Chapter 31: RLC Circuits. PHY2049: Chapter 31 1

Transcription

1 Chapter 31: RLC Circuits PHY049: Chapter 31 1

2 LC Oscillations Conservation of energy Topics Dampe oscillations in RLC circuits Energy loss AC current RMS quantities Force oscillations Resistance, reactance, impeance Phase shift Resonant frequency Power Transformers Impeance matching PHY049: Chapter 31

3 LC Oscillations Work out equation for LC circuit (loop rule) q i L 0 C t = C L Rewrite using i = q/t q q q L + = 0 + ω q= 0 C t t ω (angular frequency) has imensions of 1/t Ientical to equation of mass on spring x x m + kx= 0 + ω x= 0 t t ω = ω = 1 LC k m PHY049: Chapter 31 3

4 LC Oscillations () Solution is same as mass on spring oscillations q= q t+ max cos ( ω θ ) q max is the maximum charge on capacitor θ is an unknown phase (epens on initial conitions) Calculate current: i = q/t max ω = ( ) sin( ) i = ωq sin ωt+ θ = i ωt+ θ max k m Thus both charge an current oscillate Angular frequency ω, frequency f = ω/π Perio: T = π/ω PHY049: Chapter 31 4

5 Plot Charge an Current vs t qt ( ) ω = 1 T = π it ( ) PHY049: Chapter 31 5

6 Energy Oscillations Total energy in circuit is conserve. Let s see why i q L + = 0 t C Equation of LC circuit L i t i q q + = 0 Multiply by i = q/t C t L ( ) 1 ( i + q ) = 0 Use t C t x t = x x t t 1 Li 1 q + = C 0 1 Li 1 q + = const C U L + U C = const PHY049: Chapter 31 6

7 Oscillation of Energies Energies can be written as (using ω = 1/LC) U C max cos q q = = + C C Conservation of energy: ( ωt θ ) 1 1 q max sin max ( ) sin ( ) UL = Li = Lω q ωt+ θ = ωt+ θ C max q + UL = = const C Energy oscillates between capacitor an inuctor U Enless oscillation between electrical an magnetic energy C Just like oscillation between potential energy an kinetic energy for mass on spring PHY049: Chapter 31 7

8 U t ( ) C ( ) Plot Energies vs t U t Sum L PHY049: Chapter 31 8

9 Parameters C = 0μF L = 00 mh LC Circuit Example Capacitor initially charge to 40V, no current initially Calculate ω, f an T ω = 500 ra/s f = ω/π = 79.6 Hz T = 1/f = sec Calculate q max an i max q max = CV = 800 μc = C i max = ωq max = = 0.4 A Calculate maximum energies U C = q max/c = 0.016J U L = Li max/ = 0.016J ( 5 )( ) ω = 1/ LC = 1/ = 500 PHY049: Chapter 31 9

10 Charge an current q Energies Voltages LC Circuit Example () = cos( 500t) i = 0.4sin ( 500t) ( ) ( ) L U = 0.016cos 500t U = 0.016sin 500t C ( ) V = q/ C = 40cos 500t C max ( ) ( ) V = Li / t = Lωi cos 500t = 40cos 500t L Note how voltages sum to zero, as they must! PHY049: Chapter 31 10

11 RLC Circuit Work out equation using loop rule i q L + Ri+ = 0 t C Rewrite using i = q/t q R q q + + = t Lt LC 0 Solution slightly more complicate than LC case ( ω θ) ω ( ) tr /L q= qmaxe cos t+ = 1/ LC R/ L This is a ampe oscillator (similar to mechanical case) Amplitue of oscillations falls exponentially PHY049: Chapter 31 11

12 Charge an Current vs t in RLC Circuit qt ( ) it ( ) / e tr L PHY049: Chapter 31 1

13 Circuit parameters RLC Circuit Example L = 1mL, C = 1.6μF, R = 1.5Ω Calculate ω, ω, f an T ω = 70 ra/s ω = 70 ra/s f = ω/π = 1150 Hz T = 1/f = sec Time for q max to fall to ½ its initial value t = (L/R) * ln = s = 11.1 ms # perios = / ( )( 6 ) ω = 1/ = 70 ( ) ω = / 0.04 ω tr /L e = 1/ PHY049: Chapter 31 13

