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# lecture13 - AC Circuit Phasors Physics 102 Lecture 13 • I...

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Unformatted text preview: AC Circuit Phasors Physics 102: Lecture 13 • I = I max sin(2 π ft) • V R = I max R sin(2 π ft) • V R in phase with I • V C = I max X C sin(2 π ft- π/2 ) •V C lags I • V L = I max X L sin(2 π ft+ π/2 ) •V L leads I I t V L V C V R L R C Peak & RMS values in AC Circuits (REVIEW) L R C 5 ,max max C C V I X = ,max max R V I R = ,max max L L V I X = When asking about RMS or Maximum values relatively simple expressions 1 C X C ϖ = L X L ϖ = Time Dependence in AC Circuits Write down Kirchoff’s Loop Equation: V G (t) = V L (t) + V R (t) + V C (t) 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 I t V L V C V R Here is a problem that we will now learn how to solve: An AC circuit with R= 2 Ω , C = 15 mF, and L = 30 mH is driven by a generator with voltage V(t)=2.5 sin(8 π t) Volts. Calculate the maximum current in the circuit, and the phase angle. L R C 41 A reminder about sines and cosines Recall: y coordinates of endpoints are • a sin( θ + π /2) • a sin( θ ) • a sin( θ- π /2) θ θ+ π/ 2 θ-π/2 a a a x y I = I max sin(2 π ft ) ( θ = 2 π ft ) V L = I max X L sin(2 π ft + π/2 ) V R = I max R sin(2 π ft ) V C = I max X C sin(2 π ft- π/2 ) Graphical representation of voltages θ+ π/2 I max X L θ I max R θ-π/2 I max X C L R C Phasor Diagrams: A Detailed Example • I = I max sin(2 π ft) • V R = V R,max sin(2 π ft) π/6 V R,max sin( π/6 ) t = 1 f=1/12 2 π ft = π /6 Length of vector = V max across that component Vertical component = instantaneous value of V 10 V R , m a x π/3 Phasor Diagrams V R,max sin( π/3 ) t = 2 2 π ft = π /3 V R , m a x • I = I max sin( π /3) • V R = V R,max sin( π /3) Length of vector = V max across that component Vertical component = instantaneous value of V Phasor Diagrams...
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lecture13 - AC Circuit Phasors Physics 102 Lecture 13 • I...

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