1. Copyright R. Janow – Fall 2015 Physics 121 - Electricity and Magnetism Lecture 14-15_E - AC Circuits, Resonance Y&F Chapter 31, Sec. 3 – 8, No Y&F reference for slides on Complex analysis The Series RLC Circuit. Amplitude and Phase Relations –Time domain, Real functions Phasor Diagrams for Voltage and Current Impedance Magnitude and Phasors for Impedance Resonance Power in AC Circuits, Power Factor Examples Transformers Series LCR Circuit Summary Circuit Analysis using Complex Exponentials –Imaginary and Complex Numbers –Complex Exponential Phasors and Rotations –Phasors as Solutions of Steady State Oscillator Equations –Phasor representation applied to current versus voltage in R. L. C. –Series LCR Circuit using Complex Phasors –Parallel LCR Circuit using Complex Phasors –Transient Solution of Damped Oscillator using Complex Phasors

2. Copyright R. Janow – Fall 2015 Summary: AC Series LCR Circuit V L = I m X L +90º (  /2) Lags V L by 90º XL=dLXL=dL LInductor V C = I m X C -90º (-  /2) Leads V C by 90º X C =1/  d C CCapacitor V R = I m R0º (0 rad)In phase with V R RRResistor Amplitude Relation Phase Angle Phase of Current Resistance or Reactance SymbolCircuit Element R L C E vRvR vCvC vLvL Z ImIm EmEm  DtDt V L -V C VRVR X L -X C R sketch shows X L > X C

3. Copyright R. Janow – Fall 2015 Example 1: Analyzing a series RLC circuit A series RLC circuit has R = 425 Ω, L = 1.25 H, C = 3.50 μF. It is connected to an AC source with f = 60.0 Hz and ε m = 150 V. (A)Determine the impedance of the circuit. (B)Find the amplitude of the current (peak value). (C)Find the phase angle between the current and voltage. (D)Find the instantaneous current across the RLC circuit. (E)Find the peak and instantaneous voltages across each circuit element.

4. Copyright R. Janow – Fall 2015 A series RLC circuit has R = 425 Ω, L = 1.25 H, C = 3.50 μF. It is connected to an AC source with f = 60.0 Hz and ε m =150 V. Example 1: Analyzing a Series RLC circuit (A)Determine the impedance of the circuit. Angular frequency: Resistance: Inductive reactance: Capacitive reactance: (B) Find the peak current amplitude: Current phasor I m leads the Voltage E m Phase angle should be negative X C > X L (Capacitive) (C)Find the phase angle between the current and voltage.

5. Copyright R. Janow – Fall 2015 Example 1: Analyzing a series RLC circuit - continued V L leads V R by  /2 V C lags V R by  /2 Note that: Why not? Voltages add with proper phases: V R in phase with I m V R leads E m by |  | (E) Find the peak and instantaneous voltages across each circuit element. A series RLC circuit has R = 425 Ω, L = 1.25 H, C = 3.50 μF. It is connected to an AC source with f = 60.0 Hz and ε m =150 V. (D)Find the instantaneous current across the RLC circuit.

6. Copyright R. Janow – Fall 2015 200 HzCapacitive E m lags I m - 68.3º 8118  415  7957  3000  Frequency f Circuit Behavior Phase Angle  Impedance |Z| Reactance X L Reactance X C Resistance R 876 HzResistive Max current 0º 3000  Resonance 1817  3000  2000 HzInductive E m leads I m +48.0º 4498  4147  796  3000  R L C E vRvR vCvC vLvL ImIm EmEm  ImIm EmEm  ImIm EmEm  Why should f D make a difference? Example 2: Resonance in a series LCR Circuit: R = 3000  L = 0.33 HC = 0.10 mF E m = 100 V. Find Z and  for f D = 200 Hertz, f D = 876 Hz, & f D = 2000 Hz

7. Copyright R. Janow – Fall 2015 Resonance in a series LCR circuit R L C E Vary  D : At resonance maximum current flows & impedance is minimized inductance dominates current lags voltage capacitance dominates current leads voltage width of resonance (selectivity, “Q”) depends on R. Large R  less selectivity, smaller current at peak damped spring oscillator near resonance

8. Copyright R. Janow – Fall 2015 Power in AC Circuits Resistors always dissipate power, but the instantaneous rate varies as i 2 (t)R No power is lost in pure capacitors and pure inductors in an AC circuit –Capacitor stores energy during two 1/4 cycle segments. During two other segments energy is returned to the circuit –Inductor stores energy when it produces opposition to current growth during two ¼ cycle segments (the source does work). When the current in the circuit begins to decrease, the energy is returned to the circuit

