary coil and After removing Y, the voltage across X does
more energy is lost. not change.
(3) is incorrect. (3) is incorrect.
2 B The resistance of the secondary circuit
A steady current produces a steady magnetic increases.
field. (2) is correct.
(1) is incorrect. In the secondary circuit, by P = ,
A transformer may work with varying d.c.
(3) is incorrect. R P (V constant)
3 A Therefore, the input power of the transformer
Since there is resistance in the transmission decreases.
cables, V3 < V2 and the system is not 100% (1) is correct.
efficient. The relationship between V1 and V4 7 Output voltage = = 64 V
cannot be determined from the given
8 (a) Current in the cable
information.
4 C = = = 200 A
Let IR be the current in the transmission cable.
Power lost
Since T1 is ideal, we have IR = . = I 2R = 2002 × 0.15 = 6000 W
(b) Potential difference across cable
Power loss = = = IR = 200 × 0.15 = 30 V
Potential difference across equipment
On the other hand, = 250 – 30 = 220 V
Voltage across it is VR = V2 – V3 The equipment can operate properly.
Power loss = = 9 (a) When a changing current flows in the
primary coil, a changing magnetic field
5 D is produced in the soft-iron core. This
(1) and (3) reduces the resistance of the changing magnetic field induces an
cables, while (2) reduces the current flowing e.m.f. in the secondary coil.
in the cables. (b) Decrease the turns ratio.
6 B
10 (a) Turns ratio = = = = 22
Since the transformer is ideal, the secondary
voltage depends only on the input voltage and (b) Power = VI = 10 × 0.5 = 5 W
is not affected by the secondary circuit. Also, (c) Since the transformer is ideal, the power
taken from the mains is 5 W.
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4 Electricity and Magnetism Chapter 6 Transmission of Electrical Energy
(d) Current = = = 0.0227 A (c) The magnetic field in the core will be
weakened and the secondary voltage
11 (a) decreases.
Revision exercise 6
Concept traps (p.314)
1 F
An alternating current is one which reverses
The turns ratio of the transformer across
its direction periodically.
both A and B is 1.
2 T
VA + VB = 12
3 F
The number of turns connected to A is
A step-up transformer only increases the
the same as that connected to B.
transmitted voltage, but the transmitted power
VA = VB = 6 V
is unchanged or decreased.
The turns ratio for C is 2.
4 F
VC = 12 = 6 V
The function of the soft-iron core is to ensure
Therefore, all three bulbs work at their that all field lines from the primary coil pass
rated value, so PA = PB = PC = 2 W. through the secondary coil.
Power input of transformer 5 T
= power output = 2 + 2 + 2 = 6 W The voltage of a dry cell is steady and cannot
Current in primary coil induce an e.m.f. in the secondary coil of a
transformer.
= = = 0.5 A
(b) The turns ratio of the transformer for Multiple-choice questions (p.314)
each of the bulb is 2. 6 B
VA = VB = VC = 12 = 6 V Vrms = = = V0
All bulbs work at their rated value.
7 C
Similar to (a), the current in the primary
Since there are more turns in the secondary
coil is 0.5 A.
coil than in the primary coil, this is a step-up
(c) The turns ratio of the transformer for
transformer.
each of the bulb is . Similar to (b), the (1) is correct.
current in the primary coil is 0.5 A. The input can be a varying d.c.
(3) is incorrect.
12 (a) Voltage ratio = =
8 D
(b) Number of turns = = 350 The mains voltage 220 V is the r.m.s. value.
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(1) is incorrect.
Peak voltage across PQ = =
The power dissipated in the bulb is maximum
when the voltage across it is at a peak value.
When a 220-V voltage is applied, the bulb
dissipates a power of 60 W. R.m.s. voltage across PQ =
(2) is correct.
For sinusoidal a.c., a 5-A a.c. fuse can Therefore =6V
withstand an r.m.s. current of 5 A, which has a
peak current larger than 5 A. V0 = 339 V
(3) is correct. 13 D
9 C Voltage across PS = = 120 V
The output power is less than the input power,
so the efficiency is less than 100% and the Voltage across PQ = =6V
transformer is not ideal.
Voltage across PR = = 24 V
(1) and (3) are correct.
The turns ratio is equal to the voltage ratio as 14 (HKALE 2007 Paper 2A Q37)
long as there is no flux leakage. 15 (HKALE 2008 Paper 2A Q19)
(2) is incorrect. 16 (HKCEE 2011 Paper 2 Q42)
10 C 17 (HKDSE 2014 Paper 1A Q30)
Vrms = = 12.02 V
Conventional questions (p.316)
Irms = = = 4.16 A 18 (a) A.c. can be easily stepped-up and
stepped-down while steady d.c. cannot.
11 B
1A
Energy E dissipated in one period
Stepping up the voltage before
= transmitting power though long cables
reduces the power loss. 1A
= As a result, power transmission over
long distances has a smaller power loss
Average power = = if a.c. is used. 1A
(b) Factors affecting the efficiency (any one
12 C
of the following): 1A
Let V0 be the input peak voltage.
