1. Copyright R. Janow – Fall 2015 Physics 121 - Electricity and Magnetism Lecture 14E - AC Circuits & Resonance I – Series LCR Y&F Chapter 31, Sec. 3 – 8 The Series RLC Circuit. Amplitude and Phase Relations –Time domain, Real functions Phasor Diagrams for Voltage and Current Impedance Magnitude and Phasors for Impedance Examples Resonance Power in AC Circuits, Power Factor Transformers Series LCR Circuit Summary

2. Copyright R. Janow – Fall 2015 Summary: AC Series LCR Circuit Z ImIm EmEm  DtDt V L -V C VRVR X L -X C R sketch shows X L > X C R L C E vRvR vCvC vLvL 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

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 will 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 Power Dissipation in a Resistor Instantaneous power dissipation 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. Power pattern repeats every  radians (T/2) P av is an “RMS” quantity: “Root Mean Square” Square a quantity (positive) Average over a whole cycle Compute square root. For other rms quantities, divide peak value such as I m or E m by sqrt(2) For any R, L, or C Household power example: 120 volts RMS  170 volts peak The RMS power, current, voltage are useful, DC-like quantities Integrate P inst : 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: