Lecture 33: WED 12 NOV Electrical Oscillations, LC Circuits, Alternating Current I

Physics 2113 Jonathan Dowling

Nikolai Tesla

What are we going to learn? A road map

•  Electric charge è Electric force on other electric charges è Electric field, and electric potential

•  Moving electric charges : current •  Electronic circuit components: batteries, resistors, capacitors •  Electric currents è Magnetic field

è Magnetic force on moving charges •  Time-varying magnetic field è Electric Field •  More circuit components: inductors. •  Electromagnetic waves è light waves •  Geometrical Optics (light rays). •  Physical optics (light waves)

Energy Density in E and B Fields

uE =ε0E

2

2

uB =B2

2µ0

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.

Oscillators in Physics

We have studied that inductors and capacitors are devices that can store electromagnetic energy. In the inductor it is stored in a B field, in the capacitor in an E field.

E = K +U = ENERGY E = 12mv2 + 1

2k x2

dEdt

= 0 = 12m 2v dv

dt⎛⎝⎜

⎞⎠⎟ +

12k 2x dx

dt⎛⎝⎜

⎞⎠⎟

v = ′x (t)a = ′v (t) = ′′x (t)

→ mdvdt

+ k x = 0

)cos()( :Solution 00 φω += txtx

phase : frequency : amplitude :

0

0

φωx

mk=ω

PHYS2110: A Mechanical Oscillator

02

2

=+ xkdtxdm

Newton’s law F=ma!

The magnetic field on the coil starts to deflux, which will start to recharge the capacitor.

Finally, we reach the same state we started with (with opposite polarity) and the cycle restarts.

PHYS2113 An Electromagnetic LC Oscillator

Capacitor discharges completely, yet current keeps going. Energy is all in the B-field of the inductor all fluxed up.

Capacitor initially charged. Initially, current is zero, energy is all stored in the E-field of the capacitor.

A current gets going, energy gets split between the capacitor and the inductor.

Energy Conservation: Utot =UB +UE

Utot =12L i2 +

12qC

2

UB =12L i2      UE =

12qC

2

Utot =UB +UE Utot =12L i2 +

12qC

2

dUtot

dt= 0 = 1

2L 2i di

dt⎛⎝⎜

⎞⎠⎟+12C

2q dqdt

⎛⎝⎜

⎞⎠⎟

VL +VC = 0 = L didt

⎛⎝⎜

⎞⎠⎟+1C

q( )

i = ′q (t)′i (t) = ′′q (t)C

qdtqdL += 2

2

0

ω ≡1LC

q = q0 cos(ω t +ϕ0 )

Electric Oscillators: the Math

Or loop rule!

i = ′q (t) = −q0ω sin(ω t +ϕ0 )

′i (t) = ′′q (t) = −ω 2q0 cos(ω t +ϕ0 )

Energy Cons.

Both give Diffy-Q: Solution to Diffy-Q:

LC Frequency In Radians/Sec

Ceqa = C

Ceqb = 2C

Ceqc = C / 2

ω ≡ 1LCeq

T = 2πω

∝ LCeq

T a ∝ LC

T b ∝ L(2C)

T c ∝ L(C / 2)

T b > T a > T c

UB =12L i[ ]2 = 1

2L q0ω sin(ω t +ϕ0 )[ ]2 VL = L ′i (t) = −Lω 2q0 cos(ω t +ϕ0 )

q = q0 cos(ω t +ϕ0 )

Electric Oscillators: the Math

i(t) = ′q (t) = −q0ω sin(ω t +ϕ0 )

′i (t) = ′′q (t) = −ω 2q0 cos(ω t +ϕ0 )

UE =12q[ ]C

2

=12C

q0 cos(ω t +ϕ0 )[ ]2

Energy as Function of Time Voltage as Function of Time

VC =1C

q(t)[ ] = 1C

q0 cos(ω t +ϕ0 )[ ]

LC Circuit: At t=0 1/3 Of Energy Utotal is on Capacitor C and Two Thirds On Inductor L. Find Everything! (Phase φ0=?)

