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Chapter 31: RLC Circuits

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Chapter 31: RLC Circuits Chapter 31: RLC Circuits PHY2049: Chapter 31 1 Topics ÎLC Oscillations Conservation of energy ÎDamped oscillations in RLC circuits Energy loss ÎAC current RMS quantities ÎForced oscillations Resistance, reactance, impedance Phase shift Resonant frequency Power ÎTransformers Impedance matching PHY2049: Chapter 31 2 LC Oscillations ÎWork out equation for LC circuit (loop rule) qdi C −−L =0 L Cdt ÎRewrite using i = dq/dt dq22 q dq 1 Lq+=⇒00 +ω 2 = ω = dt22C dt LC ω (angular frequency) has dimensions of 1/t ÎIdentical to equation of mass on spring dx22 dx k mkx+=⇒00 +ω 2 x = ω = dt22 dt m PHY2049: Chapter 31 3 LC Oscillations (2) ÎSolution is same as mass on spring ⇒ oscillations k qq=+max cos()ω tθ ω = m qmax is the maximum charge on capacitor θ is an unknown phase (depends on initial conditions) ÎCalculate current: i = dq/dt iq=−ω maxsin(ωθ t +) =− i max sin( ωθ t + ) ÎThus both charge and current oscillate Angular frequency ω, frequency f = ω/2π Period: T = 2π/ω PHY2049: Chapter 31 4 Plot Charge and Current vs t ω =12T = π qt( ) it( ) PHY2049: Chapter 31 5 Energy Oscillations ÎTotal energy in circuit is conserved. Let’s see why di q L +=0 Equation of LC circuit dt C di q dq Li+=0 Multiply by i = dq/dt dt C dt 2 Ld221 d dx dx iq+=0 Use = 2x 22dt( ) C dt ( ) dt dt 2 2 dq⎛⎞ 112 q 11Li2 +=0 Li +=const ⎜⎟22 22C dt⎝⎠ C UL + UC = const PHY2049: Chapter 31 6 Oscillation of Energies ÎEnergies can be written as (using ω2 = 1/LC) 2 2 q q 2 Ut==max cos ()ω +θ C 22CC 2 2222q 2 ULiLq==11ω sin()ωθ t +=max sin () ωθ t + L 22max 2C q2 ÎConservation of energy: UU+=max =const CL2C ÎEnergy oscillates between capacitor and inductor Endless oscillation between electrical and magnetic energy Just like oscillation between potential energy and kinetic energy for mass on spring PHY2049: Chapter 31 7 Plot Energies vs t Sum UtC ( ) UtL ( ) PHY2049: Chapter 31 8 LC Circuit Example ÎParameters C = 20μF L = 200 mH Capacitor initially charged to 40V, no current initially ÎCalculate ω, f and T −5 ω = 500 rad/s ω ==×1/LC 1/() 2 10() 0.2 = 500 f = ω/2π = 79.6 Hz T = 1/f = 0.0126 sec ÎCalculate qmax and imax -4 qmax = CV = 800 μC = 8 × 10 C -4 imax = ωqmax = 500 × 8 × 10 = 0.4 A ÎCalculate maximum energies 2 2 UC = q max/2C = 0.016J UL = Li max/2 = 0.016J PHY2049: Chapter 31 9 LC Circuit Example (2) ÎCharge and current qt= 0.0008cos( 500 ) it= −0.4sin( 500 ) ÎEnergies 22 UtUtCL==0.016cos( 500) 0.016sin( 500 ) ÎVoltages VqCC ==/ 40cos( 500 t) VL ==− Ldi/ dt Lω imax cos( 500 t) =− 40cos( 500 t) ÎNote how voltages sum to zero, as they must! PHY2049: Chapter 31 10 RLC Circuit ÎWork out equation using loop rule di q LRi++=0 dt C ÎRewrite using i = dq/dt dq2 Rdq q ++=0 dt2 Ldt LC ÎSolution slightly more complicated than LC case −tR/2 L 2 qq=+=−max ecos()ωθω′′ t 1/ LCRL () / 2 ÎThis is