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# Ee1 chapter11 ac_circuits

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### Ee1 chapter11 ac_circuits

1. 1. IT2001PAEngineering Essentials (1/2)Chapter 11 – Alternating Current CircuitsLecturer Namelecturer_email@ite.edu.sg Aug 17, 2012Contact Number
2. 2. Chapter 11 – Alternating Current CircuitsLesson ObjectivesUpon completion of this topic, you should be able to: Explain and calculate the fundamentals of alternating current waveform. 2 IT2001PA Engineering Essentials (1/2)
3. 3. Chapter 11 – Alternating Current CircuitsSpecific Objectives Students should be able to :  Describe the basic construction of a simple alternator.  Describe the characteristics of an alternating quantity as fluctuating over a given period of time.  Define the following terms with reference to an alternating quantity :  Frequency  Average value  Instantaneous value  Maximum value IT2001PA Engineering Essentials (1/2)
4. 4. Chapter 11 – Alternating Current CircuitsSpecific Objectives Students should be able to :  State that the frequency of the generated emf is proportional to the speed and number of poles of the alternator.  Explain the term Root Mean Square (RMS) value with reference to an alternating current.  State the following equations for a sinusoidal waveform : RMS value = 0.707 x Maximum value Average value = 0.637 x Maximum value IT2001PA Engineering Essentials (1/2)
5. 5. Chapter 11 – Alternating Current CircuitsAlternating Waveform  Alternating Current (ac) current that is continuously reversing direction, alternately flowing in one direction and then in the other.  Alternating Voltage can be similarly described. The designation ac is normally applied to both current and voltage. 5 IT2001PA Engineering Essentials (1/2)
6. 6. Chapter 11 – Alternating Current CircuitsAlternating Waveform Vertical axis : current (I) or voltage (V) Horizontal axis : time (t) +V +I t t -V -I 6 IT2001PA Engineering Essentials (1/2)
7. 7. Chapter 11 – Alternating Current CircuitsWhy not Direct Current (DC)? Problems of using DC as +V power distribution system:  suffers from rapid power loss in the wires due to their t resistance, which dissipates energy as heat  DC power stations had -V useful ranges of about two kilometers  Once generated, DC power cannot be modified 7 IT2001PA Engineering Essentials (1/2)
8. 8. Chapter 11 – Alternating Current CircuitsPower Distribution System V=RI  1) 2) 3)To prevent risk of To avoid serious For the safety of thesparks and short heating effects, user, the voltagecircuits in the present at high needs to be low in thegenerator itself, current, it needs to be home.power needs to be transmitted at thegenerated with low lowest current V low  I is highvoltages. practical. I low  V is high  DC is not suitableV low  I is high 8 IT2001PA Engineering Essentials (1/2)
9. 9. Chapter 11 – Alternating Current CircuitsAdvantages of A.C. --- readily available from generators and power supply sockets in homes and workshops. --- can be stepped up or down by the use of a transformer. --- for transmission purpose. 9 IT2001PA Engineering Essentials (1/2)
10. 10. Chapter 11 – Alternating Current CircuitsAlternator An alternator is a machine that converts mechanical energy to electrical energy. A simple alternator basically consists of two parts: a) coils or windings b) magnetic poles 10 IT2001PA Engineering Essentials (1/2)
11. 11. Chapter 11 – Alternating Current Circuits Generation of Alternating Current An alternating quantity may be generated by a) rotating a coil in a magnetic field b) rotating a magnetic field within a stationary coilCarbon Brushes 1, 2 Slip Rings a, b 11 IT2001PA Engineering Essentials (1/2)
12. 12. Chapter 11 – Alternating Current CircuitsGeneration of Alternating Current How does AC generate a sine waveform? . . . . . x x x x x 12 IT2001PA Engineering Essentials (1/2)
13. 13. Chapter 11 – Alternating Current CircuitsGeneration of Alternating Current How does AC generate a sine waveform? 13 IT2001PA Engineering Essentials (1/2)
