Lecture 12 Electromagnetic Oscillations and Alternating Current Chp. 33PowerPoint PresentationSlide 4Impedance Z: New quantity for AC circuits. This is analogous to resistance in DC circuitsRL Circuit ExamplePower in AC circuitsAveraging over a sine curveCalculate Power lost in resistor from exampleSeries LRC circuitResonanceSlide 12Series LCR circuitLecture 12 Electromagnetic Oscillations and Alternating Current Chp. 33•Cartoon -. Opening Demo - Warm-up problem•Physlet •Topics–LC Circuit Qualitatively –Electrical and Magnetic energy oscillations–Alternating current–\Pure R and L, circuti–Series RLC circuit–Power and Transfomers•Demos–LR circuit–Series LRC circuitdAnBmˆ⋅=•rφAdBrr⋅=dAB θcos=nˆB€ ε =−dΦdt= −d(BAcosθ)dt= −BAd cosθdt= BA sinθdθdt= BAω sinθ but θ = ωt so dθdt= ω€ ε =BAω sinωt€ ε =εmsinωtWhere is the rotational angular frequency of the generator f and f= 60 HzAxis of rotationCoil of wire€ ε =εmsin(ωt − φ)Instantaneous voltageAmplitudeAngular frequencytimePhase constatphase€ εm€ ε€ ωt − φPhasor diagram€ ε =εmsinωt€ i = I sin(ωt)€ vL= Ldidt= LωI cos(ωt)€ didt= ωI cos(ωt)€ vL= LωI cos(ωt)€ vR= Ri€ R =VRI€ XL=VLI= LωVR=RIVL= XLIL or VL= (LI since I=IL L = 4.22mH€ ω=2πf€ f =1000Hz€ ε =vR+ vLImpedance Z: New quantity for AC circuits. This is analogous to resistance in DC circuits€ Z = R2+ (ωL)2€ I =εmZ€ I =εmR2+ (ωL)2€ XL= ωLRL Circuit ExampleSuppose m = 100 volts, f=1000 Hz, R=10 Ohms, L=4.22 mH,Find XL, Z, I, VR, and Vl.€ XL= ωL = 6.28 ×1000 × 0.00422H = 26.5Ω€ Z = 102+ (26.5)2= 28.3Ω€ Z = R2+ (ωL)2€ I =εmZ=10028.3= 3.53A€ VR= RI =10 × 3.53 = 35.3v€ VL= XLI = 26.5 × 3.53 = 93.5vPower in AC circuits€ P = i2R = (I sin(ωt))2RInstantaneous power doesn’t mean anythingNeed to average over time or one period of the sine wave€ Pavg=12πRdθ02π∫(I sin(θ))2=12πRI2sin202π∫θdθ = RI212= (I2)2RNote € Irms=I2€ Pavg= Irms2RAveraging over a sine curveCalculate Power lost in resistor from example€ Pavg= Irms2R€ Irms=I2=3.53A1.414= 2.50A€ Pavg= (2.50A)210 = 62.5WattsTo calculate power produced by the generator you need to take account of the phase difference between the voltage and the current. In general you can write:€ Pavg= εrmsIrmscosφFor an inductor P = 0 because the phase difference between current through the inductor and voltage across the inductor is 90 degreesSeries LRC circuit€ ε =vR+ vC+ vL€ ε =εmsinωt€ i = I sin(ωt − φ)VLVCVR€ I =εmR2+ (XL− XC)2=εmZXL=LXC=1/(C)€ Z = R2+ (ωL −1ωC)2€ R2+ (XL− XC)2€ tanφ=ωL −1ωCRELI the ICE manResonance€ XL= XCωL =1ωCω =1LCSeries LRC demo10 uF4.25 mH€ f =16.28 LC=16.28 4.25 ×10−3H ×10−6Ff = 2442HzSeries LCR
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