a damped oscillator (similar to mechanical case) Amplitude of oscillations falls exponentially PHY2049: Chapter 31 11 Charge and Current vs t in RLC Circuit qt( ) it( ) e−tR/2 L PHY2049: Chapter 31 12 RLC Circuit Example ÎCircuit parameters L = 12mL, C = 1.6μF, R = 1.5Ω ÎCalculate ω, ω’, f and T ω =×=1/ 0.012 1.6 10−6 7220 ω = 7220 rad/s ()( ) 2 2 ω’ = 7220 rad/s ω′ =−7220() 1.5/0.024 ω f = ω/2π = 1150 Hz T = 1/f = 0.00087 sec −tR/2 L ÎTime for qmax to fall to ½ its initial value e =1/2 t = (2L/R) * ln2 = 0.0111s = 11.1 ms # periods = 0.0111/.00087 ≈ 13 PHY2049: Chapter 31 13 RLC Circuit (Energy) di q LRi++=0 Basic RLC equation dt C di q dq LiRi++2 =0 Multiply by i = dq/dt dt C dt dq⎛⎞2 Collect terms 11Li22+=− i R ⎜⎟22 (similar to LC circuit) dt⎝⎠ C Total energy in circuit d 2 ()UU+=− iR decreases at rate of i2R dt LC (dissipation of energy) −tR/ L Uetot ∼ PHY2049: Chapter 31 14 Energy in RLC Circuit UtC ( ) UtL ( ) Sum e−tR/ L PHY2049: Chapter 31 15 Quiz ÎBelow are shown 3 LC circuits. Which one takes the least time to fully discharge the capacitors during the oscillations? (1) A (2) B (3) C C C C C C A B C C has smallest capacitance, therefore highest ω =1/ LC frequency, therefore shortest period PHY2049: Chapter 31 16 AC Circuits ÎEnormous impact of AC circuits Power delivery Radio transmitters and receivers Tuners Filters Transformers ÎBasic components R L C Driving emf ÎNow we will study the basic principles PHY2049: Chapter 31 17 AC Circuits and Forced Oscillations ÎRLC + “driving” EMF with angular frequency ωd ε = εωmdsin t di q LRi++=ε sinω t dt C md ÎGeneral solution for current is sum of two terms “Transient”: Falls “Steady state”: exponentially & disappears Constant amplitude Ignore ie∼ −tR/2 L cosω′ t PHY2049: Chapter 31 18 Steady State Solution ÎAssume steady state solution of form iI= mdsin(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=−mdsin()ω tφ di Substitute di q =−ωdmItcos()ωφ d LRi++=ε sinω t dt dt C m Im qt=−cos()ωd −φ ωd Im Imdω LIRttcos()ωτ d−+ φ m sin () ω d −− φ cos() ω d −= φ ε m sin ω d t ωdC PHY2049: Chapter 31 19 Steady State Solution for AC Current (2) Im Imdω LIRttcos()ωτ d−+ φ m sin () ω d −− φ cos() ω d −= φ ε m sin ω d t ωdC ÎExpand sin & cos expressions sin()ω tt−=φωφω sin cos − cos t sinφ dd dHigh school trig! cos()ωddtt−=φωφω cos cos + sin d t sinφ ÎCollect sinωdt&cosωdtterms separately ()ωddLC−−=1/ωφφ cos R sin 0 cosωdtterms ILmd()ω −+=1/ωφ d C sin IR m cos φε m sinωdtterms ÎThese equations can be solved for Im and φ (next slide) PHY2049: Chapter 31 20 Steady State Solution for AC Current (3) ω LC−−=1/ωφφ cos R sin 0 ()dd Same equations ILmd()ω −+=1/ωφ d C sin IR m cos φε m ÎSolve for φ and Im in terms of ω LCXX−−1/ω ε tanφ =≡ddLCI = m R R m Z ÎR, XL, XC and Z have dimensions of resistance Inductive “reactance” XLLd= ω Capacitive “reactance” XCCd= 1/ω 2 2 Total “impedance” ZRXX=+−()LC ÎLet’s try to understand this solution using “phasors” PHY2049: Chapter 