Lecture 29: WED 25 MAR 09 Lecture 29: WED 25 MAR 09 Ch. 31.1 Ch. - PowerPoint PPT Presentation

Physics 2102 Jonathan Dowling Lecture 29: WED 25 MAR 09 Lecture 29: WED 25 MAR 09 Ch. 31.1 Ch. 31.1– –4: Ele 4: Ele lectric ical l Oscilla illatio ions, LC Ch. 31.1 Ch. 31.1 lectric ical l Oscilla illatio ions, LC Cir ircuit its, Alt lternatin ing Current Cir ircuit its, Alt lternatin ing Current

EXAM 03: 6PM THU 02 APR 2009 The exam will cover: Ch.28 (second half) through Ch.32.1-3 (displacement current, and Maxwell's equations). The exam will be based on: HW08 – HW11 Final Day to Drop Course: FRI 27 MAR

What are we going to learn? hat are we going to learn? What are we going to learn? hat are we going to learn? A road map A road map A road map A road map • Electric charge  Electric force on other electric charges  Electric field , and electric potential • Moving electric charges : current • Electronic circuit components: batteries, resistors, capacitors • Electric currents  Magnetic field  Magnetic force on moving charges • Time-varying magnetic field  Electric Field • More circuit components: inductors. • Electromagnetic waves  light waves • Geometrical Optics (light rays). • Physical optics (light waves)

Oscillators in Physics Oscillators in Physics Oscillators in Physics Oscillators in Physics Oscillators are very useful in practical applications, for instance, to keep time, or to focus energy in a system. All oscillators can store energy in more than one way and exchange it back and forth between the different storage possibilities. For instance, in pendulums (and swings) one exchanges energy xchanges energy between kinetic and potential form. kinetic and potential We have studied that inductors and capacitors nductors and capacitors are devices that can store electromagnetic energy electromagnetic energy lectromagnetic energy. In the inductor it is lectromagnetic energy stored in a B field, in the capacitor in an E field.

PHYS2101: A Mechanical Oscillator PHYS2101: A Mechanical Oscillator PHYS2101: A Mechanical Oscillator PHYS2101: A Mechanical Oscillator 2 mv 2 + 1 U tot = 1 U tot = U kin + U pot = const 2 k x 2 � � � � dU tot = 0 = 1 2 m 2 v dv � + 1 2 k 2 x dx v = � x ( t ) � � � � � � � a = � v ( t ) = �� x ( t ) dt dt dt � m dv Newton’s law 2 d x dt + k x = 0 + = m k x 0 F=ma! 2 dt k � = = � + � Solution : x ( t ) x cos( t ) 0 0 m x : amplitude 0 � : frequency � : phase 0

PHYS210 101 A 1 An E n Ele lectr trom omagne gnetic tic LC LC O Osc scilla illator tor PHYS210 101 A 1 An E n Ele lectr trom omagne gnetic tic LC LC O Osc scilla illator tor Capacitor initially charged. Initially, current is zero, energy is all stored in the capacitor. Energy � Conservation: � U tot = U B + U E A current gets going, energy gets split between the capacitor and the inductor. 2 U B = 1 2 Li 2 ����� U E = 1 q 2 C Capacitor discharges completely, yet current keeps going. Energy is all in the inductor. The magnetic field on the coil starts to collapse, which will start to recharge the capacitor. 2 2 Li 2 + 1 U tot = 1 q 2 C Finally, we reach the same state we started with (with opposite polarity) and the cycle restarts.

Ele lectric ric Oscilla illators rs: the Math Ele lectric ric Oscilla illators rs: the Math 2 2 Li 2 + 1 U tot = 1 q U tot = U B + U E 2 C � � � � dU tot = 0 = 1 2 L 2 i di � + 1 2 C 2 q dq Energy Cons. � � � � � � � dt dt dt � � V L + V C = 0 = L di � + 1 ( ) Or loop rule! � C q � � dt Both give Diffy-Q: Solution to Diffy-Q: q = q 0 cos( � t + � 0 ) 2 i = � d q q q ( t ) = + 0 L i ( t ) = �� � 2 q ( t ) dt C i = � q ( t ) = � q 0 � sin( � t + � 0 ) LC Frequency 1 � � i ( t ) = �� � q ( t ) = � � 2 q 0 cos( � t + � 0 ) In Radians/Sec LC

