AC Circuits with RLC Components Enormous impact of AC circuits Basic components
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AC Circuits with RLC Components ÎEnormous impact of AC circuits Power delivery Radio transmitters and receivers Tuners Filters Transformers ÎBasic components R L C Driving ÎNow emf we will study the basic principles PHY2049: Chapter 31 18 AC Circuits and Forced Oscillations + “driving” EMF with angular frequency ωd ε = ε m sin ω d t ÎRLC L di q + Ri + = ε m sin ω d t dt C ÎGeneral solution for current is sum of two terms “Transient”: Falls exponentially & disappears “Steady state”: Constant amplitude Ignore i ∼ e−tR / 2 L cos ω ′t PHY2049: Chapter 31 19 Steady State Solution ÎAssume steady state solution of form i = I m sin (ω d t − φ ) Im is current amplitude φ is phase by which current “lags” the driving EMF Must determine Im and φ in solution: differentiate & integrate sin(ωt-φ) i = I m sin (ω d t − φ ) ÎPlug di = ω d I m cos (ω d t − φ ) dt Im q=− cos (ω d t − φ ) Substitute L di q + Ri + = ε m sin ω t dt C ωd Im I mω d L cos (ω dτ − φ ) + I m R sin (ω d t − φ ) − cos (ω d t − φ ) = ε m sin ω d t ωd C PHY2049: Chapter 31 20 Steady State Solution for AC Current (2) Im I mω d L cos (ω dτ − φ ) + I m R sin (ω d t − φ ) − cos (ω d t − φ ) = ε m sin ω d t ωd C ÎExpand sin & cos expressions sin (ω d t − φ ) = sin ω d t cos φ − cos ω d t sin φ cos (ω d t − φ ) = cos ω d t cos φ + sin ω d t sin φ ÎCollect sinωdt & cosωdt terms separately (ω d L − 1/ ω d C ) cos φ − R sin φ = 0 I m (ω d L − 1/ ω d C ) sin φ + I m R cos φ = ε m ÎThese High school trig! cosωdt terms sinωdt terms equations can be solved for Im and φ (next slide) PHY2049: Chapter 31 21 Steady State Solution for AC Current (3) ÎSolve for φ and Im tan φ = ÎR, ωd L − 1/ ωd C R X L − XC ≡ R Im = εm Z XL, XC and Z have dimensions of resistance X L = ωd L Inductive “reactance” X C = 1/ ωd C Capacitive “reactance” Z = R2 + ( X L − X C ) ÎLet’s 2 Total “impedance” try to understand this solution using “phasors” PHY2049: Chapter 31 22 Graphical Representation: R, XL, XC, Z X L − XC φ Z = R + ( X L − XC ) 2 2 X L − XC tan φ = R PHY2049: Chapter 31 23 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) ε ε = ε m sin ω d t i i = I m sin (ω d t − φ ) εm Im ωdt − φ Current “lags” voltage by φ PHY2049: Chapter 31 24 AC Source and Resistor Only ÎVoltage i is vR = iR = VR sin ω d t ÎRelation of current and voltage i = I R sin ω d t I R = VR / R Current ε ~ R is in phase with voltage (φ = 0) IR VR PHY2049: Chapter 31 ωdt 25 AC Source and Capacitor Only ÎVoltage vC = q / C = VC sin ω d t is ÎDifferentiate to find current q = CVC sin ωd t i = dq / dt = ωd CVC cos ωd t ÎRewrite i ε ~ C using phase i = ωd CVC sin (ωd t + 90° ) ÎRelation of current and voltage i = I C sin (ωd t + 90° ) ΓCapacitive reactance”: X C = 1/ ωd C IC VC ωdt + 90 ωdt Current “leads” voltage by 90° PHY2049: Chapter 31 26 AC Source and Inductor Only ÎVoltage vL = Ldi / dt = VL sin ω d t is ÎIntegrate di/dt to find current: di / dt = (VL / L ) sin ωd t i = − (VL / ω d L ) cos ω d t ÎRewrite i ε ~ L using phase i = (VL / ω d L ) sin (ω d t − 90° ) ÎRelation of current and voltage i = I L sin (ωd t − 90° ) ΓInductive reactance”: X L = ωd L VL ωdt Current “lags” voltage by 90° ωdt − 90 PHY2049: Chapter 31 IL 27 What is Reactance? Think of it as a frequency-dependent resistance 1 XC = ωd C Shrinks with increasing ωd X L = ωd L Grows with increasing ωd ( "XR " = R ) Independent of ωd PHY2049: Chapter 31 28 Quiz identical EMF sources are hooked to a single circuit element, a