9. Copyright R. Janow – Fall 2015 Instantaneous and RMS (average) power Instantaneous power consumption Power is dissipated in R, not in L or C Cos 2 (x) is always positive, so P inst is always positive. But, it is not constant. Pattern for power repeats every  radians (T/2) The RMS power, current, voltage are useful, DC-like quantities Integrate: P av is an “RMS” quantity: “Root Mean Square” Square a quantity (positive) Average over a whole cycle Compute square root. In practice, divide peak value by sqrt(2) for I m or E m since cos 2 (x) appears For any R, L, or C Household power example: 120 volts RMS  170 volts peak Integral = 1/2 cos 2 (  )

10. Copyright R. Janow – Fall 2015 Power factor for AC Circuits The PHASE ANGLE  determines the average RMS power actually absorbed due to the RMS current and applied voltage in the circuit. Claim (proven below): |Z| I rms E rms  DtDt X L -X C R Proof: start with instantaneous power (not very useful): Change variables: Average it over one full period  : Use trig identity:

11. Copyright R. Janow – Fall 2015 Power factor for AC Circuits - continued Recall: RMS values = Peak values divided by sqrt(2) Alternate form: If R=0 (pure LC circuit)    +/-  /2 and P av = P rms = 0 Also note:

12. Copyright R. Janow – Fall 2015 f D = 200 Hz Example 2 continued with RMS quantities: R = 3000  L = 0.33 HC = 0.10 mF E m = 100 V. R L C E VRVR VCVC VLVL Find E rms : Find I rms at 200 Hz: Find the power factor: Find the phase angle  : Find the average power: or Recall: do not use arc-cos to find 

13. Copyright R. Janow – Fall 2015 A 240 V (RMS), 60 Hz voltage source is applied to a series LCR circuit consisting of a 50- ohm resistor, a 0.5 H inductor and a 20  F capacitor. Find the capacitive reactance of the circuit: Find the inductive reactance of the circuit: The impedance of the circuit is: The phase angle for the circuit is: The RMS current in the circuit is: The average power consumed in this circuit is: If the inductance could be changed to maximize the current through the circuit, what would the new inductance L’ be? How much RMS current would flow in that case? Example 3 – Series LCR circuit analysis using RMS values  is positive since X L >X C (inductive)

14. Copyright R. Janow – Fall 2015 Transformers Devices used to change AC voltages. They have: Primary Secondary Power ratings power transformer iron core circuit symbol

15. Copyright R. Janow – Fall 2015 Transformers Faradays Law for primary and secondary: Ideal Transformer iron core zero resistance in coils no hysteresis losses in iron core all field lines are inside core Assume zero internal resistances, EMFs E p, E s = terminal voltages V p, V s Assume: The same amount of flux  B cuts each turn in both primary and secondary windings in ideal transformer (counting self- and mutual-induction) Assuming no losses: energy and power are conserved Turns ratio fixes the step up or step down voltage ratio V p, V s are instantaneous (time varying) but can also be regarded as RMS averages, as can be the power and current.

16. Copyright R. Janow – Fall 2015 Example: A dimmer for lights using a variable inductance f =60 Hz  = 377 rad/sec Light bulb R=50  E rms =30 V L Without Inductor: a) What value of the inductance would dim the lights to 5 Watts? b) What would be the change in the RMS current? Without inductor: P 0,rms = 18 W. With inductor: P rms = 5 W. Recall:

17. Copyright R. Janow – Fall 2015 Circuit Analysis using Complex Exponentials Roots of a quadratic equation az 2 – bz +c = 0 are complex if b 2 <4ac: is “pure imaginary” EE’s use j instead of i (i is for current) Complex conjugate: Re{z} and Im{z} are both real numbers Complex number in rectangular form: Addition:

18. Copyright R. Janow – Fall 2015 Representation using the Complex Plane Imaginary axis (y) Real axis (x) x y |z|  Magnitude 2 : Polar form: Argument: “Argand diagram” Picture also displays complex-valued functions

19. Copyright R. Janow – Fall 2015 Complex Exponentials: Euler’s Formula Taylor’s Series definition of exponential function: Evaluate for u = i  : Recognize series definitions of sine and cosine: Hence: Above is a complex number of magnitude 1, argument  : use i 2n+1 = i(-1) 2n Special Properties:

20. Copyright R. Janow – Fall 2015 Complex Quantities in Polar & Rectangular Form Multiply Euler’s Formula by magnitude r : Product: Magnitude of a product = product of individual magnitudes Argument of a product = sum of individual arguments Two complex entities are equal if and only if their amplitudes (magnitudes) are equal and their arguments (phases) are equal DeMoivre’s Theorem: Periodicity 2  : for integer n & k = 0,1….n-1 Sum: Polar form of arbitrary complex z Complex conjugate of z

21. Copyright R. Janow – Fall 2015 Rotation Operators, Evolution Operators, Phasors: Exponential factor e i  rotates complex number z by  in complex plane Let  =  t-t 0 ) A(t)  is often called the “time domain phasor” corresponding to A 0 cos(  t+  ) Re {A} = A 0 cos(  t+  ) is the measurable instantaneous value of A(t) Time dependent exponentials like e i  t are sometimes called evolution operators The corresponding “frequency domain phasor” A (often simply called a “phasor” by EE’s) is the complex quantity: evolves z from t 0 to t for t f > t 0 Im Re    Let a(t) be a sinusoidally varying real quantity to be pictured as a vector in the complex plane rotating (counterclockwise) at angular frequency 

22. Copyright R. Janow – Fall 2015 Complex exponentials (time domain phasors) are solutions of oscillator differential equations Derivatives:  can be complex chain rule Complex exponentials are alternative solutions to sines or cosines in circuit differential equations describing steady state oscillations. Example: Simple Harmonic Oscillator Second derivatives:set  =  t+  amplitude A 0 Solutions:

23. Copyright R. Janow – Fall 2015 Phasor Definitions, continued A phasor represents a sinusoidal, steady state signal whose amplitude A 0, phase  and frequency  are time invariant. peak value of a(t), real frequency domain phasor A, complex time domain phasor, complex sinusoidal value, time domain, real Phasor Transform P: forward - time to frequency domain inverse - frequency to time domain Advantage of phasor transform: For signals as above (sinusoidal, no transients), replace differentiation and integration in time domain differential equations by simple algebraic operations on phasors in frequency domain ( complex coefficients, common factors of e i  t cancel). Process: Replace real variables in time dependent analysis problem with variables written using complex exponentials. Cancel common factors of e i  t and solve remaining algebraic problem. Then return to real solution in time domain.

24. Copyright R. Janow – Fall 2015 Complex Exponential Representation applied to Passive Circuit Elements: Revisit AC Voltage vs. Current in L, C, and R Note: now using j = sqrt(-1) The applied voltage & current as time domain phasors: v R ( t ) i(t) equate magnitudes: equate arguments: Resistor: v L (t) L i(t) Inductor: equate magnitudes: equate arguments: i(t) v C ( t ) C Capacitor: equate magnitudes: equate arguments:

25. Copyright R. Janow – Fall 2015 Revisit AC Voltage vs. Current in L, C, and R using Complex Exponential Representation - Continued & current:Time domain applied voltage: For resistor:For inductor:For capacitor: V R & I Rm in phase Resistance V L leads I Lm by  /2 Inductive Reactance V C lags I Cm by  /2 Capacitive Reactance Individual passive circuit elements: e0e0 e +j  /2 e -j  /2 Relative phasor rotations of voltage drops relative to currents: Voltage drop phasors (time domain) relative to currents: Define (complex) impedance by z = v(t)/i(t) and |z| = V max /I max Impedances of simple circuit elements: Magnitudes of Impedances |z| = [zz*] 1/2

26. Copyright R. Janow – Fall 2015 Revisiting the series LCR circuit using phasors R L C E vRvR vCvC vLvL Kirchhoff Loop rule for series LRC: AC voltage: Current: Use preceding formulas for v R, v L, v C. Currents are the same everywhere in an essential branch (series LCR): Divide equation (1) above by I m and cancel common factor of e j  t : ImIm EmEm  DtDt VLVL VCVC VRVR time domain phasors General definition: The complex impedance (or simply impedance) is the ratio of the voltage phasor to the current phasor (time or frequency domain). Magnitude of Z: Sketch shows  > 0 for V L >V C  X L >X C At t = 0 sketch would show Phasors in frequency domain

27. Copyright R. Janow – Fall 2015 Revisiting the series LCR circuit, continued #1 Phase angle  for the circuit: Impedance z above (Equation 1) is also the sum of the impedances of the 3 series elements Impedance for Series LCR circuit section: Z  X L -X C R Re Im Sketch shows  >0 for X L > X C  V L > V C Z is complex but not a phasor, as it does not represent rotation; impedance is time independent. Previous “phasor diagrams” may have shown z rotating with E m. Generalize: the equivalent impedance for basic circuit elements in series (arbitrarily many) is the sum of the individual (complex) impedances. Resonance: As before, z is real for X L = X C (  2 = 1/LC ). |z| is minimized. Current amplitude I m is maximized at resonance

28. Copyright R. Janow – Fall 2015 Parallel LCR circuit using complex phasors Kirchhoff Rules Application (time domain): AC voltage : Instantaneous Current: All steady state voltages and currents oscillate at driving frequency  D vCvC vLvL R L C E vRvR iRiR iLiL iCiC i a b Instantaneous voltages across parallel branches have the same magnitude and phase 2 essential nodes “a” & “b” 4 essential branches, not all independent Current Rule: Loop Rule: E -R- E E -L- E E -C- E Current Amplitudes within the Branches:

29. Copyright R. Janow – Fall 2015 Parallel LCR circuit, continued #1 Instantaneous currents in branches, phases relative to voltages: in phase with E (t) lags E (t) by  /2 leads E (t) by  /2 I Rm ImIm E m, V R, V L, V C  DtDt I Cm -I Lm Re Im Sketch shows  >0 for I Lm > I Cm  X C >X L Substitute currents into junction rule equation: In current equation (2), cancel e j  t factor and multiply by e -j  Current amplitude addition rule (Pythagorean) As before, the magnitude of the complex impedance z is the ratio of peak EMF to peak current and the impedance

30. Copyright R. Janow – Fall 2015 Parallel LCR circuit, continued #2 Substitute, then divide above by E m Note that: Find 1/|z|: multiply 1/z by complex conjugate (1/z)* and take square root Phase angle  : As before,  is positive when applied voltage E m leads the current I m  > 0 for X C > X L Sketch shows  0 for 1/X C > 1/X L  I C > I L  X L > X C 1/z  1/X C -1/X L 1/R E ImIm

31. Copyright R. Janow – Fall 2015 Parallel LCR circuit, continued #3 Resonance: Maximize or minimize as function of frequency Minimum of I m when X L = X C, i.e. when  2 = 1/LC Same frequency as series LCR, but current is minimized At resonance, the current amplitudes I Lm and I Cm in the L & C branches are equal, but 180 o apart in phase. These cancel at all times at nodes a and b. Generalize: the equivalent impedance for circuit elements or branches in parallel (arbitrarily many) is the sum of the reciprocals of the individual (complex) branch impedances.

32. Copyright R. Janow – Fall 2015 Example: Use Complex Exponentials to solve a Differential Equation a L C E + b R Revisit Damped Oscillator: After “step response” to switch at ‘a’ saturates, turn switch to ‘b’. Decaying natural oscillations start for weak damping dissipation of potential energy not a steady state system solutions not simple sinusoids second order equation for Q(t) Try solution… … but if  s real solution oscillates forever Derivatives: Substitute into Equation (1): Cancel common factors of Q: “characteristic equation” Phasor-like trial solution (2) turned differential equation into polynomial equation  x and  y assumed real Assume complex frequency

33. Copyright R. Janow – Fall 2015 Use Complex Exponentials to solve Differential Equations #2 Equation (3) becomes 2 separate equations for real and imaginary terms Expand Re{Eq 3}  Substitute w y : Im{Eq 3}  Shifted natural frequency  x is real for underdamping imaginary for overdamping For real  x, underdamping: dampingoscillation For critical damping,  x = 0: No oscillations, decay only :

34. Copyright R. Janow – Fall 2015

35. Use Complex Exponentials to solve Differential Equations #3 For overdamping:  x above imaginary. Equations 4.1 & 4.2 invalid Return to Eq. 3.0, assume frequency is pure imaginary (  no oscillation) Eq. 3.0 becomes: quadratic, real coefficients Solution: two roots both pure imaginary Both roots lead to decay w/o oscillation + root:  + implies damping faster than e -rt/2L - root:  - implies damping slower than e -rt/2L but not growth Most general solution: linear combination of Q + and Q -, each of form of Eq. 2.0 Recall definition, hyperbolic cosine: Correctly reduces to critical damped Eq. 8.0