Resistance of wires
Peak voltage across PS = Magnetization and demagnetization of
the core
Induced currents in the core
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Ways to improve efficiency =
(corresponding to the above): 1A
Use thicker wires. = 67.3 W 1A
Use a soft iron core. (e) If the transformer is not used, the current
Use a laminated core. flowing in the printer will be twice its
(c) Use thicker cables. 1A rated value. 1A
This overlarge current can damage the
19 (a) By R = , 1M
printer. 1A
l= = = 89.3 m 1A 21 (a) Current through bulb = 1M
(b) Voltage across cable =
= IR = 12 × 0.5 = 6 V 1M
=2A 1A
Supply voltage = 220 – 6 = 214 V 1A
(b) Since the step-down transformer is
(c) Power supply = VI 1M
100% efficient,
= 220 × 12
= 2640 W 1A By = , 1M
(d) Power loss
= I 2R = 122 × 0.5 = 72 W 1M Ip = = = 0.1 A
Percentage power loss
Power loss = I 2R 1M
= = 2.73 % 1A
= 0.1 × 102
= 0.1 W 1A
20 (a) Turns ratio = 1M
(c) Since the step-up transformer is 100%
efficient,
=
Current drawn from input
=2 1A
= = =2A 1A
(b) Peak current = 1M
= (d) Consider the step-down transformer.
= 0.849 A 1A
VBD = = 12 = 240 V 1M
(c) Operating power = VI 1M
= 110 × 0.6 Voltage drop across cables
= 66 W 1A = IR = 0.1 × 10 = 1 V
(d) Efficiency = VAC = 240 + 1 = 241 V 1M
Consider the step-up transformer.
1M
Voltage of the input a.c.
Input power =
= = 241 = 12.1 V 1A
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Or Since the step-up transformer is 100% (ii) The average power decreases in the
efficient, case of Figure m and unchanged in
Input power = 24 + 0.1 = 24.1 W 1M the case of Figure n. 1A
Input voltage = 1M As seen from the above graphs,
energy is sometimes not dissipated
= = 12.1 V 1A in the former case 1A
but the energy dissipation is not
22 (a) The one in Figure m is an a.c. and the
affected in the later case. 1A
one in Figure n is a d.c. 1A
(b) Since the resistance and average power
are the same in the two cases, the r.m.s.
23 S is S is closed
current must be the same.
open case (i) case (ii)
Consider Figure n.
Power of X P P 4P
Irms = = 1M Voltmeter
V V V
reading
Consider Figure m.
Ammeter
Peak current = 1M I 2I 2I
reading
= 6 × 1A
=5A 1A 24 (a) Magnetic field
(c) (i) = 1M
= 0.0942 T 1A
(b) Induced e.m.f.
(I is zero for more than half a
period) 1A = 1M
(Current passes through only when
= 1M
it is greater than a certain positive
value) 1A =
= 0.113 V 1A
(c) Reduce the turns ratio. 1A
(d) Zero 1A
25 (a) A sudden change in the primary current
(Correct graph) 1A causes a sudden change in the magnetic
field in the transformer. 1A
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This induces a large e.m.f. in the while a d.c. always travels in one
secondary coil. 1A direction. 1A
The large number of turns in the
secondary coil also contributes to the
high secondary voltage. 1A
(b) Increase the rate of change of the
primary current. 1A
This will increase the rate of change of
the magnetic flux through the secondary
coil, thus increase the induced e.m.f. (Correct example of a.c.) 1A
1A (Correct example of d.c.) 1A
26 (a) (i) Maximum current (b) To transmit power at a high voltage. 1A
(c)
= 1M
= 2000 A 1A
(ii) Maximum power loss
= I 2R 1M (Correct drawing of structure) 1A
= 20002 × 0.1 (Correct labels) 1A
= 4 × 105 W 1A (Correct turns ratio) 1A
(iii) Maximum voltage drop
(d) = slope of graph ≈ =
= IR 1M
= 2000 × 0.1
The transformer is ideal. 1A
= 200 V 1A
28 (HKCEE 2009 Paper 1 Q12)
(b) (i) Turns ratio = 1M 29 (HKCEE 2011 Paper 1 Q10)
= Experiment questions (p.319)
30 (HKDSE Sample Paper 2008 Paper 1B Q13)
= 300 1A
(ii) If the voltage is higher, the current
Physics in article (p.320)
in the cables is smaller. 1A
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. 1A
so the meter’s reading is zero. 1A
27 (a) An a.c. reverses its direction periodically
(b) Both of them make use of the mutual
1A
inductance 1A
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and works only with a current that is
changing. 1A
(c) No, 1A
The coil is parallel to the magnetic field
of the a.c. 1A
The change in magnetic field through the
coil is always zero. 1A
(d) It cannot be used to measure a steady
d.c. 1A
(Or other reasonable answers)
32 (HKDSE 2012 Paper 1B Q9)
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