Prof. David R. JacksonDept. of ECE

Spring 2020

Notes 27DC Currents

ECE 3318 Applied Electricity and Magnetism

1

KCL Law

10

0

N

nn

totin

i

i

=

=

=

∑or

Wires meet at a “node”

2

Note:The “node” can be a single

point or a larger region.

A proof of the KCL law is given next.

1i

2i

3i4i

Ni

KCL Law (cont.)

totin

dvi Cdt

=

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

A = surface area of the “node.”

3

A single node is considered.

The node forms the top plate of a stray capacitor.Ground

Q+

( )v t

C

A

+

-

KCL Law (cont.)

1) In “steady state” (no time change)

2) As area of node A → 0

0

0totin

dvdti

=

=

0 00tot

in

A Ci→ ⇒ →

=

totin

dvi Cdt

=

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 λ0.

8

02.99792458 10c

f fλ ×

= =

f λ0

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 (ECE 3317)

( ) ( )sinI z A kz=

( )2 /k ω µε π λ λ= = = wavelength on line

2 fω π=

(phasor domain)

+-

+-

( )I z

( )I z

( )V z0Z

gZ

z

KCL Law (cont.)

7

Open-circuited transmission line

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

+-

+-

( )I z

( )V z0Z

gZ

z

General volume (3D) form of KCL equation:

ˆ 0outS

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

8

KCL Law (cont.)

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

nJ

S

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

1 ˆlimV

S

J J n dSV∆ →

∆

∇ ⋅ ≡ ⋅∆ ∫

ˆ 0outS

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 Eσ=

Charge-Current Formula

vJ vρ=

(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

VEL

VJ EL

VI J A AL

σ σ

σ

=

= = = =

Solve for V from the last equation:

[ ]LRAσ

= Ω

VRI

=

Hence we have

We also have that

A long narrow resistor: 0ˆ xE x E≈ Note:The electric field is constant (does not change with x)

since the current must be uniform (KCL law).

LV IAσ

=

11

σ

L

A

I

x

V

J

+ -

Joule’s Law

12

Conducting body

J Eσ=E

σ

[ ]2

2Wd

V V V

JP J E dV E dV dVσ

σ= ⋅ = =∫ ∫ ∫

The power dissipated inside the body as heat is:

(Please see one of the appendices for a derivation.)

Power Dissipation by Resistor

d x xP J E VI V VA L

IV

= ∆

= ∆

= 2d

d

P VIP RI

=

=

Hence

Note: The passive sign convention applies to the VI formula.13

ResistorPassive sign convention labeling

V AL∆ =

-

σ

+

LAI

V

x

Assume a uniform current.

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 problem”

14

( ) ( )r F rε =

A

+ -VAB

B EC

Insulating material

( ) ( )r F rσ =

A

-VAB

B ER

I

+

Conducting material

(same conductors)

C Gε σ→→

RC Analogy

Recipe for calculating resistance:

1) Calculate the capacitance of the corresponding C problem.

2) Replace ε everywhere with σ to obtain G.

3) Take the reciprocal to obtain R.

In symbolic form:

15

(Please see one of the appendices for a derivation.)

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

RC Formula

RC εσ =

16

σ = conductivity in resistor problemε = permittivity in capacitor problem

The resistance R of the resistor problem is related to the capacitance Cof the capacitor problem as follows:

(Please see one of the appendices for a derivation.)

Example

Find R

C problem:

C Gε σ→→

AChε

=

AGhσ

=

Method #1 (RC analogy or “recipe”)

hRAσ

=

AChε

=

17

Hence

σA

h

εA

h

ExampleFind R

C problem:

AChε

=

Method #2 (RC formula)

RC εσ =

hRAσ

=AR

hε ε

σ =

18

σA

h

εA

h

Example

Find the resistance

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

(not a single conductivity).

19

A

2h1h1σ

2σ

Example (cont.)

C problem:

1 2

1 1 1C C C= + 1

11

AChε

= 22

2

AChε

=

20

A

2h1h1ε

2ε

Example (cont.)

1 2

1 1 1G G G= + 1 2

1 21 2

A AG Gh hσ σ

= =

C Gε σ→→

21

→

1 2

1 1 1C C C= + 1 2

1 21 2

A AC Ch hε ε

= =

↓ ↓

A

2h1h1σ

2σ

A

2h1h1ε

2ε

1 2R R R= + 11

1

hRAσ

=

Hence, we have

22

2

hRAσ

=

Example (cont.)

22

A

2h1h1σ

2σ

Lossy Capacitor

AChε

=

hRAσ

=

This is modeled by a parallel equivalent circuit (the proof is in one of the appendices).