31 21 Understanding AC Circuits Using Phasors ÎPhasor Voltage or current represented by “phasor” Phasor rotates counterclockwise with angular velocity = ωd Length of phasor is amplitude of voltage (V) or current (I) y component is instantaneous value of voltage (v) or current (i) εm ε Im ε = εωsin t md i ωdt − φ iI=−mdsin()ω tφ Current “lags” voltage by φ PHY2049: Chapter 31 22 AC Source and Resistor Only i ÎVoltage is viRVRRd==sinω t ε ~ R ÎRelation of current and voltage iI==RdRRsinω tI VR / Current is in phase with voltage (φ = 0) IR VR ωdt PHY2049: Chapter 31 23 AC Source and Capacitor Only i ÎVoltage is vqCVCCd==/sinω t ÎDifferentiate to find current qCV= Cdsinω t ε ~ C idqdt==/cosωdC CVω d t ÎRewrite using phase iCVt=+°ωdCsin(ω d 90 ) ÎRelation of current and voltage iI=+°=Cdsin()ω t 90 I CCC V / X IC ωdt ωdt + 90 ΓCapacitive reactance”: XCCd=1/ω Current “leads” voltage by 90° VC PHY2049: Chapter 31 24 AC Source and Inductor Only i ÎVoltage is vLLd== Ldi/sin dt Vω t ÎIntegrate di/dt to find current: di//sin dt= () VLd Lω t ε ~ L iV=−()Ld/cosω Lω d t ÎRewrite using phase iV=−°()Ld/sin90ω L(ω d t ) ÎRelation of current and voltage IL iI=−°=sin()ω t 90 I VX / ω t Ld LLL V d L ωdt − 90 ΓInductive reactance”: XLLd= ω Current “lags” voltage by 90° PHY2049: Chapter 31 25 What is Reactance? Think of it as a frequency-dependent resistance ω → 0, X →∞ 1 d C X = - Capacitor looks like a break C ω →∞, X → 0 ωdC d C - Capacitor looks like a wire (“short”) ωd → 0, XL → 0 - Inductor looks like a wire (“short”) X L = ωd L ωd →∞, XL →∞ - Inductor looks like a break ("X R "= R ) Independent of ωd PHY2049: Chapter 31 26 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 Im c fd For inductor, higher frequency gives higher XLLd= ω reactance, therefore lower current PHY2049: Chapter 31 27 AC Source and RLC Circuit Im ÎVoltage is ε = εωmdsin t ÎRelation of current and voltage iI=−mdsin()ω tφ Current “lags” voltage by φ Impedance: Due to R, XC and XL ÎCalculate Im and φ using geometry εm VR See next slide VL ωdt − φ VC PHY2049: Chapter 31 28 AC Source and RLC Circuit (2) ÎRight triangle with sides VR, VL-VC and εm VV− VIR tanφ = LC Rm= εm V R VIXLmL= φ 22 2 εmR=+VVV() LC − VL−VC VR VIXCmC= ÎSolve for current: iI= mdsin(ω t−φ ) (Magnitude = Im, lags emf by phase φ) IZmm= ε / XX−−ωω L1/ C tanφ ==LC d d RR 2222 Z =+−RXX()LC =+ R()ωω d L −1/ d C PHY2049: Chapter 31 29 AC Source and RLC Circuit (3) XXLC− 2 2 tanφ = ZRXX=+−() R LC ÎOnly XL − XC is relevant, reactances cancel each other ÎWhen XL = XC, then φ = 0 Current in phase with emf, “Resonant circuit”:ωd ==ω0 1/ LC Z = R (minimum impedance, maximum current) ÎWhen XL < XC, then φ < 0 Current leads emf, “Capacitive circuit”: ωd < ω0 ÎWhen XL > XC, then φ > 0 Current lags emf, “Inductive circuit”: ωd > ω0 PHY2049: Chapter 31 30 RLC Example 1 ÎBelow are shown the driving emf and current vs time of an RLC circuit.
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