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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 NAB = 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