BEEE_unit2 for jntuk and jntuh 1st year students.pptx

BEE first year students engineering jntuk and jntuh

1. BASICS OF ELECTRICAL ENGINEERING

2. CONTENTS UNIT – II L-9 Fundamentals of A.C. Circuits : Generation of A.C. voltage - frequency, average value, R.M.S. value, form factor, peak factor for sinusoidal only; Phasor representation of alternating quantities; Analysis of simple series and parallel A.C. circuits (simple numerical problems). Balanced Three phase systems : Relation between phase and line quantities of voltages and currents in star and delta connected systems (Elementary treatment only). BASICS OF ELECTRICAL AND ELECTRONICS ENGINEERING 2

3. BASICS OF ELECTRICAL AND ELECTRONICS ENGINEERING 3 WHAT IS ALTERNATING CURRENT (A.C.)? Alternating current is the current which constantly changes in amplitude, and which reverses direction at regular intervals. Because the changes are so regular, alternating voltage and current have a number of properties associated with any such waveform. These basic properties include : Frequency : It is the number of complete cycles that occurred in one second. The frequency of the wave is commonly measured in cycles per second (cycles/sec) and expressed in units of Hertz (Hz). It is represented in mathematical equations by the letter ‘ f ’.

4. BASICS OF ELECTRICAL AND ELECTRONICS ENGINEERING 4 Alternating Voltage is the voltage which varies its magnitude as well as polarity periodically. example- v= Vm sin ωt When the current flowing in the circuit varies in magnitude as well as in direction periodically is called as an alternating current. example- i= Im sin ωt ALTERNATING VOLTAGE ALTERNATING CURRENT

5. BASICS OF ELECTRICAL AND ELECTRONICS ENGINEERING 5 WAVEFORM The shape of the curve obtained by plotting the instantaneous values of voltage or current as ordinate against time as abscissa is called its wave form or wave shape. One complete set of positive or negative values of an alternating quantity is known as cycle . One cycle spans 360 electrical degrees. CYCLE

6. BASICS OF ELECTRICAL AND ELECTRONICS ENGINEERING 6 The time taken in seconds to complete one cycle of an alternating quantity is called its time period . Units- Seconds   The number of cycles occur in one sec is called frequency . Units: Cycles/seconds (or) Hz The maximum value (positive or negative) obtained by an alternating quantity is called its amplitude ( V m or I m ).   TIME PERIOD(T) FREQUENCY(f) AMPLITUDE (OR) PEAK VALUE (OR) MAXIMUM VALUE

7. BASICS OF ELECTRICAL AND ELECTRONICS ENGINEERING 7 The arithmetical average of an alternating quantity over one cycle is called its average value. Average Value can be determined by Graphical Method or Analytical Method From fig (1) average value can calculate by using graphical method as. I 2 I 3 I 4 I n-1 I n I 1 Time (S) Current (A) Fig. 1. Current wave form       AVERAGE VALUE

8. BASICS OF ELECTRICAL AND ELECTRONICS ENGINEERING 8 The effective or RMS value of an alternating current is the steady current (D.C) which when flowing through a given resistance for a given time produces the same amount of heat as produced by a alternating current when flowing through the same resistance for the same time . RMS Value can be determined by Graphical Method or Analytical Method Current (A) Time (S) dt RMS VALUE (or) EFFECTIVE VALUE    

9. BASICS OF ELECTRICAL AND ELECTRONICS ENGINEERING 9 The peak factor of an alternating quantity is defined as the ratio of its maximum value to the RMS value.   The form factor of an alternating quantity is defined as the ratio of RMS value to the average value .   FORM FACTOR PEAK FACTOR (or) CREST FACTOR

10. BASICS OF ELECTRICAL AND ELECTRONICS ENGINEERING 10 AVERAGE VALUE OF A SINUSOIDAL WAVEFORM   The given wave form is a symmetrical wave form, consider only alternation.  

11. BASICS OF ELECTRICAL AND ELECTRONICS ENGINEERING 11 RMS VALUE OF A SINUSOIDAL WAVEFORM The given wave form is a symmetrical wave form with a time period    

12. The voltage supplied by an AC source is sinusoidal with a period T = 2p/w = 1/f. A circuit consisting of a resistor of resistance R connected to an AC source, indicated by Resistor in AC circuit D v = D v R = D V max sin w t i R = D v R /R = ( D V max /R) sin w t = I max sin w t Kirchhoff’s loop rule: D v + D v R = 0 amplitude ~

13. Fig 33-3, p.1035 Plots of the instantaneous current i R and instantaneous voltage  v R across a resistor as functions of time. The current is in phase with the voltage . At time t = T , one cycle of the time-varying voltage and current has been completed. (b) Phasor diagram for the resistive circuit showing that the current is in phase with the voltage. D v R = D V max sin w t i R = D v R /R = ( D V max /R)sin w t = I max sin w t PHASOR DIAGRAM

14. Fig 33-5, p.1037 Graph of the current (a) and of the current squared (b) in a resistor as a function of time. The average value of the current i over one cycle is zero. Notice that the gray shaded regions above the dashed line for I 2 max /2 have the same area as those below this line for I 2 max /2 . Thus, the average value of I 2 is I 2 max /2 . The root-mean-square current, I rms , is = I 2 max sin 2 w t

15. A circuit consisting of an inductor of inductance L connected to an AC source. inductive reactance Inductors in an AC circuit Kirchhoff’s law Instantaneous current in the inductor

16. Plots of the instantaneous current i L and instantaneous voltage  v L across an inductor as functions of time. (b) Phasor diagram for the inductive circuit. The current lags behind the voltage by 90°.

17. A circuit consisting of a capacitor of ca- pacitance C connected to an AC source. capacitive reactance Capacitor in an AC circuit

18. Plots of the instantaneous current i C and instantaneous voltage  v C across a capacitor as functions of time. The voltage lags behind the current by 90°. (b) Phasor diagram for the capacitive circuit. The current leads the voltage by 90°.

19. At what frequencies will the bulb glow the brightest? High, low? Or is the brightness the same for all frequencies? High freq.: low X c , high X L

20. Fig 33-13, p.1044 A series circuit consisting of a resistor, an inductor, and a capacitor connected to an AC source. (b) Phase relationships for instantaneous voltages in the series RLC circuit. The current at all points in a series AC circuit has the same amplitude and phase.

21. Phasors for (a) a resistor, (b) an inductor, and (c) a capacitor connected in series. Phasor diagram for the series RLC circuit of Fig. 33.13a. The phasor  V R is in phase with the current phasor I max , the phasor  V L leads I max by 90°, and the phasor  V C lags I max by 90°.  V max makes an angle  with I max .(b) Simplified version of part (a) of the figure.

22. Fig 33-15, p.1045 Phasor diagrams for the series RLC circuit of Fig. 33.13a. Z is the impedance of the circuit

23. Fig 33-17, p.1046 Label each part of the figure as being X L > X C , X L = X C , or X L < X C . f

24. Fig 33-18, p.1046 The phasor diagram for a RLC circuit with D V max = 120 V, f = 60 Hz, R = 200 W , and C = 4 m F. What should be L to match the phasor diagram? L =…= 0.84 H

25. A series RLC circuit has D V max =150 V, w = 377 s -1 Hz, R= 425 W , L= 1.25 H, and C = 3.5 m F. Find its Z, X C , X L , I max , the angle f between current and voltage, and the maximum and instantaneous voltages across each element. X C = 471 W, X L = 758 W Z = 513 W, I max = 0.292 A f = -34 o

26. Power in an AC Circuit The average power delivered by the source is converted to internal energy in the resistor Pure resistive load: f = 0  No power losses are associated with pure capacitors and pure inductors in an AC circuit ( ) Momentary values, average is 0

27. Introduction to three phase system Generation of three phase EMF Phase sequence Connection of three phase winding Star connection Delta connection Line and phase quantities in star connected circuit Line and phase quantities in delta connected circuit Power in three phase systems with balanced load Balanced and Unbalanced Systems Balanced Three phase systems CONTENTS

28. Introduction to three phase system Any system utilizing more than one winding is called Poly phase system Phase angle ( Φ )between adjacent phases In 3 phase system Φ = 360/3 = 120 In 4 phase system Φ = 360/4 = 90 In N phase system Φ = 360/N ( Where N = no of phases) In three phase system, 3 independent winding are displaced by 120 . R1 R V RN

29. Why three phase system ? Designing of high rated machines are complicated in single phase system. Three phase motors has good power factor and efficiency For power generation and power transmission three phase system is more convenient

30. Generation of 3- φ EMF and phasor diagram For a balanced supply RMS Value of all the phase voltages are equal emf eqns 120 120 120 V RN V YN V BN

31. 120 120 R phase B phase Y phase

32. Phase sequence The order in which the phase voltages of a three phase system attains their peak or maximum positive values is called the phase sequence The phase sequence is RYB 120 120 120 V RN V YN V BN 1 2 3 1 2 3

33. V RN V YN V BN V RN V YN V BN Phase sequence is RYB Phase sequence is RBY Determine phase sequence from the Phasor diagram

34. Connection of three phase winding Two conductors for each phase Six conductors for transferring power in three phase system Two possible connections are Star connection Delta connection Load 1 Load 2 Load 3 R R 1 V RN Y Y 1 B B 1 V BN 1 2 3 4 5 6 V YN

35. Reduced total no of conductors Neutral point or star point Star connection B B 1 V BN + - Y Y 1 V YN + - R R 1 V RN + - Neutral point I R R I Y Y I B B R R 1 V RN Y Y 1 v YN B B 1 V BN R 1 R V RN Y 1 Y v YN B 1 B V BN line voltages are phase voltages are phase currents are line currents are V RN , V YN & V BN V RY , V YB & V BR I 1 , I 2 & I 3 I R , I Y & I B I 1 I 3 I 2 V BR V RY V YB Starting point Finishing point

36. Delta connection Starting end of one phase being joined to the finishing end of another phase R R 1 V RN Y Y 1 V YN B B 1 V BN I R R B I B I Y Y here line voltages are V RY V YB V BR V RN V YN V BN here phase voltages are here phase currents are here line currents are V RN , V YN & V BN V RY , V YB & V BR I 1 , I 2 & I 3 I R , I Y & I B

37. Voltage current relation ship in an A.C circuit for different loads R R R 1 V RN R V RN V RN L R R 1 V RN C C V RN R R 1 V RN L I R I C I L I R I L I C V RN I R V RN V RN I L I C

38. Voltage current relation ship in an A.C circuit R V RN L I V RN I Most of the electric loads are RL Loads Φ

39. Line and phase quantities in star connected circuit -V RY +V RN –V YN =0 V YB = V YN –V BN V RY = V RN –V YN V BR = V BN –V RN Similarly line current = phase current B B 1 V BN + - Y Y 1 V YN + - R R 1 V RN + - Neutral point I R R I Y Y I B B I 1 I 3 I 2 V BR V RY V YB Apply KCL at node A we get A -I 1 + I R = 0 I 1 = I R Relation ship between phase current and line current Relation ship between phase voltage and line voltage here phase currents are I 1 , I 2 & I 3 here line currents are I R , I Y & I B Similarly I 2 = I Y & I 3 = I B I L = I ph + - + + - - Apply KVL in loop abcdea , we get d a b c e here line voltages are here phase voltages are V RN , V YN & V BN V RY , V YB & V BR

40. V RN V Y N V B N 120 120 120 V YB = V YN –V BN V RY = V RN –V YN V BR = V BN –V RN - V YN V RY V YB -V BN -V RN V BR 30 30 Line and phase quantities in star connected circuit

41. B A C Phasor addition (parallelogram method)

42. Line and phase quantities in star connected circuit contd.. From phasor diagram Similarly and Line Voltage = √3 times of phase voltage V Y N V YB 30 30 -V BN

43. Line and phase quantities in delta connected circuit Similarly Line voltage = phase voltage Relation ship between phase voltage and line voltage here line voltages are here phase voltages are V RN , V YN & V BN V RY , V YB & V BR Apply KVL in loop abcda , we get -V RY V RN –V YN =0 V RY V YB V BR V RN V YN V BN + - + + - - a b c d -V RY +V RN =0 V RY =V RN V YB = V YN & V BR = V BN V L = V ph Apply KCL at node ‘b’, we get Relation ship between phase current and line current here phase currents are I 1 , I 2 & I 3 here line currents are I R , I Y & I B I R – I 1 +I 3 =0 I R = I 1 -I 3 Similarly I Y = I 2 -I 1 & I B = I 3 -I 2

44. Φ Φ I 1 Line and phase quantities in delta connected circuit I R = I 1 -I 3 I B = I 3 -I 2 V RN V Y N V B N I 2 I 3 -I 3 I R Φ 30 30 -I 1 -I 2 I Y I B I Y = I 2 -I 1 Assume balanced RL load is connected

45. For balanced load phase current values are equal in magnitude, Similarly and Line current = √3 times of phase current Line and phase quantities in delta connected circuit In delta connection I 1 -I 3 I R 30 30

46. Power in three phase systems with balanced load

47. Power in three phase star connected circuit In star connected system

48. Power in three phase delta connected circuit In delta connected system

49. Types of AC Supply and AC Load Balanced supply Unbalanced supply B B 1 V BN + - Y Y 1 V YN + - R R 1 V RN + - I R I Y I B R Y B Three Phase Load Three Phase Source Balanced load Unbalanced load

50. B B 1 V BN + - Y Y 1 V YN + - R R 1 V RN + - I R I Y I B R Y B Balanced supply :- magnitude of voltage induced in each phase must be equal Angle between two adjacent phases must be 120 120 120 V RN V YN V BN 120

51. B B 1 V BN + - Y Y 1 V YN + - R R 1 V RN + - I R I Y I B R Y B Unbalanced supply :- magnitude of voltage induced in each phase may not be equal or angle between two adjacent phases may not 120 120 120 V RN V YN V BN 120 150 90 V RN V YN V BN 120

52. Balanced load:- To each phase same load is connected Three phase supply 5 Ω 5 Ω 5 Ω Three phase supply Z 1 Z 2 Z 3 Unbalanced load:- To each phase different load is connected Three phase supply 3 Ω 2 Ω 5 Ω Three phase supply Z 1 Z 2 Z 3