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AC Circuit PhasorsPhasors Physics 102: Lecture 13 Exam III L R C I = I max sin(2  ft) V R = I max R sin(2  ft) V R in phase with I V C = I max X C sin(2.

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Presentation on theme: "AC Circuit PhasorsPhasors Physics 102: Lecture 13 Exam III L R C I = I max sin(2  ft) V R = I max R sin(2  ft) V R in phase with I V C = I max X C sin(2."— Presentation transcript:

1 AC Circuit PhasorsPhasors Physics 102: Lecture 13 Exam III L R C I = I max sin(2  ft) V R = I max R sin(2  ft) V R in phase with I V C = I max X C sin(2  ft-  ) V C lags I V L = I max X L sin(2  ft+  ) V L leads I I t VLVL VCVC VRVR

2 Peak + RMS values in AC Circuits (REVIEW) L R C 5 When asking about RMS or Maximum values relatively simple expresions

3 Time Dependence in AC Circuits Write down Kirchoff’s Loop Equation: V G + V L + V R + V C = 0 at every instant of time L R C 5 However … V G,max  V L,max +V R,max +V C,max Maximum reached at different times for R,L,C

4 A reminder about sines and cosines Recall: y coordinates of endpoints are asin(  +  /2) asin(  ) asin(  -  /2)    a a a x y

5 L R C I = I max sin(2  ft) (  = 2  ft) V L = I max X L sin(2  ft +  ) V R = I max R sin(2  ft) V C = I max X C sin(2  ft -  ) Graphical representation of voltages  I max X L  I max R  I max X C

6 Phasor Diagrams I = I max sin(  /6) V R = V R,max sin(  /6)  V R,max sin(  ) t = 1 f=1/12 2  ft =  /6 Length of vector = V max across that component Vertical component = instantaneous value of V 10 V R,max

7  Phasor Diagrams V R,max sin(  ) t = 2 2  ft =  /3 V R,max I = I max sin(  /3) V R = V R,max sin(  /3) Length of vector = V max across that component Vertical component = instantaneous value of V

8 Phasor Diagrams  V R,max sin(  )=V 0 t = 3 2  ft =  /2 V R,max I = I max sin(  /2) V R = V R,max sin(  /2) Length of vector = V max across that component Vertical component = instantaneous value of V

9 Phasor Diagrams  V R,max sin(4  ) t = 4 2  ft = 4  /6 V R,max I = I max sin(4  /6) V R = V R,max sin(4  /6) Length of vector = V max across that component Vertical component = instantaneous value of V

10 Phasor Diagrams  t = 6 2  ft =  V R,max I = I max sin(  ) V R = V R,max sin(  ) V R,max sin(  )=0 Length of vector = V max across that component Vertical component = instantaneous value of V

11 Phasor Diagrams  V R,max sin(8  ) t = 8 2  ft = 8  V R,max I = I max sin(8  /6) V R = V R,max sin(8  /6) Length of vector = V max across that component Vertical component = instantaneous value of V

12 Phasor Diagrams  V R,max sin(10  ) t = 10 2  ft = 10  V R,max I = I max sin(10  /6) V R = V R,max sin(10  /6) Length of vector = V max across that component Vertical component = instantaneous value of V

13 Drawing Phasor Diagrams VLVL (2)Inductor vector: upwards Length given by V L (or X L ) VCVC (3)Capacitor vector: downwards Length given by V C (or X C ) VRVR (1) Resistor vector: to the right Length given by V R (or R) VCVC VRVR VLVL (5)Rotate entire thing counter-clockwise Vertical components give instantaneous voltage across R, C, L (4) (coming soon) 15

14 Phasor Diagrams I = I max sin(2  ft) V R = I max R sin(2  ft) I max R I max R sin(2  ft) V C = I max X C sin(2  ft-  ) = -I max X C cos(2  ft) V L = I max X L sin(2  ft+  ) = I max X L cos(2  ft) I max X L cos(2  ft) -I max X C cos(2  ft) I max X L I max X C Voltage across resistor is always in phase with current! Voltage across capacitor always lags current! Voltage across inductor always leads current! Instantaneous Values: 17