Note:The time constant is

RCτ =

23

,ε σ+ + + + + + + + + + + + + + + + + +

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

+

-

2A m

hE J( )v t( )i t

+

-

( )i t

( )v tR

C

Appendix

24

Derivation of Joule’s Law

Joule’s Law

( )

( )

( )

ABB

vA

v

v v

W Q V

V E dr

V E r

rV r E V E tt

ρ

ρ

ρ ρ

∆ = ∆

≈ ∆ ⋅

≈ ∆ ⋅∆

∆ = ∆ ∆ ⋅ = ∆ ⋅ ∆ ∆

∫

∆W = 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 cause

current to change along the conductor, violating the KCL law.)

25

Conducting body

ρv = charge density from electrons in conduction band

- -- -

- -- -

J Eσ=

r∆

A Bv

Evρ

V∆

Joule’s Law (cont.)

( )1v

W v Et V

J E

ρ∆ = = ⋅ ∆ ∆ = ⋅

power / volume

[ ]WdV

P J E dV= ⋅∫

We can also write2

2W[ ]d

V V

JP E dV dVσ

σ= =∫ ∫

( )v vrW V E t V v E tt

ρ ρ∆ ∆ = ∆ ⋅ ∆ = ∆ ⋅ ∆ ∆

The total power dissipated is then

26

Appendix

27

Derivation of RC Analogy

RC Analogy

“C problem” “R problem”

28

( ) ( )r F rε =

A

+ -VAB

B EC

Insulating material

( ) ( )r F rσ =

A

-VAB

B ER

I

+

Conducting material

(same conductors)

RC Analogy (cont.)

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

“R problem”

29

(same conductors)

“C problem”

( ) ( )r F rε =

+ -

CEA B

ABV

( ) ( )r F rσ =

+ -I

REA B

ABV

“C problem” “R problem”

( )0

0D

Eε∇ ⋅ =

∇ ⋅ = ( )0

0J

Eσ∇⋅ =

∇ ⋅ =

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

RC Analogy (cont.)Proof of “theorem”

Hence,

30

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

C RE E=

A

A

AB

sSB

A

QQCV V

dS

E dr

ρ

= =

=⋅

∫

∫

“C Problem”

RC Analogy (cont.)

ˆAS

B

A

E n dSC

E dr

ε ⋅

=⋅

∫

∫

Hence

ˆ ˆs D n E nρ ε= ⋅ = ⋅Use

31

( ) ( )r F rε =

+ -

CEA B

ABV

ˆA

AB

AB

A

S

VVRI I

E dr

J n dS

= =

⋅

=⋅

∫

∫

RC Analogy (cont.)

ˆA

B

A

S

E drR

E n dSσ

⋅

=⋅

∫

∫

Hence

“R Problem”

J Eσ=Use

32

( ) ( )r F rσ =

+ -I

REA B

ABV

Hence

( ) ( ) ( )r r F rσ ε= =

C G=

ˆAS

B

A

E n dSC

E dr

ε ⋅

=⋅

∫

∫

ˆ1 AS

B

A

E n dSG

RE dr

σ ⋅

= =⋅

∫

∫

RC Analogy (cont.)

Recall that

33

Compare:

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

RC Formula

ˆASB

A

E n dSC

E drε

⋅

=⋅

∫

∫

ˆAS

B

A

E n dSG

E drσ

⋅

=⋅

∫

∫

G C σε

=

Hence,

RC εσ =

or

34

Appendix

35

Equivalent Circuit for Lossy Capacitor

36

Lossy Capacitor

Therefore,

( ) ( )Ain x

dQi t i t J Adt

= − =

Total (net) current entering top (A) plate:

( ) xdQi t J Adt

= +

Derivation of equivalent circuit

,ε σ+ + + + + + + + + + + + + + + + + +

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

+

-

2A m

hE J( )v t( )i t

x

37

Lossy Capacitor (cont.)

( ) x

As

x

x

x

dQi t J Adt

dE A AdtdDv A A

h dtdEv A

h dtA

ρσ

σ

ε

σ

= +

= +

= +

= +

( ) xdEvi t Ah dtA

v A dvR h dtv dvCR dt

ε

σε

= +

= +

= +

,ε σ+ + + + + + + + + + + + + + + + + +

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

+

-

2A m

hE J( )v t( )i t

x

38

Lossy Capacitor (cont.)

( ) ( ) ( )v t dv ti t C

R dt= +

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

,ε σ+ + + + + + + + + + + + + + + + + +

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

+

-

2A m

hE J( )v t( )i t

x

+

-

( )i t

( )v tR

C

39

Lossy Capacitor (cont.)

We can also write the current as

( ) xx

dDi t J A Adt

= +

Conduction current Displacement current

DH Jt

∂∇× = +

∂

Ampere’s law:

Conduction current (density)

Displacement current (density)

,ε σ+ + + + + + + + + + + + + + + + + +

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

+

-

2A m

hE J( )v t( )i t

x

Note on displacement current: