19. LC and RLC Oscillators

Transcription

1 University of Rhode Island PHY 204: Eleentary Physics II Physics ourse Materials L and RL Oscillators Gerhard Müller University of Rhode Island, guller@uri.edu reative oons License This work is licensed under a reative oons Attribution-Noncoercial-Share Alike 4.0 License. Follow this and additional works at: Abstract Part nineteen of course aterials for Eleentary Physics II (PHY 204), taught by Gerhard Müller at the University of Rhode Island. Docuents will be updated periodically as ore entries becoe presentable. Soe of the slides contain figures fro the textbook, Paul A. Tipler and Gene Mosca. Physics for Scientists and Engineers, 5 th /6 th editions. The copyright to these figures is owned by W.H. Freean. We acknowledge perission fro W.H. Freean to use the on this course web page. The textbook figures are not to be used or copied for any purpose outside this class without direct perission fro W.H. Freean. Recoended itation Müller, Gerhard, "19. L and RL Oscillators" (2015). PHY 204: Eleentary Physics II. Paper 7. This ourse Material is brought to you for free and open access by the Physics ourse Materials at Digitaloons@URI. It has been accepted for inclusion in PHY 204: Eleentary Physics II by an authorized adinistrator of Digitaloons@URI. For ore inforation, please contact digitalcoons@etal.uri.edu.

2 Mechanical Oscillator law of otion: F = a, law of force: F = kx a = d2 x dt 2 d 2 x equation of otion: dt 2 = k x displaceent: x(t) = x ax cos(ωt) velocity: v(t) = ωx ax sin(ωt) angular frequency: ω = r k kinetic energy: K = 1 2 v2 k x potential energy: U = 1 2 kx2 total energy: E = K + U = const. 4/11/2015 [tsl294 1/12]

3 Electroagnetic Oscillator ( L ircuit) loop rule: Q + LdI dt = 0, I = dq dt d 2 Q equation of otion: dt 2 = 1 L Q charge on capacitor: Q(t) = Q ax cos(ωt) current through inductor: I(t) = ωq ax sin(ωt) angular frequency: ω = 1 L agnetic energy: U B = 1 2 LI2 (stored on inductor) L electric energy: U E = Q2 (stored on capacitor) 2 total energy: E = U B + U E = const. 4/11/2015 [tsl295 2/12]

4 Mechanical vs Electroagnetic Oscillations echanical oscillations position: x(t) = A cos(ωt) [red] velocity: v(t) = A sin(ωt) [green] potential energy: U(t) = 1 2 kx2 (t) [r] kinetic energy: K(t) = 1 2 v2 (t) [g] period: τ = 2π ω, ω = r k total energy: E = U(t) + K(t) = const A E t Τ A t Τ electroagnetic oscillations charge: Q(t) = A cos(ωt) [red] current: I(t) = A sin(ωt) [green] period: τ = 2π ω, ω = 1 L electric energy: U E (t) = 1 2 Q2 (t) [r] agnetic energy: U B (t) = 1 2 LI2 (t) [g] total energy: E = U E (t) + U B (t) = const 4/11/2015 [tsl497 3/12]

5 Mechanical Oscillator with Daping law of otion: F = a, law of force: F = kx bv, equation of otion: d 2 x dt 2 + b a = d2 x dt 2 v = dx dt dx dt + k x = 0 x t v t t Solution for initial conditions x(0) = A, v(0) = 0: t (a) underdaped otion: b 2 < 4k x(t) = Ae bt/2» cos(ω t) + b 2ω sin(ω t) with ω = s k b2 4 2 (b) overdaped otion: b 2 > 4k x(t) = Ae bt/2» cosh(ω t) + b 2Ω sinh(ω t) with Ω = s b k 4/11/2015 [tsl299 4/12]

6 Daped Electroagnetic Oscillator (RL ircuit) loop rule: RI + L di dt + Q equation of otion: d 2 Q dt 2 + R L = 0, I = dq dt dq dt + 1 L Q = 0 Solution for initial conditions Q(0) = Q ax, I(0) = 0: Q t I t t t (a) underdaped otion: R 2 < 4L Q(t) = Q ax e Rt/2L» cos(ω t) + R 2Lω sin(ω t) with ω = s 1 L R2 4L 2 (b) overdaped otion: R 2 > 4L Q(t) = Q ax e Rt/2L» cosh(ω t) + R 2LΩ sinh(ω t) with Ω = s R 2 4L 2 1 L 4/11/2015 [tsl300 5/12]

7 L ircuit: Application (1) Nae the L circuit with the highest and the lowest angular frequency of oscillation. L L (1) L L (2) L L L (3) L (4) 4/11/2015 [tsl296 6/12]

8 L ircuit: Application (2) At tie t = 0 a charge Q = 2 is on each capacitor and all currents are zero. (a) What is the energy stored in the circuit? (b) At what tie t 1 are the capacitors discharged for the first tie? (c) What is the current through each inductor at tie t 1? (1) (2) (3) (4) 4/11/2015 [tsl297 7/12]

9 L ircuit: Application (3) In these L circuits all capacitors have equal capacitance and all inductors have equal inductance L. Sort the circuits into groups that are equivalent. (1) (3) (5) (7) (2) (4) (6) (8) 4/11/2015 [tsl298 8/12]

10 Oscillator with Two Modes Electroagnetic: ode #1: L di dt + Q + Q + LdI dt = 0, di dt = Q L ode #2: L di dt + Q + 2Q = 0, di dt = 3Q L Mechanical: I = dq dt d2 Q dt 2 = ω2 Q, ω = 1 L I = dq dt d2 Q dt 2 = ω2 Q, ω = r 3 L a b a b ode #1: ω = ode #2: ω = r k r 3k k k k 4/11/2015 [tsl498 9/12]

11 RL ircuit: Application (1) In the circuit shown the capacitor is without charge. When the switch is closed to position a... (a) find the initial rate di/dt at which the current increases fro zero, (b) find the charge Q on the capacitor after a long tie. Then, when the switch is thrown fro a to b... (c) find the tie t 1 it takes the capacitor to fully discharge, (d) find the axiu current I ax in the process of discharging. 12V 4Ω 5H 20nF a b 4/11/2015 [tsl438 10/12]

12 RL ircuit: Application (2) In the circuit shown the capacitor is without charge and the switch is in position a. (i) When the switch is oved to position b we have an RL circuit with the current building up gradually: I(t) = (E/R)[1 e t/τ ]. Find the tie constant τ and the current I ax after a long tie. (ii) Then we reset the clock and ove the switch fro b to c with no interruption of the current through the inductor. We now have a an L circuit: I(t) = I ax cos(ωt). Find the angular frequency of oscillation ω and the axiu charge Q ax that goes onto the capacitor periodically. 34V 14Ω 6.2 µ F c 54H b a 4/11/2015 [tsl499 11/12]

13 RL ircuit: Application (3) In the circuit shown the capacitor is without charge and the switch is in position a. (i) When the switch is oved to position b we have an R circuit with the capacitor being charged up gradually: Q(t) = E[1 e t/τ ]. Find the tie constant τ and the charge Q ax after a long tie. (ii) Then we reset the clock and ove the switch fro b to c. We now have a an L circuit: Q(t) = Q ax cos(ωt). Find the angular frequency of oscillation ω and the axiu current I ax that flows through the inductor periodically. 34V 6.2 µ F 54H 14Ω b c a 4/11/2015 [tsl500 12/12]