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EEE105: CI RCUI T THEORY
CHAPTER 5: CAPACITORS AND INDUCTORS
5.1 Introduction
• Unlike resistors, which dissipate energy, capacitors
and inductors store energy.
• Thus, these passive elements are called storage
elements.
5.2 Capacitors
• Capacitor stores energy in its electric field.
• A capacitor is typically constructed as shown in
Figure 5.1.
Figure 5.1
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A capacitor consists of two conducting plates
separated by an insulator (or dielectric)
• When a voltage v is applied, the source deposits a
positive charge q on one plate and negative charge –q
on the other.
Figure 5.2
• The charge stored is proportional to the applied
voltage, v
q = Cv
(5.1)
where C is the constant of proportionality, which is
known as the capacitance of the capacitor.
• Unit for capacitance: farad (F).
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• Definition of capacitance:
Capacitance is the ratio of the charge on one plate
of a capacitor to the volatge difference between the
two plates.
• Capacitance is depends on the physical dimensions
of the capacitor.
• For parallel-plate capacitor, capacitance is given by
C=
∈A
d
(5.2)
where A is the surface area of each plate
d is the distance between the plates
∈ is the permittivity of the dielectric material
between the plates
• The symbol of capacitor:
Figure 5.3
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EEE105: CI RCUI T THEORY
• The current flows into the positive terminal when the
capacitor is being charged.
• The current flows out of the positive terminal when
the capacitors is discharging.
• Differentiating both sides of Equation 5.1,
dq
dv
=C
dt
dt
Thus,
dv
i =C
dt
(5.3)
• Capacitors that satisfy Equation 5.3 are said to be
linear.
• The voltage-current relation:
1 t
v = ∫−∞ i (t )dt
C
v=
1 t
i (t )dt + v(t0 )
∫
t0
C
(5.4)
where v (t0 ) = q (t0 ) C is the voltage across the
capacitor at time to.
• Thus, the capacitor voltage is depends on the past
history of the capacitor current – has memory.
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EEE105: CI RCUI T THEORY
• The instantaneous power given by:
dq
p = vi = Cv
dt
(5.5)
• The energy stored given by:
t
t
w = ∫−∞ pdt = C ∫−∞ v
t
dv
1
t
dt = C ∫−∞ vdv = Cv 2
dt
2
t = −∞
Note that v ( −∞) = 0 because the capacitor was
uncharged at t = −∞ .
Thus,
1 2 q2
w = Cv =
2
2C
(5.6)
• Four issues:
(i)
From Equation 5.3, when the voltage across a
capacitor is not changing with time (i.e., dc
voltage), the current through the capacitor is
zero.
A capacitor is an open circuit to dc.
(ii)
The voltage on the capacitor must be
continuous. The capacitor resists an abruot
change in the voltage across it. According to
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(iii)
(iv)
EEE105: CI RCUI T THEORY
Equation 5.3, discontinuous change in voltage
requires an infinite current, which is
physically impossible.
The ideal capacitor does not dissipate energy.
A real, nonideal capacitor has a parallel-model
linkage resistance.
Figure 5.4
• Example 1:
The voltage across a 5µF capacitor is
v(t ) = 10 cos 6000t V
Calculate the current through it.
dv
−6 d
i (t ) = C = 5 × 10
(10 cos 6000t )
dt
dt
i (t ) = −5 ×10 −6 × 6000 ×10 sin 6000t
= −0.3 sin 6000t
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EEE105: CI RCUI T THEORY
• Example 2:
An initially charged 1-mF capacitor has the current
as shown in Figure 5.5. Calculate the voltage across
it at t = 2 ms and t = 5 ms.
Figure 5.5
The current waveform can
mathematically as:
50t mV
i (t ) = 
100 mV
be
described
0<t <2
t>2
For t = 2 ms:
v=
1 t
1 2
−3
i
(
t
)
dt
+
v
(
t
)
=
50
×
10
t + v ( 0)
∫
0
− 3 ∫0
t0
C
10
−3 2 2
1 50 × 10 t
v = −3
10
2
0
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+ 0 = 100 mV
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EEE105: CI RCUI T THEORY
For t = 5 ms:
1 t
v = ∫t i (t )dt + v(t0 )
C 0
1 2
5
−3
= −3 ∫0 50 ×10 t + ∫2 100 ×10 −3 + v(0)
10
−3 2 2


1  50 × 10 t
5
v = −3
+ 100t 2  + 0 = 400 mV

10 
2
0

(
)
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EEE105: CI RCUI T THEORY
5.3 Series and Parallel Capacitors
• Consider the circuit as shown in Figure 5.6 with N
capacitors in parallel:
Figure 5.6
Applying KCL:
i = i1 + i2 + ... + iN
But,
ik = Ck
dv
dt
Hence,
dv
dv
dv
i = C1 + C2 + ... + C N
dt
dt
dt
N
dv
dv

=  ∑ Ck  = Ceq
dt
 k =1  dt
where,
Ceq = C1 + C2 + ... +C N
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(5.7)
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EEE105: CI RCUI T THEORY
• Consider the circuit as shown in Figure 5.7 with N
capacitors in series:
Figure 5.7
Applying KVL,
v = v1 + v2 + ... + v N
But,
vk =
1 t
i (t )dt + vk (t0 )
∫
t0
Ck
Thus,
1 t
1 t
v = ∫t i (t )dt + v1 (t0 ) + ∫t i (t )dt + v2 (t0 )
C1 0
C2 0
+ ... +
1 t
i (t )dt + v N (t0 )
∫
t0
CN
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EEE105: CI RCUI T THEORY
1 1
1 t
 ∫t i (t )dt
v =  + + ... +
CN  0
 C1 C2
+ v1 (t0 ) + v2 (t0 ) + ... + v N (t0 )
v=
1 t
i (t )dt + v(t0 )
∫
t0
Ceq
where,
1
1 1
1
= + + ... +
Ceq C1 C2
CN
(5.8)
• Example:
Find the equivalent capacitance seen at terminals of
the circuit in Figure 5.8.
Figure 5.8
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Ceq = (50 µ + 70 µ ) (20 µ + (60 µ 120 µ ))
= 40 µF
• Example 2:
Find the voltage across each of the capacitors in
Figure 5.9.
Figure 5.9
Ceq = (40 µ ) (20 µ + (60 µ 30 µ ))
Ceq = 20 µF
The total charge,
q = Ceq v = 20 × 10 −6 × 60 = 1.2 ×10 −3 C
q 1.2 × 10 −3
∴ v1 = =
= 30 V
−6
C1 40 ×10
Using KVL,
60V = v1 + v2
∴v2 = 30 V
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EEE105: CI RCUI T THEORY
For Loop 2:
v2 = v3 + v4
But,
q Loop 2 = C60 µ 30 µ v2
q Loop 2 = ( 20 × 10 − 6 )( 30 ) = 0 .6 × 10 C
q Loop 2
0.6 ×10 −3
v3 =
=
= 10 V
−6
C3
60 × 10
v4 = v2 − v3 = 20 V
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EEE105: CI RCUI T THEORY
5.4 Inductors
• Inductor is a pasive element designed to store energy
in its magnetic field.
• Any conductor of electric current has inductive
properties and may be regarded as an inductor.
• To enhance the inductive effect, a practical inductor
is usually formed into a cylindrical coil with many
turns of conducting wire.
Figure 5.10
An inductor consists of a coil of conducting wire.
• If the current passes through an inductor, the voltage
across the inductor is proportional to the time of
change of the current.
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EEE105: CI RCUI T THEORY
• In mathematical form:
di
v=L
dt
(5.9)
where L is the constant of proportinality called the
inductance of the inductor.
• The unit of inductance is henry (H).
Inductance is the property whereby an inductor
exhibits opposition to the charge of current
flowing through it
• The inductance depends on inductor’s physical
dimension and construction, which is given by:
N 2 µA
L=
l
(5.10)
where N is the number of turns
l is the length
A is the cross sectional area
µ
is the permeability of the core
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EEE105: CI RCUI T THEORY
• The symbol of inductor:
Figure 5.11
• The current-voltage relationship:
di =
1
vdt
L
Intergrating gives,
1 t
i = ∫−∞ v(t )dt
L
or
1 t
i = ∫t v(t ) dt + i (to )
L o
(5.11)
where i (to ) is the total current for − ∞ < t < to
i (−∞) = 0 . (Note: there must be a time in the
past when there was no current
in the inductor.)
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EEE105: CI RCUI T THEORY
• The inductor stores energy in its magnetic field.
• The power delivered to the inductor:
 di 
p = iv =  L i
 dt 
(5.12)
• The energy stored:
 di 
w = ∫−∞ pdt = ∫−∞  L idt
 dt 
1
1
t
w = L ∫−∞ idi = Li 2 (t ) − Li 2 (−∞)
2
2
since i ( −∞) = 0 ,
t
t
1 2
w = Li
2
(5.13)
• 4 issues:
(i)
From equation 5.9, the voltage across an
inductor is zero when the current is constant
(i.e. dc source).
An inductor acts like a short circuit to dc.
(ii)
An important property of the inductor is its
opposition to the change in current flowing
through it.
The current through an inductor cannot
change instantaneously.
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(iii)
(iv)
EEE105: CI RCUI T THEORY
According to Equation 5.9, discontinuous
change in current requires an infinite voltage,
which is physically impossible.
The ideal inductor does not dissipate energy.
A real, nonideal inductor has a serial-model
resistance. This resistance is called a winding
resistance, Rw.
Figure 5.12
• Example 1:
If the current through a 1 mH inductor is
i (t ) = 20 cos100t mA, find the terminal voltage
and the energy stored.
The terminal voltage,
di
= 10 −3 × 20 × 100 × − sin 100t
dt
v = −2 sin 100t mV
v=L
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EEE105: CI RCUI T THEORY
The energy stored,
1 2 1 −3
w = Li = (10 )(20 ×10 −3 cos100t ) 2
2
2
∴ w = 0.2 cos 2 100t µJ
• Example 2:
Consider the circuit as shown in Figure 5.13. under
dc conditions, find (a) i, vc and iL, (b) the energy
stored in the capacitor and inductor.
Figure 5.13
(a)
Under dc condition;
The capacitor – open circuit
The inductor – short circuit
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Figure 5.14
From Figure 5.14,
12
i = iL =
=2A
1+ 5
vc = iL R5Ω = (2)(5) = 10 V
(b)
The energy in the capacitor,
1
1
wC = CvC2 = (1)(10 2 ) = 50 J
2
2
The energy in the inductor,
1 2 1
wL = LiL = ( 2)(2 2 ) = 4 J
2
2
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EEE105: CI RCUI T THEORY
5.5 Series and Parallel Inductors
• Consider the circuit as shown in Figure 5.15 with N
inductors in series:
Figure 5.15
Applying KVL:
v = v1 + v2 + ... + v N
But,
di
vk = Lk
dt
Hence,
di
di
di
v = L1 + L2 + ... + LN
dt
dt
dt
N
di
di


