DESCRIPTION AND EXPLANATION ON AC CIRCUITS.pptx

EXPLANATION ON AC CIRCUIT

1. AC CIRCUITS CHATURVEDULA 1 @ CHATURVEDULA

2. @ CHATURVEDULA 2 Introduction Alternating current (AC) unlike Direct current (DC) flows first in one direction then in the opposite direction The most common AC waveform is a sine (or sinusoidal) waveform. See Fig.1 Fig .1

3. @ CHATURVEDULA 3 A C Fundamentals Alternating Current When the current varies in magnitude and direction periodically it is called an alternating current Cycle One complete set of all positive and negative values of an alternating quantity, voltage or current is known as a cycle or A cycle can also define in terms of angular measure 1cycle=360 electrical or 2  radians

4. @ CHATURVEDULA 4 Definitions Cycle One complete set of +ve and –ve values of an alternating quantity is known as a “Cycle” It is also defined in terms of angular measure i.e., one cycle corresponds to 360 electrical or 2 π radians

5. @ CHATURVEDULA 5 Angular velocity(  ) It is an angle turn by one cycle during the time period Unit is radians/second Wave Form The shape of the curve drawn between instantaneous voltage or current on Y- axis and time on X – axis is known as wave form A C Fundamentals (contd..)

6. @ CHATURVEDULA 6 Frequency (f) The number of cycles completed by an alternating quantity in one second is called the frequency. Frequency ,f=1/T Unit is cycles/sec or Hz (hertz) Definitions of Alternating Quantities (contd..)

7. @ CHATURVEDULA 7 Frequency The frequency of an alternating quantity is the “number of cycles completed in one second.” It is denoted by the letter “ f ” and expressed in Hertz (Hz).

8. @ CHATURVEDULA 8 Time period (T) Time taken by an alternating quantity to complete one cycle is called the time period Time period , T=1/f Unit is sec Definitions of Alternating Quantities (contd..)

9. @ CHATURVEDULA 9 Time Period Time taken by an alternating quantity to complete one cycle is known as time period ” T ”. Frequency is the reciprocal of time period T = 1/f sec Fig 2 Fig 2 shows the sine wave

10. @ CHATURVEDULA 10 Examples of waveforms representing A.C Fig.1

11. @ CHATURVEDULA 11 Instantaneous value The magnitude of an alternating quantity at any instant is known as the instantaneous value of the alternating quantity Definitions of Alternating Quantities Fig.1 Instantaneous value

12. @ CHATURVEDULA 12 Maximum value (or) Amplitude The maximum value of positive or negative of an alternating quantity is called the amplitude or peak value Definitions of Alternating Quantities (contd…) Maximum value Fig. 2 Amplitude

13. @ CHATURVEDULA 13 Average value It is defined as the steady (dc) current which when flowing through a given circuit for a given time produces the same amount of charge as produced by an alternating current when flowing through the same circuit for the same time. Definitions of Alternating Quantities (contd..)

14. @ CHATURVEDULA 14 Root Mean Square (R.M.S) Value It is defined as the steady (dc) current which when flowing through given circuit for a given time produces the same amount of heat as produced by an alternating current when flowing through the same circuit for the same time It is also called as an Effective or Virtual value Definitions of Alternating Quantities (contd..)

15. @ CHATURVEDULA 15 Form Factor It is the ratio of RMS value to average value of an alternating quantity Peak Factor It is the ratio of maximum value to the RMS value of an alternating quantity. Definitions of Alternating Quantities (contd..)

16. @ CHATURVEDULA 16 R.M.S. Value Root mean square value is that current (DC) when flows through a given circuit for a given time produces same amount of heat as produced by the alternating current (AC) when flowing through the same circuit for the same time

17. @ CHATURVEDULA 17 It is also known as Effective value or Virtual Value. Its value can be determined by 1. Mid-ordinate method 2. Analytical method R.M.S. Value

18. @ CHATURVEDULA 18 Mid-Ordinate Method If the alternating current is passed through a circuit of resistance R ohm, then I r.m.s = I = √(i 1 2 + i 2 2 + i 3 2 + …………+ i n 2 )/n E r.m.s = E = √(e 1 2 + e 2 2 + e 3 2 + ……+e n 2 )/n π I 2 π t

19. @ CHATURVEDULA 19 Analytical Method ( for Sinusoidal AC. only ) We know the equation of an alternating current is i = i m sin θ The root of the mean of the squares of the instantaneous currents over half a circle I r.m.s = I m / √2 = 0.707 I m E r.m.s = E m / √2 = 0.707 E m

20. @ CHATURVEDULA 20 I R.M.S = = Calculation Of R.M.S Value By Analytical Method (contd..)

21. @ CHATURVEDULA 21 I = = Calculation of R.M.S Value By Analytical Method (contd..)

22. @ CHATURVEDULA 22 I R.M.S = = I R.M.S = 0.707*max. value Calculation of R.M.S Value By Analytical Method (contd..)

23. @ CHATURVEDULA 23 Average Value Average value of an alternating quantity is defined as the steady current (DC) which when flows through a given circuit for a given time produces same amount of charge as produced by the alternating current (AC) when flowing through the same circuit for the same time

24. @ CHATURVEDULA 24 Average value can also be find in two ways 1) Mid-ordinate method 2) Analytical method The Average value by Mid-ordinate method I av = = (i 1 + i 2 + i 3 + …………+ i n )/n The Average value by Analytical method I ave = 2I m / π = 0.637 I m π I 2 π t

25. @ CHATURVEDULA 25 Calculation Of Average Value By Analytical Method i =Im sin I av =

26. @ CHATURVEDULA 26 = = = Calculation Of Average Value By Analytical Method (contd..)

27. @ CHATURVEDULA 27 I av = I av = = 0.637I m I av = 0.637*max value Calculation Of Average Value By Analytical Method (contd..)

28. @ CHATURVEDULA 28 Form Factor It is the ratio of r.m.s. value to average value and gives the idea of the wave shape of the current or voltage. For Sinusoidal wave form Form factor = RMS value/Average value = 0.707I m / 0.637I m Form factor = K f = 1.11

29. @ CHATURVEDULA 29 Peak Factor It is the ratio of maximum value to r.m.s value Peak Factor = Max. value/RMS value K a = I m / (I m / √2) = √2 = 1.414

30. Problem 1 The maximum values of an alternating voltage and current are 400 V and 20 A respectively. In a circuit connected to 50 Hz supply and these quantities are sinusoidal. The instantaneous values of the voltage and current are 283 and 10 A respectively at t = 0 both increasing positively. Write the expression for voltage and current at time ‘t’ @ CHATURVEDULA 30

31. Solution @ CHATURVEDULA 31 Given data v = 283 Volt, V max = 400 Volt, I max = 20A The general form of sinusoidal voltage equation v = V max sin( t+) 283 = 400sin( x 0 + ) , sin() = 283/400  = 45 o (or) /4 radians Hence v = 400 sin(100 t+ /4 ) volt

32. @ CHATURVEDULA 32 Similarly the current equation , i= I max sin( t+ ) at t = 0, 10 = 20 sin ( x 0+ ) , sin  = 10/20 Therefore  = 30 (or) /6 radain The current equation , i=20sin (100 t+ /6 ) amp Solution (contd…)

33. Problem 2 An alternating current of frequency 60 Hz has a maximum value of 120A . Write down the equation for its instantaneous value. Reckoning time from the instant the current is zero and is becoming positive , find a) The instantaneous value after 1 /360 second b) The time taken to reach 96 A for the first time. @ CHATURVEDULA 33

34. Solution for (a) Given data f = 60 Hz I max = 120 A i = 120sin2  f t = 120 sin 2  x 60 x t but t = 1/360 second i = 120sin(120 x 180 x 1/360) = 120sin60 o i =103.9 A @ CHATURVEDULA 34

35. Solution for (b) Given data f = 60Hz i = 96 A I max = 120 A i = I m sin ( t+) (general form), here  = 0 96 = 120 sin(t) sin(t) = 96/120 t = sin -1 (96/120) = 53 o t = /2  f = 53/(360 x 60) = 0.00245 sec @ CHATURVEDULA 35

36. @ CHATURVEDULA 36 Example : Calculate the RMS value, average value, form factor and peak factor for the current over one cycle of an AC. voltage e=200 sin 314t is applied to a resistor of 20 Ω ,which offer the flow of current in one direction only Given data E m = 200V, R = 20 Ω I m = E m / R = 200 / 20 = 10 A I r.m.s = I m / = 10 / = 7.07 A

37. @ CHATURVEDULA 37 I ave = 2I m / π = 2X10/ π Form factor = I r.m.s / I ave Peak factor = I max / I r.m.s Solution = 6.37 A = 7.07/6.36 = 1.11 =10 /7.07 5 = 1.414

38. @ CHATURVEDULA 38 Phase The angle turned by an alternating current or voltage from a given instant is known as Phase The phase of an alternating current may be defined as the fraction of a time period or cycle through which the alternating current has advanced from a given instant Fig(3) shows a simple sine wave Fig 3 Max value Current Or Voltage Time in seconds π 2 π

39. @ CHATURVEDULA 39 Phase Difference The angle between the phases of two alternating quantities of the same frequency as measured in degrees is known as phase difference. Fig .4 O Ф P Q P Q Ф OP & OQ are two AC vectors Ф is called as ‘phase’ angle

40. @ CHATURVEDULA 40 With reference to diagram the leading quantity is through its’ zero The lag or lead obviously depends upon the assumed direction of rotation of the vectors. The vector diagram is counter clock wise Phase Difference

41. @ CHATURVEDULA 41 If the vectors are voltage vectors leading represented by e P = E m sin ω t Lagging vector represented by e Q = E m sin ( ω t – Φ ) Phase Difference

42. @ CHATURVEDULA 42 Resistance It may be defined as “the opposition offered by a substance or body to the passage of an electric current through it”. It is represented by the letter ‘R’ and is measured in ohm ( Ω ) ( or ) “It is the property of a substance due to which it opposes the flow of electric current through it”

43. @ CHATURVEDULA 43 Inductance It may be defined as “the number of Weber-turns per ampere of current in a coil". It is represented by the letter ‘L’ and is measured in Henry (H) (or) “It is the property of a coil due to which it opposes any increase or decrease of current or flux through it”

44. @ CHATURVEDULA 44 Capacitance It may be defined as “the property of a capacitor to store electricity ” (or) “The amount of change required to create a unit potential difference between plates” Capacitance C = Charge/Potential difference = Q/V Coulomb/Volt or Fared

45. @ CHATURVEDULA 45 Pure Resistance across AC V i R i v =V m Sin ω t Fig shows a simple circuit containing a resister R and energized by alternating voltage sinusoidally as v =V m Sin ω t

46. @ CHATURVEDULA 46 Resistor has only resistive property The current flowing in the resister has magnitude which is proportional to the applied voltage at any instant If voltage is zero, current also zero at that instant Pure Resistance across AC (contd..)

47. @ CHATURVEDULA 47 When voltage reaches maximum current also acquire maximum The relation for current at any instant written as i = V/R i = V m Sin ω t / R There fore i = I m Sin ω t Where I m = Maximum value of current = V m /R Pure Resistance across AC (contd..)

48. @ CHATURVEDULA 48 If V rms and I rms are RMS( effective) voltage and current V rms = V m / √2 and I rms = I m / √2 V = V m / √2 and I = I m / √2 (assume V rms = V and I rms = I) Pure impedance Z = V∟0 / I∟0 = R Pure Resistance across AC (contd..)

49. @ CHATURVEDULA 49 Fig.A Shows Wave forms of instantaneous voltage and current in a Purely Resistive circuit Fig.B Shows Phasor diagram v = V m Sin ω t i = I m Sin ω t V ,i Y X V I Fig.A Pure Resistance across AC (contd..) Fig.B

50. @ CHATURVEDULA 50 Power Instantaneous power p = v x I watt = V m Sin ω t X I m Sin ω t ( v = Vm Sin ω t & i = Im Sin ω t) = V m I m Sin 2 ω t = V m I m [ 1-Cos 2 ω t / 2 ] = [V m I m / 2 ] – [Cos 2 ω t / 2 ] Pure Resistance across AC (contd..)

51. @ CHATURVEDULA 51 Power Constant part : V m I m / 2 Variable part :V m I m Cos 2 ω t / 2 of double frequency Average power over the whole cycle is P = V m I m / 2 = V m / x I m / = V r.m.s. I r.m.s = V.I.watt P v i t Pure Resistance across AC (contd..)

52. @ CHATURVEDULA 52 Pure Inductance across AC Let us consider a circuit of pure inductance coil If an alternating e.m.f.applied,magnetic flux set up when current flows through it, induced an alternating e.m.f.called back e.m.f due to self inductance The back e.m.f will oppose rise or fall of current through it It is equal and opposite to applied e.m.f

53. @ CHATURVEDULA 53 v = L.di / dt Where v = V m Sin ω t and L = Self Inductance in Henry Therefore L.di / dt = v = V m Sin ω t di = V m / L Sin ω t.dt Pure Inductance across AC (contd..)

54. @ CHATURVEDULA 54 Integrating both sides we have ∫ di = ∫ V m / L Sin ω t.dt i = V m / L[ -Cos ω t / ω ] i = V m / ω L[ Sin ω t – 90 ] Current ‘I’ is maximum I m when Sin ( ω t – 90 ) is unity I m = V m / ω L Hence current equation becomes I = I m Sin ( ω t – 90 ) Pure Inductance across AC (contd..)

55. @ CHATURVEDULA 55 Comparing v = V m Sin ω t & I = I m Sin ( ω t - 90 ) the current lags behind applied voltage by 90 or phase difference is π / 2 as fig (a) shown below 90 V I L i V = V m Sin ω t Fig (a) π /2 v = Vm Sin ω t I = Im Sin ( ω t – π /2) t Pure Inductance across AC (contd..)

56. @ CHATURVEDULA 56 π /2 v = Vm Sin ω t I = Im Sin ( ω t – π /2) 90 180 270 360 After reaching voltage to 90 , current starts rising from 90 Variation of V & I wave forms Pure Inductance across AC (contd..)

57. @ CHATURVEDULA 57 ‘ ω L’ plays the part of résistance, called “inductive reactance” of the coil. It is represented by “X L ” and measured in “ Ohm” X L = ω L = 2 π f L Power Instantaneous power = - V m I m / 2 Sin 2 ω t Total power for the whole cycle is P = ∫ - V m I m / 2 Sin 2 ω t dt = 0 2 π Pure Inductance across AC (contd..)

58. @ CHATURVEDULA 58 Pure Capacitance across AC Fig below shows a capacitor ‘C’ connected across the terminals of an AC source i V I C Circuit with Capacitance Only

59. @ CHATURVEDULA 59 Wave form Voltage and Current Wave Forms for pure Capacitive Circuit Pure Capacitance across AC (contd..) V m CURRENT APPLIED VOLTAGE

60. @ CHATURVEDULA 60 V(t) =1/c ∫ i(t) dt Consider the function i(t) = I m Sin ω t = I m ∟0 V(t) = 1/c ∫ I m Sin ω t = 1/ ω c I m ( - Cos ω t ) = 1/ ω c I m Sin ( ω t – 90 ) Pure Capacitance across AC (contd..)

61. @ CHATURVEDULA 61 There fore V(t) = V m Sin ( ω t – 90 ) = V m ∟- 90 Where V m = I m / ω c Z = V m ∟- 90 / I m ∟0 = I m / ω c x 1/ I m ∟0 = - j / ω c = - j X c Pure Capacitance across AC (contd..)

62. @ CHATURVEDULA 62 Where X c = 1/ ω c and is called “Capacitive Reactance” In pure Capacitor the current leads the voltage by 90 X C = 1 / ω c Pure Capacitance across AC (contd..)

63. @ CHATURVEDULA 63 Problems Problem 1 : The current flowing through pure inductor is 2.5 A when the voltage applied is 50 sin 314t volt. Find the inductance and power consumption Solution I=2.5A ; v=50sin 314t ; Find L,P V = 0.707 V max = 0.707 x 50 = 35.35 X L = V/I = 35.35/2.5 = 14.14 Ω ω = 2 π f = 314 f = 314 / 2 π =50 Hz X L = 2 π fL L = 14.14/314 = 0.045 H or 45 mH

64. @ CHATURVEDULA 64 Problem 2 : A Capacitor of 50 micro farad is connected to a 500V, 50 Hz supply. What will be the current flowing through it. Given Data V r.m.s = 500 V ; C = 50 x 10-6 F ; f = 50 Hz Solution I r.m.s = V r.m.s / 1/ ω c = ω c V r.m.s =2 π fcV r.m.s = 2 π x 50 x50 x 10-6 x 500 = 7.854 A

65. @ CHATURVEDULA 65 Problem 3 : Determine the current flowing through a pure inductor of 150mH when the frequency (a) 50Hz (b) 60Hz.The voltage applied is 230 V. Given Data L=150x 10 -3 H ; (a)50Hz & (b)60Hz ; V=230V Solution X L = 2 π fL = 2 π x 50 x 150 x 10 -3 = 47.13 Ω I = V/X L = 230/47.13 = 4.88 A. X L = 2 π x 60 x 150 x 10 -3 = 56.556 Ω I = V/X L = 230/56.556 = 4.067 A.

66. @ CHATURVEDULA 66 Problem 4 : A sinusoidal voltage of 100 sin 314t is applied across a capacitor. Find the value of capacitance and power factor when the current flowing is 1.5 A Given Data V max = 100V ; ω = 2pf = 314;I = 1.5 A ; Solution V = 0.707x V max = 0.707 x 100 = 70.7 V X C = V/I = 70.7/1.5 = 47.133 Ω X C = 1/ 2 π fC Therefore C = 1/314 x 47.133 = 0.0000675 F or 67.5 Power factor,Cos Φ = R/Z = 0

67. Difference Between Resistance and Impedance Resistance Impedance It is used in DC circuits. It is used in AC circuits. Resistance can be seen in both AC and DC circuits. Impedance can be seen only in AC circuits. It happens due to resistive elements. It happens due to reactance and resistance. It is represented by letter R. It is represented by letter Z. It is represented using real numbers such as 5.3 ohms. It is represented using real and imaginary values such as R + ik . It does not vary depending upon the frequency of DC current. It varies according to the frequency of AC current. It does not have a magnitude and phase angle. It does have a phase angle and magnitude. If kept in a electromagnetic field, it only shows the power dissipation. If kept in a electromagnetic field, it shows power dissipation and energy stored. @ CHATURVEDULA 67

68. @ CHATURVEDULA 68 Summary In this period we have discussed about Cycle: One complete set of +ve and –ve values of an alternating quantity is known as a “Cycle” Frequency: The frequency of an alternating quantity is the “number of cycles completed in in one second. Time Period: Time taken by an alternating quantity to complete one cycle is known as Time Period ”T”.

69. @ CHATURVEDULA 69 R.M.S.Value 1.Mid-Ordinate Method Ir.m.s = I = Er.m.s = E = Summary 2.Analytical Method Ir.m.s = Im / √2 = 0.707 Im Er.m.s = Em / √2 = 0.707 Em Form factor = RMS value/Average value Peak Factor = Max.value/RMS value

70. @ CHATURVEDULA 70 Phase difference The angle between the phases of two alternating quantities of the same frequency as measured in degrees is known as phase difference Phase The angle turned by an alternating current or voltage from a given instant is known as Phase Summary

71. @ CHATURVEDULA 71 Average value It is defined as the steady (D.C) current which transfers across any circuit, the same charge as transferred by the alternating current during the same time

72. @ CHATURVEDULA 72 Calculation Of Average Value By Analytical Method Average value = T Time period of the alternating quantity f(t) function of time

73. @ CHATURVEDULA 73 Root Mean Square Value The R.M.S value of an alternating quantity is given by that steady (DC current) which when flowing through a given circuit for a given line producer by the alternating current when flow up through the same circuit for same time

74. @ CHATURVEDULA 74 Calculation Of R.M.S Value By Analytical Method (contd…) R.M.S value = where T Time period of the alternating quantity f(t) Function value

75. @ CHATURVEDULA 75 Form Factor Ratio of R.M.S value to Average value of an alternating quantity R.M.S value Form factor = = Average value = 1.11

76. @ CHATURVEDULA 76 Peak Factor (or) Crest Factor(or) Amplitude Factor Peak factor is the ratio of the maximum value to the R.M.S value of the alternating quantity Maximum value Peak factor = = R.M.S value = 1.414

77. @ CHATURVEDULA 77 Sinusoidal wave form representation The wave form equation is v=V m sin  For this wave form the average value for one complete cycle is zero. Hence the average value will be calculated for a half cycle Fig.1 ∏/2 2∏ ∏ Time period

78. @ CHATURVEDULA 78 For sinusoidal wave form if ‘t’ varies from 0 to T/2  varies from 0 to  radian. Average value = = Average value of sine wave

79. @ CHATURVEDULA 79 Average value = The average value of sinusoidal voltage = The average value of sinusoidal current = Average value of sine wave (contd..)

80. @ CHATURVEDULA 80 R.M.S value The R.M.S value = = =

81. @ CHATURVEDULA 81 = = 0 R.M.S value (contd..)

82. @ CHATURVEDULA 82 = = R.M.S value (contd..)

83. @ CHATURVEDULA 83 Average value = R.M.S value = Form factor = = 1.11 R.M.S value (contd..)

84. @ CHATURVEDULA 84 Peak factor = = = 1.414 R.M.S value (contd..)

85. @ CHATURVEDULA 85 Phase It is an angular measurement that specifies the position of the alternating quantity relative to a reference

86. @ CHATURVEDULA 86 Consider the following sinusoidal wave form Fig.1  2 Time period Θ = ω t The equation of the wave form Shown is V=V m sin Θ When Θ =0, V=0(i.e wave form starts from zero) When Θ = ∏/2, V=V m (maximum value) At a given instant Θ the phase or state of the wave can be known With respect to the reference.

87. @ CHATURVEDULA 87 When the sine wave is shifted left or right with respect to the reference as follows Θ = ω t Θ = ω t Fig.2 Fig.3 Equation of this wave form is V=V m sin( Θ + Φ ) V=0 when Θ = - Φ Φ Φ Equation of this wave form is V=V m sin( Θ - Φ ) V=0 when Θ = Φ Phase

88. @ CHATURVEDULA 88 α Direction Of flux Coil A Coil B Direction Of rotation When the above system is rotated in the magnetic field α is the electrical phase angle difference. e.m.f is first induced in coil A and then in coil B. Fig. 4 Phase (contd..)

89. @ CHATURVEDULA 89 Equation of emf induced in coil A is equal to V A =V m sin Θ (Reference) Equation of emf induced in coil B is equal to V B = V m sin( Θ - α ) ‘ α ’ known as Phase difference between V A and V B V A is said to be leading V B V B is said to be lagging V A Phase Difference

90. @ CHATURVEDULA 90 When coil B is taken as reference, the equations will be as follows v B =V m sin Θ (Reference) v A = V m sin( Θ + α ) Hence phase difference is the difference of angles between two alternating quantities Phase Difference (contd..)

91. @ CHATURVEDULA 91 The above equations can be represented as follows Fig.5  2 Θ = ω t α By taking V A as reference V A V B

92. @ CHATURVEDULA 92 Definition of Vector Vector is a physical quantity, which has both the magnitude & direction 92

93. @ CHATURVEDULA 93 Graphical representation of current and voltage method 93 Fig-1

94. @ CHATURVEDULA 94 Advantage of vector method Vector method is preferred over graphical method because vector method greatly simplifies the problems in A.C circuits

95. @ CHATURVEDULA 95 Vector diagram of voltage & current circuit 95 Fig-2

96. @ CHATURVEDULA 96 Diagram of a pure resistive circuit Fig-3 Relationship between voltage and current in a pure resistive circuit

97. @ CHATURVEDULA 97 Let the applied voltage is given by the equation v = V m sin  = V m sin  t ---- eq (  ) Let R = Ohmic resistance i= instantaneous current Ohmic drop (v) = iR -------eq (  ) Relationship between voltage and current in a pure resistive circuit (contd..)

98. @ CHATURVEDULA 98 Keeping the value of v in eq( ), we get v =  V m sin  t = I R , i =  V m sin  t /R -------- eq( ) when sin  t =  then i =  V m / R here i is maximum current when sin  t =  so maximum current I m = V m / R Relationship between voltage and current in a pure resistive circuit (contd…)

99. @ CHATURVEDULA 99 Substituting the value of I m in eq( ) we get i=i m sin  t Thus in a pure resistive circuit, v =  V m sin  t i = i m sin  t Relationship between voltage and current in a pure resistive circuit (contd..)

100. @ CHATURVEDULA 100 Vector diagram of voltage & current circuit Resistive circuit Fig-2

101. @ CHATURVEDULA 101 Power in Pure Resistive Circuit Instantaneous power, p = vi = V m I m sin2  t = V m I m (1-cos2  t) = (V m I m )/2-(V m I m cos2  t) /2 The 1st part V m I m /2 is a constant The 2nd part V m I m cos2  t /2 is a fluctuating part and Depends on the frequency of the voltage or current wave for a complete cycle cos2  t is 0

102. @ CHATURVEDULA 102 Therefore power for the whole cycle, p = (V m I m )/2 = (V m /  2 )(I m /  2) i.e p = VI watts where V & I are the R.M.S value of the voltage and current respectively In a pure resistive circuit, the power in whole cycle is never zero Power in Pure Resistive Circuit (contd..)

103. @ CHATURVEDULA 103 Problem An a.c circuit consists of a pure resistance of 30 Ω and is connected across an a.c. supply of 220 V , 50 Hz. Find the current, power consumed and sinusoidal equation for voltage and current?

104. @ CHATURVEDULA 104 Solution Given Data R = 30 Ω, V=220 Volts, f = 50 Hz current = I = V/R= 220/30 = 7.3 Amps Power consumed = p = V*I =7.3*220=1606Watts V R.M.S =V = 220Volts V m =  2 V =  2 *220 = 311.08 Volts Similarly I m =  2 I =  2 *7.3 = 10.32 amps =2**f=2**50=314 rad/sec

105. @ CHATURVEDULA 105 Equation for voltage v = V m sin  t, v = 311.08 sin 314t volts Equation for current = i = I m sin  t i = 10.32sin314t amps 105 Solution (contd..)

106. @ CHATURVEDULA 106 Instantaneous AC power in a pure resistive circuit is always positive Fig-1

107. @ CHATURVEDULA 107 Inductors An inductor is usually a coil of wire Inductors oppose changes in current through them, by dropping a voltage directly proportional to the rate of change of current If current is increasing in magnitude, the induced voltage will “ push against ” the electron flow If current is decreasing, the polarity will reverse and “push with” the electron flow to oppose the decrease

108. @ CHATURVEDULA 108 A.C. Circuit with pure Inductance Fig-2                                        

109. @ CHATURVEDULA 109 Relationship Between Current & Voltage In Pure Inductive Circuit The relationship between the voltage dropped across the inductor and rate of current change through the inductor is as such: 109

110. @ CHATURVEDULA 110 Relationship Between Current & Voltage In Pure Inductive Circuit (contd..)

111. @ CHATURVEDULA 111 Relationship Between Current & Voltage In Pure Inductive Circuit (contd..)

112. @ CHATURVEDULA 112 Graphical Representation Of Voltage & current In a Pure Inductive Circuit

113. @ CHATURVEDULA 113 Inductive Reactance X L It is the ratio of the magnitudes of the voltage and current .It is measured in ohms V m = X L .I m .

114. @ CHATURVEDULA 114 Self Inductance we define the (self) inductance of the coil thus: φ B (t) = L.I(t) The voltage that appears across an inductor is due to its own magnetic field and Faraday's law of electromagnetic induction The current i(t) in the coil sets up a magnetic field, whose magnetic flux φ B is proportional to the field strength, which is proportional to the current flowing 114

115. @ CHATURVEDULA 115 Power Consumed In The Circu it Instantaneous power (p) =

116. @ CHATURVEDULA 116 Average Power In A Pure Inductive Circuit Average power consumed in the circuit over a complete cycle =

117. @ CHATURVEDULA 117 Plot of The Power For Inductive Circuit

118. @ CHATURVEDULA 118 Negative Power It means that the inductor is releasing power back to the circuit, while a positive power means that it is absorbing power from the circuit.

119. @ CHATURVEDULA 119 Problem 1) An A.C circuit consists of a pure resistance of 20 ohms and is connected across an A.C supply of 230 V, 50 Hz. Find a) Current b) Power consumed c) sinusoidal equations for voltage and current

120. @ CHATURVEDULA 120 Solution Given Data Resistance (R ) = 20 ohms Voltage (v) = 230 V Frequency (f) = 50 Hz

121. @ CHATURVEDULA 121 I = V/R = 230/20 = 11.5 A P= V* I = 230*11.5 = 2645 W Solution (contd..)

122. @ CHATURVEDULA 122 Solution (contd..)

123. @ CHATURVEDULA 123 Solution (contd..)

124. @ CHATURVEDULA 124 Pure Capacitive Circuit And Its Vector Diagram Fig.1

125. @ CHATURVEDULA 125 Relationship Between Current & Voltage In Pure Capacitive Circuit The relationship between the current “through” the capacitor and rate of voltage change across the capacitor is as such: i =

126. @ CHATURVEDULA 126 Explanation The expression dv/dt is the rate of change of instantaneous voltage (v) over time, in volts per second The capacitance (C) is in Farads, and the instantaneous current (i), is in amps capacitors oppose changes in voltage by drawing or supplying current as they charge or discharge to the new voltage level The flow of electrons “through” a capacitor is directly proportional to the rate of change of voltage across the capacitor

127. @ CHATURVEDULA 127 Let the alternating voltage applied across the circuit is v such that v = Vm sin ω t charge on the capacitance at any instant q = C * V = C * V m * sin ω t Relationship Between Current & Voltage In Pure Capacitive Circuit

128. @ CHATURVEDULA 128 Relationship Between Current & Voltage In Pure Capacitive Circuit (contd..)

129. @ CHATURVEDULA 129 The value of i will be maximum i.e (I m ) when is unity Relationship Between Current & Voltage In Pure Capacitive Circuit (contd..)

130. @ CHATURVEDULA 130 Phase relation among voltage and current 130

131. @ CHATURVEDULA 131 Capacitative Reactance Xc Capacitive reactance Xc the ratio of the magnitude of the voltage to magnitude of the current in a capacitor. It is measured in farads Xc is equal to and

132. @ CHATURVEDULA 132 Power consumed in the circuit Instantaneous power (p)=

133. @ CHATURVEDULA 133 Average Power Average power consumed in the circuit over a complete cycle =

134. @ CHATURVEDULA 134 Power Wave in capacitance circuit

135. @ CHATURVEDULA 135 Impedance of components The impedance is defined as the ratio of voltage to current

136. @ CHATURVEDULA 136 Impedance of the different components 136

137. @ CHATURVEDULA 137 Problem A 60 µf capacitor is connected across 200 V, 50Hz supply. Determine a) the maximum instantaneous charge on the capacitor and b) the maximum instantaneous energy stored in the capacitor?

138. @ CHATURVEDULA 138 Solution Data : capacitance (C ) = 60 * 10-6 farads voltage (v) =200 V frequency (f) =50 Hz EE 302.39

139. @ CHATURVEDULA 139 V m = √ 2 V rms = 282.84 Qm = C * Vm = 60 * 10 -6 * 282.84 = 16.97mC Solution (contd..)

140. @ CHATURVEDULA 140 Maximum instantaneous energy stored=E m = 0.5 * 60 * 10 -6 * (282.84) 2 = 2.4 J Solution (contd..)