Lecture 7
Capacitors and dielectric permittivity
Capacitance of parallel-plate capacitor
Capacitance per meter of coaxial cable
Magnetic permeability
Inductance
Inductance of Solenoid
Inductance of Toroid
Mutual Inductance
Coefficient of Coupling
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ECE342 Lecture 7
1
Capacitor
• Formed whenever two conductors are separated
by an insulator (dielectric) material, such as
plastic, mica, or ceramic.
• Capacitor stores electric charge.
• +Q charge is stored on one conductor.
• -Q charge is stored on the other conductor.
• Charge stored “Q” is proportional to the voltage
impressed across the capacitor’s conductors.
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ECE342 Lecture 7
2
Capacitance is a function of the geometries of the
conductors and the permittivity “ε = ε0 εR” of the dielectric.
ε0 = permittivity of free space (or air)
= 8.854 X 10-12 units of C2/N*m2 = Farads/meter
εR = relative permittivity of the medium (> 1.0)
It indicates the ability of the material to polarize, where the
positive and negative charge centers within its atoms
separate slightly, under the influence of an external E field.
This sets up an opposing internal field that acts to make
the overall E field in the material less.
EEXT
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+
EINT
Polarized Atom of
dielectric
ECE342 Lecture 7
3
Definition of Capacitance of a Capacitor
• The capacitance of a capacitor indicates its ability
to store charge.
• Definition of capacitance: If a voltage source of
strength “V” is placed across a capacitor, a positive
charge “Q” will be forced onto one conductor, and
at the same time, an equal amount of negative
charge “-Q” will necessarily be forced onto the
other conductor). Then
C=Q/V
with units of Coulombs / Volt = Farads (F)
(Note: capacitance is “Charge per unit volt”)
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ECE342 Lecture 7
4
Parallel-Plate
Capacitor
i(t)
+
+
V
Each conducting
plate is of cross
sectional area
“A”
(Transient) Charging current i(t)
+
+
+Q
+ +
+
+
d
E field
-
-
-
-
-
-
-
-
-
-Q
z
y
i(t)
(Transient) Charging current i(t)
x
C=Q/V
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ECE342 Lecture 7
5
d
V

 E dL
0
E d
(Since E field remains
vertical and approximately
constant as one moves
away from a horizontal
sheet of uniform charge
density)
Thus E = V / d, and the electric field
between the plates (E) will be
proportional to the voltage across the
plates. Since Q = CV, and E is also
proportional to the charge “Q” stored
on the plates.
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ECE342 Lecture 7
6
Intuitive observations
• For a constant voltage “V” across the capacitor plates,
the magnitude of the charge on either plate “Q” doubles
as the separation distance “d” is halved, since the
electric field between the plates E = V / d becomes
twice as strong, attracting twice as many charges onto
the plates.
• For a constant voltage across the capacitor plates, Q
increases as the plate area “A” increases, since there is
physically more room for the charges to spread out on
the plate.
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ECE342 Lecture 7
7
Capacitor current-voltage relationship
• C = Q / V (by definition)
• Q = C V (Solve for Q)
• The current “i(t)” flowing into the top conductor of the
capacitor or away from the bottom conductor of the
capacitor is the rate at which charge is transported
onto the conductor and hence is the rate at which the
stored charge “Q” changes:
• i(t) = dQ/dt = d(CV)/dt
i(t)
C
Assuming C is a constant,
i(t) = C dV/dt
+ v(t) -
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ECE342 Lecture 7
8
Relative Permittivity of Dielectric
• Recall what happens when some material other than a
vacuum with εR > 1.0 is placed between the capacitor
plates.
• The net E field between the plates is diminished, since
the material polarizes (atomic charge centers separate)
to some extent.
• Imagine a parallel-plate capacitor with an air dielectric is
charged to a specific voltage. Then with the voltage
source disconnected, a polystyrene dielectric (with εR =
2.5) is slid between the plates, with d held constant.
This makes the voltage across the capacitor, V = E / d,
decrease, since E becomes 2.5 times smaller. Because
C = Q/V, the capacitance of the parallel-plate capacitor,
must increase by a factor of 2.5!
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ECE342 Lecture 7
9
Capacitance of a parallel-plate capacitor
•
We have shown, from Gauss Law, as well as from elementary application
of Coulomb’s Law, that the E field due to an infinite sheet of uniform
surface charge density ρso is given by
E = ρso / (2ε)in
Where in is the unit vector normal to the surface of the sheet of charge.
•
Thus if the parallel plates of the capacitor carry equal and opposite
charges Q and –Q, then the E field set up by the two plates reinforces in
the region between the plates, and the E field largely cancels elsewhere if
the separation distance d is considerably less than the length or width of
either plate.
d
ρso = -Q/A
ρso = +Q/A
Where A is the plate area
Ex
x
In between the plates,
E = 2(ρso / (2ε))ix = (ρso/ε)ix = Q/(Aε)ix
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ECE342 Lecture 7
10
The capacitor plate separation "d" is kept small compared to the plate
length "L" or width "W", where the plate area A = WL. Thus, from any
vantage point between the plates (and well away from the plate edges), the
plates appear to be approximately infinitely extended, and therefore the E
field is perpendicular to the plates. Near the edges the E field "fringes"
where E is no longer perpendicular to the plates, but this is kept to a
minimum for d << W,L
Thus the voltage across the capacitor plates is
d
d

 E x dx
0
V
 
so

dx
 
0
Minor E field fringing,
d << W,L. => E ≈ Ex ix
d
d
 Q

dx
 A 
0
Q d
A 
Using the definition of capacitance,
Q
V
C

A 
d
 0  R
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REMEMBER: This simple
formula is an approximation that
holds only if d is small compared
to the plate width and length
ECE342 Lecture 7
Severe E field
fringing, since
d is not << W,L
E ≠ Ex ix
d
11
•
•
•
•
•
Capacitance of Coaxial Cable
A coaxial cable consists of an inner conductor and an outer concentric cylindrical
conductor (the shield), with an insulating dielectric (air or polystyrene) in between.
Let us calculate the capacitance of an L-meter length of coaxial cable.
A voltage V is placed across the inner and outer conductors, depositing charge
“Q” on the inner conductor and “-Q” on the inner conductor.
A closed concentric Gaussian cylindrical surface (pill bottle indicated by the dotted
line) is positioned as shown (r = constant), and Gauss’ law is applied around its
surface. Because Er is a constant along the Gaussian cylinder wall, and is
tangential to the front and rear “end caps” of the cylinder,
Gauss’ Law => The electric displacement flux emanating from the pillbox =
εEr*(2πrL) = Q => Er = Q / (2πεrL)
Let the z axis be
directed into the paper
and located at the
center of the circles.
Then by symmetry
the electric field has
only a radial outward
component: E = Er(r)ir
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for a < r < b
ix
b
r
a
iz
iy
Q
-Q
ECE342 Lecture 7
12
The voltage at the positively charged inner conductor with
respect to the negatively charged outer conductor is
b
V

 Er dr
a
b

Q

dr
 2   r L
a
b
 ln  
2   L  a 
Q
From the definition of capacitance,
C
Q
V
2   L
b
ln  
a
Example: Find the capacitance of a 1-m length of coaxial
cable with polystryrene foam dielectric with
 R  2.5
 0  8.854  10
 