=  ∑ Lk  = Leq
dt
 k =1  dt
where,
Leq = L1 + L2 + ... + L N
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(5.14)
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EEE105: CI RCUI T THEORY
• Consider the circuit as shown in Figure 5.16 with N
inductors in parallel:
Figure 5.16
Applying KCL,
i = i1 + i2 + ... + iN
But,
ik =
1 t
v(t ) dt + ik (t0 )
∫
t0
Lk
Thus,
1 t
1 t
i = ∫t vdt + i1 (t0 ) + ∫t vdt + i2 (t0 )
L1 0
L2 0
+ ... +
1 t
vdt + iN (t0 )
∫
t0
LN
1 1
1 t
i =  + + ... +  ∫t vdt
LN  0
 L1 L2
+i1 (t0 ) + i2 (t0 ) + ... + iN (t0 )
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i=
EEE105: CI RCUI T THEORY
1 t
vdt + i (t0 )
∫
t0
Leq
where,
1
1 1
1
= + + ... +
Leq L1 L2
LN
(5.15)
• Example 1:
Find the equivalent inductance of the circuit in
Figure 5.17.
Figure 5.17
Leq = 4 H + 8 H + (7 H (20 H + 12 H + 10 H ))
∴ Leq = 18 H
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EEE105: CI RCUI T THEORY
• Example 2:
For
the
circuit
in
Figure
5.18,
i (t ) = 4(2 − e −10t ) mA. If i2 (0) = −1mA, find (a)
i1 (0) , (b) v(t), v1(t) and v2(t), (c) i1(t) and i2(t).
Figure 5.18
−10t
(a)
From i (t ) = 4( 2 − e
(b)
The equivalent inductance is
) mA,
i (0) = 4( 2 − 1) = 4 mA
Since i = i1 + i2 ,
i1 (0) = i (0) − i2 (0) = 4 − (−1) = 5 mA
Leq = 2 + 4 12 = 5 H
Thus,
di
v(t ) = Leq = 5(4)(−1)(−10)e −10t mV
dt
∴ v(t ) = 200e −10t mV
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and
di
v1 (t ) = L2 H = 2(−4)(−10)e −10t mV
dt
∴ v1 (t ) = 80e −10t mV
Since v = v1 + v2 ,
v2 (t ) = v(t ) − v1 (t ) = 120e −10t mV
(c)
The current i1 is obtained as,
1 t
i1 (t ) = ∫0 v2 dt + i1 (0)
4
120 t −10t
=
e dt + 5mA
∫
0
4
= −3e −10t t0 + 5mA
= −3e −10t + 3 + 5 = 8 − 3e −10t mA
Similarly,
1 t
v dt + i2 (0)
∫
0 2
12
120 t −10t
=
e dt − 1mA
∫
0
12
= −e −10t t0 − 1mA
i2 (t ) =
= −e −10t + 1 − 1 = −e −10t mA
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EEE105: CI RCUI T THEORY
5.6 Conclusion
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