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Chapter 21: RLC Circuits Chapter 21: RLC Circuits PHY2054: Chapter 21 1 Voltage and Current in RLC Circuits ÎAC emf source: “driving frequency” f ε = εωm sin t ω = 2π f ÎIf circuit contains only R + emf source, current is simple ε ε iI==sin()ω tI =m () current amplitude R mmR ÎIf L and/or C present, current is not in phase with emf ε iI=−=sin()ωφ t I m mmZ ÎZ, φ shown later PHY2054: Chapter 21 2 AC Source and Resistor Only ÎDriving voltage is ε = εωm sin t i ÎRelation of current and voltage iR= ε / ε ~ R ε iI==sinω tI m mmR Current is in phase with voltage (φ = 0) PHY2054: Chapter 21 3 AC Source and Capacitor Only q ÎVoltage is vt==ε sinω CmC ÎDifferentiate to find current i qC= εm sinω t idqdtCV==/cosω C ω t ε ~ C ÎRewrite using phase (check this!) iCVt=+°ω C sin()ω 90 ÎRelation of current and voltage εm iI=+°=mmsin()ω t 90 I ( XC =1/ωC) XC ΓCapacitive reactance”: XCC =1/ω Current “leads” voltage by 90° PHY2054: Chapter 21 4 AC Source and Inductor Only ÎVoltage is vLm== Ldi/sin dtε ω t ÎIntegrate di/dt to find current: i di//sin dt= ()εm Lω t iLt=−()εm /cosωω ε ~ L ÎRewrite using phase (check this!) iLt=−°()εm /sin90ωω( ) ÎRelation of current and voltage εm iI=−°=mmsin()ω t 90 I ( XLL = ω ) X L ΓInductive reactance”: XLL = ω Current “lags” voltage by 90° PHY2054: Chapter 21 5 General Solution for RLC Circuit ÎWe assume steady state solution of form iI= m sin(ω t−φ ) Im is current amplitude φ is phase by which current “lags” the driving EMF Must determine Im and φ ÎPlug in solution: differentiate & integrate sin(ωt-φ) iI=−m sin()ω tφ di Substitute di q =−ωItm cos()ωφ LRi++=ε sinω t dt dt C m I qt=−m cos()ω −φ ω I I ωLtcos()ωφ−+ IRt sin () ωφ −−m cos() ωφ t −= ε sin ω t mmωC m PHY2054: Chapter 21 6 General Solution for RLC Circuit (2) I I ωLtcos()ωφ−+ IRt sin () ωφ −−m cos() ωφ t −= ε sin ω t mmωC m ÎExpand sin & cos expressions sin()ωtt−=φωφω sin cos − cos t sinφ High school trig! cos()ωtt−=φωφω cos cos + sin t sinφ ÎCollect sinωt&cosωtterms separately ()ωLC−−=1/ωφ cos R sin φ 0 cosωtterms ILmmm()ω −+=1/ωφ C sin IR cos φε sinωtterms ÎThese equations can be solved for Im and φ (next slide) PHY2054: Chapter 21 7 General Solution for RLC Circuit (3) ÎSolve for φ and Im ωLC−1/ω XX− ε tanφ =≡LCI = m R R m Z ÎR, XL, XC and Z have dimensions of resistance XLL = ω Inductive “reactance” XCC =1/ω Capacitive “reactance” 2 2 Total “impedance” ZRXX=+−()LC ÎThis is where φ, XL, XC and Z come from! PHY2054: Chapter 21 8 AC Source and RLC Circuits ε I = m Maximum current m Z XX− tanφ = LC Phase angle R Inductive reactance XLL ==ω ()ωπ2 f Capacitive reactance XCC =1/ω 2 2 Total impedance ZRXX=+−()LC φ= angle that current “lags” applied voltage PHY2054: Chapter 21 9 What is Reactance? Think of it as a frequency-dependent resistance 1 X = Shrinks with increasing ω C ωC Grows with increasing ω X L = ωL ("X R "= R ) Independent of ω PHY2054: Chapter 21 10 Pictorial Understanding of Reactance 2 2 ZRXX=+−()LC XX− tanφ = LC R R cosφ = Z PHY2054: Chapter 21 11 Summary of Circuit Elements, Impedance, Phase Angles XXLC− 2 2 tanφ = ZRXX=+−()LC R PHY2054: Chapter 21 12 Quiz ÎThree identical EMF sources are hooked to a single circuit element, a resistor, a capacitor, or an inductor. The current amplitude is then measured as a function of frequency. Which one of the following curves corresponds to an inductive circuit? (1) a (2) b a (3) c b (4) Can’t tell without more info Imax c f XL==ω ()ωπ2 f L For inductor, higher frequency gives higher IXmax= ε max / L reactance, therefore lower current PHY2054: Chapter 21 13 RLC Example 1 ÎBelow are shown the driving emf and current vs time of an RLC circuit. We can conclude the following Current “leads” the driving emf (φ<0) Circuit is capacitive (XC > XL) I ε t PHY2054: Chapter 21 14 RLC Example 2 ÎR = 200Ω, C = 15μF, L = 230mH, εmax = 36v, f = 60 Hz X L =××2π 60 0.23 = 86.7 Ω XC > XL Capacitive circuit −6 XC =×××=Ω1/() 2π 60 15 10 177 2 2 Z =+−=Ω200() 86.7 177 219 IZmax===ε max / 36/ 219 0.164A −1 ⎛⎞86.7− 177 Current leads emf φ ==−°tan⎜⎟ 24.3 (as expected) ⎝⎠200 it=+°0.164sin(ω 24.3 ) PHY2054: Chapter 21 15 Resonance ÎConsider impedance vs