Ch17 Solution
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Ch17/P.1
17 Electromagnetic Induction 17.1 Electricity from magnetism Warm-up 17.1 (p. 142) Option '2'. Current is not always produced when a wire moves in a magnetic field. A current is produced only when
− −
the wire is connected in a complete circuit, and the wire moves in a magnetic field in a direction other than that parallel to the magnetic field.
Check-point 1 (p. 148) 1 2
C Oppose, D A, B, B
Practice 17.1 (p. 149) 1 2 3 4
5
A A B (a)
(b)
(i) (ii) (iii) (iv) (v)
(a)
There is voltage induced across AB. It is because the metal rod cuts magnetic field lines. However, no current is induced because the circuit is not complete. The current will flow through the light bulb from C to D. The direction can be determined by the Fleming's right-hand rule. The brightness of the bulb can be increased by moving the metal rod faster and by increasing the magnetic field.
(b) (c)
The induced current will increase. The induced current will be zero. The induced current will flow in opposite direction. The induced current will be zero. The induced current will increase.
Ch17/P.2
17.2 Generators Warm-up 17.2 (p. 150) 1 2
B The third statement.
Check-point 2 (p. 156) 1 2
C Half, 0.1 1, 10 1, 0.1 1, 10
Practice 17.2 (p .156) 1 2 3 4 5
D C C D (a) (b)
6
A bicycle dynamo has a simpler structure, as carbon brushes and slip rings are not required. And current does not have to flow through moving contacts, thus can give a more stable output. This is because some of the kinetic energy required to drive the bicycle is converted into electrical energy by the dynamo.
(a), (b) and (c)
Ch17/P.3
17.3 Transformers Warm-up 17.3 (p. 158) The forth and the fifth statements.
Check-point 3 (p. 169) 1 2 3
B A Number of turns in secondary coil, primary voltage, 10, 22
Practice 17.3 (p .170) 1
A
2
B
Turns ratio =
3
C
For transformer with no power loss: Vp Vs
4
B
5
(a)
(b) (c) (d)
Turns ratio =
Np Ns
=
=
Is Ip
Vs
⇒
Vp Vs
=
=
110 1 = 220 2
Vp Vs
=
100 110 Is 220 and = 220 Ip Ip
100
220 ⇒ Ip = 0.909 A Ip
220 = 22 : 1 10
Np Ns
=
Is Ip
(a)
The primary voltage Vp = 1 × 5 = 5 V The secondary voltage Vs = 3.5 × 5 = 17.5 V Vp 5 2 = = Voltage ratio = Vs 17.5 7
(b)
By
(c)
110 ⇒ = 220
The turns ratio of the transformer is 22:1. By P = VI, ⇒ P = 10 × 0.5 = 5 W The power taken by the bulb is 5 W. The efficiency of the transformer is 100%, i.e. no power loss. Hence, power from the mains = power of the bulb = 5 W P 5 By I = , I = = 0.0227 A V 220 The current from the mains is 0.0227 A. (Or With no power loss,
6
Vp
Np Ns
=
Vp Vs
,
⇒
100 2 = Ns 7
22 0.5 = 1 Ip
⇒
Ip = 0.0227 A)
Ns = 350
The number of turns in the secondary coil is 350. When one of the C-cores is removed, just a small fraction of the magnetic field lines coming out of the primary coil would 'cut' through the secondary coil. This reduces the induced voltage across the secondary coil. Vp Np Hence, the voltage ratio, , would be much larger than the turns ratio, . Vs Ns
Ch17/P.4
17.4 Generating and transmitting electrical energy Warm-up 17.4 (p. 172) The third reason.
Check-point 4 (p. 178) 1 2
B P, 3.42 × 109, 1.24 × 104 A I2, (1.24 × 104)2, 769 MW
Practice 17.4 (p .179) 1 2
D C P , V By V = IR,
1000 = 4.55 A 220 voltage across the appliance = 220 − IR = 220 − 4.55 × 5 = 197 V
By I =
3
current flowing along a cable =
D At the transmission cable, voltage = V2, power = 0.9P ⇒ By P = IV,
current flowing through the cable =
0.9 P V2 2
0.9 P × R By P = I2R, power lost to transmission cable = V2
4
(a)
(b)
5
(a) (b) (c) (d)
By I =
P , V
current flowing along transmission cable =
50 × 10 3 = 200 A 250
By P = I2R, P = 2002 × 0.15 = 6 kW The power loss in the cable is 6 kW. By V = IR, supply voltage at hospital = 250 − IR = 250 − 200 × 0.15 = 220 V Hence, equipment rated at 220 V a.c. can be used properly. Steam is needed in power station to drive the steam turbine which in turn drives the alternator in the power station. It is because power station needs a huge amount of water for cooling down steam in the condenser. The power loss in transmission cable will be reduced if voltage is stepped up before transmission. It is because a.c. can be stepped up and stepped down easily for transmission.
