Prof. David R. JacksonECE Dept.

Fall 2014

Notes 27

ECE 2317 Applied Electricity and Magnetism

1

KCL Law

1

0

0

N

nn

totin

i

i

Wires meet at a “node”

i2

i3

iN

i4

i1

2

Note: The “node” can be a single

point or a larger region.

A proof of the KCL law is given next.

KCL Law (cont.)

A

C

+Q

totin

dvi C

dt

Ground

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

-

v

A = surface area of the “node.”

3

A single node is considered.

The node forms the top plate of a stray capacitor.

KCL Law (cont.)

1) In “steady state” (no time change)

2) As area of node A 0

0

0totin

dv

dt

i

0 0

0totin

A C

i

totin

dvi C

dt

Two cases for which the KCL law is valid:

4

KCL Law (cont.)

In general, the KCL law will be accurate if the size of the “node” is small compared with the wavelength 0.

8

0

2.99792458 10c

f f

f 0

60 Hz 5000 [km]

1 kHz 300 [km]

1 MHz 300 [m]

1 GHz 30 [cm]

5

Node

Currents enter a node at some frequency f.

KCL Law (cont.)

6

Example where the KCL is not valid

Open-circuited transmission line

sinI z A kz

k

2 f

(phasor domain)

Zg

Z0

+

-V(z)

I(z)

z

I z

+-

KCL Law (cont.)

7

Open-circuited transmission line

Current enters this region (“node”) but does not leave.

Zg

Z0I(z)

z

I z

+-

General volume (3D) form of KCL equation:

ˆ 0out

S

J n dS i (valid for D.C. currents)

8

KCL Law (cont.)

The total current flowing out must be zero: whatever flows in must flow out.

S

n̂J

KCL Law (Differential Form)

To obtain the differential form of the KCL law for static (D.C.) currents, start with the definition of divergence:

0J

0

1ˆlim

VS

J J n dSV

ˆ 0out

S

J n dS i

J

V

(valid for D.C. currents)

For the right-hand side:Hence

9

Important Current Formulas

Ohm’s Law

J E

Charge-Current Formula

vJ v

(This is an experimental law that was introduced earlier in the semester.)

(This was derived earlier in the semester.)

These two formulas hold in general (not only at DC).

10

Resistor Formula

0

0

x

x x

x

VE

LV

J EL

VI J A A

L

+

-V

J

L

xA

Solve for V from the last equation:

LR

A

VR

I

Hence we have

We also have that

A long narrow resistor: 0ˆ xE x E Note: The electric field is constant since the current must be uniform (KCL law).

LV I

A

11

I

Joule’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

DW = work (energy) given to a small volume of charge as it moves inside the conductor from point A to point B.

This goes to heat!

(There is no acceleration of charges in steady state, as this would violate the KCL law.)

12

V

A B v

E- -- -

- -- -

v

Conducting body

J E

r

v = charge density from electrons in conduction band

Joule’s Law (cont.)

1v

Wv E

t V

J E

power / volume

Wd

V

P J E dV We can also write

22

W[ ]d

V V

JP E dV dV

v v

rW V E t V v E t

t

The total power dissipated is then

13

Power Dissipation by Resistor

d x xP J E V

I VV

A L

IV

2

d

d

P VI

P RI

Hence

Note: The passive sign convention applies to the VI formula.

V AL

14

Resistor

-V

I

A

+

L

x

Passive sign convention labeling

RC Analogy

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

“C problem”

r F r

A

+

-VAB

B EC

“R problem”

15

r F r

A

-VAB

B ER

I

+

Insulating material Conducting material

(same conductors)

RC Analogy (cont.)

Theorem: EC = ER (same field in both problems)

“C problem”

r F r

A

+ -VAB

B EC

“R problem”

16

r F r

A

+ -VAB

B ER

I

(same conductors)

EC = ER

“C problem” “R problem”

0

0

D

E

0

0

J

E

Same differential equation since (r) = (r) Same B. C. since the same voltage is applied

RC Analogy (cont.)

Proof

Hence,

17

(They must be the same function from the uniqueness of the solution to the differential equation.)

A

A

AB

s

S

B

A

QQC

V V

dS

E dr

“C Problem”

RC Analogy (cont.)

ˆAS

B

A

E n dS

C

E dr

Hence

ˆ ˆs D n E n Use

r F r

A

+ -VAB

B EC

18

ˆA

AB

A

B

A

S

VVR

I I

E dr

J n dS

RC Analogy (cont.)

ˆA

B

A

S

E dr

RE n dS

Hence

“R Problem”

J EUse

r F r

A

+ -VAB

B ER

I

19

Hence

r r F r

C G

ˆAS

B

A

E n dS

C

E dr

ˆ1

AS

B

A

E n dS

GR

E dr

RC Analogy (cont.)

Recall that

20

Compare:

C G

RC Analogy

Recipe for calculating resistance:

1) Calculate the capacitance of the corresponding C problem.2) Replace everywhere with to obtain G.3) Take the reciprocal to obtain R.

In equation form:

21

This is a special case: A homogeneous medium of conductivity surrounds the two conductors (there is only one value of ).

RC Formula

ˆAS

B

A

E n dS

C

E dr

ˆAS

B

A

E n dS

G

E dr

G C

Hence,

RC

or

22

Example

Find R

C problem:

A

h

C G

AC

h

AG

h

Method #1 (RC analogy or “recipe”)

hR

A

AC

h

23

A

h

Hence

Example

Find R

C problem:

AC

h

Method #2 (RC formula)

RC

hR

A

A

h

A

h

AR

h

24

Example

Find the resistance

Note: We cannot use the RC formula, since there is more than one region (not a single conductivity).

25

h1

h2 2

1

A

Example

C problem:

1 2

1 1 1

C C C 1

11

AC

h

2

22

AC

h

26

h1

h2 2

1

A

Example (cont.)

1 2

1 1 1

G G G 1 2

1 21 2

A AG G

h h

C G

27

h1

h2 2

1

A

h1

h2 2

1

A

1 2

1 1 1

C C C 1 2

1 21 2

A AC C

h h

1 2R R R 11

1

hR

A

Hence, we have

22

2

hR

A

Example (cont.)

28

h1

h2 2

1

A

Lossy Capacitor

Cv (t)

R

i (t)

+

-

AC

h

hR

A

This is modeled by a parallel equivalent circuit (proof on next slides).

A [m2] Q

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

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

Ev(t)+

-

i (t)

J

Note: The time constant is

RC

29

30

Lossy Capacitor (cont.)

Therefore,

A [m2] Q

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

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

Ev(t)+

-

i (t)

Jx

Ain x

dQi t i t J A

dt

Total current entering top (A) plate:

x

dQi t J A

dt

Derivation of equivalent circuit

31

Lossy Capacitor (cont.)A [m2] Q

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

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

Ev(t)+

-

i (t)

Jx

x

As

x

x

x

dQi t J A

dt

dE A A

dtdDv

A Ah dt

dEvA

h dtA

xdEvi t A

h dtA

v A dv

R h dt

v dvC

R dt

32

Lossy Capacitor (cont.)A [m2] Q

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

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

Ev(t)+

-

i (t)

Jx

v t dv ti t C

R dt

This is the KCL equation for a resistor in parallel with a capacitor.

Cv (t)

R

i (t)

+

-

33

Lossy Capacitor (cont.)

We can also write

xx

dDi t J A A

dt

A [m2] Q

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

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

Ev(t)+

-

i (t)

Jx

Conduction current Displacement current

DH J

t

Ampere’s law:

Conduction current (density)

Displacement current (density)