frequency 2222 Z =+−RXX()LC =+− R()ωω L1/ C ÎZ is minimum when ωLC=1/ω ωω==0 1/ LC This is resonance! ÎAt resonance Impedance = Z is minimum Current amplitude = Im is maximum PHY2054: Chapter 21 16 Imax vs Frequency and Resonance ÎCircuit parameters: C = 2.5μF, L = 4mH, εmax = 10v 1/2 f0 = 1 / 2π(LC) = 1590 Hz Plot Imax vs f 2 2 Imax =+−10/RLC()ωω 1/ R = 5Ω R = 10Ω R = 20Ω Imax Resonance ff= 0 f / f0 PHY2054: Chapter 21 17 Power in AC Circuits ÎInstantaneous power emitted by circuit: P = i2R 22 PIR=−mdsin ()ω tφ Instantaneous power oscillates ÎMore useful to calculate power averaged over a cycle Use <…> to indicate average over a cycle 221 2 P =−=IRmdsin (ωφ t) 2 IR m ÎDefine RMS quantities to avoid ½ factors in AC circuits Im εm 2 Irms = εrms = Pave= IR rms 2 2 ÎHouse current Vrms = 110V ⇒ Vpeak = 156V PHY2054: Chapter 21 18 Power in AC Circuits Î 2 Power formula Pave= IR rms IIrms= max /2 εrms εrms ÎRewrite using I = PIRIave== rmsε rms rms cosφ rms Z Z Z R XX− Pave= εφ rmsI rms cos cosφ = L C Z φ R Îcosφ is the “power factor” To maximize power delivered to circuit ⇒ make φ close to zero Max power delivered to load happens at resonance E.g., too much inductive reactance (XL) can be cancelled by increasing XC (e.g., circuits with large motors) PHY2054: Chapter 21 19 Power Example 1 ÎR = 200Ω, XC = 150Ω, XL = 80Ω, εrms = 120v, f = 60 Hz 2 2 Z =+−=Ω200() 80 150 211.9 IZrms==ε rms / 120/ 211.9 = 0.566A −1 ⎛⎞80− 150 Current leads emf φ ==−°tan⎜⎟ 19.3 Capacitive circuit ⎝⎠200 cosφ = 0.944 Pave==××=ε rmsI rms cosφ 120 0.566 0.944 64.1W Same 22 PIRave== rms 0.566 ×= 200 64.1W PHY2054: Chapter 21 20 Power Example 1 (cont) ÎR = 200Ω, XC = 150Ω, XL = 80Ω, εrms = 120v, f = 60 Hz ÎHow much capacitance must be added to maximize the power in the circuit (and thus bring it into resonance)? Want XC = XL to minimize Z, so must decrease XC XfCCC =Ω=150 1/ 2π = 17.7μF XXCLnew==Ω80 C new = 33.2μF So we must add 15.5μF capacitance to maximize power PHY2054: Chapter 21 21 Power vs Frequency and Resonance ÎCircuit parameters: C = 2.5μF, L = 4mH, εmax = 10v 1/2 f0 = 1 / 2π(LC) = 1590 Hz Plot Pave vs f for different R values R = 2Ω R = 5Ω R = 10Ω P Resonance ave R = 20Ω ff= 0 f / f0 PHY2054: Chapter 21 22 Resonance Tuner is Based on Resonance Vary C to set resonance frequency to 103.7 (ugh!) Circuit response Q = 500 Other radio stations. Tune for f = 103.7 MHz RLC response is less PHY2054: Chapter 21 23 Quiz ÎA generator produces current at a frequency of 60 Hz with peak voltage and current amplitudes of 100V and 10A, respectively. What is the average power produced if they are in phase? (1) 1000 W (2) 707 W 1 (3) 1414 W Pave==2 εε peakII peak rms rms (4) 500 W (5) 250 W PHY2054: Chapter 21 24 Quiz ÎThe figure shows the current and emf of a series RLC circuit. To increase the rate at which power is delivered to the resistive load, which option should be taken? (1) Increase R (2) Decrease L (3) Increase L (4) Increase C XX− tanφ = LC R Current lags applied emf (φ > 0), thus circuit is inductive. Either (1) Reduce XL by decreasing L or (2) Cancel XL by increasing XC (decrease C). PHY2054: Chapter 21 25 Example: LR Circuit ÎVariable frequency EMF source with εm=6V connected to a resistor and inductor. R=80Ω and L=40mH. At what frequency f does VR = VL? XLRL ==⇒=ω ωπ2000 f = 2000/ 2 = 318Hz At that frequency, what is phase angle φ? tanφ ==⇒=°XRL / 1φ 45 What is the current amplitude and RMS value? 22 Imax=+==ε max / 80 80 6/113 0.053A IIrms== max / 2 0.037A it= 0.053sin(ω −° 45 ) PHY2054: Chapter 21 26 Transformers ÎPurpose: change alternating (AC) voltage to a bigger (or smaller) value Input AC voltage Changing flux in in the “primary” “secondary” turns turns produces a flux induces an emf ΔΦ ΔΦ B VN= B VNss= ppΔt Δt Ns VVsp= N p PHY2054: Chapter 21 27 Transformers ÎNothing comes for free, however! Increase in voltage comes at the cost of current. Output power cannot exceed input power! power in = power out (Losses usually account for 10-20%) iVp pss= iV i VN s ==pp iVNps s PHY2054: Chapter 21 28 Transformers: Sample Problem ÎA transformer has 330 primary turns and 1240 secondary turns.
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