Copyright R. Janow – Spring 2016 Physics Electricity and Magnetism Lecture 14 - AC Circuits, Resonance Y&F Chapter 31, Sec 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 Examples Transformers

Copyright R. Janow – Spring 2016 Current/Voltage Phases in pure R, C, and L elements Sinusoidal current i (t) = I m cos(  D t) – peak is I m Voltage drop peaks in R, L, or C loads are 0,  /2, -  /2 radians out of phase Reactances (generalized resistances) are ratios of peak voltages to peak current V R & I m in phase Resistance V L leads I m by  /2 Inductive Reactance V C lags I m by  /2 Capacitive Reactance currrent Same Phase Phases of voltages in series components are referenced to the current phasor

Copyright R. Janow – Spring 2016 Phasors applied to a Series LCR circuit R L C E vRvR vCvC vLvL ImIm EmEm   D t+  DtDt Phasors all rotate CCW at frequency  D Lengths of phasors are the peak values (amplitudes) The “x” components are instantaneous values Apply Kirchhoff Loop rule to series LRC: Applied EMF: 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  Current: Same everywhere in the single essential branch ImIm EmEm  DtDt VLVL VCVC VRVR along I m perpendicular to I m Voltage phasors for R, L, & C all rotate at  D : V C lags I m by  /2 V R has same phase as I m V L leads I m by  /2 Voltage phasors add like vectors

Copyright R. Janow – Spring 2016 R ~ 0  I m normal to E m   ~ +/-  /2  tiny losses, no power absorbed X L =X C  I m parallel to E m    Z=R  maximum current (resonance)  measures the power absorbed by the circuit: Magnitude of E m in series circuit: Reactances: Same current amplitude in each component: Z ImIm EmEm  DtDt V L -V C VRVR X L -X C R For series LRC circuit, divide each voltage in | E m | by (same) peak current peak applied voltage peak current Magnitude of Z: Applies to a single series branch with L, C, R Phase angle  : See phasor diagram Impedance magnitude is the ratio of peak EMF to peak current: Voltage addition rule for series LRC circuit

Copyright R. Janow – Spring 2016 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

Copyright R. Janow – Spring 2016 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.

Copyright R. Janow – Spring 2016 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.

Copyright R. Janow – Spring 2016 Example 1: Analyzing a series RLC circuit - continued V L leads V R by  /2 V C lags V R by  /2 Add voltages above: What’s wrong? 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.

Copyright R. Janow – Spring HzCapacitive E m lags I m º 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

Copyright R. Janow – Spring 2016 Resonance in a series LCR circuit R L C E Vary  D : At resonance maximum current, minimum impedance 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

Copyright R. Janow – Spring 2016 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

Copyright R. Janow – Spring 2016 AC Power Dissipation in a Resistor The RMS power is an AC equivalent to DC power Integrate P inst in a resistor: Integral = 1/2 Instantaneous power 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 (  /2) RMS means “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

Copyright R. Janow – Spring 2016 Power factor for an AC LCR Circuit 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:

Copyright R. Janow – Spring 2016 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: Odd integrand Even integrand

Copyright R. Janow – Spring 2016 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  directly: Find the average power: or Recall: do not use arc-cos to find 

Copyright R. Janow – Spring 2016 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)

Copyright R. Janow – Spring 2016 Transformers Devices used to change AC voltages. They have: Primary Secondary Power ratings power transformer iron core circuit symbol

Copyright R. Janow – Spring 2016 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) or RMS averages, as can be the power and current.

Copyright R. Janow – Spring 2016 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:

Copyright R. Janow – Spring 2016