The Series RLC Circuit. Amplitude and Phase Relations Phasor Diagrams for Voltage and Current

1. Physics 121 - Electricity and Magnetism Lecture 14 - AC Circuits, ResonanceY&F Chapter 31, Sec. 3 - 8 • The Series RLC Circuit. Amplitude and Phase Relations • Phasor Diagrams for Voltage and Current • Impedance and Phasors for Impedance • Resonance • Power in AC Circuits, Power Factor • Examples • Transformers • Summaries

2. Current & voltage phases in pure R, C, and L circuits VR& Im in phase Resistance VC lags Im by p/2 Capacitive Reactance VL leads Im by p/2 Inductive Reactance Same Phase currrent Current is the same everywhere in a single branch (including phase). Phases of voltages in series components are referenced to the current phasor • Apply sinusoidal current i(t) = Imcos(wDt) • For pure R, L, or C loads, phase angles for voltage drops are 0, p/2, -p/2 • Reactance” means ratio of peak voltage to peak current (generalized resistances).

3. vR E R • The current i(t) is the same everywhere in the circuit • Same frequency dependance as E (t) but...... • Current leads or lags E (t) by a constant phase angle F • Same phase for the current in E, R, L, & C L vL C vC • Phasors all rotate CCW at frequency wD • Lengths of phasors are the peak values (amplitudes) • The “x” components are the measured values. F Im Em wDt+F wDt VL Em • Plot component voltage phasors rotating at wD with phases • relative to the current phasor Im and magnitudes below : • VR has same phase as Im • VClags Im by p/2 • VLleads Im by p/2 Im F wDt Kirchoff Loop rule for potentials (measured along x) VR VC along Im perpendicular to Im Series LCR circuit driven by an external AC voltage Apply EMF: wD is the driving frequency

4. Em Magnitude of Em in series circuit: Im F VL-VC Z VR XL-XC peak applied voltage wDt R peak current For series LRC circuit, 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 diagram F measures the 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) The impedance is the ratio of peak EMF to peak current Reactances: Same current amplitude

5. vR Circuit Element Symbol Resistance or Reactance Phase of Current Phase Constant 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:

6. 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

7. Determine the impedance of the circuit. Angular frequency: Resistance: Inductive reactance: Capacitive reactance: Current phasor Im leads the Voltage Em Phase angle should 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.

8. Note that: Why not? 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 across the RLC circuit. (E) Find the peak and instantaneous voltages across each circuit element. VR in phase with Im VR leads Em by |F| VL leads VR by p/2 VC lags VR by p/2

9. Example 2: Resonance in a series LCR Circuit: 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

10. 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 Vary wD: At resonance maximum current flows & impedance is minimized

11. 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 • In a capacitor, during two segments (1/4 cycle) energy is stored and during the other two segments energy is returned to the circuit • In an inductor, the source does work against the back emf of the inductor and energy is stored in the inductor. When the current in the circuit begins to decrease, the energy is returned to the circuit

12. instantaneous power consumed by circuit • Pav 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 Im or Emsince • cos2(x) appears any component Household power example: 120 volts RMS  170 volts peak Instantaneous and RMS (average) 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. • Pattern for power repeats every p radians (T/2) The RMS power, current, voltage are useful, DC-like quantities Integrate: The integral = 1/2

13. Power factor for AC Circuits Erms Irms F Z wDt XL-XC R Proof: start with instantaneous power (not very useful): The PHASE ANGLEF determines the average RMS power actually absorbed by the RMS current and applied voltage in the circuit. We Claim (prove below): Average it over one full period t: Change variables: Use trig identity:

14. Power factor for AC Circuits - continued 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

15. VR E R L VL C Find Erms: Find Irms at 200 Hz: VC Find the power factor: Find the phase angle F: 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

16. Example 3 – LCR circuit analysis using RMS values 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?

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

18. Ideal Transformer Assume: The same flux FB cuts each turn in both primary and secondary windings of an 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 Assume no losses: energy and power are conserved Turns ratio fixes the step up or step down voltage ratio Vp, Vs are instantaneous (time varying) but can also be regarded as 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:

19. Example: A dimmer for lights using a variable inductance Light R=50 W Erms=30 V L b) What is the change in the RMS current? Prms = 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: