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Capacitors and Inductors
Chapter 6
Chap. 6, Capacitors and Inductors
•
•
•
•
•
Introduction
Capacitors
Series and Parallel Capacitors
Inductors
Series and Parallel Inductors
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6.1 Introduction
•
•
Resistor: a passive element which dissipates
energy only
Two important passive linear circuit
elements:
1) Capacitor
2) Inductor
•
Capacitor and inductor can store energy
only and they can neither generate nor
dissipate energy.
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Michael Faraday (1971-1867)
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6.2 Capacitors
• A capacitor consists of two conducting plates
separated by an insulator (or dielectric).
C
εA
d
   r 0
 0  8.854 10 12 (F/m)
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•
εA
C
d
Three factors affecting the value of
capacitance:
1. Area: the larger the area, the greater the
capacitance.
2. Spacing between the plates: the smaller the
spacing, the greater the capacitance.
3. Material permittivity: the higher the permittivity,
the greater the capacitance.
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Fig 6.4
(a) Polyester capacitor, (b) Ceramic capacitor, (c) Electrolytic capacitor
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Fig 6.5
Variable capacitors
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Fig 6.3
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Fig 6.2
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Charge in Capacitors
• The relation between the charge in plates and
the voltage across a capacitor is given below.
q  Cv
1F  1 C/V
q
Linear
Nonlinear
v
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Voltage Limit on a Capacitor
• Since q=Cv, the plate charge increases as the
voltage increases. The electric field intensity
between two plates increases. If the voltage
across the capacitor is so large that the field
intensity is large enough to break down the
insulation of the dielectric, the capacitor is out
of work. Hence, every practical capacitor has a
maximum limit on its operating voltage.
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I-V Relation of Capacitor
+
i
v
C
dq
dv
q  Cv, i 
C
dt
dt
-
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Physical Meaning
dv
iC
dt
+
v
i
C
-
• when v is a constant voltage, then i=0; a constant
voltage across a capacitor creates no current through
the capacitor, the capacitor in this case is the same as
an open circuit.
• If v is abruptly changed, then the current will have an
infinite value that is practically impossible. Hence, a
capacitor is impossible to have an abrupt change in
its voltage except an infinite current is applied.
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Fig 6.7
• A capacitor is an open circuit to dc.
• The voltage on a capacitor cannot change
abruptly.
Abrupt change
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dv
iC
dt
1 t
v(t )   idt
C 
1 t
v(t )   idt  v(to)
C to
 v()  0 
 v(to)  q(to) / C 
+
v
i
C
-
• The charge on a capacitor is an integration of
current through the capacitor. Hence, the
memory effect counts.
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Energy Storing in Capacitor
dv
p  vi  Cv
dt
v (t )
dv
1 2
w   pdt  C  v dt  C v (  ) vdv  Cv
dt
2
t
t
1
w(t )  Cv 2 (t )
2
q 2 (t )
w(t ) 
2C
Ch06 Capacitors and Inductors
( v( )  0)
+
v
v (t )
v (  )
i
C
-
17
Model of Practical Capacitor
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Example 6.1
(a) Calculate the charge stored on a 3-pF
capacitor with 20V across it.
(b) Find the energy stored in the capacitor.
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Example 6.1
Solution:
(a) Since q  Cv,
q  3  10
12
 20  60pC
(b) The energy stored is
1 2 1
12
w  Cv   3  10  400  600pJ
2
2
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Example 6.2
• The voltage across a 5- F capacitor is
v(t )  10 cos 6000t V
Calculate the current through it.
Solution:
• By definition, the current is
dv
6 d
iC
 5 10
(10 cos 6000t )
dt
dt
 5  10 6  6000 10 sin 6000t  0.3 sin 6000t A
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Example 6.3
• Determine the voltage across a 2-F capacitor if the
current through it is
i(t )  6e
3000t
mA
Assume that the initial capacitor voltage is zero.
Solution:
• Since v  1 t idt  v(0) and v(0)  0,

C 0
3
t
1
3000 t
3  10 3000t t
3
v
6
e
dt  10 
e
6 0
0
2  10
 3000
 (1  e 3000t )V
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Example 6.4
• Determine the current through a 200- F
capacitor whose voltage is shown in Fig 6.9.
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Example 6.4
Solution:
• The voltage waveform can be described
mathematically as
50t V
0  t 1

