Prof. D. Wilton ECE Dept. Notes 24 ECE 2317 Applied Electricity and Magnetism Notes prepared by the...

Prof. D. WiltonECE Dept.

Notes 24

ECE 2317 ECE 2317 Applied Electricity and MagnetismApplied Electricity and Magnetism

Notes prepared by the EM group,

University of Houston.

Stored EnergyStored Energy

1

2E v

V

U dV

1

2E

V

U D E dV

0

Note: Please see the text book or supplementary notes for a derivation.

1

2E s

S

U dS

charge formula:

electric-field formula:

(volume charge density)

(surface charge density)

v D D D

Charge form to electric field form uses

The divergence theorem

ExampleExample

+ -

r hxV0

A [m2]

Method #1

0

0 00

2

00

1

2

1

2

E

V

r

E r

V

U D E dV

VE x

h

V VD x x

h h

VU dV

h

Find the stored energy

Example (cont.)Example (cont.)

2

00

1J

2E r

VU Ah

h

20 0

1

2E r

AU V

h

Or,

20

1

2EU CV

Recall that 0 r

A AC

h h

Hence


Method #2

1

2

1

2

1 1

2 2

constant od

nan A B

A B

A B

E v

V

S S

s

S S S

A s B s

S S

U dV

dS

dS dS

(since we have surface charge in this problem)

r h x

A [m2]A

B

+ + + + + + + + + + + + +

- - - - - - - - - - - - - - - - - -


1 1

2 21 1

2 21

2

E A A B B

A A A B

A B

U Q Q

Q Q

Q

0

0 0

1

21

2

EU QV

CV V

0

1

2EU QV 20

1

2EU CV

Capacitance from Stored EnergyCapacitance from Stored Energy

2

2

1

2

2

E

E

Q Q

V

U CV

UC

V

The previous derivation assumes only that ...

there is a charge and on the two

capacitor plates

the potential difference between them is

Since the derivati

on is

it applies in general and provides another way to compute capacitance!

independent of capacitor geometry,

ExampleExample

v0 [C/m3]0 a

Method #1:1

2E

V

U D E dV

Gauss’s Law:

r < a

30

20

43

4

v aE r

r

3

0

20

43

4

v rE r

r

r > a

Find the stored energy

Solid sphere of uniform volume charge density


2 20 0

22 20

0 0 0

2 20

0

2 23 3

0 02 2

0 02 20 00

2 2

sin2

2 22

4 43 32 2

4 4

E r

V V

r

r

a v v

E

a

U E dV E dV

E r r dr d d

E r r dr

r aU r dr r dr

r r

v0 [C/m3]0 a


Result:2

50

0

4

15v

EU a

Let:3

0

0 3

4

3

3

4

v

v

Q a

Qa

as 0EU

a

2

0

1 3J

20E

QU

a

Note:

KCL LawKCL Law

1

0

0

N

nn

totin

i

i

i2

i3

iN

i4

i1

Wires meet at a “node”.

KCL Law (cont.)KCL Law (cont.)

A

C

+Q

totin

dvi C

dt

ground

C is the stray capacitance between the “node” and ground

+

-

v

KCL Law (cont.)KCL Law (cont.)

1) In “steady state” (no time change)

2) As A (area of node) 0, C 0

0

0totin

dv

dt

i

0totini

totin

dvi C

dt

Two cases for which the KCL law is valid:

KCL Law (Differential Form)KCL Law (Differential Form)

1

0 or 0N

totn in

n

i i

Differential form:

0J

0

1limV

S

J J n dSV

0

out

S

in

J n dS i

i

J

V

(valid for D.C. currents)

For D.C. : (circuit form)

Integral form:

DC Current FormulasDC Current Formulas

Ohm’s Law

J E

Charge-Current Formula

vJ v

(an experimental law)

(this was derived earlier in the semester)

Resistor FormulaResistor Formula

x

x x

x

VE

LV

J EL

VI J A A

L

LV I

A

+ -V

J

L

xA

Hence

LR

A

I

Joule’s LawJoule’s Law

AB

B

v

A

v

v v

W Q V

V E dr

V E r

rV r E V E t

t

W = work done on a small volume of charge as it moves a small distance inside the conductor from point A to point B.

This goes to heat!

V

A B

v

E

+ ++ +

+ ++ +

v

conducting bodyJ E

Joule’s Law (cont.)Joule’s Law (cont.)

Power v

WV v E

tV J E

3[W/m ]

W

d

Vd

V

P J E

P J E dV

Since, we also have

22 3[W/m ]d

JP E

v v

rW V E t V v E t

t

Power/unit volume d

WP J E

V t

J E

Power Dissipation by ResistorPower Dissipation by Resistor

Wvolume

Vd x xP J E LA

I

A

V

L

L

A

IV2

Vd

Vd

P VI

P RI

Hence,

Note: passive sign convention appears in the final result.

Resistor -V

I

A

+

L

x

RC AnalogyRC Analogy

Goal: Assuming we know how to solve the C problem (find C), can we solve the R problem (find R)?

“C Problem”

r F r

A

+ -VAB

B EC

“R Problem”

r F r

A

+ -VAB

B ER

I

RC Analogy (cont.)RC Analogy (cont.)

Theorem: EC = ER (same field)

“C Problem”

r F r

A

+ -VAB

B EC

“R Problem”

r F r

A

+ -VAB

B ER

IA

Hence: EC = ER

“C Problem” “R Problem”

0 0

0

C

C

D E

E

0 0

0

R

R

J E

E

• Same D. E. since (r) = (r)• Same B. C. since the same voltage is applied


A

A

AB

s

S

B

A

s

QQC

V V

ds

E dr

D n E n

“C Problem”

r F r

A

+ -VAB

B EC


AS

B

A

E n ds

C

E dr

AS

B

A

E n ds

G

E dr

“R Problem”

r F r

A

+ -VAB

B ER

IA


1

A

A

AB

A

B

A

S

B

A

S

VVR

I I

E dr

J n ds

J E

E dr

RRE n ds

Hence

r r F r

C G

AS

B

A

E n ds

C

E dr

1

AS

B

A

E n ds

GR

E dr


C G

RC AnalogyRC Analogy

Recipe for calculating resistance:

1. Calculate the capacitance of the C problem.

2. Replace everywhere with to obtain G.

3. Take the reciprocal to obtain R.

In equation form:

Special case: A single homogeneous medium of conductivity surrounds the two conductors.

RC FormulaRC Formula

AS

B

A

E n ds

C

E dr

AS

B

A

E n ds

G

E dr

Hence,

RC

orC G

geometry(GF)

factor

ExampleExample

A

h Find R

C problem:

A

h

C G

AC

h

A

Gh

Method #1 (RC analogy)

hR

A

ExampleExample

Find R

C problem:

AC

h

Method #2 (RC formula)

AC

h

RC

hR

A

A

h

r

A

h

ExampleExampleFind resistance

h1

h2 2

1

Note: We cannot use the RC formula, since there is more than one region.

ExampleExample

h1

h22

1

C problem:

1 2

1 1 1

C C C 1

11

AC

h

2

22

AC

h


h1

h2 2

1

1 2

1 1 1

G G G 1

11

AG

h

2

22

AG

h

C G

h1

h2 2

1

1 2R R R 11

1

hR

A

Hence

22

2

hR

A


ExampleExampleFind equivalent circuit

Cv (t)

R

i (t)

+

-

AC

h

hR

A

A Q

h, + + + + + + + + + + + + + + + + + +

- - - - - - - - - - - - - - - - - - - - - - - - - -

Ev(t)+

-

i (t)

Prof. D. Wilton ECE Dept. Notes 24 ECE 2317 Applied Electricity and Magnetism Notes prepared by the...

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