RLC Circuits PHY2049: Chapter 31 1

ω = ω = Work out equation for LC circuit (loop rule) C dt C − − L = 0 LC Oscillations Work out equation for LC circuit (loop rule) q di C dt C − − L = 0 L Rewrite using i = dq/dt 2 2 + = 0 ⇒ +ω2q = 0 dt2 C dt2 1 LC d q q d q ω = L ω (angular frequency) has dimensions of 1/t Identical to equation of mass on spring 2 2 k m d x d x ω = +kx = 0 ⇒ +ω x = 0 2 m dt dt PHY2049: Chapter 31 3

ω = Solution is same as mass on spring ⇒ oscillations LC Oscillations (2) Solution is same as mass on spring ⇒ oscillations q = qmax cos(ωt +θ) k m ω = qmax is the maximum charge on capacitor θ is an unknown phase (depends on initial conditions) Calculate current: i = dq/dt i = −ωqmax sin(ωt +θ) = −imax sin(ωt +θ) Thus both charge and current oscillate Angular frequency ω, frequency f = ω/2π Period: T = 2π/ω 4

Plot Charge and Current vs t q(t) i(t) PHY2049: Chapter 31 5

(i2)+ (q2) = 0 Total energy in circuit is conserved. Let’s see why Energy Oscillations Total energy in circuit is conserved. Let’s see why di q dt C + = 0 L Equation of LC circuit di q dq dt C dt L i + = 0 Multiply by i = dq/dt (i2)+ 1 d 2C dt (q2) = 0 2 = 2x dt dt Ld 2 dt dx dx Use ⎛ q2 ⎞ 2 q2 d 1 2 ⎜ ⎜ ⎟ ⎟ = 0 Li2 + 1 2 1 Li + = const 1 2 2 C UL + UC = const dt ⎝ ⎠ C PHY2049: Chapter 31 6

ωt +θ Energies can be written as (using ω2 = 1/LC) Oscillation of Energies Energies can be written as (using ω2 = 1/LC) q2 2C 2 2C qmax cos2(ωt +θ) UC = = 2 2C UL = 1 Li2 = 1 Lω2qmax sin2( ) = qmax 2 sin2(ωt +θ) ωt +θ 2 2C Energy 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 UL(t) PHY2049: Chapter 31 UC(t) Sum 8

( Parameters Calculate ω, f and T ω =1/ LC =1/ Calculate qmax and imax LC Circuit Example Parameters C = 20μF L = 200 mH Capacitor initially charged to 40V, no current initially Calculate ω, f and T ω = 500 rad/s ( 2×10−5)(0.2) = 500 ω =1/ LC =1/ f = ω/2π = 79.6 Hz T = 1/f = 0.0126 sec Calculate qmax and imax qmax = CV = 800 μC = 8 × 10-4 C imax = ωqmax = 500 × 8 × 10-4 = 0.4 A Calculate maximum energies UC = q2max/2C = 0.016J UL = Li2max/2 = 0.016J PHY2049: Chapter 31 9

Note how voltages sum to zero, as they must! LC Circuit Example (2) Charge and current q = 0.0008cos(500t) i = −0.4sin(500t) Energies UC = 0.016cos2(500t) UL = 0.016sin2(500t) Voltages VC = q/C = 40cos(500t) VL = Ldi/dt = −Lωimax cos(500t) = −40cos(500t) Note how voltages sum to zero, as they must! PHY2049: Chapter 31 10

Work out equation using loop rule RLC Circuit Work out equation using loop rule di q dt C + Ri+ = 0 L Rewrite using i = dq/dt 2 d q R dq q L dt LC + + = 0 dt Solution slightly more complicated than LC case −tR/2L cos(ω′t +θ) ω′ = 1/LC −(R/2L) 2 q = qmaxe This is a damped oscillator (similar to mechanical case) Amplitude of oscillations falls exponentially PHY2049: Chapter 31 11

e i(t) Charge and Current vs t in RLC Circuit q(t) −tR/2L PHY2049: Chapter 31 12

(0.012)(1.6×10−6) = 7220 Circuit parameters Calculate ω, ω’, f and T RLC Circuit Example Circuit parameters L = 12mL, C = 1.6μF, R = 1.5Ω Calculate ω, ω’, f and T ω = 7220 rad/s (0.012)(1.6×10−6) = 7220 ω =1/ ω’ = 7220 rad/s f = ω/2π = 1150 Hz ω′= 72202 −(1.5/0.024) 2 ω T = 1/f = 0.00087 sec Time for qmax to fall to ½ its initial value e−tR/2L =1/2 t = (2L/R) * ln2 = 0.0111s = 11.1 ms # periods = 0.0111/.00087 ≈ 13 PHY2049: Chapter 31 13