2    0  R L
C 
b

ln  
a
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 12 F

m
L  1  m
C  1.11  10
 10
F
ECE342 Lecture 7
a  0.2  cm b  0.7  cm
= 111 pF
13
Magnetic Permeability
•
A material’s magnetic permeability μ = μRμ0 is a measure of
– The existence of magnetic dipoles (loops of spinning and orbiting
charges) inside the atoms of the material.
– How easy it is for these magnetic dipoles to re-orient themselves along
the direction of an externally applied magnetic field.
•
A nonmagnetic material (such as air, polystyrene, glass) has no available,
moveable magnetic dipoles. For these materials μR = 1.0
Thus in non-magnetic materials, such as air
B = μ0H, where μ0 = 4π*10-7 with units of Wb/(A*m) = H/m
Due to random thermal vibration, a magnetic material (a material with μR >
1.0) will normally (with no magnetic field present) have its microscopic
magnetic dipoles randomly oriented in all directions, so there is no net
macroscopic magnetic field set up in the material.
•
•
But if an external magnetic field intensity “H” is applied across such a
magnetic material, it will cause the randomly oriented dipoles to “line up” to
some extent, producing an enhanced magnetic flux field B in that material.
(Think of “H” as the stimulus and “B” as the response, or resulting flow field.)
B = μR μ0H
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where μR > 1.0
ECE342 Lecture 7
14
Current Carrying Loops Are
Magnetic Dipoles:
B
B
i(t)
The magnetic flux flows upward through the center of
the loop, then it curls back downward outside of the loop
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ECE342 Lecture 7
15
Ferromagnetic Materials
In ferromagnetic materials (Fe, Co, Ni, Gd (Gadolinium), and Dy (Dysprosium)
and their alloys) a special effect called exchange coupling occurs between
groups of neighboring atoms (domains), that causes all of their magnetic
dipole moments to line up in parallel and point in a single direction.
The domains in a ferromagnetic material can be seen under a microscope
using a colloidal suspension of finely powered iron oxide.
With no magnetic field present, each domain points in a random direction due
to thermal agitation of the molecules, so there is no net magnetic field in the
material.
When an H field is applied, the domain boundaries (walls) shift, allowing the
domains pointing along the direction of the applied H field to get bigger, while
the domains pointing in other directions shift.
This results in a VERY BIG enhancement in the resulting magnetic flux (B), or
flow of the magnetic field through a ferromagnetic material, thus μR can be
quite large (on the order 100 – 300) for ferromagnetic materials.
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ECE342 Lecture 7
16
Ferromagnetic domains in an
unmagnetized sample of iron with
no applied H field
With no applied
magnetic field, the
net macroscopic
magnetic field = 0,
since the
microscopic
domain magnetic
dipole moments
cancel out.
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With applied magnetic
field, the domain walls
shift such that the
domains pointing
along the magnetic
field get bigger,
resulting in an
enhancement of the
magnetic flux in the
region. If the applied
magnetic field is
strong enough, the
domains can be
“rotated into
alignment”
ECE342 Lecture 7
17
Steps in Magnetization of
Ferromagnetic Material
a) No applied H field (unmagnetized state)
b) Weak applied H field (domain wall motion)
c) Strong applied H field (domain rotation)
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ECE342 Lecture 7
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Nonlinear B vs H curve (Hysteresis Curve)
of a ferromagnetic material
If the applied H field is varied over
a large range, we can no longer
say that B = μH except over a
small range of H variation, since
the relationship between B and H
is not linear. This is because
eventually all of the domains get
rotated into alignment with the
applied H field, and “that’s all
there is, and their aint no more!”
The material is said to be
“magnetically saturated.” Thus
the slope of the B vs. H curve
levels off at higher H values,
meaning that there is less B field
enhancement for higher H values.
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ECE342 Lecture 7
19
Magnetized Sample (Permanent Magnet)
Note as H is increased from 0 in the
initially unmagnetized sample (Point a)
up to a maximum value at Point b, the
material approaches magnetic
saturation. Then, as the H field is
decreased back to zero, B remains at a
nonzero positive B field value (Point c).
This is because the domain walls move
with friction. The walls do NOT move
back all the way in order to return to
their original unmagnetized value.
Instead, the sample retains a residual B
field. The sample has been
“magnetized”. The same principle
allows the sample to be magnetized in
reverse. (See Point f). This is the
principle used in magnetic recording
media, such as a PC’s hard drive
read/write head!
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ECE342 Lecture 7
20
Principle used in digital magnetic recording media (tape /
magnetic strip on credit cards / hard drive / floppy drive, etc.)
READ: a left-to-right or right-to-left pointing magnetic
field is picked up at air gap and transmitted through
the ferrite core and pickup coil. By Faraday’s law of
induction, v(t) = -d/dt(Flux cutting coil), the domain
transitions pick up a positive or negative-going
voltage pulse
WRITE: Magnetic domains in magnetic
recording medium (iron-oxide coated
plastic surface) are left aligned either leftto-right (Logic 1), or right-to-left (Logic 0),
depending on direction of the current
pulse through read/write head coil, which
results in a magnetic field across the
narrow air gap
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ECE342 Lecture 7
21
Use of ferromagnetic materials to
conduct and guide magnetic flux
• In much the same way that copper wire, which
has a high conductivity compared to air, is used
to conduct electrical current in an electrical
circuit, ferromagnetic material, which has a
relatively high permeability compared to air, is
used to conduct and guide magnetic field flux
lines in a magnetic device such as a hard drive
read/write head, toroidal inductor core, electric
motor, loudspeaker, D’Arsonval meter
movement, etc.
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ECE342 Lecture 7
22
Inductance
• If two coils are near each other, a time-varying current in one coil will
set up a time-varying magnetic flux that cuts the surface enclosed by
the second coil, and hence by Faraday’s Law of Induction, a timevarying voltage will be induced in the second coil. This is called
“transformer action”.
• However, two coils are not needed to demonstrate Faraday’s Law of
Induction. If a single coil carries a time-varying current, a self-induced
magnetic flux will be set up that cuts the surface enclosed by this coil,
which by Faraday’s Law of Induction, will set up a time-varying
voltage across that coil that is given by
v  t
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d
  N    t  where N is the number of turns in the coil
and N is called the "flux linkages"
dt
ECE342 Lecture 7
23
• We have found that the magnetic field intensity due to current flowing
in a wire
Hø = i(t)/2πr
is always proportional to the current that creates it, i(t).
• Thus the resulting magnetic flux Φ must be proportional to i(t), since
B = µH = µ0µRH (for a linear magnetic medium)
and Φ is the integral of B·ds
• Exception: ferromagnetic (nonlinear) medium, where the the B vs. H
curve is nonlinear, unless i(t) varies over a suitably small range, in
which case an approximate µR can still be assigned.
• Let this proportionality constant be defined as the “self-inductance” of
the coil, “L” where
N   ( t)
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L  i ( t)
We will take this as the definition
of the inductance (L) of a coil,
where the units are in
Weber-turns / Ampere
Volt*second / Ampere = Henrys
ECE342 Lecture 7
24
Inductor voltage – current relationship
• Applying Faraday’s law to our defining equation for inductance,
we may determine the voltage across the inductor v(t) in terms of
the current through the inductor i(t).
v ( t)
d
  N   ( t) 
dt
N   ( t)
v ( t)
where N is the number of turns in the coil
and N is called the "flux linkages"
+
L  i ( t)
d 
N   ( t) 
dt
v(t)
d

