2. L C   R

3. Today... RLC circuits with a sinusoidal drive Phase difference between current & voltage for Resistors, Capacitors, and Inductors. Reactance Phasors Application to frequency filters (high-, low-pass)

4. Driven LRC circuits Last time we discovered that an LC circuit was a natural oscillator: L C ++ - - R However, the resistance of any real inductor will cause oscillations to damp out, unless we can supply energy at the rate the resistor dissipates it! How? A sinusoidally varying emf (AC generator) will sustain sinusoidal current oscillations! Very useful → TV, radio, computer clocks, …   o sin  t Generic problem: we are given an “ac voltage source” “driving” a circuit We need to find the current that flows: I(t)=I o sin(  t -  ) It is also sinusoidal at the same frequency,  possibly with some “phase” angle relative to the voltage source

5. Preview Our goal is to understand how an AC LRC circuit works. Physical picture of each object: – Source: produces an oscillating voltage (supplies whatever current the rest of the circuit “requires”) – Resistor: causes a voltage drop when a current flows through. As soon as the voltage changes, so does the current  always in phase. – Capacitor: resists change in charge Q  resists change in voltage. voltage across capacitor lags behind (90˚) the current (charge leaving & entering the plates). – Inductor: resists change in magnetic flux  resists change in current. current always lags voltage (90˚). ~

6. AC Circuits Series LRC Statement of problem: Given  =  m sin  t, find I(t). Everything else will follow. We could solve this equation with tons of algebra involving sin(  t) and cos(  t) ; or with simple complex algebra. We will do neither, but start by considering simple circuits with one element ( R, C, or L ) in addition to the driving emf. L C   R Procedure: start with loop equation

7.  R Circuit R only: Loop eqn gives:   I R R  Note: this is always, always, true in LRC circuit, always… 0 0 t I R 0 0 V R t Voltage across R in phase with current through R  But voltage across R is not always in phase with source!

8.  C Circuit Now consider C only: Loop eqn gives:    C I C  Voltage across C lags current through C by one-quarter cycle (90  ). Is this always true? YES! 0 0 t VCVC t 0 0 ICIC

9.  L Circuit Now consider L : Loop eqn gives:   Voltage across L leads current through L by one- quarter cycle (90  ). I L   L Yes, yes, but how to remember? t 0 0 VLVL t 0 0 ILIL

10. V L leads I L V C lags I C Hi kids, I’m Eli and I’ll help you learn physics ! …we’ll see about that Introducing...

11. Summary thus far… LRC Circuit –Given: –Assume solution for current: I(t) = I m sin(  t -  ) L C   R –Note that in all cases, though there may be a phase shift: XCXC XLXL reactance

12. What is reactance? You can think of it as a frequency-dependent resistance. For high ω, χ C ~0 - Capacitor looks like a wire (“short”) For low ω, χ C  ∞ - Capacitor looks like a break For low ω, χ L ~0 - Inductor looks like a wire (“short”) For high ω, χ L  ∞ - Inductor looks like a break (inductors resist change in current)

13. Filter Example #1 ~ Consider the AC circuit shown. For very high frequencies, is V out big or small? Recall: capacitor resists change in voltage. High frequency  more change  smaller reactance  smaller V C,  V out pulled to ground Low frequency  capacitor has time to charge up  larger V C,  no current flows  no voltage drop across R V out ~ ε What is ω 0 ? Use dimensional analysis. So, this is a circuit that only passes low frequencies: “low-pass” filter  Bass knob on radio ε V out

14. Lecture 19, Act 1 A driven RC circuit is connected as shown. –For what frequencies  of the voltage source is the current through the resistor largest? (a)  small (b)  large (c) C   R 

15. More Filters ~ V out a. b. c. ω=0No current V out ≈ 0 ω=∞Capacitor ~ wire V out ≈ ε ~ V out ω = ∞ No current V out ≈ 0 ω = 0Inductor ~ wire V out ≈ ε ω = 0 No current because of capacitor ω = ∞ No current because of inductor (Conceptual sketch only) High- pass filter Low- pass filter Band- pass filter ~ V out

16. AC Circuits, Quantitative Our goal is to calculate the voltages across the various elements, and also the current flowing through the circuit. If these were just three resistors in series, we could calculate the net resistance simply by adding the individual resistances. Because current and voltage are 90  out of phase for the capacitor and the inductor, a straight sum does not work (unless you use complex numbers, which we don’t in P212). Instead we use “phasors”, a geometric way to visualize an oscillating function (avoiding nasty trig. or complex numbers). A phasor is a rotating “vector” whose magnitude is the maximum value of a quantity (e.g., V or I ). –The instaneous value is the projection on the y-axis. –The phasor rotates at the drive frequency . (Appendix) L C   R

17. “Beam me up Scotty – It ate my phasor!”

18. Phasors Problem: Given V drive = ε m sin(ωt), find V R, V L, V C, i R, i L, i C Strategy: We will use Kirchhoff’s voltage law that the (phasor) sum of the voltages V R, V C, and V L must equal V drive. L C   R

19. Phasors, cont. 1.Draw V R phasor along x -axis (this direction is chosen for convenience). Note that since V R = i R R, this is also the direction of the current phasor i R. Because of Kirchhoff’s current law, i L = i C = i R ≡ i (i.e., the same current flows through each element). V R, i R R L C   R Problem: Given V drive = ε m sin(ωt), find V R, V L, V C, i R, i L, i C

20. Phasors, cont. L C   R Problem: Given V drive = ε m sin(ωt), find V R, V L, V C, i R, i L, i C 2.Next draw the phasor for V L. Since the inductor current i L always lags V L  draw V L 90˚ further counterclockwise. The length of the V L phasor is i L X L = i  L V R = i R V L = i X L

21. Phasors, cont. 3.The capacitor voltage V C always lags i C  draw V C 90˚ further clockwise. The length of the V C phasor is i C X C = i /  C V R = i R V L = i X L V C = i X C The lengths of the phasors depend on R, L, C, and ω. The relative orientation of the V R, V L, and V C phasors is always the way we have drawn it. Memorize it! Problem: Given V drive = ε m sin(ωt), find V R, V L, V C, i R, i L, i C L C   R

22. Phasors, cont. The phasors for V R, V L, and V C are added like vectors to give the drive voltage V R + V L + V C = ε m : From this diagram we can now easily calculate quantities of interest, like the net current i, the maximum voltage across any of the elements, and the phase between the current the drive voltage (and thus the power). VRVR VLVL VCVC εmεm Problem: Given V drive = ε m sin(ωt), find V R, V L, V C, i R, i L, i C L C   R

23. A series RC circuit is driven by emf  =  m sin  t. Which of the following could be an appropriate phasor diagram? Lecture 19, Act 2 (a)(c)(b) VRVR VLVL VCVC εmεm VRVR VCVC εmεm ~ 2A For this circuit which of the following is true? (a)The drive voltage is in phase with the current. (b)The drive voltage lags the current. (c)The drive voltage leads the current. 2B VRVR εmεm VCVC

24. RC Circuit, quantitative: V R = iR εmεm  V C = iX C ~  m sin  t C R

25. RC Circuit, cont. Ex.: C = 1 μF, R = 1Ω High-pass filter Note: this is ω, V out ~  C R

26. LRC Circuits, quantitative L C   R  The phasor diagram gives us graphical solutions for  and I m : where...  ImRImR I m X L I m X C εmεm  εmεm ImRImR I m X L I m X C

27. Summary Reactances  ~ frequency-dependent resistance –Capacitors »look like wires for high frequencies »look like breaks for low frequencies »Voltage lags current by 90˚ –Inductors »look like breaks for high frequencies »look like wires for low frequencies »Current lags voltage by 90˚ –Filters (low-pass, high-pass, band-pass) LRC Circuit –Apply KVL using phasors ImRImR I m X L I m X C εmεm 

28. Ex. Source = y-component of the V-phasor Appendix: Phasors A phasor is a “vector” whose magnitude is the maximum value of a quantity (e.g., V ) and which rotates counterclockwise in a 2- d plane with angular velocity . Recall uniform circular motion: The projections of r (on the vertical y axis) execute sinusoidal oscillation.  x y y 1 2 4 ωt=0 V=0 ωt=90˚ V=εmV=εm ωt=270˚ V=-εmV=-εm ωt=45˚ 3

29. When the current through the circuit is maximum, what is the potential difference across the inductor? a) V L = 0 b) V L = V Lmax /2 c) V L = V Lmax When the capacitor is fully charged, what is the magnitude of the voltage across the inductor? a) V L = 0 b) V L = V Lmax /2 c) V L = V Lmax The phasor picture corresponds to a snapshot at some time t. The projections of the phasors along the vertical axis are the actual values of the voltages at the given time. One can draw the phasors at different times, simply by rotating the entire diagram. With this understanding, other questions can be easily answered…

30. Explanation: Since the current and V L are 90 degrees out of phase, when the current is at a maximum, V L will be at 0. When the capacitor is fully charged, the current through the circuit will be 0, and the magnitude of L will be at a maximum. V L will actually be at a minimum because V C and V L are 180 degrees out of phase. Current = max VRVR VLVL VCVC VRVR VLVL VCVC Charge on C = max