Physics 212 Lecture 20, Slide 1 Physics 212 Lecture 20 AC Circuits Phasors

Actual current I (t) is the projection of the “phasor” onto the y-axis at time t. R

L V L = LdI/dt Inductor phasor 90 o Inductive “reactance”

90 o C

Physics 212 Lecture 20, Slide 5 R V max = R I max V R in phase with I L V max = X L I max V L 90 o ahead of I X L =  L C V max = X C I max V C 90 o behind I X C = 1/  C Current comes first since it charges capacitor Summary Like a short circuit at high  Like a short circuit at low 

Physics 212 Lecture 20, Slide 6 Add phasor voltages around the loop. These add like vectors. L R C If phasors satisfy Khirchoff’s Law, then so will their projections. The projections are the actual voltages.

Physics 212 Lecture 20, Slide 7 Add component by component. I max X L I max X C I max R L R C  max I max R I max (X L -X C ) I max X L I max X C I max R

Physics 212 Lecture 20, Slide 8 R (X L -X C )  Impedance Phasors Define total circuit impedance Z(  )  max = I max Z I max R I max (X L -X C )

Physics 212 Lecture 20, Slide 9 Summary: I max =  max / Z R  X L -X C )   max = I max Z V Cmax = I max X C V Lmax = I max X L V Rmax = I max R I max X L I max X C I max R L R C  max

Physics 212 Lecture 20, Slide 10 Example: RL Circuit X c =0 L R I max X L I max R  max I max X L I max R  max

Physics 212 Lecture 20, Slide 11 An RL circuit is driven by an AC generator as shown in the figure. An RL circuit is driven by an AC generator as shown in the figure. For what driving frequency  of the generator will the current through the resistor be largest ? A)  large B) Current through R doesn’t depend on  C)  small L R RL ACT

Physics 212 Lecture 20, Slide 12 ABCABC Checkpoint 1a Draw Voltage Phasors I max X L I max R  max

Physics 212 Lecture 20, Slide 13 ABCABC Checkpoint 1b Draw Voltage Phasors I max X L I max R  max

Physics 212 Lecture 20, Slide 14 I max X L I max R  max Checkpoint 1c The CURRENT is THE CURRENT ABCDABCD  is the phase between generator and current 

Physics 212 Lecture 20, Slide 15 ABCABC Checkpoint 2a IX c IR  IX L

Physics 212 Lecture 20, Slide 16 Checkpoint 2b ABCABC IX c IR  IX L

Physics 212 Lecture 20, Slide 17 Checkpoint 2c ABCABC IX c = V c = Q/C IR  IX L = V L

Physics 212 Lecture 20, Slide 18Calculation Conceptual Analysis –The maximum voltage for each component is related to its reactance and to the maximum current. –The impedance triangle determines the relationship 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 LCR L V ~ 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. What is X L, the reactance of the inductor, at this frequency?

Physics 212 Lecture 20, Slide 19Calculation Compare X L and X C at this frequency: (A) X L X C (D) Not enough information This information is determined from the phase –Current leads voltage V L = I max X L V C = I max X C V R (phase of current) VLVL VCVC V leads 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. What is X L, the reactance of the inductor, at this frequency?CR L V ~ IRV  

Physics 212 Lecture 20, Slide 20Calculation What is Z, the total impedance of the circuit? (A) (B) (C) (D)CR L V ~ 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. What is X L, the reactance of the inductor, at this frequency? 35.4 k  50 k  21.1 k  70.7 k 

Physics 212 Lecture 20, Slide 21Calculation What is R? (A) (B) (C) (D) D etermined from impedance triangleR Z=50k  (X C -X L )   CR L V ~ 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. What is X L, the reactance of the inductor, at this frequency? Z = 50k  sin(45)=.707 cos(45)= k  50 k  21.1 k  70.7 k  = 50 k  x = 35.4 k  R = Z cos(45 o )

Physics 212 Lecture 20, Slide 22CalculationCR L V ~ (A) (B) (C) (D) 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. What is X L, the reactance of the inductor, at this frequency? Z = 50k  R = 35.4k  R Z (X C -X L )   We start with the impedance triangle: What is X C ? 35.4 k  50 k  21.1 k  70.7 k  X L = X C - R V Cmax = I max X C X L = 56.5 k  – 35.4 k 