resistor, a capacitor, or an inductor. The current amplitude is then measured vs frequency. Which curve corresponds to an inductive circuit? ÎThree (1) a (2) b (3) c (4) Can’t tell without more info a b Im c fd X L = ωd L For inductor, higher frequency gives higher reactance, therefore lower current PHY2049: Chapter 31 29 AC Source and RLC Circuit Im ÎVoltage is ε = ε m sin ω d t R, L, C all present ÎRelation of current and voltage i = I m sin (ω d t − φ ) “lags” voltage by φ Impedance: Due to R, XC and XL Current ÎCalculate Im See VL εm V R I ωdt − φ and φ next slide VC PHY2049: Chapter 31 30 AC Source and RLC Circuit (2) ÎSolution using trig (or geometry) Magnitude = Im, lags emf by phase φ) i = I m sin (ω d t − φ ) Im = εm / Z X L − X C ωd L − 1/ ωd C tan φ = = R R Z = R + ( X L − X C ) = R + (ωd L − 1/ ωd C ) 2 2 2 PHY2049: Chapter 31 2 31 AC Source and RLC Circuit (3) X L − XC tan φ = R ÎWhen Z = R2 + ( X L − X C ) 2 XL = XC, then φ = 0 in phase with emf, “Resonant circuit”: ω d Z = R (minimum impedance, maximum current) Current ÎWhen XL < XC, then φ < 0 Current ÎWhen = ω0 = 1/ LC leads emf, “Capacitive circuit”: ωd < ω0 XL > XC, then φ > 0 Current lags emf, “Inductive circuit”: ωd > ω0 PHY2049: Chapter 31 32 Graphical Representation: VR, VL, VC, εm εm VL − VC φ εm = 2 VR + (VL − VC ) 2 X L − X C VL − VC = tan φ = R VR PHY2049: Chapter 31 33 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) ωd < ω0 I ε t PHY2049: Chapter 31 34 Quiz ÎWhich one of these phasor diagrams corresponds to an RLC circuit dominated by inductance? (1) Circuit 1 (2) Circuit 2 (3) Circuit 3 Inductive: Current lags emf, φ>0 εm 1 εm 2 εm 3 PHY2049: Chapter 31 35 Quiz ÎWhich one of these phasor diagrams corresponds to an RLC circuit dominated by capacitance? (1) Circuit 1 (2) Circuit 2 (3) Circuit 3 Capacitive: Current leads emf, φ<0 εm 1 εm 2 εm 3 PHY2049: Chapter 31 36 RLC Example 2 ÎR = 200Ω, C = 15μF, L = 230mH, εm = 36v, fd = 60 Hz ωd = 120π = 377 rad/s Natural XL frequency ω 0 = 1/ ( 0.230 ) (15 ×10−6 ) = 538rad/s = 377 × 0.23 = 86.7Ω ( ) X C = 1/ 377 × 15 × 10−6 = 177Ω Z = 2002 + ( 86.7 − 177 ) = 219Ω I m = ε m / Z = 36 / 219 = 0.164 A XL < XC Capacitive circuit 2 −1 ⎛ 86.7 − 177 ⎞ φ = tan ⎜ ⎝ 200 ⎟ = −24.3° ⎠ PHY2049: Chapter 31 Current leads emf (as expected) 37 RLC Example 2 (cont) Î VL = I m X L = 0.164 × 86.7 = 14.2 V Î VC = I m X C = 0.164 × 177 = 29.0 V Î VR = I m R = 0.164 × 200 = 32.8V Î εm φ VL − VC 32.8 + (14.2 − 29.0 ) = 36.0 = ε m 2 2 PHY2049: Chapter 31 38 RLC Example 3 ÎCircuit parameters: C = 2.5μF, L = 4mH, εm = 10v ω0 = (1/LC)1/2 = 104 rad/s Plot Im vs ωd / ω0 R = 5Ω R = 10Ω R = 20Ω Im Resonance ω d = ω0 ωd / ω0 PHY2049: Chapter 31 39 Power in AC Circuits ÎInstantaneous power emitted by circuit: P = i2R P = I m2 R sin 2 (ωd t − φ ) ÎMore Instantaneous power oscillates useful to calculate power averaged over a cycle Use <…> to indicate average over a cycle P = I m2 R sin 2 (ω d t − φ ) = 12 I m2 R ÎDefine RMS quantities to avoid ½ factors in AC circuits I rms = ÎHouse Vrms Im 2 ε rms = εm 2 2 Pave = I rms R current = 110V ⇒ Vpeak = 156V PHY2049: Chapter 31 40 Power in AC Circuits (2) 2 P = I ÎRecall power formula ave rms R ÎRewrite Pave R ⎛ ε rms ⎞ =⎜ I rms R = ε rms I rms = ε rms I rms cos φ ⎟ Z ⎝ Z ⎠ Pave = ε rms I rms cos φ Îcosφ R cos φ = Z is the “power factor” Z φ X L − XC R maximize power delivered to circuit ⇒ make φ close to zero Most power delivered to load happens near resonance E.g., too much inductive reactance (XL) can be cancelled by increasing XC (decreasing C) To PHY2049: Chapter 31 41
