1 / 28

AC Circuit Phasors

L. R. C. Physics 102: Lecture 13. AC Circuit Phasors. I = I max sin(2 p ft) V R = I max R sin(2 p ft) V R in phase with I. I. V R. V C = I max X C sin(2 p ft- p/2 ) V C lags I. t. V L. V L = I max X L sin(2 p ft+ p/2 ) V L leads I. V C. L. R. C.

janettt
Download Presentation

AC Circuit Phasors

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. L R C Physics 102:Lecture 13 AC Circuit Phasors • I = Imaxsin(2pft) • VR = ImaxR sin(2pft) • VR in phase with I I VR • VC = ImaxXC sin(2pft-p/2) • VC lags I t VL • VL = ImaxXL sin(2pft+p/2) • VL leads I VC

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

  3. L R C Time Dependence in AC Circuits Write down Kirchoff’s Loop Equation: VG(t) = VL(t) + VR(t) + VC(t)at every instant of time I VR • However … • VG,max  VL,max+VR,max+VC,max • Maximum reached at different times for R, L, C t VL VC 5

  4. L R C Here is a problem that we will now learn how to solve: Example An AC circuit with R= 2 W, C = 15 mF, and L = 30 mH is driven by a generator with voltage V(t)=2.5 sin(8pt) Volts. Calculate the maximum current in the circuit, and the phase angle. 41

  5. q+p/2 a a q q-p/2 a A reminder about sines and cosines y • Recall: y coordinates of endpoints are • asin(q + p/2) • asin(q) • asin(q - p/2) x

  6. L R C q+p/2 ImaxR ImaxXL q q-p/2 ImaxXC Graphical representation of voltages I = Imaxsin(2pft) (q = 2pft) VL = ImaxXLsin(2pft + p/2) VR = ImaxR sin(2pft) VC = ImaxXC sin(2pft - p/2)

  7. Phasor Diagrams: A Detailed Example t = 1 f=1/12 2pft = p/6 • I = Imaxsin(2pft) • VR = VR,maxsin(2pft) VR,max VR,maxsin(p/6) p/6 Length of vector = Vmax across that component Vertical component = instantaneous value of V 10

  8. VR,max Phasor Diagrams t = 2 2pft = p/3 • I = Imaxsin(p/3) • VR = VR,maxsin(p/3) VR,maxsin(p/3) p/3 Length of vector = Vmax across that component Vertical component = instantaneous value of V

  9. VR,max Phasor Diagrams t = 3 2pft = p/2 • I = Imaxsin(p/2) • VR = VR,maxsin(p/2) VR,maxsin(p/2)=V0 p/2 Length of vector = Vmax across that component Vertical component = instantaneous value of V

  10. Phasor Diagrams t = 4 2pft = 4p/6 • I = Imaxsin(4p/6) • VR = VR,maxsin(4p/6) VR,max VR,maxsin(4p/6) 4p/6 Length of vector = Vmax across that component Vertical component = instantaneous value of V

  11. VR,max Phasor Diagrams t = 6 2pft = p • I = Imaxsin(p) • VR = VR,maxsin(p) p VR,maxsin(p)=0 Length of vector = Vmax across that component Vertical component = instantaneous value of V

  12. Phasor Diagrams t = 8 2pft = 8p/6 • I = Imaxsin(8p/6) • VR = VR,maxsin(8p/6) 8p/6 VR,max VR,maxsin(8p/6) Length of vector = Vmax across that component Vertical component = instantaneous value of V

  13. Phasor Diagrams t = 10 2pft = 10p/6 • I = Imaxsin(10p/6) • VR = VR,maxsin(10p/6) 10p/6 VR,maxsin(10p/6) VR,max Length of vector = Vmax across that component Vertical component = instantaneous value of V

  14. Resistor vector: to the right • Length given by VR,max (or R) VL,max VR,max • (2) Inductor vector: upwards • Length given by VL,max (or XL) VC,max • (3) Capacitor vector: downwards • Length given by VC,max (or XC) VL(t) VR(t) • (5) Rotate entire thing counter-clockwise • Vertical components give instantaneous voltage across R, C, L VC(t) Drawing Phasor Diagrams (4) Generator vector (coming soon) 15

  15. ImaxXL ImaxR ImaxXL cos(2pft) ImaxR sin(2pft) ImaxXC -ImaxXC cos(2pft) Phasor Diagrams Instantaneous Values: • I = Imaxsin(2pft) • VR = ImaxR sin(2pft) • VC = ImaxXC sin(2pft-p/2) • = -ImaxXC cos(2pft) • VL = ImaxXL sin(2pft+ p/2) • = ImaxXL cos(2pft) Voltage across resistor is always in phase with current! Voltage across capacitor always lags current! Voltage across inductor always leads current! 17

  16. Phasor Diagram Practice Example 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 Inductor Leads Capacitor Lags VR VL R: It has largest vertical component VC Decreasing, spins counter clockwise Inductor, it has longest line. 21

  17. “phase angle” Kirchhoff: Impedance Triangle • Instantaneousvoltage across generator (Vgen) must equal sum of voltage across all of the elements at all times: ImaxXL=VL,max Vgen,max=ImaxZ Vgen (t) = VR (t)+VC (t)+VL (t) Imax(XL-XC) f Vgen,max = Imax Z ImaxR=VR,max ImaxXC=VC,max 25

  18. ImaxZ Imax 2pft + f 2pft Phase angle f I = Imaxsin(2pft) Vgen = ImaxZ sin(2pft + f) f is positive in this particular case.

  19. Resistor vector: to the right • Length given by VR,max (or R) VL,max VR,max • (2) Capacitor vector: Downwards • Length given by VC,max (or XC) Vgen,max VC,max • (3) Inductor vector: Upwards • Length given by VL,max (or XL) • (4) Generator vector: add first 3 vectors • Length given by Vgen,max (or Z) Vgen Drawing Phasor Diagrams VL VR • (5) Rotate entire thing counter-clockwise • Vertical components give instantaneous voltage across R, C, L VC Rotates Counter Clockwise 27

  20. Vgen VR Vgen VR VR VR Vgen Vgen time 4 time 3 time 1 time 2 ACTS 13.1, 13.2, 13.3 When does Vgen = 0 ? time 2 When does Vgen = VR ? time 3 30

  21. f time 4 time 3 time 1 time 2 ACTS 13.1, 13.2, 13.3 When does Vgen = 0 ? time 2 When does Vgen = VR ? time 3 The phase angle is: (1) positive (2) negative (3) zero? negative Look at time 1: Vgen is below VR 31

  22. 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 Imax VR,max = ½ Imax Vgen,max cos(f) = Irms Vgen,rms cos(f) 34

  23. AC Summary Resistors: VR,max=Imax R In phase with I Capacitors: VC,max =Imax XC Xc = 1/(2pf C) Lags I Inductors: VL,max=Imax XL XL = 2pf L Leads I Generator: Vgen,max=Imax Z Z = √R2 +(XL -XC)2 Can lead or lag I tan(f) = (XL-XC)/R Power is only dissipated in resistor: P = ½ImaxVgen,max cos(f) 37

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

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

  26. ACT/Preflight 13.1 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 40% 28% 32% ImaxXL=VL,max Vgen,max Imax(XL-XC) Rotates Counter Clockwise f ImaxR Vgen(t) = VL(t)+VR(t)+VC(t) at all times! Vrms ≠ VL,rms+VR,rms+VC,rms ImaxXC = VC,max 43

  27. Imax XL Vgen,max f Imax R Imax XC ACT: Voltage Phasor Diagram At this instant, the voltage across the generator is maximum. What is the voltage across the resistor at this instant? 1) VR = ImaxR 2) VR = ImaxR sin(f) 3) VR = ImaxR cos(f) 46

  28. See You Monday!

More Related