2. Induction - Spring 20061 Time Varying Circuits April 10, 2006

3. Induction - Spring 20062 What is going on? There are only 7 more classes and the final is 3 weeks away. There are only 7 more classes and the final is 3 weeks away. Scotty, beam me somewhere else! Scotty, beam me somewhere else! Exam Issues Exam Issues Look Ashamed! Look Ashamed! Inductor Circuits Inductor Circuits Quiz Friday Quiz Friday AC Next week & Following Monday AC Next week & Following Monday

4. Induction - Spring 2006 3 Second Problem a h Both Currents are going into the page. B   Other Wire

5. Induction - Spring 2006 4 First Problem VV

6. Induction - Spring 20065 The Last two Problems were similar to WebAssigns that were also reviewed in class. Circular Arc – Easy Biot-Savart Moving Rod

7. Induction - Spring 2006 6 Question What about these problems was “unfair”? Why so many blank or completely wrong pages?

8. Induction - Spring 2006 7 And Now ….. From the past

9. Induction - Spring 2006 8 Max Current Rate of increase = max emf V R =iR ~current

10. Induction - Spring 2006 9 Solve the loop equation.

11. Induction - Spring 2006 10 We also showed that

12. Induction - Spring 2006 11 LR Circuit i Steady Source

13. Induction - Spring 2006 12 Time Dependent Result:

14. Induction - Spring 2006 13 RLRL

15. Induction - Spring 2006 14 At t=0, the charged capacitor is now connected to the inductor. What would you expect to happen??

16. Induction - Spring 2006 15 The math … For an RLC circuit with no driving potential (AC or DC source):

17. Induction - Spring 2006 16 The Graph of that LR (no emf) circuit..

18. Induction - Spring 2006 17

19. Induction - Spring 2006 18 Mass on a Spring Result Energy will swap back and forth. Add friction  Oscillation will slow down  Not a perfect analogy

20. Induction - Spring 2006 19

21. Induction - Spring 2006 20 LC Circuit High Q/C Low High

22. Induction - Spring 2006 21 The Math Solution (R=0):

23. Induction - Spring 2006 22 New Feature of Circuits with L and C These circuits produce oscillations in the currents and voltages Without a resistance, the oscillations would continue in an un-driven circuit. With resistance, the current would eventually die out.

24. Induction - Spring 2006 23 Variable Emf Applied emf Sinusoidal DC

25. Induction - Spring 2006 24 Sinusoidal Stuff “Angle” Phase Angle

26. Induction - Spring 2006 25 Same Frequency with PHASE SHIFT 

27. Induction - Spring 2006 26 Different Frequencies

28. Induction - Spring 2006 27 Note – Power is delivered to our homes as an oscillating source (AC)

29. Induction - Spring 2006 28 Producing AC Generator x x x x x x x x x x x x x x x x x x x x x x x

30. Induction - Spring 2006 29 The Real World

31. Induction - Spring 2006 30 A

32. Induction - Spring 2006 31

33. Induction - Spring 2006 32 The Flux:

34. Induction - Spring 200633 April 12, 2006

35. Induction - Spring 200634 Schedule Today Today Finish Inductors Finish Inductors Friday Friday Quiz on this weeks material Quiz on this weeks material Some problems and then AC circuits Some problems and then AC circuits Monday Monday Last FULL week of classes Last FULL week of classes Following Monday is last day of class Following Monday is last day of class FINAL IS LOOMING! FINAL IS LOOMING!

36. Induction - Spring 200635 Some Problems

37. Induction - Spring 2006 36 14. Calculate the resistance in an RL circuit in which L = 2.50 H and the current increases to 90.0% of its final value in 3.00 s.

38. Induction - Spring 2006 37 16. Show that I = I 0 e – t/τ is a solution of the differential equation where τ = L/R and I 0 is the current at t = 0.

