B tech ee ii_ eee_ u-2_ ac circuit analysis_dipen patel
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This document provides an overview of AC circuit analysis and three-phase systems. It discusses: 1. The basics of AC circuits including sinusoidal waveforms, impedance, and Ohm's law for AC circuits. 2. Three-phase systems including how the three voltages are phase-shifted by 120 degrees, derivation of line voltages, and generation of three-phase voltages using a three-phase generator. 3. Different connections for three-phase systems including star, delta, 4-wire and 3-wire systems and the implications of each.
![AC circuits -- Impedance
• Impedance and Ohm’s Law for AC:
– Impedance is Z = R + jX,
where j = -1, and X is the reactance in [].
– Ohm’s AC Law in s domain: v = i Z
• Resistance R dissipates power as heat.
• Reactance X stores and returns power.
– Inductors have positive reactance Xl=L
– Capacitors have negative reactance Xc=-1/C](https://image.slidesharecdn.com/btecheeiieeeu-2accircuitanalysisdipenpatel-150317005057-conversion-gate01/85/B-tech-ee-ii_-eee_-u-2_-ac-circuit-analysis_dipen-patel-14-320.jpg)
B tech ee ii_ eee_ u-2_ ac circuit analysis_dipen patel
- 1. Unit 2 AC Circuit Analysis Course : B.Tech Branch : EE Semester : II Subject : Elements of Electrical Engineering
- 2. AC circuit • In alternating current (AC, also ac), the flow of electric charge periodically reverses direction. In direct current (DC, alsodc), the flow of electric charge is only in one direction. The abbreviations AC and DC are often used to mean simply alternating and direct, as when they modify current or voltage. • AC is the form in which electric power is delivered to businesses and residences. The usual waveform of an AC power circuit is a sine wave. In certain applications, different waveforms are used, such as triangular or square waves. Audio and radio signals carried on electrical wires are also examples of alternating current. In these applications, an important goal is often the recovery of information encoded (or modulated) onto the AC signal.
- 3. Generation of Sinusoidal Waveforms Fig 1
- 4. Cont.. • However, if the conductor moves in parallel with the magnetic field in the case of points A and B, no lines of flux are cut and no EMF is induced into the conductor, but if the conductor moves at right angles to the magnetic field as in the case of points C and D, the maximum amount of magnetic flux is cut producing the maximum amount of induced EMF. • Also, as the conductor cuts the magnetic field at different angles between points A and C, 0 and 90o the amount of induced EMF will lie somewhere between this zero and maximum value. Then the amount of emf induced within a conductor depends on the angle between the conductor and the magnetic flux as well as the strength of the magnetic field.
- 5. Cont.. • to convert a mechanical energy such as rotation, into electrical energy, a Sinusoidal Waveform. A simple generator consists of a pair of permanent magnets producing a fixed magnetic field between a north and a south pole. Inside this magnetic field is a single rectangular loop of wire that can be rotated around a fixed axis allowing it to cut the magnetic flux at various angles as shown below.
- 6. Fig 2
- 7. Cont.. • As the coil rotates anticlockwise around the central axis which is perpendicular to the magnetic field, the wire loop cuts the lines of magnetic force set up between the north and south poles at different angles as the loop rotates. The amount of induced EMF in the loop at any instant of time is proportional to the angle of rotation of the wire loop. • As this wire loop rotates, electrons in the wire flow in one direction around the loop. Now when the wire loop has rotated past the 180o point and moves across the magnetic lines of force in the opposite direction, the electrons in the wire loop change and flow in the opposite direction. Then the direction of the electron movement determines the polarity of the induced voltage.
- 8. Cont.. • So we can see that when the loop or coil physically rotates one complete revolution, or 360o, one full sinusoidal waveform is produced with one cycle of the waveform being produced for each revolution of the coil. As the coil rotates within the magnetic field, the electrical connections are made to the coil by means of carbon brushes and slip- rings which are used to transfer the electrical current induced in the coil. • The amount of EMF induced into a coil cutting the magnetic lines of force is determined by the following three factors. • • Speed – the speed at which the coil rotates inside the magnetic field. • • Strength – the strength of the magnetic field. • • Length – the length of the coil or conductor passing through the magnetic field.
- 9. Cont.. • We know that the frequency of a supply is the number of times a cycle appears in one second and that frequency is measured in Hertz. As one cycle of induced emf is produced each full revolution of the coil through a magnetic field comprising of a north and south pole as shown above, if the coil rotates at a constant speed a constant number of cycles will be produced per second giving a constant frequency. So by increasing the speed of rotation of the coil the frequency will also be increased. Therefore, frequency is proportional to the speed of rotation, ( ƒ ∝ Ν ) where Ν = r.p.m.
- 10. Cont.. • Also, our simple single coil generator above only has two poles, one north and one south pole, giving just one pair of poles. If we add more magnetic poles to the generator above so that it now has four poles in total, two north and two south, then for each revolution of the coil two cycles will be produced for the same rotational speed. Therefore, frequency is proportional to the number of pairs of magnetic poles, ( ƒ ∝ P ) of the generator where P = is the number of “pairs of poles”. • Then from these two facts we can say that the frequency output from an AC generator is:
- 11. Cont..
- 12. Instataneous voltage. • The EMF induced in the coil at any instant of time depends upon the rate or speed at which the coil cuts the lines of magnetic flux between the poles and this is dependant upon the angle of rotation, Theta ( θ ) of the generating device. Because an AC waveform is constantly changing its value or amplitude, the waveform at any instant in time will have a different value from its next instant in time. • For example, the value at 1ms will be different to the value at 1.2ms and so on. These values are known generally as the Instantaneous Values, or Vi Then the instantaneous value of the waveform and also its direction will vary according to the position of the coil within the magnetic field as shown below.
- 13. Cont.. • The instantaneous values of a sinusoidal waveform is given as the “Instantaneous value = Maximum value x sin θ ” and this is generalized by the formula. • Where, Vmax is the maximum voltage induced in the coil and θ = ωt, is the angle of coil rotation. • If we know the maximum or peak value of the waveform, by using the formula above the instantaneous values at various points along the waveform can be calculated. By plotting these values out onto graph paper, a sinusoidal waveform shape can be constructed. In order to keep things simple we will plot the instantaneous values for the sinusoidal waveform at every 45o and assume a maximum value of 100V.
- 14. AC circuits -- Impedance • Impedance and Ohm’s Law for AC: – Impedance is Z = R + jX, where j = -1, and X is the reactance in []. – Ohm’s AC Law in s domain: v = i Z • Resistance R dissipates power as heat. • Reactance X stores and returns power. – Inductors have positive reactance Xl=L – Capacitors have negative reactance Xc=-1/C
- 15. 15 1.2: Three-Phase System In a three phase system the source consists of three sinusoidal voltages. For a balanced source, the three sources have equal magnitudes and are phase displaced from one another by 120 electrical degrees. A three-phase system is superior economically and advantage, and for an operating of view, to a single- phase system. In a balanced three phase system the power delivered to the load is constant at all times, whereas in a single-phase system the power pulsates with time.
- 16. 16 16 v(t) t vR vY vB The instantaneous e.m.f. generated in phase R, Y and B: eR = EmR sin t eY = EmY sin (t -120o) eB = EmB sin (t -240o) = EmBsin (t +120o)
- 17. 17 1.3: Generation of Three-Phase Three separate windings or coils with terminals R-R’, Y-Y’ and B-B’ are physically placed 120o apart around the stator. Y’ BY B’ Stator Rotor Y R B R R’ N S
- 19. The instantaneous e.m.f. generated in phase R, Y and B: eR = EmR sin ωt eY = EmY sin (ωt -120o) eB = EmB sin (ωt -240o) = EmBsin (ωt +120o) In phasor domain: ER = ERrms 0o EY = EYrms -120o EB = EBrms 120o Phase voltage 120o -120o 0o ERrms = EYrms = EBrms = Ep Magnitude of phase voltage 19
- 20. ER Three-phase Load Three-phase AC generator VY IR VR EY EB ZR IY IB ZB ZY VB IN Line voltage ERY ERY = ER - EY
- 21. 21 Line voltage ERY = ER - EY 120o -120o 0o -EY ERY = Ep 0o - Ep -120o = 1.732Ep ERY 30o= √3 Ep = EL 30o 30o
- 22. ER Three-phase Load Three-phase AC generator VY IR VR EY EB ZR IY IB ZB ZY VB IN Line voltage EYB EYB = EY - EB
- 23. Line voltage EYB = EY - EB 120o -120o 0o -EB EYB = Ep -120o - Ep 120o = 1.732Ep EYB -90o -90o= √3 Ep = EL -90o
- 24. ER Three-phase Load Three-phase AC generator VY IR VR EY EB ZR IY IB ZB ZY VB IN Line voltage EBR EBR = EB - ER
- 25. 25 25 Line voltage EBR = EB - ER 120o -120o 0o -ER EBR = Ep 120o - Ep 0o = 1.732Ep EBR 150o 150o= √3 Ep = EL 150o For star connected supply, EL= √3 Ep
- 26. 26 120o -120o 0o Phase voltages ER = Ep 0o EY = Ep -120o EB = Ep 120o Line voltages ERY = EL 30o EYB = EL -90o EBR = EL 150o It can be seen that the phase voltage ER is reference.
- 27. 27 27 Phase voltages ER = Ep -30o EY = Ep -150o EB = Ep 90o Line voltages ERY = EL 0o EYB = EL -120o EBR = EL 120o Or we can take the line voltage ERY as reference.
- 28. ER Three-phase Load Three-phase AC generator VY IR VR EY EB ZR IY IB ZB ZY VB ERY Delta connected Three-Phase supply ERY = ER = Ep 0o
- 29. ER Three-phase Load Three-phase AC generator VY IR VR EY EB ZR IY IB ZB ZY VB EYB EBR For delta connected supply, EL= Ep Delta connected Three-Phase supply
- 30. 30 Connection in Three Phase System 4-wire system (neutral line with impedance) 3-wire system (no neutral line ) 4-wire system (neutral line without impedance) Star-Connected Balanced Loads a) 4-wire system b) 3-wire system 3-wire system (no neutral line ), delta connected load Delta-Connected Balanced Loads a) 3-wire system
- 31. ER Three-phase Load Three-phase AC generator VY IR VR EY EB ZR IY IB ZB ZY VB IN ZN VN 4-wire system (neutral line with impedance) VN = INZNVoltage drop across neutral impedance: 1.1
- 32. ER Three-phase Load Three-phase AC generator VY IR VR EY EB ZR IY IB ZB ZY VB IN ZN VN 4-wire system (neutral line with impedance) IR + IY + IB= IN Applying KCL at star point 1.2
- 33. ER Three-phase Load Three-phase AC generator VY IR VR EY EB ZR IY IB ZB ZY VB IN ZN VN 4-wire system (neutral line with impedance) Applying KVL on R-phase loop 33
- 34. ER Three-phase Load Three-phase AC generator IR VR ZR IN ZN VN Applying KVL on R-phase loop ER – VR – VN = 0 ER – IRZR – VN = 0 IR = Thus ER – VN ZR 1.3 4-wire system (neutral line with impedance) 34
- 35. 35 ER Three-phase Load Three-phase AC generator VY IR VR EY EB ZR IY IB ZB ZY VB IN ZN VN 4-wire system (neutral line with impedance) Applying KVL on Y-phase loop
- 36. 36 Three-phase Load Three-phase AC generator VY EY IY ZY IN ZN VN Applying KVL on Y-phase loop 4-wire system (neutral line with impedance) EY – VY – VN = 0 EY – IYZY – VN = 0 IY = Thus EY – VN ZY 1.4
- 37. Three-phase Load Three-phase AC generator EB IB ZB VB IN ZN VN 4-wire system (neutral line with impedance) Applying KVL on B-phase loop EB – VB – VN = 0 EB – IBZB – VN = 0 IB = Thus EB – VN ZB 1.5 37
- 38. 38 4-wire system (neutral line with impedance) IR + IY + IB= IN Substitute Eq. 1.2, Eq.1.3, Eq. 1.4 and Eq. 1.5 into Eq. 1.1: = EB – VN ZB + EY – VN ZY ER – VN ZR + VN ZN ER – VN EY – VN + + EB – VN = VN ZNZR ZR ZY ZY ZB ZB ER ZR + EY ZY + EB ZB = 1 ZN + 1 ZR + 1 ZY VN + 1 ZB
- 39. 39 4-wire system (neutral line with impedance) ER ZR + EY ZY + EB ZB = 1 ZN + 1 ZR + 1 ZY VN + 1 ZB VN = ER ZR + EY ZY + EB ZB 1 ZN + 1 ZR + 1 ZY + 1 ZB 1.6
- 40. 4-wire system (neutral line with impedance) VN = ER ZR + EY ZY + EB ZB 1 ZN + 1 ZR + 1 ZY + 1 ZB 1.6 VN is the voltage drop across neutral line impedance or the potential different between load star point and supply star point of three-phase system. We have to determine the value of VN in order to find the values of currents and voltages of star connected loads of three-phase system. 40
- 41. Example ER Three-phase Load ZY= 2 Ω IR VR EY EB ZR = 5 Ω IY IB ZB = 10 Ω VB IN ZN =10 Ω VN Find the line currents IR ,IY and IB. Also find the neutral current IN. EL = 415 volt
- 42. ER Three-phase Load Three-phase AC generator VY IR VR EY EB ZR IY IB ZB ZY VB IN ZN VN 3-wire system (no neutral line )
- 43. ER Three-phase Load Three-phase AC generator VY IR VR EY EB ZR IY IB ZB ZY VB VN 3-wire system (no neutral line ) No neutral line = open circuit , ZN = ∞ 43
- 44. Example ER Three-phase Load ZY= 2 Ω IR VR EY EB ZR = 5 Ω IY IB ZB = 10 Ω VB VN EL = 415 volt Find the line currents IR ,IY and IB . Also find the voltages VR, VY and VB.
- 45. 3-wire system (no neutral line ),delta connected load ER Three-phase Load Three-phase AC generator VY IR VR EY EB ZR IY IB ZB ZY VB
- 46. 3-wire system (no neutral line ),delta connected load ER Three-phase Load Three-phase AC generator IR EY EB IY IB VRY ZRYZBR ZYB VYB VBR Ir Ib Iy
- 47. 3-wire system (no neutral line ),delta connected load ER Three-phase Load Three-phase AC generator IR EY EB IY IB VRY ZRYZBR ZYB VYB VBR Ir Ib Iy ERY =VRY EYB =VYB EBR =VBR
- 48. 48 3-wire system (no neutral line ),delta connected load Phase currents 30o Ir = VRY ZRY = ERY ZRY = EL ZRY -90o Iy = VYB ZYB = EYB ZYB = EL ZYB 150o Ib = VBR ZBR = EBR ZBR = EL ZBR
- 49. 3-wire system (no neutral line ),delta connected load ER Three-phase Load Three-phase AC generator IR EY EB IY IB VRY ZRYZBR ZYB VYB VBR Ir Ib Iy ERY =VRY EYB =VYB EBR =VBR Line currents IR = Ir Ib- = EL ZRY 30o - 150oEL ZBR IY = Iy Ir- = EL ZYB -90o - 30oEL ZRY
- 50. 3-wire system (no neutral line ),delta connected load ER Three-phase Load Three-phase AC generator IR EY EB IY IB VRY ZRYZBR ZYB VYB VBR Ir Ib Iy ERY =VRY EYB =VYB EBR =VBR Line currents IB = Ib Iy- = EL ZBR 150o - -90oEL ZYB
- 51. ZRY ZBR ZYB ZR ZB ZY Star to delta conversion ZRY = ZRZY + ZYZB + ZBZR ZB ZYB = ZRZY + ZYZB + ZBZR ZR ZBR = ZRZY + ZYZB + ZBZR ZY
- 52. Example ER Three-phase Load ZY= 2 Ω IR VR EY EB ZR = 5 Ω IY IB ZB = 10 Ω VB VN Find the line currents IR ,IY and IB . EL = 415 volt Use star-delta conversion.
- 53. ER Three-phase Load Three-phase AC generator VY IR VR EY EB ZR IR IB ZB ZY VB IN ZN VN 4-wire system (neutral line without impedance) = 0 Ω VN = INZN = IN(0) = 0 volt 53
- 54. 54 4-wire system (neutral line without impedance) For 4-wire three-phase system, VN is equal to 0, therefore Eq. 1.3, Eq. 1.4, and Eq. 1.5 become, IB = EB ZB 1.5 EB – VN IY = EY ZY 1.4 EY – VN IR = ER ZR 1.3 ER – VN
- 55. Example ER Three-phase Load ZY= 2 Ω IR VR EY EB ZR = 5 Ω IY IB ZB = 10 Ω VB IN VN Find the line currents IR ,IY and IB . Also find the neutral current IN. EL = 415 volt
- 56. 56 v(t) t vR vY vB The instantaneous e.m.f. generated in phase R, Y and B: eR = EmR sin t eY = EmY sin (t -120o) eB = EmB sin (t -240o) = EmBsin (t +120o)
- 57. 57 1.4: Phase sequences RYB and RBY 120o -120o 120o VR VY VB o )rms(RR 0VV o )rms(YY 120VV o )rms(B o )rms(BB 120V 240VV VR leads VY, which in turn leads VB. This sequence is produced when the rotor rotates in the counterclockwise direction. (a) RYB or positive sequence
- 58. 58 (b) RBY or negative sequence 120o -120o 120o VR VB VY o )rms(RR 0VV o )rms(BB 120VV o rmsY o rmsYY V V 120 240 )( )( V VR leads VB, which in turn leads VY. This sequence is produced when the rotor rotates in the clockwise direction.
- 59. 59 1.5: Connection in Three Phase System R Y B ZR Z Y Z B 1.5.1: Star Connection a) Three wire system
- 60. 60 Star Connection b) Four wire system VRN VBN VYN ZR Z Y Z B R B N Y
- 61. 61 Wye connection of Load Z1 Z3 Z2 R B Y N Load Z3 Z 1 Z 2 R Y B Load N
- 63. 63 Delta connection of load Zc Za Zb R B Y Load Z c Zb Za R Y B Load
- 64. 64 1.6: Balanced Load Connection in 3- Phase System
- 65. 65 Example ER Three-phase Load ZY= 20 Ω IR VR EY EB ZR = 20 Ω IY IB ZB = 20 Ω VB VN EL = 415 volt Find the line currents IR ,IY and IB . Also find the voltages VR, VY and VB. Wye-Connected Balanced Loads b) Three wire system
- 66. 66 Wye-Connected Balanced Loads b) Three wire system VN = = 0 volt VR = ER VY = EY VB = EB
- 67. 67 Example ER Three-phase Load ZY= 20 Ω IR VR EY EB ZR = 20 Ω IY IB ZB = 20 Ω VB IN VN Find the line currents IR ,IY and IB . Also find the neutral current IN. EL = 415 volt 1.6.1: Wye-Connected Balanced Loads a) Four wire system
- 68. 68 VRN VBN Z1 Z 2 Z 3 R B N Y VYN IR IY IB IN BYRN IIII For balanced load system, IN = 0 and Z1 = Z2 = Z3 3 o BN B 2 o YN Y 1 o RN R Z 120V I Z 120V I Z 0V I BNYNRNphasa phasaBN phasaYN phasaRN VVVVwhere 120VV 120VV 0VV 1.6.1: Wye-Connected Balanced Loads a) Four wire system
- 69. 69 Wye-Connected Balanced Loads b) Three wire system R Y B Z1 Z 2 Z 3 IR IY IB VRY VYB VBR S 0III BYR 3 o BS B 2 o YS Y 1 o RS R Z 120V I Z 120V I Z 0V I BSYSRSphasa phasaBS phasaYS phasaRS VVVVwhere 240VV 120VV 0VV
- 70. 70 1.6.2: Delta-Connected Balanced Loads Z Z Z R Y B VRY VYB VBR IR IRY IBR IYB IB IY Phase currents: 3 o BR BR 2 o YB YB 1 o RY RY Z 120V I Z 120V I Z 0V I Line currents: YBBRB RYYBY BRRYR III III III lineBYR phasaBRYBRY IIIIand IIIIwhere
- 72. 72 1.7.1: Wye-Connected Unbalanced Loads Four wire system VRN VBN Z1 Z 2 Z 3 R B N Y VYN IR IY IB IN BYRN IIII For unbalanced load system, IN 0 and Z1 Z2 Z3 3 o BN B 2 o YN Y 1 o RN R Z 120V I Z 120V I Z 0V I 120VV 120VV 0VV phasaBN phasaYN phasaRN
- 73. 73 1.7.2: Delta-Connected Unbalanced Loads Z Z Z R Y B VRY VYB VBR IR IRY IBR IYB IB IY Phase currents: 3 o BR BR 2 o YB YB 1 o RY RY Z 120V I Z 120V I Z 0V I Line currents: YBBRB RYYBY BRRYR III III III 120VV 120VV 0VV phasaBN phasaYN phasaRN
- 74. 74 1.8 Power in a Three Phase System
- 75. 75 Power Calculation The three phase power is equal the sum of the phase powers P = PR + PY + PB If the load is balanced: P = 3 Pphase = 3 Vphase Iphase cos θ
- 76. 76 1.8.1: Wye connection system: I phase = I L and Real Power, P = 3 Vphase Iphase cos θ Reactive power, Q = 3 Vphase Iphase sin θ Apparent power, S = 3 Vphase Iphase or S = P + jQ phaseLL VV 3 WattIV LLL cos3 VARIV3 LLL sin VAIV3 LLL
- 77. 77 1.8.2: Delta connection system VLL= Vphase P = 3 Vphase Iphase cos θ phaseL I3I WattIV LLL cos3
- 78. References - images • http://www.electronicstutorials.ws/accircuits/a cp15.gif http://s.eeweb.com/members/andrew_carter/ blog/2012/02/18/Fundamental-Concepts-in- Generation-of-Sinusoidal-Waveform- 1329577510.jpg
- 79. REFERENCE- BOOK • B.L.Theraja, “Electrical Technology Vol.1”, S.Chand Publication. • D.P.Kothari, “Basic Electrical Engineering”, Tata McGraw-Hill publication. • U.A.Patel “Circuits and Networks”. WEB REFRENCE WWW.SCRIBD.COM WWW.AUTHORSTREAM.COM