1. Chapter 28: Alternating Current Phasors and Alternating Currents  Alternating current (AC current) Current which varies sinusoidally in time is called alternating current (AC) as opposed to direct current (DC). One example of AC current source is a coil of wire rotating with constant angular velocity in a magnetic field. The symbol~is used to denote an AC source. In general a source means either a source of alternating current or voltage. In the U.S. and Canada, commercial electric-power distribution system uses a frequency of f = 60 Hz, corresponding to  = 377 rad/s. In much of the rest of the world uses f = 50 Hz. In Japan, however, the country is divided in two regions with f = 50 Hz and 60 Hz.

2. Phasors and Alternating Currents  Phasors O tt IPIP  I=I P sin  t A convenient way to express a quantity varying sinusoidally with time is by a phasor in phasor diagram as shown. phasor  Rectifier and rectified current +- -+

3. Phasors and Alternating Currents  Rectifier and rectified current (cont’d)

4. Phasors and Alternating Currents  Root-mean-square current and voltage Root-mean-square current of a sinusoidal current time averaged Root-mean-square voltage of a sinusoidal voltage For 120-volt AC, V=170 V.

5. Reluctance  Resistance, inductance, capacitance and reactance Resistor in an AC circuit   I R R  0 0 t I R 0 0 V R t Voltage across R in phase with current through R I VRVR RR IRIR tt Given: At time t I=  m /R

6.  Resistance, inductance, capacitance and reactance Inductor in an AC circuit   Voltage across L leads current through L by one-quarter cycle (90°). I L   L t 0 0 VLVL t 0 0 ILIL I VLVL ILIL tt Given: Reluctance At time t I=  m /(  L) mm 

7.  Resistance, inductance, capacitance and reactance Capacitor in an AC circuit    C I C  Voltage across C lags current through C by one-quarter cycle (90°). 0 0 t VCVC t 0 0 ICIC I VCVC mm ICIC tt Given: Reluctance At time t tt I=  C  m 

8.  LRC series circuit and reluctance XCXC XLXL reactance Given: Assume the solution for current: LRC circuit summary amplitude Reluctance (See derivation later)

9. 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) You can think of it as a frequency-dependent resistance. What is reactance?  LRC series circuit and reluctance (cont’d) Reluctance f 

10. L C   R     I m R I m  L I m  C  m  Assume: Given:  This picture corresponds to a snapshot at t=0. The projections of these phasors along the vertical axis are the actual values of the voltages at the given time.  LRC series circuit (cont’d) amplitude LRC Circuits ImIm

11. Problem: Given V drive = ε m sin ωt, find V R, V L, V C, I R, I L, I C Strategy: 1. Draw V drive phasor at t=0 2. Guess i R phasor 3. Since V R = i R R, this is also the direction for the V R phasor. -φ-φ 4. Realize that due to Kirchhoff’s current law, i L = i C = i R (i.e., the same current flows through each). (No L or C → f = 0) L C   R  LRC series circuit (cont’d)   LRC Circuits

12. -φ-φ V R = I R V L = I X L V C = I X C 5.The inductor current I L always lags V L  draw V L 90˚ further counterclockwise. 6.The capacitor voltage V C always lags I C  draw V C 90˚ further clockwise. 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.  is determined such that V R + V L + V C = ε (Kirchhoff’s voltage rule) These are added like vectors.  LRC series circuit (cont’d)  LRC Circuits

13.  Phasor diagrams for LRC circuits: Example y x ε VCVC IR ~ V out y x ε LRC Circuits amplitude of current

14.  Filters : Example Ex.: C = 1 μF, R = 1Ω High-pass filter Note: this is ω, ~ V out LRC Circuits

15.  Filters ~ V out ~ 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 = 0No current because of capacitor ω = ∞ No current because of inductor (Conceptual sketch only) High- pass filter Low- pass filter Band-pass filter LRC Circuits

16.  Phasor diagrams for LRC circuits: Example 2  I m R  m  I m (X(X L -X-X C )    I m R  m I m X C I m X L  Impedance Z amplitude LRC Circuits Reluctance for inductor Reluctance for capacitor

17. LRC Circuits  Phasor diagrams for LRC circuits: Tips This phasor diagram was drawn as a snapshot of time t=0 with the voltages being given as the projections along the y-axis. From this diagram, we can also create a triangle which allows us to calculate the impedance Z: f ImRImR ImXLImXL ImXCImXC mm “Full Phasor Diagram” Sometimes, in working problems, it is easier to draw the diagram at a time when the current is along the x-axis (when I=0). I m R mm I m X C I m X L y x f f f “Impedance Triangle” f | |     

18. Resonance in Alternating Current Circuits  Resonance For fixed R, C, L the current I m will be a maximum at the resonant frequency w 0 which makes the impedance Z purely resistive. Note that this resonant frequency is identical to the natural frequency of the LC circuit by itself! At this frequency, the current and the driving voltage are in phase! i.e.: reaches a maximum when: XX C = L This condition is obtained when:  L C ~  R resonance frequency

19. Resonance in Alternating Current Circuits  Resonance (cont’d) ImIm 0 0 oo  0 R m  R=RoR=Ro R=2R o R XLXL XCXC  Z X L - X C Plot the current versus , the frequency of the voltage source: → cos  R I m m  

20. Resonance in Alternating Current Circuits  Resonance (cont’d) On Resonance: L C ~  R On resonance, the voltage across the reactive elements is amplified by Q! Necessary to pick up weak radio signals, cell phone transmissions, etc.  and Z=R

21. Power in Alternating Current Circuits  Power The instantaneous power (for some frequency, w) delivered at time t is given by: The most useful quantity to consider here is not the instantaneous power but rather the average power delivered in a cycle. To evaluate the average on the right, we first expand the sin(  t-  ) term.

22. Power in Alternating Current Circuits  Power sin  tcos  t  t 0  0 +1 (Product of even and odd function = 0) sin 2  t  t 0  0 +1 0 1/2 Expanding, Taking the averages, Generally: Putting it all back together again,

23. Power in Alternating Current Circuits  Power Power delivered depends on the phase, f, the “power factor” Phase depends on the values of L, C, R, and  Therefore... This result is often rewritten in terms of rms values:

24. Power in Alternating Current Circuits  Power We can write this in the following manner (which we won’t try to prove): …introducing the curious factors Q and x... Power, as well as current, peaks at  =  0. The sharpness of the resonance depends on the values of the components. Recall:

25. Resonance in Alternating Current Circuits  Power and resonance where U max is max energy stored in the system and  U is the energy dissipated in one cycle A parameter “Q” is often defined to describe the sharpness of resonance peaks in both mechanical and electrical oscillating systems. “Q” is defined as For RLC circuit, U max is Losses only come from R: This gives And for completeness, note period

26. Resonance in Alternating Current Circuits  Power and resonance 0 0 oo  0 2 R rms  R=RoR=Ro R=2R o Q=3 FWHM For Q > few, FWHM Full Width at Half Maximum Q Quality of the peak Higher Q = sharper peak = better quality

27. Transformers  Transformers AC voltages can be stepped up or stepped down by the use of transformers. The AC current in the primary circuit creates a time-varying magnetic field in the iron The iron is used to maximize the mutual inductance. We assume that the entire flux produced by each turn of the primary is trapped in the iron. 2 1 (primary) (secondary) ~  N N iron V2V2 V1V1 This induces an emf on the secondary windings due to the mutual inductance of the two sets of coils.

28. Transformers  Ideal transformer without a load No resistance losses All flux contained in iron Nothing connected on secondary   N 2 N 1 (primary) (secondary) iron V 2 V 1 The primary circuit is just an AC voltage source in series with an inductor. The change in flux produced in each turn is given by: The change in flux per turn in the secondary coil is the same as the change in flux per turn in the primary coil (ideal case). The induced voltage appearing across the secondary coil is given by: Therefore, N 2 > N 1  secondary V 2 is larger than primary V 1 (step-up) N 1 > N 2  secondary V 2 is smaller than primary V 1 (step-down) Note: “no load” means no current in secondary. The primary current, termed “the magnetizing current” is small!

29. Transformers  Ideal transformer with a load   N 2 N 1 (primary) (secondary) iron V 2 V 1 R What happens when we connect a resistive load to the secondary coil? Changing flux produced by primary coil induces an emf in secondary which produces current I 2 This current produces a flux in the secondary coil µ N 2 I 2, which opposes the change in the original flux -- Lenz’s law This induced changing flux appears in the primary circuit as well; the sense of it is to reduce the emf in the primary, to “fight” the voltage source. However, V 1 is assumed to be a voltage source. Therefore, there must be an increased current I 1 (supplied by the voltage source) in the primary which produces a flux µ N 1 I 1 which exactly cancels the flux produced by I 2.

30. Transformers  Ideal transformer with a load (cont’d) Power is dissipated only in the load resistor R.   N 2 N 1 (primary) (secondary) iron V 2 V 1 R = = The primary circuit has to drive the resistance R of the secondary. Where did this power come from? It could come only from the voltage source in the primary:

31. Exercises  Exercise 1 Suppose  m = 100 volts, f=1000 Hz, R=10 Ohms, L=4.22 mH, Find X L, Z, I, V R, and V l.

32. Exercises  Exercise 2: Calculate power lost in R in Exercise 1 To calculate power produced by the generator you need to take account of the phase difference between the voltage and the current. In general you can write: For an inductor P = 0 because the phase difference between current through the inductor and voltage across the inductor is 90 degrees