U B (0)

U E 0( ) =12

L q0ω sin(ϕ0 )⎡⎣ ⎤⎦2

12C

q0 cos(ϕ0 )⎡⎣ ⎤⎦2=

U total / 32U total / 3

UB (t) =12L q0ω sin(ω t +ϕ0 )[ ]2

UE t( ) = 12C

q0 cos(ω t +ϕ0 )[ ]2UB (0) =

12L q0ω sin(ϕ0 )[ ]2 =U total / 3

UE 0( ) = 12C

q0 cos(ϕ0 )[ ]2 = 2U total / 3

LC q0ω sin(ϕ0 )⎡⎣ ⎤⎦2

q0 cos(ϕ0 )⎡⎣ ⎤⎦2

=12

ω = 1 / LCq0 =VC

tan(ϕ0 ) =12

ϕ0 = arctan 1 / 2( ) = 35.3°

q = q0 cos(ω t +ϕ0 )i(t) = −q0ω sin(ω t +ϕ0 )′i (t) = −ω 2q0 cos(ω t +ϕ0 )

VL (t) = −q0Ccos(ω t +ϕ0 )

VC (t) =q0Ccos(ω t +ϕ0 )

02

2

=+ xkdtxdm

Analogy Between Electrical And Mechanical Oscillations

q→ x 1 /C→ ki→ v L→ m

LC1=ω

)cos()( 00 φω += txtx

mk=ω

Cq

dtqdL += 2

2

0

q = q0 cos(ω t +ϕ0 )

i = ′q (t) = −q0ω sin(ω t +ϕ0 )

′i (t) = ′′q (t) = −ω 2q0 cos(ω t +ϕ0 )

v = ′x (t) = −ωx0 sin(ω t +ϕ0 )

a = ′′x (t) = −ω 2x0 cos(ω t +ϕ0 )

Charqe q -> Position x Current i=q’ -> Velocity v=x’ Dt-Current i’=q’’-> Acceleration a=v’=x’’

-1.5

-1

-0.5

0

0.5

1

1.5

Time

ChargeCurrent

)cos( 00 φω += tqq)sin( 00 φωω +−== tq

dtdqi

UB =12Li2 = 1

2Lω 2q0

2 sin2 (ω t +ϕ0 )

0

0.2

0.4

0.6

0.8

1

1.2

Time

Energy in capacitorEnergy in coil

UE =12qC

2

=12C

q02 cos2 (ω t +ϕ0 )

LCxx 1 and ,1sincos

that,grememberin And

22 ==+ ω

Utot =UB +UE =12C

q02

The energy is constant and equal to what we started with.

LC Circuit: Conservation of Energy

-1.5

-1

-0.5

0

0.5

1

1.5

Time

ChargeCurrent

)cos( 00 φω += tqq

)sin( 00 φωω +−== tqdtdqi

LC Circuit: Phase Relations

Take ϕ0 = 0 as origin of time.

q ∝ cos(ω t)

i ∝−sin(ω t)

Trigamarole: − sin(ω t −π / 2) = cos(ω t)

The current runs 90° out of phase with respect to the charge.

-1.5

-1

-0.5

0

0.5

1

1.5

Time

ChargeCurrent

ω ≡1LC

T = 2πω

t = 0 ×T t = Tt = T / 4 t = T / 2 t = 3T / 4

(a) T / 2

(b) T

(c) T / 2

-1.5

-1

-0.5

0

0.5

1

1.5

Time

ChargeCurrent

t = 0

t = T

t = T / 2

t = T / 4

t = 3T / 4(d) T / 4

Vc = q /C

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?

(a) KLSU 91.1 (b) WRKF 89.3 (c) Eagle 98.1 WDGL

FM radio stations: frequency is in MHz.

ω =1LC

=1

1×10−6 × 3.18 ×10−12rad/s

= 5.61×108 rad/s

f = ω2π

= 8.93×107Hz= 89.3 MHz

Example 2 •  In an LC circuit,

L = 40 mH; C = 4 µF •  At t = 0, the current is a

maximum; •  When will the capacitor

be fully charged for the first time?

ω =1LC

=1

16x10−8rad/s

•  ω = 2500 rad/s •  T = period of one complete cycle • T = 2π/ω = 2.5 ms •  Capacitor will be charged after T=1/4 cycle i.e at •  t = T/4 = 0.6 ms -1.5

-1

-0.5

0

0.5

1

1.5

Time

ChargeCurrent

t = 0 t = Tt = T / 2t = T / 4 t = 3T / 4

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 ωq0 of the resulting oscillating current.

•  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 =

ωq0 =1

(1mH)(1µF)(10µC) = 0.316 A

)sin( 00 φωω +−= tqi

b a

E=10 V 1 mH 1 mF

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?

)cos( 00 φω += tqq)sin( 00 φωω +−== tq

dtdqi

Maximum current is i0=ωq0 Maximum charge: q0=i0/ω��� Angular frequency w=1/√LC=(1mH 10 mF)–1/2 = (10-8)–1/2 = 104 rad/s Maximum charge is q0=i0/ω = 1A/104 rad/s = 10–4 C