2. Induction1 Time Varying Circuits 2008

3. Induction2 A look into the future We have one more week after today (+ one day) We have one more week after today (+ one day) Time Varying Circuits Including AC Time Varying Circuits Including AC Some additional topics leading to waves Some additional topics leading to waves A bit of review if there is time. A bit of review if there is time. There will be one more Friday morning quiz. There will be one more Friday morning quiz. I hope to be able to return the exams on Monday at which time we will briefly review the solutions. I hope to be able to return the exams on Monday at which time we will briefly review the solutions.

4. Induction 3 The Final Exam 8-10 Problems similar to (or exactly) Web- Assignments 8-10 Problems similar to (or exactly) Web- Assignments Covers the entire semester’s work Covers the entire semester’s work May contain some short answer questions. May contain some short answer questions.

5. Induction 4 Max Current Rate of increase = max emf V R =iR ~current

6. Induction 5 We Solved the loop equation.

7. Induction 6 We also showed that

8. Induction 7 At t=0, the charged capacitor is now connected to the inductor. What would you expect to happen??

9. Induction 8 The math … For an RLC circuit with no driving potential (AC or DC source):

10. Induction 9 The Graph of that LR (no emf) circuit.. I

11. Induction 10

12. Induction 11 Mass on a Spring Result Energy will swap back and forth. Add friction  Oscillation will slow down  Not a perfect analogy

13. Induction 12

14. Induction 13 LC Circuit High Q/C Low High

15. Induction 14 The Math Solution (R=0):

16. Induction 15 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.

17. Induction 16 Variable Emf Applied emf Sinusoidal DC

18. Induction 17 Sinusoidal Stuff “Angle” Phase Angle

19. Induction 18 Same Frequency with PHASE SHIFT 

20. Induction 19 Different Frequencies

21. Induction 20 Note – Power is delivered to our homes as an oscillating source (AC)

22. Induction 21 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

23. Induction 22 The Real World

24. Induction 23 A

25. Induction 24

26. Induction 25 The Flux:

27. Induction26 problems …

28. Induction 27 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.

29. Induction 28 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?

30. Induction 29 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?

31. Induction 30 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.

32. Induction 31 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?

33. Induction 32 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?

34. Induction 33 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?

35. Induction 34 Source Voltage:

36. Induction 35 Average value of anything: Area under the curve = area under in the average box T h

37. Induction 36 Average Value For AC:

38. Induction 37 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

39. Induction 38 Average Value

40. Induction 39 RMS

41. Induction 40 Usually Written as:

42. Induction 41 Example: What Is the RMS AVERAGE of the power delivered to the resistor in the circuit: E R ~

43. Induction 42 Power

44. Induction 43 More Power - Details

45. Induction 44 Resistive Circuit We apply an AC voltage to the circuit. Ohm’s Law Applies

46. Induction 45 Consider this circuit CURRENT AND VOLTAGE IN PHASE

47. Induction 46

48. Induction47 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

49. Induction48  vv  V(t) tt VpVp -V p V = V P sin (  t -  v ) Phase Term

50. Induction49 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

51. Induction50 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.

52. Induction51 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.

53. Induction52 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.

54. Induction53 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).

55. Induction54 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 

56. Induction55 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.

57. Induction56 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)

58. Induction57 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

59. Induction58 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.

60. Induction59

61. Induction60 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.

62. Induction61 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.

63. Induction62 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.

64. Induction63 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

65. Induction64 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

66. Induction65 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”

67. Induction66 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.

68. Induction67 Reactance - Phasor Diagrams VpVp IpIp  t VpVp IpIp VpVp IpIp ResistorCapacitor Inductor

69. Induction68 “Impedance” of an AC Circuit R L C ~ The impedance, Z, of a circuit relates peak current to peak voltage: (Units: OHMS)

70. Induction69 “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.)

71. Induction70 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.

72. Induction71 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.

73. Induction72 Phasors for a Series RLC Circuit IpIp V Rp (V Cp - V Lp ) VPVP  V Cp V Lp

74. Induction73 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

75. Induction74 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

76. Induction75 Impedance of an RLC Circuit Solve for the current: R L C ~

77. Induction76 Impedance of an RLC Circuit Solve for the current: Impedance: R L C ~

78. Induction77 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

79. Induction78 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

80. Induction79 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

81. Induction80 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.)

82. Induction81 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 )

83. Induction82 Power in an AC Circuit Now:

84. Induction83 Power in an AC Circuit Now:

85. Induction84 Power in an AC Circuit Use: and: So Now:

86. Induction85 Power in an AC Circuit Use: and: So Now: which we usually write as

87. Induction86 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.

88. Induction 87 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.

89. Induction 88 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?

90. Induction 89 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?

91. Induction 90 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?