AC Circuits, Phasors, Forced Oscillations

1. Physics 121 - Electricity and Magnetism Lecture 14 - AC Circuits, ResonanceY&F Chapter 31, Sec. 1 - 8 • AC Circuits, Phasors, Forced Oscillations • Phase Relations for Current and Voltage in Simple Resistive, Capacitive, Inductive Circuits • Phasor Diagrams for Voltage and Current • The Series RLC Circuit. Amplitude and Phase Relations • Impedance and Phasors for Impedance • Resonance • Power in AC Circuits, Power Factor • Transformers

2. load the driving frequency of EMF STEADY STATE RESPONSE (after transients die away): . Current in load is sinusoidal, has same frequency wD as source ... but . Current may be retarded or advanced relative to E by “phase angle” F for the whole circuit(due to inertia L and stiffness 1/C). resonant frequency w0, in general R C L In a Series LCR Circuit: • Everything oscillates at driving frequency wD • Current is the same everywhere (including phase) • At “resonance”: • F is zero at resonance  circuit acts purely resistively. • Otherwise F can be positive (current lags applied EMF) • or negative (current leads applied EMF) Sinusoidal AC Source driving a circuit USE KIRCHHOFF LOOP & JUNCTION RULES, INSTANTANEOUS QUANTITIES Application of KIRCHHOFF RULES: . In a series branch, current has the same amplitude and phase everywhere. . Across parallel branches, voltage drops have the same amplitudes and phases . At nodes (junctions), instantaneous net currents in & out are conserved

3. Peak amplitudes  lengths of the phasors. • Measurable, real, instantaneous values of i( t ) and • E( t ) are projections of the phasors onto the x-axis. F “phase angle”, the angle from current phasor to EMF phasor in the driven circuit. wDt and are independent Real X • Current is the same (phase included) everywhere • in a single essential branch of any circuit • Current vector is reference for series LCR circuit. • Applied EMF Emax(t) leads or lags the current by a • phase angle F in the range [-p/2, +p/2] Relative phases for individual series circuit elements (to be shown): • Voltage across R is in phase with the current. • Voltage across C lags the current by 900. • Voltage across L leads the current by 900. VR VC VL XC>XL XL>XC represent currents and potentials as vectors rotating at frequency wD in complex plane Phasors Imaginary Y

4. Peak voltage and current amplitudes are just the coefficients out front • represented by lengths of rotating phasors • Simplistic time averaging of periodic quantities gives zero (and useless) results). • example: integrate over a whole number of periods – one t is enough (wt=2p) Integrand is odd over a full cycle The meaningful types of quantities in AC circuits can be instantaneous, peak, or average (RMS) • Instantaneous voltages and currents refer to a specific time: • oscillatory, depend on time through argument “wt” • possibly advanced or retarded relative to each other by phase angle • represented by x-component of rotating “phasors” – see slides below So how should “average” be defined?

5. “RMS averages” are used the way instantaneous quantities were in DC circuits: • “RMS” means “root, mean, squared”. NOTE: Integrand is positive on a whole cycle after squaring Prescription: RMS Value = Peak value / Sqrt(2) Definitions of Average AC Quantities • “Rectified average values” are useful for DC but seldom used in AC circuits: • Integrate |cos(wt)| over one full cycle or cos(wt) over a positive half cycle

6. Current/Voltage Phasing in pure R, C, and L circuit elements • Sinusoidal current i(t) = Imcos(wDt). Peak is Im • vR(t) ~ i(t) vL(t) ~ di(t)/dt • Peak voltage drops across R, L, or C lead/lag current by 0, p/2, -p/2 radians • Reactances (generalized resistances) are ratios of peak voltages to peak currents Phases of voltages in a series branch are referenced to the current phasor VR& Im in phase Resistance VC lags Im by p/2 Capacitive Reactance VL leads Im by p/2 Inductive Reactance Same Phase currrent

7. i(t) vR( t ) Applied EMF: From Kirchhoff loop rule: Time dependent current: The ratio of the AMPLITUDES (peaks) VRto Im is the resistance: The phases of vR(t) and i(t) coincide Peak current and peak voltage are in phase across a resistance, rotating at the driving frequency wD AC current i(t) and voltage vR(t) in a resistor are in phase From voltage drop across R:

8. i(t) Applied EMF: L vL(t) From Kirchhoff loop rule: Time dependent current: Note: Definition: inductive reactance The RATIO of the peaks (AMPLITUDES) VL to Im is the inductive reactance XL: • Inductive Reactance • limiting cases • w 0: Zero reactance. • Inductor acts like a wire. • w  infinity: Infinite reactance. • Inductor acts like a broken wire. AC current i(t) lags the voltage vL(t) by 900 in an inductor Sine from derivative From voltage drop across L (Faraday Law): so...phase angle F = + p/2 for inductor Voltage phasor leads current by +p/2 in inductive part of a circuit (F positive)

9. Applied EMF: From Kirchhoff loop rule: i vC ( t ) C Time dependent current: Note: Definition: capacitive reactance The RATIO of the AMPLITUDES VC to Im is the capacitive reactance XC: • Capacitive Reactance • limiting cases • w 0: Infinite reactance. DC blocked. C acts like broken wire. • w  infinity: Reactance is zero. Capacitor acts like a simple wire AC current i(t) leads the voltage vC(t) by 900 in a capacitor From voltage drop across C (proportional to Q(t)): Sine from integral so...phase angle F = - p/2 for capacitor Voltage phasor lags the current by p/2 in a pure capacitive circuit branch (F negative)

10. Applied EMF: vR E R Same current everywhere in the single branch L vL • Refer voltage phasors to current phasor • Same frequency dependance as E (t) • Same phase for the current in E, R, L, & C, but...... • Current leads or lags E (t) by a constant phase angle F C vC • Recall: • Phasors rotate CCW at frequency wD • Lengths of phasors are the peak values (amplitudes) • Instantaneous values are the “x” components F Im Em wDt+F wDt Apply Loop Rule to instantaneous voltages: VL Em Im • Voltage phasors for VR, VL, & VC all rotate at wD : • VClags Im by p/2 • VR has same phase as Im • VLleads Im by p/2 F wDt VR Voltage phasor magnitudes add like vectors VC along Im perpendicular to Im Phasors applied to a Series LCR circuit Current:

11. Voltage addition rule for series LRC circuit Magnitude of Em in series circuit: Em Im F VL-VC Z XL-XC VR wDt R Define: Impedance is the ratio of peak EMF to peak current. peak applied voltage peak current Divide each voltage in |Em| by (same) peak current: Applies to a single series branch with L, C, R Magnitude of Z: Phase angle F: See phasor diagram F measures peak power absorbed by the circuit: • R ~ 0  tiny losses, no power absorbed  Im normal to Em  F ~ +/- p/2 • XL=XC Im parallel to Em  F = 0  Z=R  maximum current (resonance) Reactances: Same current amplitude in each component:

12. vR Circuit Element Symbol Resistance or Reactance Phase of Current Phase Angle Amplitude Relation E R Resistor R R In phase with VR 0º (0 rad) VR = ImR L vL C Capacitor C XC=1/wdC Leads VC by 90º -90º (-p/2) VC = ImXC vC Inductor L XL=wdL Lags VL by 90º +90º (p/2) VL = ImXL Em Im F VL-VC Z VR XL-XC wDt R sketch shows XL > XC Summary: AC Series LCR Circuit

13. 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 εm= 150 V. Determine the impedance of the circuit. Find the amplitude of the current (peak value). Find the phase angle between the current and voltage. Find the instantaneous current across the RLC circuit. Find the peak and instantaneous voltages across each circuit element. Example 1: Analyzing a series RLC circuit

14. Determine the impedance of the circuit. Angular frequency: Resistance: Inductive reactance: Capacitive reactance: Current phasor Im leads the Voltage Em Phase angle will be negative XC > XL (Capacitive) (B) Find the peak current amplitude: • Find the phase angle between the current and voltage. 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 εm=150 V.

15. Add voltages above: What’s wrong? Voltages add with proper phases: Example 1: Analyzing a series RLC circuit - continued 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 εm=150 V. • Find the instantaneous current in the RLC circuit. (E) Find the peak and instantaneous voltages in each circuit element. VR in phase with Im VR leads Em by |F| = 0 VL leads VR by p/2 VC lags VR by p/2

16. Resonance in a series LCR Circuit: Example 2: Em = 100 V. R = 3000 W L = 0.33 H C = 0.10 mF vR Find |Z| and F for fD = 200 Hertz, fD = 876 Hz, & fD = 2000 Hz E R Why should fD make a difference? L vL C vC Frequency f Resistance R Reactance XC Reactance XL Impedance |Z| Phase Angle F Circuit Behavior 200 Hz 3000 W 7957 W 415 W 8118 W - 68.3º Capacitive Em lags Im 876 Hz 3000 W 1817 W 1817 W 3000 W Resonance 0º Resistive Max current 2000 Hz 3000 W 796 W 4147 W 4498 W +48.0º Inductive Em leads Im Em Im Im Em F < 0 F > 0 Im F=0 Em

17. E R L C width of resonance (selectivity, “Q”) depends on R. Large R  less selectivity, smaller current at peak inductance dominates current lags voltage capacitance dominates current leads voltage damped spring oscillator near resonance Resonance in a series LCR circuit Vary wD: At resonance maximum current, minimum impedance

18. Power in AC Circuits • Resistors always dissipate power, but the instantaneous rate varies as i2(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

19. Instantaneous power • Power is dissipated in R, not in L or C • cos2(x) is always positive, so Pinst is always • positive. But, it is not constant. • Power pattern repeats every p radians (t/2) The RMS power is an AC equivalent to DC power Integrate Pinst in resistor over t: Integral = 1/2 • RMS means “Root Mean Square” • Square a quantity (positive) • Average over a whole cycle • Compute square root. • COMPUTING RMS QUANTITIES: • For any RMS quantity divide peak • value such as Im or Em by sqrt(2) For any R, L, or C Household power example: 120 volts RMS  170 volts peak AC Power Dissipation in a Resistor

20. Power factor for an AC LCR Circuit The PHASE ANGLEF determines the average RMS power actually absorbed due to the RMS current and applied voltage in the circuit. Claim (proven below): Erms Irms F |Z| wDt XL-XC R Proof: Start with instantaneous power (not very useful): Average it over one full period t: Change variables: Use trig identity:

21. Power factor for AC Circuits - continued Odd integrand Even integrand Recall: RMS values = Peak values divided by sqrt(2) Also note: Alternate form: If R=0 (pure LC circuit)F +/- p/2 and Pav = Prms = 0

22. VR E R L VL C Find Erms: Find Irms at 200 Hz: VC Find the power factor: Find the phase angle F directly: Find the average power: or Example 2 continued with RMS quantities: Em = 100 V. R = 3000 W L = 0.33 H C = 0.10 mF fD = 200 Hz Recall: do not use arc-cos to find F

23. Example 3 – Use RMS values inSeries LCR circuit analysis F is positive since XL>XC (inductive) 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 mF 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?

24. Transformers power transformer • Devices used to change • AC voltages. They have: • Primary winding • Secondary winding • Power ratings iron core circuit symbol

25. Ideal Transformer Assume: The same amount of flux FB cuts each turn in both primary and secondary windings in ideal transformer (counting self- and mutual-induction) iron core • zero resistance in coils • no hysteresis losses in iron core • all field lines are inside core Assuming no losses: energy and power are conserved Turns ratio fixes the step up or step down voltage ratio Vp, Vs are instantaneous (time varying) or RMS averages, as can be the power and current. Transformers Assume zero internal resistances, EMFs Ep, Es = terminal voltages Vp, Vs Faradays Law for primary and secondary:

26. Example: A dimmer for lights using a variable inductance Light bulb R=50 W Erms=30 V L b) What would be the change in the RMS current? P0,rms = 18 W. Without inductor: Prms = 5 W. With inductor: f =60 Hz w = 377 rad/sec Without Inductor: a) What value of the inductance would dim the lights to 5 Watts? Recall: