2. Induction1 Time Varying Circuits 2009

3. Induction 2 The Final Exam Approacheth 8-10 Problems similar to Web-Assignments 8-10 Problems similar to Web-Assignments Covers the entire semester’s work Covers the entire semester’s work May contain some short answer (multiple choice) questions. May contain some short answer (multiple choice) questions.

4. Induction 3 Spring 2009 Final Exam Schedule Tuesday, April 28 - Monday, May 4 No Exams on Sunday EXAM TIMES Class Meeting Times EXAM DAY 1 TUES 4/28 EXAM DAY 2 WED 4/29 EXAM DAY 3 THURS 4/30 EXAM DAY 4 FRI 5/1 EXAM DAY 5 SAT 5/2 EXAM DAY 6 MON 5/4 7:00 a.m. - 9:50 a.m. 7:30-10:20 T 9:00- 10:15 TR (all a.m.) 7:30-10:20 W 8:30- 9:20 MWF 9:00- 10:15 MW (all a.m.) 7:30-8:45 TR 7:30- 10:20 R (all a.m.) 7:30-10:20 F 9:00- 10:15 M/ 7:30-8:45 F 9:30- 10:20 MWF (all a.m.) Finals For Saturday Classes Are Held During Regular Class Meeting Times 7:30-8:20 MWF 7:30-8:45 MW 7:30- 10:20 M (all a.m.) 10:00 a.m. - 12:50 p.m. 10:30-11:45 TR 10:30-1:20 T 10:30-1:20 W 11:30-12:20 MWF 12:00-1:15 MW 12:00-1:15 WF 10:30-1:20 R 12:00- 1:15 TR 10:30-1:20 F 12:30- 1:20 MWF 12:00- 1:15 M/ 10:30-11:45 F Finals For Saturday Classes Are Held During Regular Class Meeting Times 10:30-11:20 MWF 10:30-11:45 MW 10:30-1:20 M 1:00 p.m. - 3:50 p.m. 1:30-2:45 TR 1:30- 4:20 T 1:30-4:20 W 2:30- 3:20 MWF 3:00- 4:15 WF 3:00-4:15 MW 1:30-4:20 R 3:00- 4:15 TR 1:30-4:20 F 3:30- 4:20 MWF 3:00- 4:15 M/ 1:30-2:45 F FREE PERIOD 1:30-2:20 MWF 1:30-2:45 MW 1:30- 4:20 M 4:00 p.m. - 6:50 p.m. 6:00-7:15 TR and Alternate Time 6:00-7:15 MW and Alternate Time 4:30-5:45 TR and Alternate Time 4:30-7:15 F and Alternate Time FREE PERIOD 4:30-5:45 MW and Alternate Time 7:00 p.m. - 9:50 p.m. 6:00-8:50 T 7:30- 10:20 T (all p.m.) 6:00-8:50 W 7:00- 7:50 MWF 7:30- 8:45 MW 7:30- 10:20 W (all p.m.) 6:00-8:50 R 7:30- 8:45 TR 7:30-10:20 R (all p.m.) 6:00-8:50 F 8:00- 8:50 MWF (all p.m.) FREE PERIOD 6:00-6:50 MWF 6:00-8:50 M 7:30- 10:20 M (all p.m.)

5. HowjaDo?? Induction4 A. I done good B. I done ok C. I done not so ok D. I screwed up major

6. The Test Itself Was Induction5 A. Fair B. Not so fair. C. Really Unfair. D. The worst kind of unfair in the entire universe.

7. Back to Circuits for a bit …. Sort of like RC circuit issues.

8. Definition Current in loop produces a magnetic field in the coil and consequently a magnetic flux. If we attempt to change the current, an emf will be induced in the loops which will tend to oppose the change in current. This this acts like a “resistor” for changes in current!

9. Remember Faraday’s Law Lentz

10. Look at the following circuit:  Switch is open  NO current flows in the circuit.  All is at peace!

11. Close the circuit…  After the circuit has been close for a long time, the current settles down.  Since the current is constant, the flux through the coil is constant and there is no Emf.  Current is simply E/R (Ohm’s Law)

12. Close the circuit…  When switch is first closed, current begins to flow rapidly.  The flux through the inductor changes rapidly.  An emf is created in the coil that opposes the increase in current.  The net potential difference across the resistor is the battery emf opposed by the emf of the coil.

13. Close the circuit…

14. Moving right along …

15. Definition of Inductance L UNIT of Inductance = 1 henry = 1 T- m 2 /A   is the flux near the center of one of the coils making the inductor

16. Consider a Solenoid n turns per unit length l

17. So…. Depends only on geometry just like C and is independent of current.

18. Inductive Circuit  Switch to “a”.  Inductor seems like a short so current rises quickly.  Field increases in L and reverse emf is generated.  Eventually, i maxes out and back emf ceases.  Steady State Current after this. i

19. THE BIG INDUCTION  As we begin to increase the current in the coil  The current in the first coil produces a magnetic field in the second coil  Which tries to create a current which will reduce the field it is experiences  And so resists the increase in current.

20. Back to the real world… i Switch to “a”

21. Solution (See textbook)

22. Switch position “b”

23. Max Current Rate of increase = max emf V R =iR ~current

24. Solve the loop equation.

25. IMPORTANT QUESTION  Switch closes.  No emf  Current flows for a while  It flows through R  Energy is conserved (i 2 R) WHERE DOES THE ENERGY COME FROM??

26. For an answer Return to the Big C  We move a charge dq from the (-) plate to the (+) one.  The (-) plate becomes more (-)  The (+) plate becomes more (+).  dW=Fd=dq x E x d +q -q E=  0 A/d +dq

27. The calc The energy is in the FIELD !!!

28. What about POWER?? power to circuit power dissipated by resistor Must be dW L /dt

29. So Energy stored in the Capacitor

30. WHERE is the energy?? l

31. Remember the Inductor??

32. So …

33. ENERGY IN THE FIELD TOO!

34. IMPORTANT CONCLUSION  A region of space that contains either a magnetic or an electric field contains electromagnetic energy.  The energy density of either is proportional to the square of the field strength.

35. Solution (From Before)

36. At t=0, the charged capacitor is now connected to the inductor. What would you expect to happen?? Induction35

37. The math … Induction36 For an RLC circuit with no driving potential (AC or DC source):

38. The Graph of that LR (no emf) circuit Induction37 I

39. Induction38

40. Mass on a Spring Result Induction39  Energy will swap back and forth.  Add friction  Oscillation will slow down  Not a perfect analogy

41. Induction40

42. LC Circuit Induction41 High Q/C Low High

43. The Math Solution (R=0): Induction 42

44. New Feature of Circuits with L and C Induction43  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.

45. Variable Emf Applied Induction44 emf Sinusoidal DC

46. Sinusoidal Stuff Induction45 “Angle” Phase Angle

47. Induction46 Same Frequency with PHASE SHIFT 

48. Different Frequencies Induction47

49. Note – Power is delivered to our homes as an oscillating source (AC) Induction48

50. Producing AC Generator Induction49 x x x x x x x x x x x x x x x x x x x x x x x

51. The Real World Induction50

52. Induction51 A

53. Induction52

54. The Flux: Induction53

55. problems … Induction54

56. Induction 55 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.

57. Induction 56 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?

58. Induction 57 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?

59. Induction 58 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.

60. Induction 59 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?

61. Induction 60 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?

62. Induction 61 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?

63. Source Voltage: Induction62

64. Induction63 Average value of anything : Area under the curve = area under in the average box T h

65. Average Value Induction64 For AC:

66. So … Induction65  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

67. Average Value Induction66

68. RMS Induction67

69. Usually Written as: Induction68

70. Example: What Is the RMS AVERAGE of the power delivered to the resistor in the circuit: Induction69 E R ~

71. Power Induction70

72. More Power - Details Induction71

73. Resistive Circuit  We apply an AC voltage to the circuit.  Ohm’s Law Applies Induction 72

74. Consider this circuit Induction73 CURRENT AND VOLTAGE IN PHASE

75. Induction74

76. Induction75 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

77. Induction76  vv  V(t) tt VpVp -V p V = V P sin (  t -  v ) Phase Term

78. Induction77 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

79. Induction78 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.

80. Induction79 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.

81. Induction80 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.

82. Induction81 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).

83. Induction82 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 

84. Induction83 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.

85. Induction84 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)

86. Induction85 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

87. Induction86 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.

88. Induction87

89. Induction88 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.

90. Induction89 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.

91. Induction90 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.

92. Induction91 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

93. Induction92 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

94. Induction93 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”

95. Induction94 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.

96. Induction95 Reactance - Phasor Diagrams VpVp IpIp  t VpVp IpIp VpVp IpIp ResistorCapacitor Inductor

97. Induction96 “Impedance” of an AC Circuit R L C ~ The impedance, Z, of a circuit relates peak current to peak voltage: (Units: OHMS)

98. Induction97 “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.)

99. Induction98 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.

100. Induction99 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.

101. Induction100 Phasors for a Series RLC Circuit IpIp V Rp (V Cp - V Lp ) VPVP  V Cp V Lp

102. Induction101 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

103. Induction102 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

104. Induction103 Impedance of an RLC Circuit Solve for the current: R L C ~

105. Induction104 Impedance of an RLC Circuit Solve for the current: Impedance: R L C ~

106. Induction105 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

107. Induction106 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

108. Induction107 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

109. Induction108 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.)

110. Induction109 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 )

111. Induction110 Power in an AC Circuit Now:

112. Induction111 Power in an AC Circuit Now:

113. Induction112 Power in an AC Circuit Use: and: So Now:

114. Induction113 Power in an AC Circuit Use: and: So Now: which we usually write as

115. Induction114 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.

116. Induction 115 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.

117. Induction 116 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?

118. Induction 117 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?

119. Induction 118 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?