Physics 2102

Jonathan Dowling

Lecture 30: FRI 27 MAR 09Lecture 30: FRI 27 MAR 09

Ch. 31.4–7: Electrical Ch. 31.4–7: Electrical Oscillations, LC Circuits, Oscillations, LC Circuits,

Alternating CurrentAlternating CurrentQuickTime™ and a

decompressorare needed to see this picture.

QuickTime™ and a decompressor

are needed to see this picture.

EXAM 03: 6PM THU 02 APR LOCKETT 5 EXAM 03 REVIEW: 6PM WED 01 APR NICHOLSON 130

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: TODAY FRI 27 MAR

The formula sheet for the exam can be found here:http://www.phys.lsu.edu/classes/spring2009/phys2102/formulasheet.pdf

You can see examples of old exam IIIs here:http://www.phys.lsu.edu/classes/spring2009/phys2102/Test3.oldtests.pdf

Damped LCR OscillatorDamped LCR Oscillator

Ideal LC circuit without resistance: oscillations go

on for ever; = (LC)–1/2

Real circuit has resistance, dissipates energy:

oscillations die out, or are “damped”

Math is complicated! Important points:

– Frequency of oscillator shifts away from

= (LC)-1/2

– Peak CHARGE decays with time constant =

QLCR=2L/R

– For small damping, peak ENERGY decays with

time constant

ULCR= L/R 0 4 8 12 16 200.0

0.2

0.4

0.6

0.8

1.0

UE

time (s)

Umax =Q2

2Ce−

RtL

C

RL

2

2

If we add a resistor in an circuit (see figure) we must

modify the energy equation, because now energy is

being dissipated on the resistor: .

E B

RL

dUi R

dt

qU U U

=−

= + =

Damped Oscillations in an CircuitRCL

22

2 2Li dU q dq di

Li i RC dt C dt dt+ → = + =−

( )

2

2

2

/2

2

2 2

10. This is the same equation as that

of the damped harmonics osc 0, which has theillator:

The angular f

solution

re( ) c que

:

os

.bt mm

d x dxm b kxdt dt

x

dq di d q d q dqi L R qdt dt dt dt dt

t x e t

C

ω φ−

= → = → + + =

+ + =

′= +

( )2

/22

2

2ncy

For the damped circuit the solution is:

The angular freque1

( ) cos . .4

ncy

.

4

Rt L Rq

k b

RC

t Qe t

m

L

L

C L

m

ω φ ω

ω

− ′ ′ −

−

+

=

= =

′

(31-6)

/ 2Rt LQe−

/ 2Rt LQe−

( )q tQ

Q−

( )q t ( )/ 2( ) cosRt Lq t Qe tω φ− ′= +

2

2

1

4

R

LC L′= −

/2

2

2

The equations above describe a harmonic oscillator with an exponentially decaying

amplitude . The angular frequency of the damped oscillator

1 is always smaller than the angular

4

Rt LQe

R

LC L

−

′= −

2

2

1frequency of the

1undamped oscillator. If the term we can use the approximation .

4

LCRL LC

=

′≈=

RC = RC                τ RL = L / R            τ RCL = 2L / R

SummarySummary• Capacitor and inductor combination

produces an electrical oscillator,

natural frequency of oscillator is

=1/√LC

• Total energy in circuit is conserved:

switches between capacitor (electric

field) and inductor (magnetic field).

• If a resistor is included in the

circuit, the total energy decays (is

dissipated by R).

Alternating Current:

To keep oscillations going we need to drive the circuit with an external emf that produces a current that goes back and forth.Notice that there are two frequencies involved: one at which the circuit wouldoscillate “naturally”. The other is the frequency at which we drive theoscillation.

However, the “natural” oscillation usually dies off quickly (exponentially) with time. Therefore in the long run, circuits actually oscillate with the frequency at which they are driven. (All this is true for the gentleman trying to make the lady swing back and forth in the picture too).

We have studied that a loop of wire,spinning in a constant magnetic fieldwill have an induced emf that oscillates with time,

€

E =Em sin(ωd t)

That is, it is an AC generator.

Alternating Current:

AC’s are very easy to generate, they are also easy to amplify anddecrease in voltage. This in turn makes them easy to send in distributiongrids like the ones that power our homes.

Because the interplay of AC and oscillating circuits can be quitecomplex, we will start by steps, studying how currents and voltagesrespond in various simple circuits to AC’s.

AC Driven Circuits:

1) A Resistor:

0=− Rvemf

€

vR = emf = Em sin(ωd t)

€

iR =vRR

=Em

Rsin(ωd t)

Resistors behave in AC very much as in DC, current and voltage are proportional (as functions of time in the case of AC),that is, they are “in phase”.