39. Induction - Spring 2006 38 17. Consider the circuit in Figure P32.17, taking ε = 6.00 V, L = 8.00 mH, and R = 4.00 Ω. (a) What is the inductive time constant of the circuit? (b) Calculate the current in the circuit 250 μs after the switch is closed. (c) What is the value of the final steady-state current? (d) How long does it take the current to reach 80.0% of its maximum value?

40. Induction - Spring 2006 39 18. In the circuit shown in Figure P32.17, let L = 7.00 H, R = 9.00 Ω, and ε = 120 V. What is the self-induced emf 0.200 s after the switch is closed?

41. Induction - Spring 2006 40 27. A 140-mH inductor and a 4.90-Ω resistor are connected with a switch to a 6.00-V battery as shown in Figure P32.27. (a) If the switch is thrown to the left (connecting the battery), how much time elapses before the current reaches 220 mA? (b) What is the current in the inductor 10.0 s after the switch is closed? (c) Now the switch is quickly thrown from a to b. How much time elapses before the current falls to 160 mA?

42. Induction - Spring 2006 41 32. At t = 0, an emf of 500 V is applied to a coil that has an inductance of 0.800 H and a resistance of 30.0 Ω. (a) Find the energy stored in the magnetic field when the current reaches half its maximum value. (b) After the emf is connected, how long does it take the current to reach this value?

43. Induction - Spring 2006 42 52. The switch in Figure P32.52 is connected to point a for a long time. After the switch is thrown to point b, what are (a) the frequency of oscillation of the LC circuit, (b) the maximum charge that appears on the capacitor, (c) the maximum current in the inductor, and (d) the total energy the circuit possesses at t = 3.00 s?

44. Induction - Spring 200643 Back to Variable Sources

45. Induction - Spring 2006 44 Source Voltage:

46. Induction - Spring 2006 45 Average value of anything: Area under the curve = area under in the average box T h

47. Induction - Spring 2006 46 Average Value For AC:

48. Induction - Spring 2006 47 So … Average value of current will be zero. Power is proportional to i 2 R and is ONLY dissipated in the resistor, The average value of i 2 is NOT zero because it is always POSITIVE

49. Induction - Spring 2006 48 Average Value

50. Induction - Spring 2006 49 RMS

51. Induction - Spring 2006 50 Usually Written as:

52. Induction - Spring 2006 51 Example: What Is the RMS AVERAGE of the power delivered to the resistor in the circuit: E R ~

53. Induction - Spring 2006 52 Power

54. Induction - Spring 2006 53 More Power - Details

55. Induction - Spring 200654 AC Circuits April 17, 2006

56. Induction - Spring 200655 Last Days … If you need to, file your taxes TODAY! If you need to, file your taxes TODAY! Due at midnight. Due at midnight. Note: File on Web has been updated. Note: File on Web has been updated. This week This week Monday & Wednesday – AC Circuits followed by problem based review Monday & Wednesday – AC Circuits followed by problem based review Friday – Review problems Next Week Friday – Review problems Next Week Monday – Complete Problem review. Monday – Complete Problem review.

57. Induction - Spring 200656 Final Examination Will contain 8-10 problems. One will probably be a collection of multiple choice questions. Will contain 8-10 problems. One will probably be a collection of multiple choice questions. Problems will be similar to WebAssign problems. Class problems may also be a source. Problems will be similar to WebAssign problems. Class problems may also be a source. You have 3 hours for the examination. You have 3 hours for the examination. SCHEDULE: MONDAY, MAY 1 @ 10:00 AM SCHEDULE: MONDAY, MAY 1 @ 10:00 AM http://pegasus.cc.ucf.edu/%7Eenrsvc/registrar/calend ar/exam/ http://pegasus.cc.ucf.edu/%7Eenrsvc/registrar/calend ar/exam/ http://pegasus.cc.ucf.edu/%7Eenrsvc/registrar/calend ar/exam/ http://pegasus.cc.ucf.edu/%7Eenrsvc/registrar/calend ar/exam/

58. Induction - Spring 2006 57 Back to AC

59. Induction - Spring 2006 58 Resistive Circuit We apply an AC voltage to the circuit. Ohm’s Law Applies

60. Induction - Spring 2006 59 Consider this circuit CURRENT AND VOLTAGE IN PHASE

61. Induction - Spring 2006 60

62. Induction - Spring 200661 Alternating Current Circuits  is the angular frequency (angular speed) [radians per second]. Sometimes instead of  we use the frequency f [cycles per second] Frequency  f [cycles per second, or Hertz (Hz)]  f V = V P sin (  t -  v ) I = I P sin (  t -  I ) An “AC” circuit is one in which the driving voltage and hence the current are sinusoidal in time.  vv  V(t) tt VpVp -V p

63. Induction - Spring 200662  vv  V(t) tt VpVp -V p V = VP sin (wt -  v ) Phase Term

64. Induction - Spring 200663 V p and I p are the peak current and voltage. We also use the “root-mean-square” values: V rms = V p / and I rms =I p /  v and  I are called phase differences (these determine when V and I are zero). Usually we’re free to set  v =0 (but not  I ). Alternating Current Circuits V = V P sin (  t -  v ) I = I P sin (  t -  I )  vv  V(t) tt VpVp -V p V rms I/I/ I(t) t IpIp -Ip-Ip I rms

65. Induction - Spring 200664 Example: household voltage In the U.S., standard wiring supplies 120 V at 60 Hz. Write this in sinusoidal form, assuming V(t)=0 at t=0.

66. Induction - Spring 200665 Example: household voltage In the U.S., standard wiring supplies 120 V at 60 Hz. Write this in sinusoidal form, assuming V(t)=0 at t=0. This 120 V is the RMS amplitude: so V p =V rms = 170 V.

67. Induction - Spring 200666 Example: household voltage In the U.S., standard wiring supplies 120 V at 60 Hz. Write this in sinusoidal form, assuming V(t)=0 at t=0. This 120 V is the RMS amplitude: so V p =V rms = 170 V. This 60 Hz is the frequency f: so  =2  f=377 s -1.

68. Induction - Spring 200667 Example: household voltage In the U.S., standard wiring supplies 120 V at 60 Hz. Write this in sinusoidal form, assuming V(t)=0 at t=0. This 120 V is the RMS amplitude: so V p =V rms = 170 V. This 60 Hz is the frequency f: so  =2  f=377 s -1. So V(t) = 170 sin(377t +  v ). Choose  v =0 so that V(t)=0 at t=0: V(t) = 170 sin(377t).

69. Induction - Spring 200668 Review: Resistors in AC Circuits E R ~ EMF (and also voltage across resistor): V = V P sin (  t) Hence by Ohm’s law, I=V/R: I = (V P /R) sin(  t) = I P sin(  t) (with I P =V P /R) V and I “In-phase” V tt  I 

70. Induction - Spring 200669 This looks like I P =V P /R for a resistor (except for the phase change). So we call X c = 1/(  C) the Capacitive Reactance Capacitors in AC Circuits E ~ C Start from: q = C V [V=V p sin(  t)] Take derivative: dq/dt = C dV/dt So I = C dV/dt = C V P  cos (  t) I = C  V P sin (  t +  /2) The reactance is sort of like resistance in that I P =V P /X c. Also, the current leads the voltage by 90 o (phase difference). V tt   I V and I “out of phase” by 90º. I leads V by 90º.

71. Induction - Spring 200670 I Leads V??? What the **(&@ does that mean?? I V Current reaches it’s maximum at an earlier time than the voltage! 1 2 I = C  V P sin (  t +  /2)  Phase= - (  /2)

72. Induction - Spring 200671 Capacitor Example E ~ C A 100 nF capacitor is connected to an AC supply of peak voltage 170V and frequency 60 Hz. What is the peak current? What is the phase of the current? Also, the current leads the voltage by 90o (phase difference). I=V/X C

73. Induction - Spring 200672 Again this looks like I P =V P /R for a resistor (except for the phase change). So we call X L =  L the Inductive Reactance Inductors in AC Circuits L V = V P sin (  t) Loop law: V +V L = 0 where V L = -L dI/dt Hence: dI/dt = (V P /L) sin(  t). Integrate: I = - (V P / L  cos (  t) or I = [V P /(  L)] sin (  t -  /2) ~ Here the current lags the voltage by 90 o. V tt   I V and I “out of phase” by 90º. I lags V by 90º.

74. Induction - Spring 200673

75. Induction - Spring 200674 Phasor Diagrams VpVp IpIp  t Resistor A phasor is an arrow whose length represents the amplitude of an AC voltage or current. The phasor rotates counterclockwise about the origin with the angular frequency of the AC quantity. Phasor diagrams are useful in solving complex AC circuits. The “y component” is the actual voltage or current. A phasor is an arrow whose length represents the amplitude of an AC voltage or current. The phasor rotates counterclockwise about the origin with the angular frequency of the AC quantity. Phasor diagrams are useful in solving complex AC circuits. The “y component” is the actual voltage or current.

76. Induction - Spring 200675 Phasor Diagrams VpVp IpIp  t VpVp IpIp ResistorCapacitor A phasor is an arrow whose length represents the amplitude of an AC voltage or current. The phasor rotates counterclockwise about the origin with the angular frequency of the AC quantity. Phasor diagrams are useful in solving complex AC circuits. The “y component” is the actual voltage or current. A phasor is an arrow whose length represents the amplitude of an AC voltage or current. The phasor rotates counterclockwise about the origin with the angular frequency of the AC quantity. Phasor diagrams are useful in solving complex AC circuits. The “y component” is the actual voltage or current.

77. Induction - Spring 200676 Phasor Diagrams VpVp IpIp  t VpVp IpIp VpVp IpIp ResistorCapacitor Inductor A phasor is an arrow whose length represents the amplitude of an AC voltage or current. The phasor rotates counterclockwise about the origin with the angular frequency of the AC quantity. Phasor diagrams are useful in solving complex AC circuits. The “y component” is the actual voltage or current. A phasor is an arrow whose length represents the amplitude of an AC voltage or current. The phasor rotates counterclockwise about the origin with the angular frequency of the AC quantity. Phasor diagrams are useful in solving complex AC circuits. The “y component” is the actual voltage or current.

78. Induction - Spring 200677 Steady State Solution for AC Current (2) Expand sin & cos expressions Collect sin  d t & cos  d t terms separately These equations can be solved for I m and  (next slide) High school trig! cos  d t terms sin  d t terms

79. Induction - Spring 200678 Steady State Solution for AC Current (2) Expand sin & cos expressions Collect sin  d t & cos  d t terms separately These equations can be solved for I m and  (next slide) High school trig! cos  d t terms sin  d t terms

80. Induction - Spring 200679 Solve for  and I m in terms of R, X L, X C and Z have dimensions of resistance Let’s try to understand this solution using “phasors” Steady State Solution for AC Current (3) Inductive “reactance” Capacitive “reactance” Total “impedance”

81. Induction - Spring 200680 REMEMBER Phasor Diagrams? VpVp IpIp  t VpVp IpIp VpVp IpIp ResistorCapacitor Inductor A phasor is an arrow whose length represents the amplitude of an AC voltage or current. The phasor rotates counterclockwise about the origin with the angular frequency of the AC quantity. Phasor diagrams are useful in solving complex AC circuits. A phasor is an arrow whose length represents the amplitude of an AC voltage or current. The phasor rotates counterclockwise about the origin with the angular frequency of the AC quantity. Phasor diagrams are useful in solving complex AC circuits.

82. Induction - Spring 200681 Reactance - Phasor Diagrams VpVp IpIp  t VpVp IpIp VpVp IpIp ResistorCapacitor Inductor

83. Induction - Spring 200682 “Impedance” of an AC Circuit R L C ~ The impedance, Z, of a circuit relates peak current to peak voltage: (Units: OHMS)

84. Induction - Spring 200683 “Impedance” of an AC Circuit R L C ~ The impedance, Z, of a circuit relates peak current to peak voltage: (Units: OHMS) (This is the AC equivalent of Ohm’s law.)

85. Induction - Spring 200684 Impedance of an RLC Circuit R L C ~ E As in DC circuits, we can use the loop method: E - V R - V C - V L = 0 I is same through all components.

86. Induction - Spring 200685 Impedance of an RLC Circuit R L C ~ E As in DC circuits, we can use the loop method: E - V R - V C - V L = 0 I is same through all components. BUT: Voltages have different PHASES  they add as PHASORS.

87. Induction - Spring 200686 Phasors for a Series RLC Circuit IpIp V Rp (V Cp - V Lp ) VPVP  V Cp V Lp

88. Induction - Spring 200687 Phasors for a Series RLC Circuit By Pythagoras’ theorem: (V P ) 2 = [ (V Rp ) 2 + (V Cp - V Lp ) 2 ] IpIp V Rp (V Cp - V Lp ) VPVP  V Cp V Lp

89. Induction - Spring 200688 Phasors for a Series RLC Circuit By Pythagoras’ theorem: (V P ) 2 = [ (V Rp ) 2 + (V Cp - V Lp ) 2 ] = I p 2 R 2 + (I p X C - I p X L ) 2 IpIp V Rp (V Cp - V Lp ) VPVP  V Cp V Lp

90. Induction - Spring 200689 Impedance of an RLC Circuit Solve for the current: R L C ~

91. Induction - Spring 200690 Impedance of an RLC Circuit Solve for the current: Impedance: R L C ~

92. Induction - Spring 200691 The circuit hits resonance when 1/  C-  L=0:  r =1/ When this happens the capacitor and inductor cancel each other and the circuit behaves purely resistively: I P =V P /R. Impedance of an RLC Circuit The current’s magnitude depends on the driving frequency. When Z is a minimum, the current is a maximum. This happens at a resonance frequency:  The current dies away at both low and high frequencies. rr L=1mH C=10  F

93. Induction - Spring 200692 Phase in an RLC Circuit IpIp V Rp (V Cp - V Lp ) VPVP  V Cp V Lp We can also find the phase: tan  = (V Cp - V Lp )/ V Rp or; tan  = (X C -X L )/R. or tan  = (1/  C -  L) / R

94. Induction - Spring 200693 Phase in an RLC Circuit At resonance the phase goes to zero (when the circuit becomes purely resistive, the current and voltage are in phase). IpIp V Rp (V Cp - V Lp ) VPVP  V Cp V Lp We can also find the phase: tan  = (V Cp - V Lp )/ V Rp or; tan  = (X C -X L )/R. or tan  = (1/  C -  L) / R More generally, in terms of impedance: cos  R/Z

95. Induction - Spring 200694 Power in an AC Circuit V(t) = V P sin (  t) I(t) = I P sin (  t) P(t) = IV = I P V P sin 2 (  t) Note this oscillates twice as fast. V tt   I tt  P  = 0 (This is for a purely resistive circuit.)

96. Induction - Spring 200695 The power is P=IV. Since both I and V vary in time, so does the power: P is a function of time. Power in an AC Circuit Use, V = V P sin (  t) and I = I P sin (  t+  ) : P(t) = I p V p sin(  t) sin (  t+  ) This wiggles in time, usually very fast. What we usually care about is the time average of this: (T=1/f )

97. Induction - Spring 200696 Power in an AC Circuit Now:

98. Induction - Spring 200697 Power in an AC Circuit Now:

99. Induction - Spring 200698 Power in an AC Circuit Use: and: So Now:

100. Induction - Spring 200699 Power in an AC Circuit Use: and: So Now: which we usually write as

101. Induction - Spring 2006100 Power in an AC Circuit  goes from -90 0 to 90 0, so the average power is positive) cos(  is called the power factor. For a purely resistive circuit the power factor is 1. When R=0, cos(  )=0 (energy is traded but not dissipated). Usually the power factor depends on frequency.