Electricity & Magnetism Lecture 20

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1 Electricity & Magnetism Lecture 20 Today s Concept: AC Circuits Maximum currents & voltages Phasors: A Simple Tool Electricity & Magne?sm Lecture 20, Slide 1

2 Other videos: Prof. W. Lewin, MIT Open Courseware Mechanical Universe, Driven LRC circuits

3 Resistors ε = V max sin(ωt) R I = V R /R = V max /R sin(ωt) Amplitude = V max /R Electricity & Magne?sm Lecture 20, Slide 3

4 Capacitors Q = CV = CV max sin(ωt) ε = V max sin(ωt) C I = dq/dt I = V max ωc cos(ωt) 90 o Amplitude = V max /X C where X C = 1/ωC is like the resistance of the capacitor X C depends on ω Electricity & Magne?sm Lecture 20, Slide 4

5 Inductors di/dt = V L = V max sin(ωt) ε = V max sin(ωt) L I = (V max /ωl) cos(ωt) 90 o Amplitude = V max /X L where X L = ωl is like the resistance of the inductor X L depends on ω Electricity & Magne?sm Lecture 20, Slide 5

6 RL Clicker Question An RL circuit is driven by an AC generator as shown in the figure. L X L = ωl As ω 0, so does X L R As ω 0, impedance resistance of circuit R current gets bigger For what driving frequency ω of the generator will the current through the resistor be largest A) ω large B) Current through R doesn t depend on ω C) ω small Electricity & Magne?sm Lecture 20, Slide 6

7 Summary R I max = V max /R V R in phase with I Because resistors are simple C I max = V max /X C V C 90 o behind I X C = 1/ωC Current comes first since it charges capacitor Like a wire at high ω L I max = V max /X L V L 90 o ahead of I X L = ωl Opposite of capacitor Like a wire at low ω Electricity & Magne?sm Lecture 20, Slide 7

8 Makes sense to write everything in terms of I since this is the same everywhere in a one- loop circuit: Phasors make this simple to see I max X L V max = I max X C V 90 o behind I ε max V max = I max X L V 90 o ahead of I I max X C V max = V in phase with I Do you have any fancy- schmancy simula?ons for to show me? Always looks the same. Only the lengths will change Electricity & Magne?sm Lecture 20, Slide 8

9 The Voltages still Add Up But now we are adding vectors: I max X L I max X L ε max I max X L I max X C I max X C I max X C ε max Electricity & Magne?sm Lecture 20, Slide 9

10 Make this Simpler I max X L I max X L ε max I max X C I max X C Electricity & Magne?sm Lecture 20, Slide 10

11 Make this Simpler I max X L ε max = I max Z I max (X L X C ) I max X C Electricity & Magne?sm Lecture 20, Slide 11

12 Make this Simpler ε max = I max Z I max (X L X C ) Electricity & Magne?sm Lecture 20, Slide 12

13 Make this Simpler ε max = I max Z φ I max (X L X C ) φ R (X L X C ) Impedance Triangle Electricity & Magne?sm Lecture 20, Slide 13

14 Summary V Cmax = I max X C V Lmax = I max X L V Rmax = ε max = I max Z I max = ε max / Z φ R (X L X C ) Electricity & Magne?sm Lecture 20, Slide 14

15 Example: RL Circuit X c = 0 I max X L ε max Electricity & Magne?sm Lecture 20, Slide 15

16 CheckPoint 2 Draw Voltage Phasors I max X L ε max 60 A B C Electricity & Magne?sm Lecture 20, Slide 16 1

17 CheckPoint 4 Draw Voltage Phasors I max X L ε max 60 A B C Electricity & Magne?sm Lecture 20, Slide 17

18 CheckPoint 6 The CURRENT is THE CURRENT I max X L φ ε max φ is the phase between generator and current 50 A B C D Electricity & Magne?sm Lecture 20, Slide 18

19 CheckPoint A B C 1 IX L ε IR ε IR What does the voltage phasor diagram look like when the current is a maximum? IX L IX c IX c Electricity & Magne?sm Lecture 20, Slide 19

20 CheckPoint IX L IX c ε IR IR ε IX L IX c A B0 C 1 What does the voltage phasor diagram look like when the capacitor is fully charged? Electricity & Magne?sm Lecture 20, Slide 20

21 CheckPoint 12 IX L IX c ε 50 IR IR ε IX c IX L 13 0 A B C 1 What does the voltage phasor diagram look like when the voltage across capacitor is at its posi?ve maximum? Electricity & Magne?sm Lecture 20, Slide 21

22 Calculation Consider the harmonically driven series LCR circuit shown. V max = 100 V C I max = 2 ma V Cmax = 113 V The current leads generator voltage by 45 o L and R are unknown. V ~ R L What is X L, the reactance of the inductor, at this frequency? Conceptual Analysis The maximum voltage for each component is related to its reactance and to the maximum current. The impedance triangle determines the rela?onship between the maximum voltages for the components Strategic Analysis Use V max and I max to determine Z Use impedance triangle to determine R Use V Cmax and impedance triangle to determine X L Electricity & Magne?sm Lecture 20, Slide 22

23 Calculation Consider the harmonically driven series LCR circuit shown. V max = 100 V I max = 2 ma V Cmax = 113 V The current leads generator voltage by 45 o L and R are unknown. V ~ C R L What is X L, the reactance of the inductor, at this frequency? Compare X L and X C at this frequency: A) X L < X C B) X L = X C C) X L > X C D) Not enough informa?on V L = I max X L V C = I max X C V L This informa?on is determined from the phase Current leads voltage V R (phase of current) V C V leads V IR 45 ο Electricity & Magne?sm Lecture 20, Slide 23

24 Calculation Consider the harmonically driven series LCR circuit shown. V max = 100 V C I max = 2 ma V Cmax = 113 V The current leads generator voltage by 45 o L and R are unknown. V ~ R L What is X L, the reactance of the inductor, at this frequency? What is Z, the total impedance of the circuit? 70.7 kω 50 kω 35.4 kω 21.1 kω A) B) C) D) Electricity & Magne?sm Lecture 20, Slide 24

25 Calculation Consider the harmonically driven series LCR circuit shown. V max = 100 V I max = 2 ma V Cmax = 113 V The current leads generator voltage by 45 o L and R are unknown. V ~ C R L What is X L, the reactance of the inductor, at this frequency? What is R? A) 70.7 kω B) C) 35.4 kω D) 50 kω 21.1 kω Z = 50 kω sin(45 ) =0.707 cos(45 ) =0.707 Determined from impedance triangle R 45 ο R = Z cos(45 (X C X L ) o ) = 50 kω = 35.4 kω Z = 50kΩ Electricity & Magne?sm Lecture 20, Slide 25

26 Calculation Consider the harmonically driven series LCR circuit shown. V max = 100 V I max = 2 ma V Cmax = 113 V The current leads generator voltage by 45 o L and R are unknown. V ~ C R L What is X L, the reactance of the inductor, at this frequency? A) 70.7 kω B) C) 35.4 kω D) We start with the impedance triangle: R 45 ο (X C X L ) Z 50 kω 21.1 kω X L = X C R What is X C? V Cmax = I max X C Z = 50 kω R = 35.4kΩ X L = 56.5 kω 35.4 kω Electricity & Magne?sm Lecture 20, Slide 26

27 Practical Test Hints You will have 4 banana plug wires, ~4 alligator clips 1 Scope Probe (x1/x10) 1 BNC wire adaptors: Tee, BNC- Male Banana, BNC- Female Banana proto- board (Use it!) scope, func?on generator, DMM We will have two 60 min sessions: 12:30 13:30 13:40 14:30

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