14 L i t RLC Circuit (Energy) q + Ri+ = 0 Basic RLC equation C i q q L i+ Ri + = t C t 0 Multiply by i = q/t t 1 1 q Li + = C ( U L + U C) = i R t / Utot e tr L i R Collect terms (similar to LC circuit) Total energy in circuit ecreases at rate of i R (issipation of energy) PHY049: Chapter 31 14

15 Energy in RLC Circuit U ( ) C t U L ( ) Sum t / e tr L PHY049: Chapter 31 15

16 Quiz Below are shown 3 LC circuits. Which one takes the least time to fully ischarge the capacitors uring the oscillations? (1) A () B (3) C C C C C C A B C ω =1/ LC C has smallest capacitance, therefore highest frequency, therefore shortest perio PHY049: Chapter 31 16

17 AC Circuits Enormous impact of AC circuits Power elivery Raio transmitters an receivers Tuners Filters Transformers Basic components R L C Driving emf Now we will stuy the basic principles PHY049: Chapter 31 17

18 AC Circuits an Force Oscillations RLC + riving EMF with angular frequency ω ε = ε sin ω m t i q L Ri msin t t + + C = ε ω General solution for current is sum of two terms Transient : Falls exponentially & isappears Steay state : Constant amplitue Ignore tr / L i e cosω t PHY049: Chapter 31 18

19 Steay State Solution Assume steay state solution of form I m is current amplitue φ is phase by which current lags the riving EMF Must etermine I m an φ i = I sin( ω t φ ) Plug in solution: ifferentiate & integrate sin(ωt-φ) i = I sin( ω t φ ) m i = ωimcos( ωt φ) t Im q= cos( ωt φ ) ω Substitute Im ImωLcos( ωτ φ) + ImRsin ( ωt φ) cos( ωt φ) = εmsinωt ω C PHY049: Chapter m i q L Ri m sin t t + + C = ε ω

20 Steay State Solution for AC Current () Im ImωLcos( ωτ φ) + ImRsin ( ωt φ) cos( ωt φ) = εmsinωt ω C Expan sin & cos expressions ( ) ( ) sin ω t φ = sinω t cosφ cosω tsinφ cos ω t φ = cosω t cosφ + sinω t sinφ Collect sinω t&cosω tterms separately High school trig! ( ) ( ) ω L 1/ ω C cosφ Rsinφ = 0 I ω L 1/ ω C sinφ + I Rcosφ = ε m m m cosω tterms sinω tterms These equations can be solve for I m an φ (next slie) PHY049: Chapter 31 0

21 Steay State Solution for AC Current (3) ( ) ( ) ω L 1/ ω C cosφ Rsinφ = 0 I ω L 1/ ω C sinφ + I Rcosφ = ε m m m Same equations Solve for φ an I m in terms of ωl 1/ ωc XL XC ε tanφ = m I R R m = Z R, X L, X C an Z have imensions of resistance X X L C = ω L = 1/ ω C Inuctive reactance Capacitive reactance ( ) Z = R + X X L C Total impeance Let s try to unerstan this solution using phasors PHY049: Chapter 31 1

22 Unerstaning AC Circuits Using Phasors Phasor Voltage or current represente by phasor Phasor rotates counterclockwise with angular velocity = ω Length of phasor is amplitue of voltage (V) or current (I) y component is instantaneous value of voltage (v) or current (i) ε = ε sin ω i i = I sin( ω t φ ) m m t ε ε m I m ω t φ Current lags voltage by φ PHY049: Chapter 31

23 Voltage is AC Source an Resistor Only v = ir= V sinω t R R i Relation of current an voltage i = I sin ω t I = V / R R R R ε ~ R Current is in phase with voltage (φ = 0) I R V R ω t PHY049: Chapter 31 3

24 AC Source an Capacitor Only Voltage is C C Differentiate to fin current q= CV sinω t C i = q/ t =ω CV cosω t Rewrite using phase v = q/ C = V sinω t C Relation of current an voltage Capacitive reactance : C Current leas voltage by 90 C ( ω ) i = ω CV sin t+ 90 ( ω ) i = I sin t+ 90 I = V / X C C C C X = 1/ ω C i ε ~ C I C ω t V C ω t + 90 PHY049: Chapter 31 4

25 AC Source an Inuctor Only v = Li / t = V sinω t Voltage is L L Integrate i/t to fin current: = ( L ) ( ω ) i / t V / L sinω t i = V / L cosω t L Rewrite using phase ( / ω ) sin( ω 90 ) i = V L t L Relation of current an voltage ( ω ) Inuctive reactance : L Current lags voltage by 90 ε ~ i = I sin t 90 I = V / X ω t L L L L X = ω L V L i L I L ω t 90 PHY049: Chapter 31 5

26 What is Reactance? Think of it as a frequency-epenent resistance X C = 1 ω C ω 0, X C - Capacitor looks like a break ω, X C 0 - Capacitor looks like a wire ( short ) X L = ω L ω 0, X L 0 - Inuctor looks like a wire ( short ) ω, X L - Inuctor looks like a break (" X " = R ) R Inepenent of ω PHY049: Chapter 31 6

27 Quiz Three ientical EMF sources are hooke to a single circuit element, a resistor, a capacitor, or an inuctor. The current amplitue is then measure as a function of frequency. Which one of the following curves correspons to an inuctive circuit? (1) a () b (3) c (4) Can t tell without more info I m a b c f X L = ω L For inuctor, higher frequency gives higher reactance, therefore lower current PHY049: Chapter 31 7

28 Voltage is AC Source an RLC Circuit ε = ε sin ω m t I m Relation of current an voltage i = I sin( ω t φ ) m Current lags voltage by φ Impeance: Due to R, X C an X L Calculate I m an φ using geometry See next slie V L ε m V R ω t φ V C PHY049: Chapter 31 8

29 AC Source an RLC Circuit () Right triangle with sies V R, V L -V C an ε m VL VC tanφ = VR = ImR VR V ε ( ) L = ImXL m = VR + VL VC V = I X C m C V L V C ε m φ V R Solve for current: i = I sin( ω t φ ) m (Magnitue = I m, lags emf by phase φ) I m = ε m / Z XL XC ωl 1/ ωc tanφ = = R R ( ) ( ω 1/ ω ) L C Z = R + X X = R + L C PHY049: Chapter 31 9

30 AC Source an RLC Circuit (3) XL XC tanφ = R ( ) Z = R + X X L C Only X L X C is relevant, reactances cancel each other When X L = X C, then φ = 0 Current in phase with emf, Resonant circuit : Z = R (minimum impeance, maximum current) When X L < X C, then φ < 0 ω = ω0 = 1/ LC Current leas emf, Capacitive circuit : ω < ω 0 When X L > X C, then φ > 0 Current lags emf, Inuctive circuit : ω > ω 0 PHY049: Chapter 31 30

31 RLC Example 1 Below are shown the riving emf an current vs time of an RLC circuit. We can conclue the following Current leas the riving emf (φ<0) Circuit is capacitive (X C > X L ) ω < ω 0 I ε t PHY049: Chapter 31 31

32 Quiz Which one of these phasor iagrams correspons to an RLC circuit ominate by inuctance? (1) Circuit 1 () Circuit (3) Circuit 3 Inuctive: Current lags emf, φ>0 ε m ε m ε m 1 3 PHY049: Chapter 31 3

33 Quiz Which one of these phasor iagrams correspons to an RLC circuit ominate by capacitance? (1) Circuit 1 () Circuit (3) Circuit 3 Capacitive: Current leas emf, φ<0 ε m ε m ε m 1 3 PHY049: Chapter 31 33

34 RLC Example R = 00Ω, C = 15μF, L = 30mH, ε m = 36v, f = 60 Hz ω = 10π = 377 ra/s Natural frequency ( )( 6 ) ω 0 = 1/ = 538ra/s X = = 86.7Ω L X C ( 6 ) = 1/ = 177Ω ( ) Z = = 19Ω X L < X C Capacitive circuit I = ε / Z = 36/ 19 = 0.164A m m φ = tan = Current leas emf (as expecte) PHY049: Chapter 31 34

35 RLC Example 3 Circuit parameters: C =.5μF, L = 4mH, ε m = 10v ω 0 = (1/LC) 1/ = 10 4 ra/s Plot I m vs ω / ω 0 I m Resonance ω = ω 0 R = 5Ω R = 10Ω R = 0Ω PHY049: Chapter 31 35

