prof. david r. jackson ece dept. fall 2014 notes 27 ece 2317 applied electricity and magnetism 1
TRANSCRIPT
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)