Ch17/P.5
Revision exercise 17 Multiple-choice (p. 182) 1 2 3 4 5 6 7 8 9 10 11
B D D D D D (HKCEE 1999 Paper II Q31) E (HKCEE 2001 Paper II Q34) A (HKCEE 2001 Paper II Q35) A (HKCEE 2002 Paper II Q38) C B
Conventional (p. 184) 1
(a)
(b) (c)
(d)
2
(a)
Correct shape and direction of the field lines Correct separation between field lines If an a.c. flows along the wire, the direction of the magnetic field changes alternately while the shape of the field lines remains the same. Since the coil intercepts a changing magnetic field produced by the wire connected to an a.c. power supply, a current is induced in the coil by Lenz's law and the light bulb gives out light. Any one of the following: − To determine whether a current is a.c. or d.c. − To compare the size of the a.c. flowing in different wires.
(1A) (1A) (1A) (1A) (1A) (1A) (1A) (1A)
When the bar magnet spins, the pole of the magnet facing the solenoid changes alternately. (1A) Then by the Lenz's law, current is induced in the coil to oppose the motion of the magnet and the light bulb gives out light. (1A)
(b)
Correct labelled axes Correct curve showing alternating positive and negative voltage
(1A) (1A)
Ch17/P.6
Correct positions of the magnet (the poles can be interchanged)
(1A)
Ch17/P.7
(c)
(d)
3
(a)
(b)
Any two of the following: (2 × 1A) − Use a solenoid with more turns. − Use thicker wire for the solenoid. − Use a stronger magnet. − Spin the magnet at a faster speed. − Use more elastic threads/stiffer elastic threads. I don't agree. (1A) When the magnet spins, a current is induced in the solenoid to oppose the motion of the magnet, i.e. a force acts on the magnet by the solenoid. (1A) Therefore, Newton's first law is not valid in this case and the magnet will slow down and stop. (1A) (Or I don't agree. 1A By the conservation of energy, the elastic potential energy of elastic threads changes into the kinetic energy of the spinning magnet, and in turn the kinetic energy of the magnet is converted into the light energy of the light bulb. 1A The magnet will stop spinning when all elastic potential energy of the elastic thread is converted into light energy. 1A) P , V 220 000 I= = 1000 A 220 The current in the cables is 1000 A. (ii) By P = I2R, P = 10002 × 0.1 = 105 W The power loss in cables is 105 W. P 220 000 − 100 000 (iii) By V = , V= = 120 V I 1000 The voltage available at the village is 120 V. (Or By V = IR, voltage available at the village = 220 − IR = 220 − 1000 × 0.1 = 120 V (i) A step-down transformer with turns ratio 300:1 should be connected to the transmission cable. (i)
By I =
Correct circuit diagram Correct labels
(1M) (1A) (1M) (1A) (1A)
1A) (1A) (1A)
(1A) (1A)
Ch17/P.8
(ii) By I = I=
(c)
P , V 220 000 66 000
= 3.33 A
The current flowing in cables is 3.33 A. By P = I2R, power loss in cables = (3.33)2 × 0.1 = 1.11 W 66 000 Voltage available at village = 300 = 220 V Power should be transmitted at 66 kV. Any one of the following: − Shorten the length of transmission cables. − Reduce the resistance of cables.
4
(HKCEE 1999 Paper I Q4)
5
(HKCEE 2000 Paper I Q10)
(1A)
(1A)
(1A) (1A) (1A)
Ch17/P.9
6
(a)
(b)
1.7 × 10 8 (1M1A) 25 000 = 6800 A (1A) The current in one generator is 6800 A. (ii) Since a large current is generated in a generator, the heating effect of a current flowing through the wires is huge. (1A) Hence, water is needed to cool down the generator. (1A) (i) The voltage is stepped up by a step-up transformer, (1A) which has turns ratio 1:16. (1A) (ii) By P = IV, if electricity is transmitted at high voltage, the current flowing along transmission cables would be reduced. (1A) 2 Then, by P = I R, the power loss in transmission cables can be reduced. (1A) (i)
By I =
P , V
⇒
I=
(c)
Overhead
Advantage Any one of the following:
Disadvantage Any one of the following: (1A)
− Cheaper installation − No coolant required − Easier to maintain Underground Any one of the following:
− Ugly − Affected by extreme weather condition Any one of the following: (1A)
− Out of sight − Does not take up land
(d)
(1A)
Any three of the following: − The water flow of the rivers is not fast enough. − There are only a few suitable rivers in the area. − Wind is not always available.
(1A) − Expensive installation (high cost of burying cables) − Need to handle electrical and thermal insulation (3 × 1A)
Ch17/P.10
−
A large area is needed to accommodate a lot of wind turbines to generated electricity.
17 Electromagnetic Induction 17.1 Electricity from magnetism Warm-up 17.1 (p. 142) Option '2'. Current is not always produced when a wire moves in a magnetic field. A current is produced only when
− −
the wire is connected in a complete circuit, and the wire moves in a magnetic field in a direction other than that parallel to the magnetic field.
Check-point 1 (p. 148) 1 2
C Oppose, D A, B, B
Practice 17.1 (p. 149) 1 2 3 4
5
A A B (a)
(b)
(i) (ii) (iii) (iv) (v)
(a)
There is voltage induced across AB. It is because the metal rod cuts magnetic field lines. However, no current is induced because the circuit is not complete. The current will flow through the light bulb from C to D. The direction can be determined by the Fleming's right-hand rule. The brightness of the bulb can be increased by moving the metal rod faster and by increasing the magnetic field.
(b) (c)
The induced current will increase. The induced current will be zero. The induced current will flow in opposite direction. The induced current will be zero. The induced current will increase.
Ch17/P.2
17.2 Generators Warm-up 17.2 (p. 150) 1 2
B The third statement.
Check-point 2 (p. 156) 1 2
C Half, 0.1 1, 10 1, 0.1 1, 10
Practice 17.2 (p .156) 1 2 3 4 5
D C C D (a) (b)
6
A bicycle dynamo has a simpler structure, as carbon brushes and slip rings are not required. And current does not have to flow through moving contacts, thus can give a more stable output. This is because some of the kinetic energy required to drive the bicycle is converted into electrical energy by the dynamo.
(a), (b) and (c)
Ch17/P.3
17.3 Transformers Warm-up 17.3 (p. 158) The forth and the fifth statements.
Check-point 3 (p. 169) 1 2 3
B A Number of turns in secondary coil, primary voltage, 10, 22
Practice 17.3 (p .170) 1
A
2
B
Turns ratio =
3
C
For transformer with no power loss: Vp Vs
4
B
5
(a)
(b) (c) (d)
Turns ratio =
Np Ns
=
=
Is Ip
Vs
⇒
Vp Vs
=
=
110 1 = 220 2
Vp Vs
=
100 110 Is 220 and = 220 Ip Ip
100
220 ⇒ Ip = 0.909 A Ip
220 = 22 : 1 10
Np Ns
=
Is Ip
(a)
The primary voltage Vp = 1 × 5 = 5 V The secondary voltage Vs = 3.5 × 5 = 17.5 V Vp 5 2 = = Voltage ratio = Vs 17.5 7
(b)
By
(c)
110 ⇒ = 220
The turns ratio of the transformer is 22:1. By P = VI, ⇒ P = 10 × 0.5 = 5 W The power taken by the bulb is 5 W. The efficiency of the transformer is 100%, i.e. no power loss. Hence, power from the mains = power of the bulb = 5 W P 5 By I = , I = = 0.0227 A V 220 The current from the mains is 0.0227 A. (Or With no power loss,
6
Vp
Np Ns
=
Vp Vs
,
⇒
100 2 = Ns 7
22 0.5 = 1 Ip
⇒
Ip = 0.0227 A)
Ns = 350
The number of turns in the secondary coil is 350. When one of the C-cores is removed, just a small fraction of the magnetic field lines coming out of the primary coil would 'cut' through the secondary coil. This reduces the induced voltage across the secondary coil. Vp Np Hence, the voltage ratio, , would be much larger than the turns ratio, . Vs Ns
Ch17/P.4
17.4 Generating and transmitting electrical energy Warm-up 17.4 (p. 172) The third reason.
Check-point 4 (p. 178) 1 2
B P, 3.42 × 109, 1.24 × 104 A I2, (1.24 × 104)2, 769 MW
Practice 17.4 (p .179) 1 2
D C P , V By V = IR,
1000 = 4.55 A 220 voltage across the appliance = 220 − IR = 220 − 4.55 × 5 = 197 V
By I =
3
current flowing along a cable =
D At the transmission cable, voltage = V2, power = 0.9P ⇒ By P = IV,
current flowing through the cable =
0.9 P V2 2
0.9 P × R By P = I2R, power lost to transmission cable = V2
4
(a)
(b)
5
(a) (b) (c) (d)
By I =
P , V
current flowing along transmission cable =
50 × 10 3 = 200 A 250
By P = I2R, P = 2002 × 0.15 = 6 kW The power loss in the cable is 6 kW. By V = IR, supply voltage at hospital = 250 − IR = 250 − 200 × 0.15 = 220 V Hence, equipment rated at 220 V a.c. can be used properly. Steam is needed in power station to drive the steam turbine which in turn drives the alternator in the power station. It is because power station needs a huge amount of water for cooling down steam in the condenser. The power loss in transmission cable will be reduced if voltage is stepped up before transmission. It is because a.c. can be stepped up and stepped down easily for transmission.
Ch17/P.5
Revision exercise 17 Multiple-choice (p. 182) 1 2 3 4 5 6 7 8 9 10 11
B D D D D D (HKCEE 1999 Paper II Q31) E (HKCEE 2001 Paper II Q34) A (HKCEE 2001 Paper II Q35) A (HKCEE 2002 Paper II Q38) C B
Conventional (p. 184) 1
(a)
(b) (c)
(d)
2
(a)
Correct shape and direction of the field lines Correct separation between field lines If an a.c. flows along the wire, the direction of the magnetic field changes alternately while the shape of the field lines remains the same. Since the coil intercepts a changing magnetic field produced by the wire connected to an a.c. power supply, a current is induced in the coil by Lenz's law and the light bulb gives out light. Any one of the following: − To determine whether a current is a.c. or d.c. − To compare the size of the a.c. flowing in different wires.
(1A) (1A) (1A) (1A) (1A) (1A) (1A) (1A)
When the bar magnet spins, the pole of the magnet facing the solenoid changes alternately. (1A) Then by the Lenz's law, current is induced in the coil to oppose the motion of the magnet and the light bulb gives out light. (1A)
(b)
Correct labelled axes Correct curve showing alternating positive and negative voltage
(1A) (1A)
Ch17/P.6
Correct positions of the magnet (the poles can be interchanged)
(1A)
Ch17/P.7
(c)
(d)
3
(a)
(b)
Any two of the following: (2 × 1A) − Use a solenoid with more turns. − Use thicker wire for the solenoid. − Use a stronger magnet. − Spin the magnet at a faster speed. − Use more elastic threads/stiffer elastic threads. I don't agree. (1A) When the magnet spins, a current is induced in the solenoid to oppose the motion of the magnet, i.e. a force acts on the magnet by the solenoid. (1A) Therefore, Newton's first law is not valid in this case and the magnet will slow down and stop. (1A) (Or I don't agree. 1A By the conservation of energy, the elastic potential energy of elastic threads changes into the kinetic energy of the spinning magnet, and in turn the kinetic energy of the magnet is converted into the light energy of the light bulb. 1A The magnet will stop spinning when all elastic potential energy of the elastic thread is converted into light energy. 1A) P , V 220 000 I= = 1000 A 220 The current in the cables is 1000 A. (ii) By P = I2R, P = 10002 × 0.1 = 105 W The power loss in cables is 105 W. P 220 000 − 100 000 (iii) By V = , V= = 120 V I 1000 The voltage available at the village is 120 V. (Or By V = IR, voltage available at the village = 220 − IR = 220 − 1000 × 0.1 = 120 V (i) A step-down transformer with turns ratio 300:1 should be connected to the transmission cable. (i)
By I =
Correct circuit diagram Correct labels
(1M) (1A) (1M) (1A) (1A)
1A) (1A) (1A)
(1A) (1A)
Ch17/P.8
(ii) By I = I=
(c)
P , V 220 000 66 000
= 3.33 A
The current flowing in cables is 3.33 A. By P = I2R, power loss in cables = (3.33)2 × 0.1 = 1.11 W 66 000 Voltage available at village = 300 = 220 V Power should be transmitted at 66 kV. Any one of the following: − Shorten the length of transmission cables. − Reduce the resistance of cables.
4
(HKCEE 1999 Paper I Q4)
5
(HKCEE 2000 Paper I Q10)
(1A)
(1A)
(1A) (1A) (1A)
Ch17/P.9
6
(a)
(b)
1.7 × 10 8 (1M1A) 25 000 = 6800 A (1A) The current in one generator is 6800 A. (ii) Since a large current is generated in a generator, the heating effect of a current flowing through the wires is huge. (1A) Hence, water is needed to cool down the generator. (1A) (i) The voltage is stepped up by a step-up transformer, (1A) which has turns ratio 1:16. (1A) (ii) By P = IV, if electricity is transmitted at high voltage, the current flowing along transmission cables would be reduced. (1A) 2 Then, by P = I R, the power loss in transmission cables can be reduced. (1A) (i)
By I =
P , V
⇒
I=
(c)
Overhead
Advantage Any one of the following:
Disadvantage Any one of the following: (1A)
− Cheaper installation − No coolant required − Easier to maintain Underground Any one of the following:
− Ugly − Affected by extreme weather condition Any one of the following: (1A)
− Out of sight − Does not take up land
(d)
(1A)
Any three of the following: − The water flow of the rivers is not fast enough. − There are only a few suitable rivers in the area. − Wind is not always available.
(1A) − Expensive installation (high cost of burying cables) − Need to handle electrical and thermal insulation (3 × 1A)
Ch17/P.10
−
A large area is needed to accommodate a lot of wind turbines to generated electricity.