 100  50t V
1 t  3
v(t )  
 200  50t V
3t  4

0 otherwise

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Example 6.4
• Since i = C dv/dt and C = 200 F, we take the
derivative of to obtain
0  t  1  10mA
0  t 1
 50
 50
1  t  3  10mA
1 t  3
6
i (t )  200  10  

50
3t  4
10mA
3t  4


0 otherwise
 0 otherwise 
• Thus the current waveform is shown in
Fig.6.10.
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Example 6.4
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Example 6.5
• Obtain the energy stored in each capacitor in
Fig. 6.12(a) under dc condition.
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Example 6.5
Solution:
• Under dc condition, we replace each capacitor
with an open circuit. By current division,
3
i
(6mA )  2mA
3 2 4
 v1  2000 i  4 V, v 2  4000i  8 V
1
1
2
3
2
 w1  C1v1  (2  10 )(4)  16mJ
2
2
1
1
2
3
2
w2  C2 v2  (4  10 )(8)  128mJ
2
2
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Fig 6.14
Ceq  C1  C2  C3  ....  C N
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6.3 Series and Parallel Capacitors
i  i1  i2  i3  ...  iN
dv
dv
dv
dv
i  C1  C2  C3  ...  C N
dt
dt
dt
dt
N
dv
dv


   CK   Ceq
dt
 k 1  dt
Ceq  C1  C2  C3  ....  CN
• The equivalent capacitance of N parallelconnected capacitors is the sum of the
individual capacitance.
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Fig 6.15
1
1
1
1
1
 
  ... 
Ceq C1 C2 C3
CN
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Series Capacitors
v(t )  v1 (t )  v2 (t )  ...  vN (t )
1
Ceq
1
1
1
1 t
id  ( C1  C2  C3  ...  CN )id
t
q(t ) q(t ) q(t )
q(t )



Ceq
C1
C2
CN
• The equivalent capacitance of seriesconnected capacitors is the reciprocal of the
sum of the reciprocals of the individual
capacitances.
C1C2
1
1 1
Ceq 
 
C1  C2
Ceq C1 C2
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Summary
• These results enable us to look the capacitor in
this way: 1/C has the equivalent effect as the
resistance. The equivalent capacitor of
capacitors connected in parallel or series can
be obtained via this point of view, so is the Y△ connection and its transformation
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Example 6.6
• Find the equivalent capacitance seen between
terminals a and b of the circuit in Fig 6.16.
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Example 6.6
Solution:
 20  F and 5  F capacitors are in series :
20  5

 4F
20  5
 4  F capacitor is in parallel with the 6  F
and 20  F capacitors :
 4  6  20  30 F
 30  F capacitor is in series with
the 60  F capacitor.
30  60
Ceq 
F  20F
30  60
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Example 6.7
• For the circuit in Fig 6.18, find the voltage
across each capacitor.
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Example 6.7
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Example 6.7
Solution:
• Two parallel capacitors:
 Ceq 
• Total charge
1
1 1 1
 
60 30 20
mF  10mF
3
q  Ceq v  10  10  30  0.3 C
• This is the charge on the 20-mF and 30-mF capacitors,
because they are in series with the 30-v source. ( A
crude way to see this is to imagine that charge acts
like current, since i = dq/dt)
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Example 6.7
q
0
.
3
• Therefore,
v1  
 15 V,
3
C1 20  10
q
0.3
v2 

 10 V
3
C2 30  10
• Having determined v1 and v2, we now use KVL to
determine v3 by
v3  30  v1  v2  5V
• Alternatively, since the 40-mF and 20-mF capacitors
are in parallel, they have the same voltage v3 and their
combined capacitance is 40+20=60mF.
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q
0.3
 v3 