36 Power in AC Circuits Instantaneous power emitte by circuit: P = i R sin ( ω φ ) P= I t m More useful to calculate power average over a cycle Use < > to inicate average over a cycle ( ω φ) 1 m sin m P = I R t = I R Instantaneous power oscillates Define RMS quantities to avoi ½ factors in AC circuits I m ε I rms = ε rms = m House current V rms = 110V V peak = 156V ave rms P = I R PHY049: Chapter 31 36

37 Power in AC Circuits () Recall power formula ave rms P = I R Rewrite Base on P ave εrms Pave = IrmsR=εrmsIrms Z cos φ V m cos R I R φ = = = ε I Z rms I rms cos cosφ is the power factor m m To maximize power elivere to circuit make φ close to zero Most power elivere to loa happens near resonance E.g., too much inuctive reactance (X L ) can be cancelle by increasing X C (ecreasing C) R Z = ε φ cosφ = R Z PHY049: Chapter 31 37

38 Power Example 1 R = 00Ω, X C = 150Ω, X L = 80Ω, ε rms = 10v, f = 60 Hz ω = 10π = 377 ra/s ( ) Z = = 11.9Ω I = ε / Z = 10/ 11.9 = 0.566A rms rms φ = tan = cosφ = Current leas emf Capacitive circuit P = ε I cos φ = = 64.1W ave rms rms Pave = IrmsR= = 64.1W Same PHY049: Chapter 31 38

39 Power Example 1 (cont) R = 00Ω, X C = 150Ω, X L = 80Ω, ε rms = 10v, f = 60 Hz How much capacitance must be ae to maximize the power in the circuit (an thus bring it into resonance)? Want X C = X L to minimize Z, so must ecrease X C X = 150Ω= 1/ ω C C = 17.7μF C Cnew X = X = 80Ω C = 33.μF L new So we must a 15.5μF capacitance to maximize power PHY049: Chapter 31 39

40 Q Factor We can quantify low amping situations using Q factor Define using values at resonance Energy in circuit Q π Energy lost per cycle 1 LI m 1 0 ω0l Q = π = I R R ( π / ω ) m Can also be expresse (using resonance values) as X L X Q = = C R R PHY049: Chapter 31 40

41 Power Example Circuit parameters: C =.5μF, L = 4mH, ε m = 10v ω 0 = (1/LC) 1/ = ra/s Plot Power vs ω / ω 0 for ifferent R values, using Q P ave Resonance ω = ω 0 R = Ω Q = 0 R = 5Ω Q = 8 R = 10Ω Q = 4 R = 0Ω Q = PHY049: Chapter 31 41

42 Q Factor (cont) εrmsr ave = rms = P I R ( ω 1/ ω ) L C + R For Q > 3 or so, can easily show that Q ω0 FWHM FWHM = Full With at Half Maximum PHY049: Chapter 31 4

43 Raio Tuner Set RLC tuner to (ugh!) Other raio stations. RLC response is less Circuit response Q = 500. Maximize for f = PHY049: Chapter 31 43

44 Quiz A generator prouces current at a frequency of 60 Hz with peak voltage an current amplitues of 100V an 10A, respectively. What is the average power prouce? (1) 1000 W () 707 W (3) 1414 W (4) 500 W (5) 50 W 1 ave = ε peak peak = εrms rms P I I PHY049: Chapter 31 44

45 Quiz The figure shows the current an emf of a series RLC circuit. To increase the rate at which power is elivere to the resistive loa, which option shoul be taken? (1) Increase R () Decrease L (3) Increase L (4) Increase C Current lags applie emf (peak occurs later, φ > 0), thus circuit is inuctive. Max power is at φ = 0, so nee to either (1) reuce X L by ecreasing L or () cancel X L by increasing X C (ecrease C). PHY049: Chapter 31 45

46 RLC Circuit Example If you wante to increase the power elivere to this RLC circuit, which moification(s) woul work? (a) increase R (b) increase C (c) increase L () ecrease R (e) ecrease C (f) ecrease L Again, current lags emf. See previous page for etails. PHY049: Chapter 31 46

47 Example (Prob ) Variable frequency EMF source with ε m =6V connecte to a resistor an inuctor. R=80Ω an L=40mH. At what frequency f oes V R = V L? ω L f = R ω = = ω / π = 318Hz 000ra/s At that frequency, what is phase angle φ? tanφ = 1 φ = 45 What is current amplitue? I m m = ε / = 6 /113 = 0.053A ε m I V m L V R PHY049: Chapter 31 47