Chapter 7. Electrodynamics (전기동력학 -...

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Chapter 7. Electrodynamics (전기동력학)

Lecture Note #7A

7.1 Electromotive Force (기전력)

7.2 Electromagnetic Induction (전자기유도)

7.3 Maxwell’s Equations

The Ohm’s Law :

The electrical power delivered by the current I in the electric potential V :

The electromotive force (emf) of the circuit:

EJ

RIV

where is the conductivity of the medium (전도도).

RIIVP 2

: the Joule heating law

ldfldf s

q

Ff

Electric Generator (발전기) = a motional emfdt

d

Eddy current

Electrostatics

(정전기학)

Magnetostatics

(정자기학)

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7.1 Electromotive Force

For the force pushing charges f (= the force per unit charge), the current density becomes

7.1.1 Ohm’s Law

fJ

(7.1)

where is the conductivity(전도도) of the medium.

1

The resistivity(비저항) of the medium is written as

(Do not confuse this with the surface charge density)

(Do not confuse this with the volume charge density)

For metals

0

For insulators

0

q

Ff

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For an electromagnetic force pushing charges

7.1.1 Ohm’s Law

BvEJ

(7.2)

When the velocity of the charges is sufficiently small, the 2nd term can be ignored:

EJ

: true in most metals or solids

- continued

(7.3): Ohm’s Law

: Not true in plasmas where the magnetic contribution to the force is important.

(Solution)

[Example 7.1]

VL

A

L

VAAEAJI

The field inside the resistor is uniform, and

the current density is uniform.

IRA

LIV

A

L

A

LR

BvE

q

BvEq

q

Ff

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7.1.1 Ohm’s Law - continued

(Solution)

[Example 7.2]

ss

E ˆ2 0

The field between the cylinder is

where is the charge per unit length on the inner cylinder.

The current becomesLadEadJI

0

(The integral is over any surface enclosing the inner cylinder.)

The potential difference between the cylinders is

a

bdss

ldEaVbVb

a

b

aln

22 00

V

abln

2 0

Thus, the current is written as

V

ab

LLI

ln

2

0

IR

L

abIV

2

ln

L

abR

2

ln

V

a

bV ln

2 0

where is the

conductivity(전도도)

of the medium.

0enclosed

S

qadE

0

2

LsLE

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L

I



where R is the “resistance”.

The total current flowing from one electrode to the other (potential difference between them)

: a function of the geometry of the arrangement and

the conductivity of the medium.

RIV

L

abR

2

ln

(7.4)

: Ohm’s Law (traditional version)

Unit : ohm ()

A

L

A

LR

a

b

For steady currents and uniform conductivity,

Eq. (7.3) 01

JE

(7.5) Eq. (5.33) 0 J

for a steady current density

Eq. (5.26) vJ

0 00

E

from the Gauss’s law

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(Solution)

[Example 7.3]

0V V00ˆ nJ

On the cylindrical surface,

0

n

V

0ˆ nE

02 VInside the cylinder,

EJ

Boundary conditions: L

zVzV 0

Thus, the electric field becomes zL

V

L

zV

zzzVE ˆˆ 00

: constant along the z-direction

If there is no conducting material,

The electric field becomes more complicated.

For steady currents and uniform conductivity,

01

JE

VE

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0V V0

l

there is a force acting on the charge q due to the electric field.

v

dt

dl

dt

ld

dt

dqI

E

a

lt

2

Inside the cylinder with an applied voltage V,

EqamF

: acceleration acting on the charge

2

2

1atl

The time needed for a charged particle to travel a distance l with an acceleration a

before a collision is

The average velocity of the particle is

The time between collisions due to the thermal motion is

q

m

Eqa

Does the current keep increasing with time?

E

q

22

1 laatvave

The thermal velocity (= drift velocity) of the particle = vthermal = vdrift

thermalv

lt

thermal

avev

alatv

22

1

2

02

1v attdd o

atvv o

atvvvave2

1

2

10

(l : mean free path of the charged particle)

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0V V0

E

If there are n molecules per unit volume, and f free electrons per

molecule, each with charge q and mass m, the current density is

Emv

lnfq

m

F

v

nfql

v

lanfqvnfqJ

thermalthermalthermal

222

2

: the Joule heating law

Due to the collisions, the work done by the electrical force is converted into heat

in the resistor.

q

volume

electronsfree

molecule

electronsfreef

volume

moleculesn

As temperature increases, vthermal increases, and thus

the current density J decreases.

RIIVP 2

J

The electrical power delivered by the current (charge flow per unit time) I in the

electric potential V is

I (A, ampere), R (, ohm) P (J/s, joules/second)

(7.6)

(7.7)

l

E

q

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The total force pushing charges f (= the force per unit charge) into the circuit is

7.1.2 Electromotive Force

Eff s

(7.8)

where fs is the source force confined to one part like a battery.

There are two forces involved in driving current around a circuit.

- A chemical force in a battery

since

E is an electrostatic force causing the smooth current flow in the circuit.

- Mechanical pressure converted into an electrical impulse in a piezoelectric crystal

- A temperature gradient in a thermocouple.

- The electrons loaded onto a conveyer belt in a Van de Graaff generator

The net effect determined over the entire circuit :

ldfldf s

.0 ldE

(7.9)

: electromotive force (emf) of the circuit: 회로의기전력

(Do not confuse this with the electric permitivity)

chargeunit

force a of integral

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For an ideal source of emf (ex., a resistanceless battery),

7.1.2 Electromotive Force

0 Eff s

because

The potential difference between the terminals a and b is

because outside the source.

ldfldfldEV s

b

as

b

a

(Inside the battery, fs derives current in the direction

opposite to E)

- continued

Eq. (7.1) 0

Jf

,01

Eq. (7.8) sfE

ab

0sf

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B

uq


Electric Generator (발전기) = a motional emf

7.1.3 Motional emf (발전력)

vBq

Ff

mag

mag

(7.11)

When a wire moves across a magnetic field, an induced current

flow along the wire.

- The charges in segment ab experience a magnetic force :

The charged particle moves in the direction of the resultant

velocity w.

The magnetic force acting on flowing charges per unit charge

over the wire section ab =

The work done per unit charge is :

vBhldfmag

(to the left-had direction)

qvBFmag

induces current in the clockwise direction.

- The emf is

uBq

Ff

mag

mag

pullmag fuBf

The magnetic force induces the emf, but no work.

The pulling force does provide the work.

vBhh

uBldf pull sincos

( = emf )

(⊥the emotion of charges)

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B

uq


Another method using the magnetic flux :

7.1.3 Motional emf

Bhx

Eq. (7.11)

Let be the flux of B through the loop:

For the rectangular loop in the magnetic field,

As the loop moves, the flux decreases:

vBhldfmag

adB

: the flux rule for motional emf.

- continued

Bhvdt

dxBh

dt

d

dt

d

(7.12)

(7.13)

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The flux rule can be applied to nonrectangular loops moving in arbitrary directions

through nonuniform magnetic fields; in fact, the loop need not even maintain a fixed shape.

7.1.3 Motional emf

dtldvad

For a time-varying loop of wire at time t and t+dt,

the change in flux is

ribbon

ribbon adBtdttd

(Proof)

- continued

ldvBdt

d

Thus,

The point P moves to P’ in time dt.

Let v : the velocity of the wire.

u : the velocity of a charge down the wire

w = v+ u : the resultant velocity of a charge at P.

The infinitesimal element of area on the ribbon becomes

Since w = v+ u and u is parallel to dl, we get

ldwBdt

d

ldBwldwB

BACACBCBA

From Chapter 1.

Since

ldBwdt

d ,magfBw

ldf

dt

dmag

dt

d

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7.1.3 Motional emf - continued

The magnetic flux rule is a convenient method to calculate the motional electromotive force.

The magnetic flux rule paradox

B

However, some care must be taken to avoid any error or ambiguity.

When the switch is thrown from a to b in the figure shown below,

the magnetic flux increases twice.

Do you expect any motional emf induced during the switching process?

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s


(Solution)

The speed of a point on the disk at a distance s from the axis is

ssBBvfq

Fmaf

mag ˆ

sv

The force per unit charge is

The emf is2

2

00

BadssBdsf

aa

mag

The current isR

Ba

RI

2

2

“Faraday disk”

“Faraday dynamo”

Since the current spreads out over the whole disk, it is not easy

to use the flux rule.

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7.1.3 Motional emf - continued

In the case of a chunk of aluminum is shaken

around in a nonuniform magnetic field,

Eddy current

Currents (eddy current) will be generated in

the material.

A chunk of

aluminum

A kind of “viscous drag” will be felt as though

the block is pulled through molasses (fpull).

Eddy currents are very difficult to calculate.

http://saudisico.com/index_files/TubeTestingECTMFLRFTIRIS.htm

Applications for the detection and

sizing of metal discontinuities

such as corrosion, erosion, wear,

pitting, baffle cuts, wall losses,

and cracks in nonferrous

materials.

http://www.olympus-ims.com/en/ndt-tutorials/eca-tutorial/intro/

http://isites.harvard.edu/ic

b/icb.do?keyword=k1694

0&pageid=icb.page91941

&pageContentId=icb.page

content216732&view=vie

w.do&viewParam_name=

indepth.html

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Next Class

Chapter 7. Electrodynamics


7.2 Electromagnetic Induction


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Chapter 7. ElectrodynamicsLecture Note #7B




(전자기유도)

The Faraday’s Law :

The Lentz’s law :

(Electromagnetic Induction)

Mutual Inductance :

dt

d

Self Inductance :

adt

BldE

t

BE

dt

dldE

r

ldldM 210

214

dt

dIM

dt

d 122

IL dt

dIL

dt

d

Energy stored in the magnetic field :2

2

1LIW

spacealldBW

2

0

v 2

1

IL

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Michael Faraday’s 1831 experiment

7.2.1 Faraday’s Law [Michael Faraday (22 Sept. 1791 – 25 Aug. 1867),

an English scientist]

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From the Michael Faraday’s experiment,

7.2.1 Faraday’s Law - continued

dt

d

A changing magnetic field induces an electric field.

dt

dldE

(7.14)

Thus,

adt

BldE

(7.15)

LoopS

ldad

vv

From Stokes’ Theorem

(Curl Theorem)

: Faraday’s law (in integral form)

t

BE

: Faraday’s law (in differential form)

(7.16)

In the static case, ,0 ldE

0 E

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What is the main difference between these two cases for electromagnetic induction?

B

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(Solution)

[Example 7.5]

7.22

Let us choose the z axis along the direction of M.

0 MJb

ˆ MnMKb

This is just like a long solenoid with surface current MKb

Thus, the magnetic field inside the cylinder, except near the ends, is MB

0

The magnetic flux through the cylinder changes from zero to the maximum of 2

0max aM

dt

d

“-” sign indicates the “Lenz’s law”.

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AC (alternating current) Generator

Source: http://www.electronics-tutorials.ws/accircuits/sinusoidal-waveform.html

http://www.explainthatstuff.com/generators.html

DC (direct current) Generator

http://www.ustudy.in/node/3985

http://www.wisc-online.com/objects/index_tj.asp?objID=IAU13108

http://www.wisc-online.com/objects/index_tj.asp?objID=IAU13108

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dt

d

“-” sign indicates the “Lenz’s law”.

Lenz’s law :

Heinrich Lenz• a Russian physicist of Baltic German ethnicity.

• Born (1804-02-12), Dorpat, Russian Empire

• Died (1865-02-10) (aged 60), Rome, Italy

http://en.wikipedia.org/wiki/Heinrich_Lenz

Nature abhors a change in flux.

http://hyperphysics.phy-

astr.gsu.edu/hbase/electric/farlaw.html

http://en.wikipedia.org/wiki/File:Heinrich_Friedrich_Emil_Lenz.jpg

http://en.wikipedia.org/wiki/File:Heinrich_Friedrich_Emil_Lenz.jpg

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(Solution)

As the current on the solenoid is turned on,

the induced current on the ring flows un such a direction that

its field tends to cancel the generating flux by the solenoid

according to the Lenz’s law.

[Example 7.6]

http://www.youtube.com/watch?v=Pl7KyVIJ1iE

http://www.youtube.com/watch?v=Pl7KyVIJ1iE

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Faraday’s Law generalizes the electrostatic rule to the time-dependent regime.

7.2.2 The Induced Electric Field

0 E

t

BE

The Gauss’s law holds for both cases.

0

E

If the electric field is a pure Faraday field (due to exclusively to a changing B, with = 0),

,0 E

t

BE

For the magnetostatics,

,0 B

JB

0

The Biot-Savart law is (5.42)

The analog to Biot-Savart law is

v

ˆ

4

1v

ˆ

4

122

dr

rB

td

r

rtBrE

(7.18)

dt

dldE

(7.19)

.0 encIdlB

Ampere’s lawFaraday’s law (integral form)

'ˆ'J

4 2

0 dvr

rrrB

adB

(7.12)

(5.28)

S

adJI

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The induced electric field points in the circumferential direction.

