# Lecture 27: FRI 20 MAR - LSU Physics 2113 Jonathan Dowling Lecture 34: FRI 13 NOV Electrical Oscillations, LC Circuits, Alternating Current I Nikolai Tesla What are we going to learn? 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) Energy Density in E and B Fields e0E uE = 2 2

2 B uB = 2 m0 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 between kinetic and potential form. We have studied that inductors and capacitors are devices that can store electromagnetic energy. energy In the inductor it is stored in a B field, in the capacitor in an E field. PHYS2110: A Mechanical Oscillator E =K +U =ENERGY 1 1 2 2 E = mv + kx 2 2

dE 1 dv 1 dx =0 = m 2v + k 2x dt 2 dt 2 dt dv m + kx =0 dt Solution : v= x(t) a =v(t) = x(t) Newtons law d2 x m 2 + k x =0 F=ma! dt k x(t) = x0 cos(w t + f0 ) w = m x0 : amplitude w : f0 : frequency

phase PHYS2113 An Electromagnetic LC Oscillator Capacitor initially charged. Initially, current is zero, energy is all stored in the E-field of the capacitor. Energy Conservation: Utot =U B +U E A current gets going, energy gets split between the capacitor and the inductor. 1 2 1 q2 UB = L i UE = 2 2C Capacitor discharges completely, yet current keeps going. Energy is all in the B-field of the inductor all fluxed up. The magnetic field on the coil starts to deflux, which will start to recharge the capacitor. 1 2 1 q2 Utot = L i + 2 2C Finally, we reach the same state we started with (with opposite polarity) and the cycle restarts. Electric Oscillators: the Math Utot =U B +U E 1 2 1 q2 Utot = L i + 2

2C dUtot 1 di 1 dq =0 = L 2i + 2q dt 2 dt 2C dt Energy Cons. di 1 VL +VC =0 =- L - ( q) dt C Or loop rule! Both give Diffy-Q: Solution to Diffy-Q: d2 q q 0 =L 2 + dt C

q =q0 cos(w t + j 0 ) w 1 LC i =q(t) i (t) =q(t) LC Frequency In Radians/Sec i =q(t) =- q0w sin(w t + j 0 ) i(t) =q(t) =- w 2 q0 cos(w t + j 0 ) T a LC T b L(2C) T c L(C / 2) b a T >T >T c 1 w LCeq 2p

T= LCeq w a Ceq =C b Ceq =2C c Ceq =C / 2 Electric Oscillators: the Math q =q0 cos(w t + j 0 ) i(t) =q(t) =- q0w sin(w t + j 0 ) i(t) =q(t) =- w 2 q0 cos(w t + j 0 ) Energy as Function of Time Voltage as Function of Time 1 1 2 2 U B = L [ i ] = L [ q0w sin(w t + j 0 )] VL =Li (t) =- Lw 2 q0 cos(w t + j 0 ) 2 2 2 1 [ q]

1 2 UE = = [ q0 cos(w t + j 0 )] 2 C 2C 1 1 VC = [ q(t)] = [ q0 cos(w t + j 0 )] C C LC Circuit: At t=0 1/3 Of Energy Utotal is on Capacitor C and Two Thirds On Inductor L. Find Everything! (Phase 0=?)) 1 2 U B (t) = L [ q0w sin(w t + j 0 )] 2 1 2 U E ( t) = [ q0 cos(w t + j 0 )] 2C 22 1 L q w sin( j

) 0 U total / 3 U BB(0) 2 00 = = total 22 1 2U total /3 U EE 0 total q00 cos(j 00 ) 2C ( ) tan(j 0 ) = 2 LC q0w sin(j 0 ) 1 w =1 / LC = 2 2 q0 =VC q0 cos(j 0 ) q =q0 cos(w t + j 0 ) i(t) =- q0w sin(w t + j 0 ) 1 2

( 1 2 U B (0) = L [ q0w sin(j 0 )] =U total / 3 2 1 2 U E ( 0 ) = [ q0 cos(j 0 )] =2U total / 3 2C ) j 0 =arctan 1 / 2 =35.3 i (t) =- w 2 q0 cos(w t + j 0 ) q VL (t) =- 0 cos(w t + j 0 ) C q VC (t) = 0 cos(w t + j 0 ) C Analogy Between Electrical And Mechanical Oscillations d2 q q 0 =L 2 + dt C 1

w= LC d2 x m 2 + k x =0 dt k w= m q =q0 cos(w t + j 0 ) x(t) = x0 cos(w t + f0 ) i =q(t) =- q0w sin(w t + j 0 ) v= x(t) =- w x0 sin(w t + j 0 ) 2 i (t) =q(t) =- w q0 cos(w t + j 0 ) q x i v 1/C k L m a = x(t) =- w 2 x0 cos(w t + j 0 ) Charqe q -> Position x

Current i=q -> Velocity v=x Dt-Current i=q-> Acceleration a=v=x LC Circuit: Conservation of Energy q =q0 cos(w t + f0 ) 1.5 1 0.5 0 Time -0.5 Charge Current -1 dq i = =- w q0 sin(w t + f 0 ) dt 1 1 U B = Li 2 = Lw 2 q02 sin 2 (w t + j 0 ) 2 2 1 q2 1 UE = = q02 cos 2 (w t + j 0 )

2C 2C -1.5 1.2 And remembering that, 1 0.8 0.6 0.4 0.2 0 Time Energy in capacitor Energy in coil 1 cos x + sin x =1, and w = LC 2 2 1 2 Utot =U B + U E = q0 2C The energy is constant and equal to what we started with.

LC Circuit: Phase Relations q =q0 cos(w t + f0 ) 1.5 1 0.5 Charge 0 Time -0.5 Current -1 dq i = =- w q0 sin(w t + f 0 ) dt Take j 0 =0 as origin of time. -1.5 q cos(w t) i - sin(w t) Trigamarole: - sin(w t - p / 2) =cos(w t) current runs 90 out of phase with respect to the c

1.5 1 w LC 1 0.5 Charge 0 Time -0.5 Current 2p T= w -1 -1.5 t =0 T t =T / 4 t =T / 2 t =3T / 4

t =T 1.5 1 0.5 Charge 0 Time -0.5 Current -1 -1.5 (a) T / 2 (b) T (c) T / 2 (d) T / 4 t =0 Vc =q/ C t =T / 4 t =T t =T / 2 t =3T / 4

Example 1 : Tuning a Radio Receiver The inductor and capacitor in my car radio have one program at L = 1 mH & C = 3.18 pF. Which is the FM station?) (b) WRKF 89.3 What is wavelength of radio wave?) How about for WJBO 1150 AM?) FM radio stations: frequency is in MHz. 1 w= LC = 1 -6 1 10 3.18 10 =5.61 10 8 rad/s w f= 2p =8.93 10 7 Hz =89.3 MHz - 12 rad/s

Example 2 In an LC circuit, L = 40 mH; C = 4 mF At t = 0, the current is a maximum; When will the capacitor be fully charged for the first time?) 1.5 1 0.5 Charge 0 Time -0.5 Current -1 -1.5 t =0 t =T / 4 t =T / 2 t =3T / 4 t =T 1

1 w= = rad/s LC 16x10 - 8 w = 2500 rad/s T = period of one complete cycle T = pw = 2.5 ms Capacitor will be charged after T=1/4 cycle i.e at t = T/4 = 0.6 ms Example 3 In the circuit shown, the switch is in position a for a long time. It is then thrown to position b. Calculate the amplitude wq0 of the resulting oscillating current. 1 mH 1 mF b E=10 V

a i =- w q0 sin(w t + f0 ) Switch in position a: q=CV = (1 mF)(10 V) = 10 mC Switch in position b: maximum charge on C = q0 = 10 mC So, amplitude of oscillating current = 1 w q0 = (10mC) = 0.316 A (1mH)(1m F) Example 4 In an LC circuit, the maximum current is 1.0 A. If L = 1mH, C = 10 mF what is the maximum charge q0 on the capacitor during a cycle of oscillation? q =q0 cos(w t + f0 ) dq i = =- w q0 sin(w t + f 0 ) dt Maximum current is i0=wq0 Maximum charge: q0=i0/w Angular frequency w=1/LC=(1mH 10 LC=(1mH 10 mF)1/2 = (10-8)1/2 = 104 rad/s Maximum charge is q0=i0/w = 1A/104 rad/s = 104 C