Ele lectric ric Oscilla illators rs: the Math Ele lectric ric Oscilla illators rs: the Math q = q 0 cos( � t + � 0 ) i = � q ( t ) = � q 0 � sin( � t + � 0 ) i ( t ) = �� � q ( t ) = � � 2 q 0 cos( � t + � 0 ) Voltage as Function of Time Energy as Function of Time 2 = 1 U B = 1 [ ] [ ] 2 2 2 L q 0 � cos( � t + � 0 ) V L = L � i ( t ) = � 2 q 0 sin( � t + � 0 ) � � 2 L i � � [ ] 2 U E = 1 q = 1 [ ] V C = 1 ] = 1 [ ] [ 2 C q 0 cos( � t + � 0 ) 2 C q 0 cos( � t + � 0 ) C q ( t ) 2 C

Analogy Between Electrical And Mechanical Oscillations 2 d q q 1 k 2 d x = + � = 0 L � = + = m k x 0 2 LC dt C m dt 2 q = q 0 cos( � t + � 0 ) = � + � x ( t ) x cos( t ) 0 0 v = � x ( t ) = � x 0 � sin( � t + � 0 ) i = � q ( t ) = � q 0 � sin( � t + � 0 ) a = �� x ( t ) = � � 2 x 0 cos( � t + � 0 ) i ( t ) = �� � q ( t ) = � � 2 q 0 cos( � t + � 0 ) Charqe q -> Position x q � x 1/ C � k Current i=q’ -> Velocity v=x’ i � v L � m D-Current i’=q’’-> Acceleration a=v’=x’’

LC Circ ircuit it: Conserv rvatio ion of Energ rgy LC Circ ircuit it: Conserv rvatio ion of Energ rgy = � + � q q cos( t ) 0 0 1.5 dq = = � � � + � i q sin( t ) 1 0 0 dt 0.5 Charge 0 2 Li 2 = 1 U B = 1 Current 2 sin 2 ( � t + � 0 ) 2 L � 2 q 0 Time -0.5 -1 2 U E = 1 q = 1 -1.5 2 cos 2 ( � t + � 0 ) 2 C q 0 2 C 1.2 And rememberin g that, 1 1 0.8 + = � = cos 2 x sin 2 x 1 , and Energy in capacitor 0.6 LC Energy in coil 0.4 U tot = U B + U E = 1 2 0.2 2 C q 0 0 Time The energy is constant and equal to what we started with.

Example le 1 : Tunin ing a Radio io Receiv iver Example le 1 : Tunin ing a Radio io Receiv iver FM radio stations: frequency is in MHz. The inductor and capacitor in my car radio are usually 1 � = set at L = 1 mH & C = 3.18 LC pF. Which is my favorite FM 1 station? = 1 � 10 � 6 � 3.18 � 10 � 12 rad/s = 5.61 � 10 8 rad/s (a) KLSU 91.1 f = � (b) WRKF 89.3 2 � (c) Eagle 98.1 WDGL = 8.93 � 10 7 Hz = 89.3 � MHz

Example 2 2 Example • In an LC circuit, 1 1 L = 40 mH; C = 4 µ F � = = 16 x 10 � 8 rad/s LC • At t = 0, the current is a maximum; • ω = 2500 rad/s • When will the capacitor • T = period of one be fully charged for the complete cycle first time? •T = 2 π / ω = 2.5 ms 1.5 • Capacitor will be 1 charged after T=1/4 0.5 Charge 0 Current cycle i.e at Time -0.5 -1 • t = T/4 = 0.6 ms -1.5

Example le 3 Example le 3 • In the circuit shown, the switch is in position “a” for a E =10 V long time. It is then thrown 1 mH 1 µ F to position “b.” • Calculate the amplitude ω q 0 a b of the resulting oscillating current. = � � � + � i q sin( t ) 0 0 • Switch in position “a”: q=CV = (1 µ F)(10 V) = 10 µ C • Switch in position “b”: maximum charge on C = q 0 = 10 µ C • So, amplitude of oscillating current = 1 � q 0 = (10 µ C) = 0.316 A (1mH)(1 µ F)

Example 4 Example 4 In an LC circuit, the maximum current is 1.0 A. If L = 1mH, C = 10 µ F what is the maximum charge q 0 on the capacitor during a cycle of oscillation? = � + � q q cos( t ) 0 0 dq = = � � � + � i q sin( t ) 0 0 dt Maximum current is i 0 = ω q 0 → Maximum charge: q 0 =i 0 / ω Angular frequency ω =1/ √ LC=(1mH 10 µ F) –1/2 = (10 -8 ) –1/2 = 10 4 rad/s Maximum charge is q 0 =i 0 / ω = 1A/10 4 rad/s = 10 –4 C