Therefore,

2 dt

dBsE

dt

dBstBs

dt

d

dt

dsEldE 22ˆ2ˆ

(Solution)

[Example 7.7]

- continued

7.25.

For an Amperian loop of radius s,

If B is increasing with time, E runs clockwise, as viewed from above.

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The changing magnetic field will induce an electric field.

Therefore, 2

2

dt

dB

b

aE

dt

dBatBa

dt

d

dt

dbEldE 22ˆ 2ˆ

(Solution)

For an Amperian loop of radius b,

→ This the motion is counterclockwise, as viewed from above.

→ This electric field exerts a force on the charges at the rim, and the

wheel starts to turn.

→ According to Lenz’s law, it will rotate in such a direction that its field

tends to restore the upward flux.

The torque on a segment of length dl is Fdrd

dlEbEdqbd

The total torque on the wheel is

dt

dBabb

dt

dBa-dl

dt

dB

b

a-bdlEbd 2

22

22

2

Since ,dt

dL the angular momentum imparted to the wheel is

0

20

22

0

bBadBabdtdt

dBabdtL

B

: does not depend on

the switching speed.

(Direction: upward B0 direction)

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For the case of magnetostatics where the magnetic field is constant,

- those magnetostatics laws can be used to calculate the magnetic field.

“quasistatic”

- continued

Even for cases of varying magnetic field,

: Biot-Savart law

2

0

2

0ˆ'

4'

ˆI

4 r

rldIdl

r

rrB

.0 encIdlB

: Ampere’s law

- The error is usually negligible, unless the field fluctuates extremely rapidly,

or you are interested in points very far from the source.

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In the quasistatic approximation,

Therefore, zKsdt

dIsE ˆ ln

2

0

000

0 lnln2

''

1

2 0

ssdt

dIlds

sl

dt

dIadB

dt

dlSElSEldE

s

s

(Solution)

For the rectangular “Amperian loop” in the figure, Faraday’s law gives :

→ Then, the induced electric field E runs parallel to the wire.

If we assume that K is a constant, E → as s → . (This is not correct !!)

K : independent of s

: is still a function of time t.

the magnetic field around the wire iss

IB

2

0

0

00

000

0 ln2

ln2

lnln2

sdt

dISEs

dt

dISEss

dt

dISE

The quasistatic approximation is valid only when tcs

if “t” is the time it takes I to change substantially.

valid only

for small s.

(7.21)

(7.20)

tc

s

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Suppose that we have two loops of wire at rest.

7.2.3 Inductance

The magnetic field

generated by the loop 1 =

r

The Biot-Savart law becomes

1

2

11

0

ˆ

4

I

r

rldIrB

Thus,

212 adB

(7.22)

(7.23)

1B

→ purely geometrical quantity depending on the size,

shapes, and relative position of the two loops.

: Neumann formula

The magnetic flux of B1 through the loop 2 =

1212 IM

where M21 is the constant of proportionality (“Mutual Inductance of the two loop).

Using the vector potential A and the Stokes’ theorem, 2121212 ldAadAadB

Eq. (5.66) →

r

ldIA 110

14

2110

2124

ldr

ldIldA

r

ldldM 210

214

Therefore,

1221 MM (7.24)

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(Solution)

- For the inner short solenoid,

it is very difficult to calculate the magnetic field

- Using the equality property of the mutual inductance, we can

calculate the magnetic field of the long solenoid, and determine the flux through the short solenoid.

The magnetic field inside the long solenoid is InB 20

The magnetic flux through a single loop of the short solenoid is2

20

2

1 aInaBs

Since there are n1l turns in all of the short solenoid, the total flux through the inner short solenoid is

lInnalnaIns 21

2

01

2

20

= the magnetic flux through the long solenoid caused by a current I in the short solenoid

The mutual inductance of this case is lnna

IM 21

2

0

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When the current in loop 1 is changed,

7.2.3 Inductance

The magnetic flux through loop 2

varies, and this changing flux

induces an emf in loop 2:

rdt

dIM

dt

d 122

(7.26)

(7.27)

Unit of L : H (henries) = Vs/A

A changing current not only induces

an emf in any nearby loops, it also

induces an emf in the source loop

itself.

IL

where L is the constant of proportionality [“Self Inductance (or simple the “inductance”) of

the loop].

If the current changes, the emf induced in the loop is

dt

dIL

- continued

(7.25)

The greater L is, the harder it is to change the current.

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(Solution)

The magnetic field inside the toroid is

hs

INB

2

0

The magnetic flux through a single turn of the toroid is

The total magnetic flux through the N-turns of the toroid is

The self-inductance of this toroid is

[Example 7.11]

ab

s

Eq. (5.60) →

a

bhINds

sh

INadB

b

aln

2

1

2

001

a

bhINNtotal ln

2

2

01

a

bhN

IL total ln

2

2

0

(7.28)

http://www.isomati

c.co.uk/smtraf.htm

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(Solution)

The total emf in this circuit =

: the first-order differential equation for I

The time constant :

[Example 7.12]

(7.29)

inductance thefrombarrery thefrom0dt

dIL

7.35

According to Ohm’s law, IRdt

dIL 0

tLReR

tI 0 1

(See [Appendix 1] for derivation of the solution)

R

L

teR

tI 10

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The work done on a unit charge, against the back emf, in one trip around the circuit = -

7.2.4 Energy in Magnetic Fields

- An amount of energy to start a current flowing in a circuit

Thus, (7.30)

dt

dILII

dt

dW

where the line integral is around the perimeter of the loop.

ldAadAadB

- The work we must do against the back emf to get the current flowing in a circuit

- A fixed amount which can be recoverable during the current turned off.

- Energy stored in the magnetic field.

The amount of charge per unit time passing down the wire = I

The total work done per unit time :

dILIdW

2

2

1LIW

Let us find out how long it takes to get the final current I.

ldALISince ,LI

IL

dt

dIL

(7.26)

(7.27)

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Therefore,(7.31)

(7.34)

0 for all space because A and B are almost zero for a large space.

: Energy stored in the magnetic field

,0JB

dlIAldAIILIW 2

1

2

1

2

1

2

1 2

For the volume current density J,

(7.35)

Using the vector product rule in Chapter 1, we get

spaceall

dBW

2

0

v 2

1

- continued

v

v 2

1dJAW

(7.32)

Since v

0

v 2

1dBAW

(7.33)

The vector product rule (IV), Chapter 1

BAABBA

BAABBA

Thus, Eq. (7.33) becomes

SadBAdB

dBAdABW

v

2

0

vv0

v 2

1

v v 2

1

SVaddV

vv

The Gauss’s Theorem (Green’s Theorem,

Divergence Theorem)

Thus, we finally get

0

2

2

B: the energy density (= energy per unit volume)

JA

2

1

: the energy density (energy

stored in the current

distribution per unit volume)

SS

adJdaJI

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The energy stored in the magnetic fields cannot do work, because the magnetic fields

themselves do no work.

Comparison between the magnetic energy and the electric energy in the electrostatics

Eqs. (2.43) & (2.45)

- continued

v

2

0v

v 2

1v

2

1dBdJAWmag

v2

v2

1

v

20

v dEVdWelec

Eqs. (7.32) & (7.35)

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7.2.4 Energy in Magnetic Fields - continued

(Solution)

According to Ampere’s law,

ˆ 2

0

s

IB

the field between the cylinders =

the field elsewhere = 0

The energy per unit volume =22

2

0

2

0

00

2

8 22

1

2 s

I

s

IB

s

ds

The energy in a cylindrical shell of length l, radius s, and thickness ds =

s

dslIdsls

s

I

4 2

8

20

22

20

The total magnetic energy stored in the dual cylinders of length l =

a

blI

s

dslIdWW ln

44

2

0

2

0

,2

1 2LIW Since the magnetic energy stored in an inductance L is

a

blILI ln

42

1 2

02

a

blL ln

2

0

: an easy way to calculate the self-inductance.

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Next Class

Chapter 7. Electrodynamics




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Solution to the 1st-order Differential Equation

IRdt

dIL 0

LI

L

R

dt

dI 0

dtIL

R

LdI

0

dt

IL

R

L

dI

0

10ln

1CtI

L

R

LLR

10ln C

L

Rt

L

RI

L

R

L

tLRCeIL

R

L

0

tLRCe

LR

LI 0

tLReR

LC

RtI 0

The initial condition :

,0tAt .0tI

00 0 R

LC

RI

LC 0

tLRtLR eR

eLR

L

RtI 0 00 1

[Appendix 1]

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