15 Phasor Diagram Practice Label the vectors that corresponds to the resistor, inductor and capacitor. Which element has the largest voltage across it at the instant shown? 1) R 2) C 3) L Is the voltage across the inductor 1)increasing or 2) decreasing? Which element has the largest maximum voltage across it? 1) R 2) C 3) L VLVL VCVC VRVR Inductor Leads Capacitor Lags R: It has largest vertical component Decreasing, spins counter clockwise Inductor, it has longest line. 21

16 I max (X L -X C ) KVL: Impedance Triangle Instantaneous voltage across generator (V gen ) must equal sum of voltage across all of the elements at all times: I max X L =V L,max I max X C =V C,max I max R=V R,max V max,gen =I max Z  V gen (t) = V R (t) +V C (t) +V L (t) “phase angle” V gen,max = I max Z 25

17 Phase angle  2  ft I max I = I max sin(2  ft) V gen = I max R sin(2  ft +  ) I max R 2  ft +   is positive in this particular case.

18 Drawing Phasor Diagrams VLVL (2)Capacitor vector: Downwards Length given by V C (or X C ) VCVC (3)Inductor vector: Upwards Length given by V L (or X L ) VRVR (1) Resistor vector: to the right Length given by V R (or R) (4)Generator vector: add first 3 vectors Length given by V gen (or Z) V gen VCVC VRVR VLVL (5)Rotate entire thing counter-clockwise Vertical components give instantaneous voltage across R, C, L V gen 27

19 time 1time 2 time 3 time 4 ACTS 13.1, 13.2, 13.3 When does V gen = V R ? When does V gen = 0 ? V gen VRVR VRVR VRVR VRVR time 2 time 3 30

20 time 1time 2 time 3 time 4 ACTS 13.1, 13.2, 13.3 The phase angle is: (1) positive (2) negative (3) zero? When does V gen = V R ? When does V gen = 0 ? Look at time 1: V gen is below V R  negative time 3 time 2 31

21 Power P=IV The voltage generator supplies power. –Resistor dissipates power. –Capacitor and Inductor store and release energy. P = IV so sometimes power loss is large, sometimes small. Average power dissipated by resistor: P = ½ I max V R,max = ½ I max V gen,max cos(  ) = I rms V rms cos(  ) 34

22 AC Summary Resistors: V Rmax =I R In phase with I Capacitors: V Cmax =I X C X c = 1/(2  f C) Lags I Inductors: V Lmax =I X L X L = 2  f L Leads I Generator: V gen,max =I ZZ= sqrt(R 2 +(X L -X C ) 2 ) Can lead or lag I tan(  ) = (X L -X C )/R Power is only dissipated in resistor: P = ½I max V gen,max cos(  ) 37

23 Problem Time! An AC circuit with R= 2 , C = 15 mF, and L = 30 mH is driven by a generator with voltage V(t)=2.5 sin(8  t) Volts. Calculate the maximum current in the circuit, and the phase angle. L R C 41

24 Problem Time! An AC circuit with R= 2 , C = 15 mF, and L = 30 mH is driven by a generator with voltage V(t)=2.5 sin(8  t) Volts. Calculate the maximum current in the circuit, and the phase angle. I max = 2.5/2.76 =.91 Amps I max = V gen,max /Z L R C 41

25 I max (X L -X C ) Preflight 13.1 I max X L =V L,ma x I max X C = V C,max I max R V gen,max  Rotates Counter Clockwise V gen =V L +V R +V C at all times. V rms does not! 33% 32% 35% The statement that the voltage across the generator equals the sum of the voltages across the resistor, capacitor and inductor is true for: (1) instantaneous voltages only (2) rms voltages only (3) both rms and instantaneous 43

26 ACT: Voltage Phasor Diagram I max X L I max X C I max R V gen,max  At this instant, the voltage across the generator is maximum. What is the voltage across the resistor at this instant? 1) V R = I max R 2) V R = I max R sin(  )3) V R = I max R cos(  ) 46

27 See You Monday!


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