14. 14. Chapter 11 – Alternating Current CircuitsGeneration of Alternating Current How does AC generate a sine waveform? x x . . 14 IT2001PA Engineering Essentials (1/2)
15. 15. Chapter 11 – Alternating Current CircuitsGeneration of Alternating Current AC generator consists of a magnet and a loop of wire which rotates in the magnetic field of the magnet. As the wire rotates in the magnetic field, the changing strength of the magnetic field through the wire produces a force which drives the electric charges around the wire. The force initially generates an electric current in one direction along the wire. Then as the loop rotates through 180 degrees the force reverses to give an electric current in the opposite direction along the wire. Every time the loop rotates through 180 degrees the direction of the force and therefore the current changes. The changing direction of the force after every 180 degrees of rotation gives the alternating current. AC generator also has slip rings which make sure that the ends of the wire are always connected to the same side of the electric circuit. This makes sure that the direction of the current changes every half revolution of the wire. 15 IT2001PA Engineering Essentials (1/2)
16. 16. Chapter 11 – Alternating Current CircuitsGeneration of Alternating emf The magnitude of the emf is given by, e = Em sin θ 16 IT2001PA Engineering Essentials (1/2)
17. 17. Chapter 11 – Alternating Current CircuitsSine Wave  A common type of ac. Also referred as sinusoidal waveform sinusoid Symbol for a sine wave voltage source : OR 17 IT2001PA Engineering Essentials (1/2)
18. 18. Chapter 11 – Alternating Current CircuitsWhat is an Alternating Quantity? An alternating quantity is one which acts in alternate directions and whose magnitude undergoes a definite cycle of changes in definite interval of time. 18 IT2001PA Engineering Essentials (1/2)
19. 19. Chapter 11 – Alternating Current CircuitsPolarity  When the voltage changes polarity at its zero crossing,the current correspondingly changes direction. +V Positive Alternation - VS R t + -V I +V Negative Alternation + I VS R t - -V 19 IT2001PA Engineering Essentials (1/2)
20. 20. Chapter 11 – Alternating Current CircuitsAmplitude  The maximum value of the sine wave. +V Positive Maximum (peak) t Negative Maximum (peak) -V 20 IT2001PA Engineering Essentials (1/2)
21. 21. Chapter 11 – Alternating Current CircuitsWhat is Cycle of a Waveform? One complete waveform is known as one cycle. Each cycle consists of two half-cycles. During 1st half-cycle, the quantity acts in one direction and during the second half-cycle, in the opposite direction. Each cycle : consists of 2 alternations. consists of 2 peaks. reaches maximum amplitude 2 times. 21 IT2001PA Engineering Essentials (1/2)
22. 22. Chapter 11 – Alternating Current CircuitsCycle  A sine wave repeats itself in identical cycles. +V t -V 1st cycle 2nd cycle 3rd cycle 22 IT2001PA Engineering Essentials (1/2)
23. 23. Chapter 11 – Alternating Current CircuitsPeriod  The time required for a given sine wave to complete one full cycle. symbol - T unit - second (s) Time taken is the same for each cycle; thus fixed value for a given sine wave. 23 IT2001PA Engineering Essentials (1/2)
24. 24. Chapter 11 – Alternating Current CircuitsPeriod Measurement  From one zero crossing to the corresponding zero crossing in the next cycle positive zero crossing to positive zero crossing. +V t -V period period period 24 IT2001PA Engineering Essentials (1/2)
25. 25. Chapter 11 – Alternating Current CircuitsPeriod Measurement  From one peak to the corresponding peak in the next cycle positive peak to positive peak. +V t -V period period 25 IT2001PA Engineering Essentials (1/2)
26. 26. Chapter 11 – Alternating Current CircuitsPeriod Measurement  From any point to the corresponding point in the next cycle. +V period t -V period 26 IT2001PA Engineering Essentials (1/2)