For time dependent periodic situations it is useful torepresent magnitudes using “phasors”. These are vectorsthat rotate at a frequency d , their magnitude is equalto the amplitude of the quantity in question and theirprojection on the vertical axis represents the instantaneous value of the quantity under study.

AC Driven Circuits:

2) Capacitors:

€

vC = emf = Em sin(ωd t)

€

qC =Cemf =CEm sin(ωd t)

€

iC =dqCdt

=ωdCEm cos(ωd t)

€

iC =ωdCEm sin(ωd t + 900)

€

iC =Em

Xsin(ωd t + 900)

reactance"" 1

whereC

Xd

=

€

im =Em

X

€

looks like i =V

RCapacitors “oppose a resistance” to AC (reactance) of frequency-dependent magnitude 1/d C (this idea is true only for maximum amplitudes, the instantaneous story is more complex).

AC Driven Circuits:

3) Inductors:

€

vL = emf = Em sin(ωd t)

Ldtv

idtid

Lv LL

LL

∫=⇒=

€

iL = −Em

Lωdcos(ωd t)

€

=Em

Lωdsin(ωd t − 900)

€

iL =Em

Xsin(ωd t − 900)

€

im =Em

X dLX = where

Inductors “oppose a resistance” to AC (reactance) of frequency-dependent magnitude d L(this idea is true only for maximum amplitudes, the instantaneous story is more complex).

Power Station

Transmission lines Erms =735 kV , I rms = 500 A Home

110 V

T1T2

Step-up transformer

Step-down transformer

R = 220Ω

1000 km=l

2heat rms

The resistance of the power line . is fixed (220 in our example).

Heating of power lines . This parameter is also fixed

(55 MW in our ex

R RA

P I R

ρ= Ω

=

Energy Transmission Requirements

l

trans rms rms

heat trans

heat rms

ample).

Power transmitted (368 MW in our example).

In our example is almost 15 % of and is acceptable.

To keep we must keep as low as possible. The onl

P I

P P

P I

=E

rms rms

y way to accomplish this

is by . In our example 735 kV. To do that we need a device

that can change the amplitude of any ac voltage (either increase or decrease).

=increasing E E

(31-24)

Ptrans=iV

= “big”

Pheat=i2R

= “small”

Solution:

Big V!

The transformer is a device that can change

the voltage amplitude of any ac signal. It

consists of two coils with a different number

of turns wound around a common iron core.

The Transformer

The coil on which we apply the voltage to be changed is called the " " and

it has turns. The transformer output appears on the second coil, which is known

as the "secondary" and has turnsP

S

N

N

primary

. The role of the iron core is to ensure that the

magnetic field lines from one coil also pass through the second. We assume that

if voltage equal to is applied across the primary then a voltagPV e appears

on the secondary coil. We also assume that the magnetic field through both coils

is equal to and that the iron core has cross-sectional area . The magnetic flux

through the primary

S

P

V

B A

Φ ( ).

The flux through the secondary ( ).

PP P P

SS S S S

d dBN BA V N A

dt dtd dB

N BA V N Adt dt

Φ= → =− =−

ΦΦ = → =− =−

eq. 1

eq. 2(31-25)

( )

( )

If we divide equation 2 by equati

.

on 1 we get:

S P

S P

PP P P P

SS S S S

SS S

P PP

d dBN BA V N A

dt dtd dB

N BA V N Adt dt

dBN AV Ndt

dBV NN Adt

V V

N N

ΦΦ = → =− =−

ΦΦ = → =− =−

−= =−

=→

eq. 1

eq. 2

The voltage on the secondary .

If 1 , we have what is known as a " " transformer.

If 1 , we have what is known as a " " transformer.

Both type

SS P

P

SS P S P

P

SS P S P

P

NV V

N

NN N V V

N

NN N V V

N

=

> → > → >

< → < → <

step up

step down

s of transformers are used in the transport of electric power over large distances.

S P

S P

V V

N N=

(31-26)

PI SI

If we close switch S in the figure we have in addition to the primary current

a current in the secondary coil. We assume that the transformer is " "

i.e., it suffers no losses due to heatin

P

S

I

I ideal,

g. Then we have: (eq. 2).

If we divide eq. 2 with eq. 1 we get:

In a step-up transformer ( ) we have that .

In a step-down transformer

.

(

P

P P S S

S SP P

P S S P

PS

P S S

PS

S P S P

I N

V I V I

V IV I

V N V N

N

I N

I IN

N N I I

N

=

= →

<

=

=

>) we have that .S P S PN I I< >

We have that:

(eq. 1).

S P

S P

S P P S

V V

N N

V N V N

=

→ =

S P

S P

V V

N N=

S S P PI N I N=

(31-27)