RLC Circuit (Energy) (UL C) = −i R di q dt C + Ri + = 0 Basic RLC equation L L i + Ri2 + = 0 di q dq dt C dt Multiply by i = dq/dt d 1 1 q2 ⎞ ⎛ Collect terms (similar to LC circuit) Li2 + 2 ⎜ ⎜ 2 ⎟ = −i R 2 dt ⎝ ⎠ ⎟ C Total energy in circuit decreases at rate of i2R (dissipation of energy) d 2 dt Utot ∼ e−tR/L (UL C) = −i R +U PHY2049: Chapter 31 14

Energy in RLC Circuit UC(t) UL(t) Sum −tR/L e PHY2049: Chapter 31 15

time to fully discharge the capacitors during the oscillations? Quiz Below 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 A B C has smallest capacitance, therefore highest frequency, therefore shortest period ω =1/ LC PHY2049: Chapter 31 16

Enormous impact of AC circuits Power delivery Radio transmitters and receivers Tuners Filters Transformers Basic components R L C Driving emf Now we will study the basic principles PHY2049: Chapter 31 17

RLC + “driving” EMF with angular frequency ωd AC Circuits and Forced Oscillations RLC + “driving” EMF with angular frequency ωd ε =εmsinωdt di q dt C + Ri+ =εm dt sinω L General solution for current is sum of two terms “Transient”: Falls exponentially & disappears Ignore i ∼ e−tR/2L cosω′t PHY2049: Chapter 31 “Steady state”: Constant amplitude 18

ωd Assume steady state solution of form i = Imsin(ωdt −φ) Im is current amplitude φ is phase by which current “lags” the driving EMF Must determine Im and φ Plug in solution: differentiate & integrate sin(ωt-φ) i = Imsin(ωdt −φ) Substitute di dt =ωdIm cos(ωdt −φ) di q dt C + Ri+ =εmsinωt L Im ωd cos(ωdt −φ) q =− ImωdLcos(ωdτ −φ)+ ImRsin(ωdt −φ)− Im ωdC cos(ωdt −φ) =εmsinωdt PHY2049: Chapter 31 19

Steady State Solution for AC Current (3) (ωdL−1/ωdC)cosφ − Rsinφ = 0 Same equations Im(ωdL−1/ωdC)sinφ + ImRcosφ =εm Solve for φ and Im in terms of tanφ = ≡ R R ωdL−1/ωdC XL − XC εm Z Im = R, XL, XC and Z have dimensions of resistance XL =ωdL XC =1/ωdC Inductive “reactance” Capacitive “reactance” Z = R +(XL C) 2 2 − X Total “impedance” Let’s try to understand this solution using “phasors” PHY2049: Chapter 31 21

ωdC XC = ("X R"= R ) Think of it as a frequency-dependent resistance What is Reactance? Think of it as a frequency-dependent resistance ωd → 0, XC → ∞ - Capacitor looks like a break ωd → ∞, XC → 0 - Capacitor looks like a wire (“short”) 1 ωdC XC = ωd → 0, XL → 0 - Inductor looks like a wire (“short”) ωd → ∞, XL → ∞ - Inductor looks like a break Independent of ωd X L =ωdL ("X R"= R ) PHY2049: Chapter 31 26

AC Source and RLC Circuit (2) Right triangle with sides VR, VL-VC and εm VR mR VR VL = ImXL 2 2 2 VC = ImXC VL −VC = I εm φ tanφ = εm =VR +(VL −VC) VL−VC VR Solve for current: i = Imsin(ωdt −φ) (Magnitude = Im, lags emf by phase φ) Im =εm /Z XL − XC ωdL−1/ωdC R R Z = R2 +( L C) = R2 +( d dC) 2 ω L−1/ω 2 X − X PHY2049: Chapter 31 29

AC Source and RLC Circuit (3) XL − XC R tanφ = Z = R2 +(XL − XC) 2 Only XL − XC is relevant, reactances cancel each other When XL = XC, then φ = 0 Current in phase with emf, “Resonant circuit”:ωd Z = R (minimum impedance, maximum current) When XL < XC, then φ < 0 Current leads emf, “Capacitive circuit”: ωd < ω0 When XL > XC, then φ > 0 Current lags emf, “Inductive circuit”: ωd > ω0 PHY2049: Chapter 31 =ω0 = 1/LC 30

ε t an RLC circuit. We can conclude the following I RLC Example 1 Below 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 ε I t PHY2049: Chapter 31 31

εm εm εm RLC circuit dominated by inductance? (3) Circuit 3 Quiz Which 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 εm εm 3 1 2 PHY2049: Chapter 31 32

εm εm εm RLC circuit dominated by capacitance? (1) Circuit 1 Quiz Which 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 εm εm 3 1 2 PHY2049: Chapter 31 33

Im Circuit parameters: C = 2.5μF, L = 4mH, εm = 10v Resonance RLC Example 3 Circuit 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 ωd =ω0 PHY2049: Chapter 31 35