CAPACITANCE,
INDUCTANCE,
AND
MUTUAL INDUCTANCE
C.T. Pan
1
6.1 The Capacitor
6.2 The Inductor
6.3 Series-Parallel Combinations of Capacitance
and Inductance
6.4 Mutual Inductance
C.T. Pan
2
6.1 The Capacitor
In this chapter, two new and important passive
linear components are introduced.
They are ideal models.
Resistors dissipate energy but capacitors and
inductors are energy storage components.
C.T. Pan
3
6.1 The Capacitor
Circuit symbol and component model.
q @ C ⋅ vC , q(t)= ∫ iC (τ ) dτ
t
q : charge
C.T. Pan
C : capacitance , in F(Farad)
4
6.1 The Capacitor
1
1 t0
1 t
iC (τ ) dτ = ∫ iC (τ ) dτ + ∫ iC (τ ) dτ
∫
C
C −∞
C t0
1 t
vC (t ) @ vC (t0 ) + ∫ iC (τ ) dτ .......... (A)
C t0
dv
dq
iC (t ) =
=C C
..................... (B)
dt
dt
vC (t ) =
1 farad = 1 Coulomb/Volt
The unit of capacitance is chosen to be farad in honor of
the English physicist , Michael Faraday(1791-1867).
C.T. Pan
5
6.1 The Capacitor
Example 1 : A parallel-plate capacitor
ε
A
d
ε : the permittivity of the dielectric
material between the plates
C=ε
C.T. Pan
6
6.1 The Capacitor
Example 1 : (cont.)
If vc > 0 and ic > 0 , or vc < 0 and ic < 0
the capacitor is being charged.
If vc ⋅ ic < 0 , the capacitor is discharging.
p(t)= vC (t ) iC (t ) = vC (t ) C
dvc ( t )
dt
ε
Energy in a capacitor
w = ∫ p (τ ) dτ =
t
−∞
1
q2
C vc 2 ( t ) =
≥0
2
2C
C.T. Pan
7
6.1 The Capacitor
(a) When vC is constant , then ic = 0.
ie. , equivalent to open circuit
(b) vC (t ) is a continuous function
if there is only finite strength energy sources
inside the circuit.
1 t +ε
iC (τ )dτ
C ∫t
lim vC (t + ε ) = vC (t ) , vC (0+ ) = vC (0− )
Q vC (t + ε ) = vC (t ) +
ε →0
i.e. vC (t) can not change abruptly for finite iC (t)
C.T. Pan
8
6.1 The Capacitor
(c) An ideal capacitor does not dissipate energy . It stores
energy in the electrical field .
(d) A nonideal capacitor has a leakage resistance
ESR : equivalent series resistance
C.T. Pan
9
6.2 The Inductor
Circuit symbol and component model.
λ
0
λ @ L ⋅ iL , λ ( t ) =
C.T. Pan
∫
t
iL
v L (τ ) d τ
λ : flux linkage , in web - turns
L : inductance , in H (Henry)
10
6.2 The Inductor
1 t
1 t0
1 t
vL (τ ) dτ = ∫ vL (τ ) dτ + ∫ vL (τ ) dτ
∫
L −∞
L −∞
L t0
1 t
iL ( t ) = iL ( t0 ) + ∫ vL (τ ) dτ − − − − − − − − − − ( A)
L t0
di ( t )
dλ
vL ( t ) =
=L L
− − − − − − − − − − − − − ( B)
dt
dt
di ( t )
p ( t ) = vL iL = LiL ( t ) L
dt
Energy in an inductor
t
1
w ( t ) = ∫ p (τ ) dτ = LiL 2 ( t ) ≥ 0
2
iL =
C.T. Pan
11
6.2 The Inductor
Example 2 : An Inductor
H : magnetic field intensity
B : flux density
l
i
A
N
φ :
∫
ur ur
B d A flu x
uuv r
Ñ∫ H d l = N i
Hl = Ni , H =
C.T. Pan
B = µH =
Ni
l
µ Ni
, µ perm eability of the core
l
12
6.2 The Inductor
Example 2 : (cont.)
µ ANi
l
µ A N 2i
λ = Nφ =
, f lu x lin k a g e
l
λ
µ AN 2
∴ L =
=
i
l
φ = BA =
l
i
A
N
C.T. Pan
13
6.2 The Inductor
The unit of inductance is the henry (H) , named in honor
of the American inventor Joseph Henry (1797-1878) .
1H = 1 volt-second / ampere
(a) when iL is constant , then vL=0 .
i.e. , equivalent to short circuit
(b) iL(t) is a continuous function if there is only finite
strength source inside the circuit .
1 t +∈
vL (τ ) dt
L ∫t
lim iL ( t + ∈) = iL ( t ) , or iL ( 0+ ) = iL ( 0− )
iL ( t + ∈) = iL ( t ) +
∈→0
i.e. iL ( t ) can not change abruptly for finite vL ( t ) .
C.T. Pan
14
6.2 The Inductor
(c) An ideal inductor does not dissipate energy .
It stores energy in the magnetic field .
(d) A nonideal inductor contains winding resistance and
parasitic capacitance .
C.T. Pan
15
6.2 The Inductor
Example 3 : Under dc and steady state conditions,
find (a) I , VC & IL , (b) WC and WL
12
= 2A
1+ 5
VC = 5I L = 10V
I = IL =
1
WC = ×1× 102 = 50 J
2
1
WL = × 2 × 22 = 4 J
2
C.T. Pan
16
6.2 The Inductor
(a) The capacity of C and L to store energy makes them
useful as temporary voltage or current sources , i.e. ,
they can be used for generating a large amount of
voltage or current for a short period of time.
(b) The continuity property of VC(t) and iL(t) makes
inductors useful for spark or arc suppression and
for converting pulsating voltage into relatively
smooth dc voltage.
C.T. Pan
17
6.2 The Inductor
(c) The frequency sensitive property of L and C makes
them useful for frequency discrimination.
(eg. LP , HP , BP filters)
C.T. Pan
18
6.3 Series-Parallel Combinations of
Capacitance and Inductance
N capacitors in parallel
Q i = i1 + i2 + L + iN
i
v
c1
i1
c2
i2
L
cN
iN
dv
dv
dv
+ c2
+ L + cN
dt
dt
dt
N
dv

 dv
=  ∑ ck 
= Ceq
dt
 k =1  dt
= c1
v1 (0) = v2 (0) = L = vN (0)
∴ Ceq = c1 + c2 + L + cN
v1 = v2 = L =vN = v
v(0) = vk (0)
C.T. Pan
19
6.3 Series-Parallel Combinations of
Capacitance and Inductance
N capacitors in series
Q vk ( t ) =
L
L
t
1
i (τ ) dτ + vk ( t0 )
ck t∫0
N
 N 1 t
∴ v =  ∑  ∫ i (τ ) dτ + ∑ vk (to )
k =1
 k =1 ck  t0
t
i1 = i2 = L = iN
1
i (τ ) dτ + v ( t0 )
=
Ceq t∫0
∴
1
1
1
1
= +
+ L +
Ceq C1 C2
CN
v ( t0 ) = v1 ( t0 ) + v2 ( t0 ) + L + vN ( t0 )
C.T. Pan
20
6.3 Series-Parallel Combinations of
Capacitance and Inductance
N inductors in series
Qv = v1 + v2 + L + vN
L
di
di
di
+ L2 + L + LN
dt
dt
dt
N
di

 di
=  ∑Lk 
= Leq
dt
 k =1  dt
= L1
i1 (0) = i2 (0) = L = iN (0)
i1 (t ) = i2 (t)= L =iN (t ) = i (t ) ∴Leq = L1 + L2 + L + LN
i(0) = i1(0) = i2 (0)L= iN (0)
C.T. Pan
21
6.3 Series-Parallel Combinations of
Capacitance and Inductance
N inductors in parallel
i =i1 ( t0 ) +
L
Q v1 = v2 = L = vN = v
∴ i = i1 + i2 + L + iN
C.T. Pan
1 t
1 t
1 t
vdτ +i2 ( t0 ) + ∫ vdτ +L+iN ( t0 ) + ∫ vdτ
∫
t
t
L1 0
L2 0
LN t0
N
N 1 t
=∑ ∫ vdτ +∑ik ( t0 )
k=1
 k=1 Lk  t0
t
1
= ∫ vdτ +i( t0 )
Leq t0
1 1 1
1
∴ = + +L+
Leq L1 L2
LN
i( t0 ) =i1 ( t0 ) +i2 ( t0 ) +L+iN ( t0 )
22
6.3 Series-Parallel Combinations of
Capacitance and Inductance
In summary
Resistor
V-I
V = RI
I-V
I=
P or W
P=
Capacitor
1
i dt
C ∫t0
dv
i=C
dt
1
W = Cv 2
2
CC
Ceq = 1 2
C1 + C 2
v = v (t 0 ) +
1
V
R
V2
= I 2R
R
Req = R1 + R2
series
parallel
R1 R2
R1 + R2
same
Req =
dc case
t
Inductor
di
dt
1 t
i = i ( t 0 ) + ∫ v dτ
L t0
1
W = Li 2
2
v=L
Leq = L1 + L2
L1 L2
L1 + L2
Ceq = C1 + C 2
Leq =
open circuit
short circuit
C.T. Pan
23
6.4 Mutual Inductance
Circuit symbol and model of coupling inductors
+
v1
M 12
i1
L1
−
i2
L2
+
v2
L1 , L2 : self inductances
M : mutual inductance
unit : in H
−
di1
di
+M 2
dt
dt
di
di
v2 = M 1 + L 2 2
dt
dt
v1 = L 1
C.T. Pan
24
6.4 Mutual Inductance
Example 4 : Mutual inductance
Apply I1 , with i2=0
ur v
H
∫ ⋅ dl=N1 ⋅ I1
C.T. Pan
25
6.4 Mutual Inductance
Assume uniform magnetic field intensity H
N 1I 1
, ∴ B = µH =
l
µ A N 1I 1
φ = ∫ B ⋅ dA =
l
µ A N 12 I 1
λ 1 = N 1φ =
; λ2
l
λ1
µ A N 12
L1 @
=
, M 21 @
I1
l
H =
C.T. Pan
µ N 1I 1
l
µ A N 1N 2I 1
l
µ A N 1N 2
=
l
= N 2φ =
λ2
I1
26
6.4 Mutual Inductance
Dot convention for mutually coupled inductors:
When the reference direction for a current enters the
dotted terminal of a coil , the polarity of the induced
voltage in the other coil is positive at its corresponding
dotted terminal.
C.T. Pan
27
6.4 Mutual Inductance
V1>0
V2
i1↗ , v1>0 , φ↗ , ( i2=0 )
Another dot of coil 2 should be placed in terminal c.
C.T. Pan
28
6.4 Mutual Inductance
V1>0
V2
Conceptually , one can connect a resistor at cd terminals.
Then i2 will be negative.
The generated flux of i2 will oppose the increasing of φ
due to increasing i1 ( Lentz law ).
Hence , another dot should be placed at c terminal.
C.T. Pan
29
6.4 Mutual Inductance
V1>0
V2
In case , the other dot is placed at d terminal , then i2
will be positive.
Hence , the generated flux of i2 will be added to the
increasing φ due to i1.
C.T. Pan
30
6.4 Mutual Inductance
V1>0
V2
Then the induced voltage at coil two will increase
and so will i2.
This will violate the conservation of energy.
C.T. Pan
31
6.4 Mutual Inductance
The procedure for determining dot markings
Step1 Assign current direction references for the coils.
Step2 Arbitrarily select one terminal of one coil and
mark it with a dot.
Step3 Use the right-hand rule to determine the direction
of the magnetic flux due to the current of the other
coil.
C.T. Pan
32
6.4 Mutual Inductance
Step4 If this flux direction has the same direction as that
of the first dot terminal current , then the second
dot is placed at the terminal where the second
current enters. Otherwise, the second dot should
be placed at the terminal where the second current
leaves.
C.T. Pan
33
6.4 Mutual Inductance
Example 5 : Determining dot markings
Step 1 : Assign i1 , i2 , i3 directions.
C.T. Pan
34
6.4 Mutual Inductance
Example 5 : (cont.)
Step 2 :
For coils 1 and 2, choose
first dot as follows
C.T. Pan
35
6.4 Mutual Inductance
Example 5 : (cont.)
For coils 1 and 3
C.T. Pan
36
6.4 Mutual Inductance
Example 5 : (cont.)
For coils 2 and 3
C.T. Pan
37
6.4 Mutual Inductance
Example 5 : (cont.)
Step 3 :
Check the relative flux directions and determine the
dot position at the other coil
For coil 1 and 2
φ 1 and φ 2 are in the same direction
L1
L2
M12
C.T. Pan
38
6.4 Mutual Inductance
Example 5 : (cont.)
For coil 1 and 3
φ1 and φ3 are in opposite direction
L1
C.T. Pan
L3
39
6.4 Mutual Inductance
Example 5 : (cont.)
For coil 2 and 3
φ2 and φ3 are in opposite direction
C.T. Pan
40
6.4 Mutual Inductance
Example 5 : (cont.)
In summary
di
di1
di
+ M 12 2 − M 13 3
dt
dt
dt
di
di1
di2
v2 = M 12
+ L2
− M 23 3
dt
dt
dt
di
di
di
v3 = − M 13 1 − M 23 2 + L3 3
dt
dt
dt
v1 = L1
+
v1
_
M12
i1
L1
i2
+
v2
_
L2
M23
M13
i3
L3
C.T. Pan
+
v3
_
41
6.4 Mutual Inductance
If the physical arrangement of the coils are not known,
the relative polarities of the magnetically coupled coils
can be determined experimentally. We need
a dc voltage source VS
a resistor R : to limit the current
a switch S
a dc voltmeter
C.T. Pan
42
6.4 Mutual Inductance
Step 1 : Assign current directions and arbitrarily assign
one dot at coil one.
L1
L2
C.T. Pan
43
6.4 Mutual Inductance
Step 2 : Connect the setup as follows.
L1
C.T. Pan
L2
44
6.4 Mutual Inductance
Step 3 : Determine the voltmeter deflection when the
switch is closed.
If the momentary deflection is upscale,
then
L1
L2
If the momentary deflection is downscale,
then
L1
L2
C.T. Pan
45
6.4 Mutual Inductance
The reciprocal property of the mutual inductance can be
proved by considering the energy relationship .
Step 1 : i2=0 , i1 increased from zero to I .
+
v1
−
M 12
i1
+
L1
L2
di1
dt
di
v2 = M 21 1
dt
input power p1 = v1i1 + v2i2
v1 = L1
i2
v2
−
= L1 (
di1
)i1
dt
input energy w1 = ∫ p1dτ
t
0
C.T. Pan
= ∫ L1i1di1 =
I1
0
L1 2
I1
2
46
6.4 Mutual Inductance
Step2 : keep i1=I , i2 is increased from zero to I2
di
di
dI
+ M 12 2 = M 12 2
dt
dt
dt
di
di
dI
v2 = M 21 + L2 2 = L2 2
dt
dt
dt
v1 = L1
Input power
p2 = v1 I1 + v2 i2
= M 12 I1
Input energy
di2
di
+ ( L2 2 )i2
dt
dt
w2 = ∫ p2 dτ
= ∫ M 12 I1di2 + ∫ L2i2 di2
I2
0
C.T. Pan
I2
0
1
= M 12 I1 I 2 + L2 I 22
2
47
6.4 Mutual Inductance
Total energy when i1=I1 , i2=I2
w = w1 + w2 =
1
1
L1 I12 + L2 I 22 + I1 I 2 M 12
2
2
Similarly , if we reverse the procedure , by first
increasing i2 from zero to I2 and then increasing i1 from
zero to I1, the total energy is
w = w1 + w2 =
1
1
L1 I12 + L2 I 22 + I1 I 2 M 21
2
2
Hence , M12=M21=M
C.T. Pan
48
6.4 Mutual Inductance
Definition of coefficient of coupling k
M
L1 L2
k@
0 ≤ k ≤1
1
0 < k < , loosely coupling
2
1
≤ k < 1 , closely coupling
2
k =1 ,
unity coupling
C.T. Pan
49
6.4 Mutual Inductance
According to dot convention chosen , the total energy
stored in the coupled inductors should be
w=
1 2 1 2
L1i1 + L2i2 ± Mi1i2 ≥ 0
2
2
In particular , consider the limiting case
1 2 1
L1i1 + L2 i22 − Mi1i2 = 0
2
2
The above equation can be put into the following form
(
C.T. Pan
L1
L
i1 − 2 i2 ) 2 + i1i2 ( L1 L2 − M ) = 0
2
2
50
6.4 Mutual Inductance
Thus , w(t) ≧0 only if
L1 L2 ≥ M
when i1 and i2 are either both positive or both negative
Hence , k≦1
C.T. Pan
51
6.4 Mutual Inductance
Example 6 : Finding mesh-current equations for a circuit
with magnetically coupled coils.
i4H
i1 5Ω
ig
20Ω
i2
ig
Three meshes and one
current source , only need
two unknowns , say i1 and
60Ω i2
i16 H
Note : current i4H = i1(t)
current i16H = ig - i2
C.T. Pan
52
6.4 Mutual Inductance
Due to existence of mutual inductance M=8H , there
are two voltage terms for each coil .
One can use dependence source to eliminate the
coupling relation as follows.
8
5Ω
ig
d
i16 H
dt
i1
20Ω
ig
i2
di
8 1
dt
C.T. Pan
60Ω
53
6.4 Mutual Inductance
8
4H
i1
5Ω
ig
d
i16 H
dt
+20Ω
16H
ig
+- 8 di1
dt
i2
60Ω
Hence , for i1 mesh
di1
d
+ 8 (ig − i2 ) + 20(i1 − i2 ) + 5(i1 − ig ) = 0
dt
dt
For i2 mesh
4
C.T. Pan
20(i2 − i1 ) + 60i2 + 16
d
d
(i2 − ig ) − 8 i1 = 0
dt
dt
54
Summary
n
n
n
Objective 1 : Know and be able to use the component
model of an inductor
Objective 2 : Know and be able to use the component
model of a capacitor.
Objective 3 : Be able to find the equivalent inductor
(capacitor) together with it equivalent
initial condition for inductors (capacitors)
connected in series and in parallel.
C.T. Pan
55
Summary
n
Objective 4 : Understand the component model of
coupling inductors and the dot convention
as well as be able to write the mesh equations
for a circuit containing coupling inductors.
Chapter Problems : 6.4
6.19
6.21
6.26
6.40
6.41
Due within one week.
C.T. Pan
56