27. 27. Chapter 11 – Alternating Current CircuitsExample 11-1  What is the period of the sine wave? V 0 t(s) 2 4 6 1 cycle takes 2s to complete. Therefore the period is 2s. 27 IT2001PA Engineering Essentials (1/2)
28. 28. Chapter 11 – Alternating Current CircuitsExample 11-2  What is the period of the sine wave? V 0 t(s) 12 5 cycles takes 12s to complete 1 cycle takes (12/5)s to complete time taken 1 cycle takes 2.4s to complete T = no. of cycles Therefore the period is 2.4s. 28 IT2001PA Engineering Essentials (1/2)
29. 29. Chapter 11 – Alternating Current CircuitsFrequency  The number of cycles a sine wave can complete in 1 second. symbol : f unit - Hertz (Hz) Examples : 160Hz - 160 complete cycles in 1s. 50Hz - 50 complete cycles in 1s. 29 IT2001PA Engineering Essentials (1/2)
30. 30. Chapter 11 – Alternating Current CircuitsFrequency vs Period  Reciprocal relationship : 1 1 f = T = T f More cycles in 1s Lesser cycles in 1s - higher frequency. - lower frequency. - shorter period. - longer period. V T V T 0 t t 0 1s 1s 30 IT2001PA Engineering Essentials (1/2)
31. 31. Chapter 11 – Alternating Current CircuitsExample 11-3  Which sine wave has the higher frequency? Determine the period and the frequency of both waveforms. V V 0 t 0 t 1s 1s Waveform A Waveform B 31 IT2001PA Engineering Essentials (1/2)
32. 32. Chapter 11 – Alternating Current CircuitsExample 11-3  continue … V V 0 t 0 t 1s 1s Waveform A Waveform B Waveform B complete more cycles in 1s. Therefore Waveform B has the higher frequency. 32 IT2001PA Engineering Essentials (1/2)
33. 33. Chapter 11 – Alternating Current CircuitsExample 11-3  continue … 3 cycles in 1s -- frequency is 3HzV period (T) = 1/f = 1/3Hz t = 0.333s 0 1s Waveform A 3 cycles take 1s OR -- 1 cycles takes (1s/3) = 0.333s -- period is 0.333s frequency (f) = 1/T = 1/0.333s = 3Hz 33 IT2001PA Engineering Essentials (1/2)
34. 34. Chapter 11 – Alternating Current CircuitsExample 11-3  continue … 5 cycles in 1s -- frequency is 5Hz V period (T) = 1/f = 1/5Hz = 0.2s t 0 1s OR Waveform B 5 cycles take 1s -- 1 cycles takes (1s/5) = 0.2s -- period is 0.2s frequency (f) = 1/T = 1/0.2s = 5Hz 34 IT2001PA Engineering Essentials (1/2)
35. 35. Chapter 11 – Alternating Current CircuitsExample 11-4  The period of a sine wave is 10ms. What is the frequency? period (T) = 10ms frequency (f) = 1/T = 1/10ms = 100Hz 35 IT2001PA Engineering Essentials (1/2)
36. 36. Chapter 11 – Alternating Current CircuitsExample 11-5  The frequency of a sine wave is 60Hz. What is the period? frequency (f) = 60Hz period (T) = 1/f = 1/60Hz = 16.7ms 36 IT2001PA Engineering Essentials (1/2)
37. 37. Chapter 11 – Alternating Current CircuitsVoltage Sources If an ac voltage is applied to a circuit, an ac current flows. The voltage and current will have the same frequency. +V +I I t t -V -I 37 IT2001PA Engineering Essentials (1/2)
38. 38. Chapter 11 – Alternating Current CircuitsSine Wave Angles  The horizontal axis can be replaced by angular measurement (degrees). + V ½ cycle one cycle angle 0o 90 o 180 o 270 o 360 o -V peak zero peak zero 38 IT2001PA Engineering Essentials (1/2)
39. 39. Chapter 11 – Alternating Current CircuitsSine Wave Values  Ways to express and measure the value of a sine wave : 1) instantaneous value. 2) peak value. 3) peak-to-peak value. 4) root-mean-square value. 5) average value. 39 IT2001PA Engineering Essentials (1/2)
40. 40. Chapter 11 – Alternating Current CircuitsInstantaneous Value  The voltage or current value of a waveform at a given instant in time. Symbol : voltage - v current - i 40 IT2001PA Engineering Essentials (1/2)
41. 41. Chapter 11 – Alternating Current CircuitsInstantaneous Value  Different at different points of time. +V +I 8.5 5.5 5 t3 3 t3 t t t1 t2 t1 t2 -6 -4.5 -V -I 41 IT2001PA Engineering Essentials (1/2)
42. 42. Chapter 11 – Alternating Current CircuitsInstantaneous Value  Different at different points of angle. +V +I Vp Ip v i degree degree θ θ -V v = Vmax sin θ -I i = Imax sin θ v = Vmax sin 2πft i = Imax sin 2πft 42 IT2001PA Engineering Essentials (1/2)
43. 43. Chapter 11 – Alternating Current CircuitsExample 11-6  A current sine wave has a maximum value of 200A. How much is the current at the instant of 30 degrees of the cycle? i = Imax sin θ i = 200 sin 30 o = 200 x 0.5 = 100A 43 IT2001PA Engineering Essentials (1/2)
44. 44. Chapter 11 – Alternating Current CircuitsPeak Value  The voltage or current value of a waveform at its maximum positive or negative points. Symbol : voltage - Vp or Vmax current - Ip or Imax 44 IT2001PA Engineering Essentials (1/2)
45. 45. Chapter 11 – Alternating Current CircuitsPeak Value  The peaks are equal for a sine wave and is characterized by a single peak value. +V Positive Maximum (peak) t Negative Maximum (peak) -V 45 IT2001PA Engineering Essentials (1/2)
46. 46. Chapter 11 – Alternating Current CircuitsPeak-to-Peak Value  The voltage or current value of a waveform measured from its minimum to its maximum points. Symbol : voltage - Vpp current - Ipp 46 IT2001PA Engineering Essentials (1/2)
47. 47. Chapter 11 – Alternating Current CircuitsPeak-to-Peak Value +V Vpp = 2Vmax Vpp = 2Vmax Vp t 1 Vpp Vp 2 Vpp = Vmax 1 Vmax = Vpp -V 2 Vmax = 0.5Vpp 47 IT2001PA Engineering Essentials (1/2)
48. 48. Chapter 11 – Alternating Current CircuitsRoot Mean Square Value  Equal to the dc voltage that produces the same amount of heat in a resistance as does the AC (sine wave) voltage. + + VS R Vdc R - I radiated - I same heat amount of Vmax = 10 V radiated heat. Vrms = 7.07V = Vdc 48 IT2001PA Engineering Essentials (1/2)
49. 49. Chapter 11 – Alternating Current CircuitsRoot Mean Square Value  Also known as the effective value. Symbol : voltage - Vrms current - Irms The voltage and current values given are usually as rms unless otherwise stated. 49 IT2001PA Engineering Essentials (1/2)
50. 50. Chapter 11 – Alternating Current CircuitsRoot Mean Square Value +V Vrms Vrms = 0.707Vmax Vp t 1 Vrms = Vmax √2 √2 Vrms = Vmax -V Vmax = √2 Vrms Vmax = 1.414 Vrms 50 IT2001PA Engineering Essentials (1/2)
51. 51. Chapter 11 – Alternating Current CircuitsAverage Value The average of a sine wave over half- cycle. The average value taken over a complete cycle is always zero. Symbol : voltage - Vavg current - Iavg 51 IT2001PA Engineering Essentials (1/2)
52. 52. Chapter 11 – Alternating Current CircuitsAverage Value +V Vavg = 0.637 Vmax Vavg 1 Vp t 0.637 V = V avg max 1 Vmax = 0.637 Vavg -V Vmax = 1.57 Vavg 52 IT2001PA Engineering Essentials (1/2)
53. 53. Chapter 11 – Alternating Current CircuitsForm & Peak Factor for Sine Wave Form factor = Rms value Avg value 0.707 X Max value = 0.637 X Max value = 1.11 Max value Peak factor = Rms value Max value = 0.707 X Max value = 1.414 53 IT2001PA Engineering Essentials (1/2)
54. 54. Chapter 11 – Alternating Current CircuitsExample 11-7  Determine Vmax when : Vpp = 3V ; Vrms = 5V ; Vavg = 4V. Vmax = 0.5 Vpp Vmax = 1.414 Vrms = 0.5 (3) = 1.414 (5) = 1.5V = 7.07V Vmax = 1.57 Vavg = 1.57 (4) = 6.28V 54 IT2001PA Engineering Essentials (1/2)
55. 55. Chapter 11 – Alternating Current CircuitsExample 11-8  Determine Vrms when Vp = 20V ; Vpp = 10V ; Vavg = 30V. Vrms= 0.707 Vmax Vmax= 0.5 Vpp Vmax = 1.57 Vavg = 0.707 (20) = 0.5 (10) = 1.57 (30) =14.14V = 5V = 47.1V Vrms = 0.707 Vmax Vrms= 0.707 Vmax = 0.707 (5) = 0.707 (47.1) = 3.535V = 33.33V 55 IT2001PA Engineering Essentials (1/2)
56. 56. Chapter 11 – Alternating Current CircuitsExample 11-9  Determine Vavg when Vmax = 37V ; Vpp = 28V ; Vrms = 46V. Vavg= 0.637 Vmax Vmax= 0.5 Vpp Vmax=1.414 Vrms = 0.637 (37) = 0.5 (28) = 1.414 (46) = 23.569V = 14V = 65. V Vavg= 0.637 Vp Vavg= 0.637 Vmax = 0.637 (14) = 0.637 (65.) = 8.918V = 41.5V 56 IT2001PA Engineering Essentials (1/2)
57. 57. Chapter 11 – Alternating Current CircuitsSine Wave Values Vpp = 2Vmax Vmax = 0.5Vpp Vrms = 0.707Vmax Vmax = 1.414 Vrms Vavg = 0.637 Vmax Vmax = 1.57 Vavg Note: 1 1 0.707 = 1.414 = 1.57 0.637 57 IT2001PA Engineering Essentials (1/2)
58. 58. Chapter 11 – Alternating Current CircuitsSummary 58 IT2001PA Engineering Essentials (1/2)
59. 59. Chapter 11 – Alternating Current CircuitsNext Lesson 59 IT2001PA Engineering Essentials (1/2)