 L  i ( t) 
 dt

L 
-
di ( t)
dt
i(t)
XB
Recall that the polarity of v(t) depends on the direction that we traverse the
closed contour when calculating the magnetic flux linking the turns of the coil.
This direction is dictated by the direction of the flux that is set up by the coil
current, i(t) according to the right-hand rule. In this case, the right -hand rule
dictates that v(t) be referenced positive w.r.t. the terminal that current leaves
the coil. Thus if we instead choose v(t) to be referenced positive w.r.t. the
terminal that current enters the coil, we may drop the (-) sign, and we have
v(t) now referenced
positive at the terminal
where the current enters
the inductor.
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v ( t)
L
di ( t)
dt
ECE342 Lecture 7
25
Inductance of a long, thin solenoidal coil
•
•
•
Consider a long, thin solenoidal coil with length “l”, N turns, cross sectional area A,
carrying a current i.
By definition, L = N Ф / i
From symmetry considerations, the B field inside the thin solenoid is directed along the
axis of the solenoid, and is approximately uniform, as can be seen from Ampere’s
circuital law, picking the Amperian contour shown below
b
Amperian contour normal
to surface of coil, with
top and bottom sides
parallel to axis, and half
inside and half outside
the solenoidal coil.
a
z axis
Js=Ni/l
H = Hziz
Length l
The only contribution to


  H dl 




around the
closed Amperian contour is along the bottom. Thus
we can find the H field = Hz  iz inside the solenoid.
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ECE342 Lecture 7
26

 H dl

Hz

Js
b
b

Hz   1 dz
0

 Hz dz
0
Ni
l

 B ds

Hz  b
Js  b
(Constant around the inside of
the solenloid, well away from its
ends)

   Hz ds


  Hz   1 ds

  Hz  A

Ni
A
l
Thus the inductance of the long solenoid becomes
L
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N 
i
N    

Ni 
 A
l

i
2
N  
ECE342 Lecture 7
A
l
27
Inductance of a toroidal inductor
Powdered Iron or
ferrite core (μR>>1)
Amperian
contour of
radius r
b
r
z
a
B
Thickness “h”
iФ
i(t)
N Turns
From symmetry, we see that B and H will be directed in the i
direction, pointing along the Amperian contour, thus
H  2    r
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N i
ECE342 Lecture 7
28
Thus
H
N i
Inside the toroidal core, for a < r < b
 2  r
The magnetic flux that circulates around the toroidal core
and cuts each turn of the coil must be
h

b
 
    H dr dz
0 a

h  N i 
h



0
b

   N  i dr dz

 2  r
a
b
ln  
a
2
By definition of inductance, the inductance of the
toroid becomes:
L
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N 
i
 N2  h   b 
 
  ln  
 2    a 
ECE342 Lecture 7
29
Numerical example
 R  250
With a  5  cm b  10  cm
h  1  cm
7 H
And
N  50 turns
 0  4    10 
m
 N2  h   b 
L   0   R  
  ln  
 2    a 
Consider a toroid with a ferrite core
L  8.664  10
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4
H
ECE342 Lecture 7
30
Mutual Inductance
• When two coils are near each other, some of the
magnetic flux created by one coil links (passes through
the surface enclosed by the other coil).
• The two coils are said to be “magnetically coupled”
• A time-varying current in one coil will induced a timevarying voltage in the other coil by Faraday’s Law of
Induction.
• When the two coils are tightly coupled (with nearly every
flux line produced by one coil linking the other coil as
well, we have a transformer.
• Mutual inductance is a measure of the degree of
magnetic coupling between two coils.
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ECE342 Lecture 7
31
Mutual Inductance
i1(t)
B flows out of paper
inside 2nd coil
B
v2(t)
+
Coil 1
N1 turns
Coil 2
N2 turns
The mutual inductance between Loop 1 and Loop 2 “M12” is defined as the
constant of proportionality between the magnetic flux linking Coil 2 due to the
current flowing in neighboring Coil 1. Note that Coil 2 is open-circuited.
N2Ф2 = M12*(i1)
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=>
M12 = N2Ф2 / i1


Differentiating with respect to time:
d
N2   2
dt
By Faraday's Law of Induction v2
d
M12  i1 => M12
dt
ECE342 Lecture 7
M12 
d
i1
dt
v2
d
i1
dt
32
Experimental measurement of Mutual Inductance between two coils
1.
2.
3.
4.
Measure the self-inductance of each of the two inductors, L1 and L2
Connect a sinusoidal ac voltage source across L1
Measure the amplitude of v1(t), Vm1, and the open-circuit value of v2(t), Vm2.
From the expressions for self-inductance and mutual inductance we find that
v1  t
d
L1  i1  t
dt
=>
M12
v2
d
i1
dt
v2
 L1
v1
5. Thus we may calculate M12 = (Vm2/Vm1)L1
6. The roles of coils L1 and L2 may be reversed and the steps repeated to
find M21. With the sinusoidal voltage source placed across L2,
M21 = (Vm1/Vm2)L2
5. However, due to the symmetry of the situation, you should find that
M12 = M21 = “M”. (There is no need to retain subscripts)
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ECE342 Lecture 7
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Coefficient of Coupling
M
i1(t)
i2(t)
+
v1(t)
-
L1
L2
RL
+
V2(t)
-
Consider two magnetically coupled coils where currents are flowing in both
coils (the second one is no longer open-circuited). Now the voltage across
each coil has a self-induced and a mutually induced component:
v1
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L1 
di1
dt
 M
di2
dt
and v2
ECE342 Lecture 7
L2 
di2
dt
 M
di1
dt
34
If the second coil is short-circuited (RL = 0), then v2(t) = 0, allowing us
to solve for di2/dt in terms of di1/dt
0
=>
L 2
di 2
dt
di 2
dt
 M
di 1
dt
M di 1

L 2 dt
Now the voltage across Coil 1 can be written purely in terms of di1/dt,
And the equivalent inductance “LEQ” across Coil 1 can be found:
v1
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2  di

M
1
L 1 

L 2  dt

ECE342 Lecture 7
L EQ 
di 1
dt
35
2
L EQ
M
L1
L2
But the stored energy in an inductor:
2
WL
1
2
 L EQ  i 1
2
M
L1
L2
2
i 1
2
can never be allowed to go negative, since that would
violate the passive sign convention and allow the
inductor to act as a source of electrical energy.
Therefore LEQ must remain positive all times.
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ECE342 Lecture 7
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Thus the maximum possible value of M is found by setting LEQ = 0
L1 
MMAX
2
L2
0
=>
MMAX
L1  L2
This mutual inductance occurs when both coils are magnetically coupled
tightly
as possible,
all the flux
createdoccurs
by onewhen
coil linking
This maximum
valuewith
of mutual
inductance
the twothe other co
coils are placed in such close proximity, or linked together by a
magnetic material of very high µR, such that all of the flux produced
by one coil completely links the other coil as well.
The coefficient of coupling “k” is defined to indicate the degree of
mutual coupling:
k
M
MMAX
M
L1  L2
where 0 < k < 1
100% magnetic coupling between two coils (M = MMAX, or k = 1) is not
possible in practice, however values of k as high as 0.98 are attainable.
7/14/2017
ECE342 Lecture 7
37