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The Last Leg The Ups and Downs of Circuits Chapter 31.

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Presentation on theme: "The Last Leg The Ups and Downs of Circuits Chapter 31."— Presentation transcript:

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2 The Last Leg The Ups and Downs of Circuits Chapter 31

3 The End is Near! Examination #3 – Friday (April 15 th ) Taxes Due – Friday (April 15 th ) Watch for new WebAssigns Final Exam is Wednesday, April 27 th. Grades will be submitted as quickly as possible.

4 That’s Two Weeks. That’s Two Weeks. That’s Two Weeks.

5 So far we have considered DC Circuits  Resistors  Capacitors  Inductors We looked at  Steady State DC behaviors  Transient DC behaviors. We have not looked at sources that varied with time.

6 Example LR Circuit i Steady Source

7 Time Dependent Result:

8 RLRL

9 Variable Emf Applied emf Sinusoidal DC

10 Sinusoidal Stuff “Angle” Phase Angle

11 Same Frequency with PHASE SHIFT 

12 Different Frequencies

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

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15 The Math Solution:

16 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 The Graph

18 Note – Power is delivered to our homes as an oscillating source (AC)

19 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

20 The Real World

21 A

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23 The Flux:

24 OUTPUT WHAT IS AVERAGE VALUE OF THE EMF ??

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

26 Average Value For AC:

27 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

28 Average Value

29 RMS

30 Usually Written as:

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

32 Power

33 More Power - Details

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

35 Consider this circuit CURRENT AND VOLTAGE IN PHASE

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37 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

38  vv  V(t) tt VpVp -V p V = VP sin (wt -  v ) Phase Term

39 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

40 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.

41 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.

42 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.

43 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).

44 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 

45 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.

46 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) 

47 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).

48 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.

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50 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.

51 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.

52 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.

53 + + i + + + + i i i i i LC Circuit time

54 Analyzing the L-C Circuit Total energy in the circuit: Differentiate : N o change in energy

55 Analyzing the L-C Circuit Total energy in the circuit: Differentiate : N o change in energy

56 Analyzing the L-C Circuit Total energy in the circuit: Differentiate : N o change in energy

57 Analyzing the L-C Circuit Total energy in the circuit: Differentiate : N o change in energy The charge sloshes back and forth with frequency  = (LC) -1/2 The charge sloshes back and forth with frequency  = (LC) -1/2


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