 5V
3
60mF 60  10
39
Joseph Henry (1979-1878)
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6.4 Inductors
• An inductor is made of a coil of conducting wire
N A
L
l
2
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Fig 6.22
N 2 A
L
l
   r 0
 0  4 10 7 (H/m)
N : number of turns.
l :length.
A:cross  sectional area.
 : permeability of the core
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Fig 6.23
(a) air-core
(b) iron-core
(c) variable iron-core
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Flux in Inductors
• The relation between the flux in inductor and
the current through the inductor is given below.
  Li
ψ
1H  1 Weber/A
Linear
Nonlinear
i
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Energy Storage Form
• An inductor is a passive element designed to
store energy in the magnetic field while a
capacitor stores energy in the electric field.
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I-V Relation of Inductors
• An inductor consists of
a coil of conducting
d
wire.
di
v
L
dt
dt
i
+
v
L
-
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Physical Meaning
d
di
v
L
dt
dt
• When the current through an inductor is a constant,
then the voltage across the inductor is zero, same as a
short circuit.
• No abrupt change of the current through an inductor
is possible except an infinite voltage across the
inductor is applied.
• The inductor can be used to generate a high voltage,
for example, used as an igniting element.
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Fig 6.25
• An inductor are like a short circuit to dc.
• The current through an inductor cannot change
instantaneously.
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1
di  vdt
L
1 t
i   v(t )dt
L 
1 t
i   v(t )dt  i (to )
L to
+
v
L
-
The inductor has memory.
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Energy Stored in an Inductor
 di 
P  vi   L  i
 dt 
+
v
L
di


w   pdt    L idt
 dt 
i (t )
1 2
1 2
 L i (  ) i di  Li (t )  Li () i ()  0,
2
2
t
t
• The energy stored in an inductor
1 2
w(t )  Li (t )
2
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Model of a Practical Inductor
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Example 6.8
• The current through a 0.1-H inductor is i(t) =
10te-5t A. Find the voltage across the inductor
and the energy stored in it.
Solution:
di
Since v  L and L  0.1H,
dt
d
v  0.1 (10te 5t )  e 5t  t (5)e 5t  e 5t (1  5t )V
dt
The energy stored is
1 2 1
w  Li  (0.1)100t 2 e 10t  5t 2 e 10t J
2
2
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Example 6.9
• Find the current through a 5-H inductor if the
voltage across it is
2
30t , t  0
v(t )  
t0
 0,
Also find the energy stored within 0 < t < 5s.
Assume i(0)=0.
Solution:
1 t
Since i  t v(t )dt  i (t0 ) and L  5H.
L
3
t
1
t
2
i  0 30t dt  0  6   2t 3 A
3
5
0
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Example 6.9
The power p  vi  60t 5 , and the energy stored is then
6
t 5
w   pdt  0 60t dt  60
 156.25 kJ
60
Alternatively, we can obtain the energy stored using
Eq.(6.13), by writing
1 2
1
w(5)  w(0)  Li (5)  Li (0)
2
2
1
3 2
 (5)(2  5 )  0  156.25 kJ
2
as obtained before.
5
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54
Example 6.10
• Consider the circuit in
Fig 6.27(a). Under dc
conditions, find:
(a) i, vC, and iL.
(b) the energy stored
in the capacitor and
inductor.
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Example 6.10
Solution:
(a ) Under dc condition : capacitor  open circuit
inductor  short circuit
12
i  iL 
 2 A, vc  5i  10 V
1 5
(b )
1
1
2
wc  Cvc  (1)(10 )  50J,
2
2
1 2 1
wL  Li  (2)(2 2 )  4J
2
2
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56
Inductors in Series
Leq  L1  L2  L3  ...  LN
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Inductors in Parallel
1
1 1
1
  
Leq L1 L2
LN
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6.5 Series and Parallel Inductors
• Applying KVL to the loop,
v  v1  v2  v3  ...  vN
• Substituting vk = Lk di/dt results in
di
di
di
di
v  L1  L2  L3  ...  LN
dt
dt
dt
dt
di
 ( L1  L2  L3  ...  LN )
dt
N
di
di


   LK   Leq
dt
 K 1  dt
Leq  L1  L2  L3  ...  LN
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Parallel Inductors
• Using KCL, i  i1  i2  i3  ...  iN
• But ik  1 tt vdt  ik (t0 )
Lk
1
i 
Lk
t
t
0
o
1
1 t
vdt  i1 (t0 )  t vdt  is (t0 )  ... 
LN
L2
0
t
t vdt  iN (t0 )
0
1 1
1 t
    ...   t vdt  i1 (t0 )  i2 (t0 )  ...  iN (t0 )
LN 
 L1 L2
0
N
N 1t
1 t
    t vdt   ik (t0 ) 
vdt  i (t0 )

t
Leq
k 1
 k 1 Lk 
0
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60
• The inductor in various connection has the
same effect as the resistor. Hence, the Y-Δ
transformation of inductors can be similarly
derived.
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Table 6.1
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Example 6.11
• