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AC Networks

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Published in: Electrical | Electronics
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Presentation on AC Networks

Trinity A / Chandigarh

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  1. Chapter 2 AC Networks
  2. Wave Description Amplitude (V ) - maximum displacement peak from the zero line Period (T) - the time required for one cycle Frequency (f) - the number of cycles that appear in a time span 1 cycle per second = 1 Hertz T=l/f
  3. Angular Velocity (o) - rotation rate of a vector (or generator shaft) o is measure in radians per second. New Waveform Equation: v=V sinot = 2TCf
  4. Sinusoidal Voltage Relationships v = V sin 0 peak v = V sin ot peak v = V sin 2Ttt/T peak v = V sin 21tft peak Which form you use depends on what you know
  5. Phase Relationship = phase angle For initial intersections before 0 0 v= Vpsin( ot+") For initial intersections after 0 0 a v = Vpsin( ot-Ø)
  6. Resistive Circuit Only for a resistive circuit can current and voltage be in phase. If the current and voltage go through zero at different times, they are said to be out of phase
  7. Sinusoidal Plot Phase Angle max ax smo 13 51' Phasor Angle e=ot max 3 60 Ö
  8. Example Problems Example 4.7: o Determine the phase relationship between the following two waveforms: i=o.12 sin(300t+100)
  9. 4.4 Effective (RMS) value These mean the same thing for AC circuits: o "equivalent DC voltage" o "effective voltage" o "rms voltage' RMS = root mean square Draw the "relating effective values " plots.
  10. 4.5 Average Value For a sine wave, the average value is zero. For other waves area(under waveform) G(average value) T(period)
  11. Example Problems Example 4.9: Write the sinusoidal expression for a voltage having an rms value of 40 mV, a frequency of 500 Hz and and initial phase shift of +40 0. O — m V) = 56.56 m v o = 2M = kHz) = 3.142 X 103 rad/s v = 56.56 mV sin(3142t + 400)
  12. Example Problems Example 4.10: Calculate the average value of the waveform below over one full cycle. 12 10
  13. What are the following for a wall outlet? eff — peak ave eff — peak ave rms rms f = 60 Hz veff = vrms = 120 V = 170 V peak v eff — peak ave Irms = limited by breaker rms
  14. Effective Values Average values for I or V are zero If we use IV to measure power with a dc meter, we will get zero o Yet the resistor gets hot — power is dissipated o I and V are not always zero — you can get shocked from ac just like dc We need a new definition for power
  15. Use Effective (rms) Values PR = = IR2R = VR2/R Look just like the dc forms of power o But we use rms values instead
  16. 4.7 The R, L, and C Elements Ohm's Law for Peak Values Resistors: V -1 R peak— peak Capacitors: V -1 X peak— peak C Inductors: V -1 X peak— peak L
  17. Inductive Reactance c Recall the form of inductance At Since I for ac is constantly changing, there is a constant opposition to the flow of current o Inductive Reactance XL = OL = 2TtfL o Measured in ohms o Energy is stored in the coil, not dissipated like in the resistor.
  18. Example An inductor of 400 mH is connected across a 120 V, 60 Hz ac source. Find XL and the current through the coil a. b. = 21tfL = 2n(60 H) = 151 Q 1 = V/XL = 120 v/151 Q
  19. XL vs. f graph for ideal inductor = 21tfL o Should give a straight line graph of slope 2TtL o Higher the L the greater the slope
  20. V and I for an Inductor From v = -Lt-; we see that voltage is greatest when 1 075 0.5 0.25 .1 -075 -0.5 -0.75 Current the change of current is greatest. Highest positive v Zero here Most negative here
  21. Voltage and Current (Coil) Current Voltage Voltage leads current by 900
  22. Example Problems Example 4.13: o The current iL through a 10-mH inductor is 5 sin(200t +30 0). Find the voltage VL across the inductor.
  23. Capacitive Reactance The effect of a capacitor is to prevent changes in voltage o Keep the potential difference from building up in the circuit Since the build up of potential is minimized, the flow of current is reduced. o Capacitors act like a resistance in an ac circuit.
  24. Capacitive Reactance V is changing in ac circuits o But Q = VC, so the charge changes also AQ , current is greatest when change in o Since At voltage is greatest. 1 0.75 0.5 0.25 1 -0.25 -115 -0.75 Current Voltage I leads V by 900
  25. CIVIL CIVIL is a memory aid. For a capacitor C, current I leads the voltage V by 90 degrees. For an inductor L, voltage V leads the current I by 90 degrees.
  26. Capacitive Reactance 1 Xc also measured in ohms x 2m f C What is the capacitive reactance of a 50 Elf capacitor when an alternating current of 60 Hz is applied? x x 1 2nfC = 530 1
  27. 4.28) Form of Reactance (Fig. 3000 300 250 200 150 100 50 C=IPIF C=2PlF 1 ooo 2000
  28. Example Problems Example 14: o The voltage across a 2-gF capacitor is 4mV (rms) at a phase angle of -600. If the applied frequency is 100kHz, find the sinusoidal expression for the current ic
  29. 3.6 Phasors and Complex Numbers A vector is a quantity has both magnitude and direction. A scalar quantity has only a magnitude. Examples: o Scalar: 50mph o Vector: 50mph North
  30. A phasor is a complex number used to represent a sine wave's amplitude and phase. A complex number can be written as where C is a complex number, A and B are real numbers and
  31. Imaginary C=A+Bj Real What is the magnitude of C? What is the direction of C? B -1 O = tan
  32. Phasor Diagrams We make the real axis the resistance The imaginary axis is reactance o Inductive reactance is plotted on the +y axis o Capacitive reactance is plotted on the —y axis The vector addition of R and X is Z
  33. Phasors Form Reactance First X = - X Form Impedance Z = R2 + X 2 Calculate phase angle = Tan-I( X
  34. Example Series R-L Circuit A coil has a resistance of 2.4 Q and a inductance of 5.8 mH. If it is connected to a 120-V, 60 Hz ac source, find the reactance and the current
  35. Solution = 2nfL = 2n(60 X 10-3 H) = 2.19 Q z = 3.250 R) = 42.40 z x 2.190 = Tan ( 240)
  36. Solution (cont'd) 01 = V/Z .11 120 V/3.25 Q = 36.9 A Notice that if f = 120 Hz, then e = 2n(120 H) = 4.37 Q Z = 4.99Q +61.20 I = 24 A
  37. Example Series R-C Circuit What is the current flow through a circuit with 120 V, 60 Hz source, an 80 uf capacitor, and a 24 Q resistor? c
  38. Solution x 27 f c = 330 z = 40.902 330 tan = 54. I o 2m(60Hz )( 80 X 10-6 f ) (240)2 = 1.38 v nov = 2.93A z 40.90
  39. Solution (cont'd) What happens is the frequency doubles? 16.60 Z = 29.2 Q + = 34.60
  40. Example - Series RLC Circuit A 240 V, 60 Hz line is connected to a circuit containing a resistor of 15 Q, an inductor of 0.08 H, and a capacitor of 80 Pif. Compute the circuit impedance and current.
  41. Solution x x x - f L - 2m(60Hz O.08H ) = 30.20 1 60Hz 80x10 1 2m f C = 33.20 f) = 30.20-33.20 = -3. OO v -240V (150 f = 15.30 = 15.7 A 15.30
  42. Power Factor In dc circuits we had P = IV and this still holds in purely resistive circuits with ac In general, I and V are out of phase in ac circuits. o At times V and I are negative and power is drawn from the circuit. o Thus the true power must be less than IV
  43. Power Factor i = I sin ot v = Vp sin( ot + is the phase between i and v p = iv = IpVp sin ot sin( ot + ) true power(P) Power Factor = P = IV( power factor) Sum of p over one cycle V COSØ rms rms
  44. Power Factor Power Factor = cos x R cos z R x For purely inductive or purely capacitive circuits, = ±900 and cos = 0 - no power is dissipated
  45. Example An industrial plant service line has an impressed power of 100 KV-A on a 60 Hz line at 440 V. An installed motor offers a total ohmic resistance of 350 Q and an inductance of I H. Find. a) b) c) d) e) f) The inductive reactance The impedance The current The power factor The power used by the motor The size capacitor required to change PF to 95%.
  46. Solution a) x L H) = 377 c) I —V = 440V = O.855A d) PF R -3500 = o. 68 514.40
  47. Solution (cont'd) PF = 0.95 = R = 368.420 0.95 Z2 —R2 = (XL ( / 3500 / x Z2 -R2 = x c = 3770- 27K 60 2620) = 2620 = 10.1gF
  48. Series Resonance If XL = X c, then Z = R and the circuit is purely resistive. This will occur at a certain frequency 1 1 res 2n LC By altering L or C, the circuit can be made to resonate at any frequency. This condition makes X = 0 V and I will be in phase minimum Z and maximum I
  49. Parallel RC Circuit General Scheme o Form impedance for each branch lov Hz o From Kirchhoff's Current Law i] = —sin( o t + "l — sin( o t + The possibility of different phase angles means parallel impedances do not add as simply as for dc circuits
  50. Total Current and Definitions i = V — sin ( o t + ) + — CD t + ) R 12 + X 2 Define G R 22 -k X 22 and B = x G is called the conductance B is called the susceptance
  51. Rule for Adding Parallel Impedances tan 0T Note: by convention, Xc always carries a negative sign in computing the impedances. Y = I/ZT is called the admittance
  52. Example
  53. Resistive Elements If parallel impedance contains only resistors, we have 0 G=I/R And the equation of parallel resistances results.
  54. Example lov 60 Hz 4 c lov 60 Hz = 0.0377S — = —2m f C + ( 0.0377 ( 50 = 0.204S
  55. Parallel Resonance Add L and C in parallel 4 4 X
  56. Parallel Resonance 1 4 1 (B2 +B3)2 = (B2 +B3) 1 Resonance occurs when the denominator becomes zero
  57. Parallel Resonance 1 res 2n LC Same relationship as for series resonance o In series resonance the impedance is a minimum (Z=R) o In parallel resonance the impedance is a maximum (Y=O)
  58. Resonance Comparisons Y=I/Z v=1Z Series Min Max Max Min Parallel Max Min Min Max
  59. Quality (Q) Factor At resonance, 00 , define The current in the circuit is... At resonance (series) Z = R, so we can write the current at resonance as 1M = V/R
  60. Q-Factor Using the definition and a little manipulation, we can derive... 2
  61. Q-Factor Q-Factor 0/00

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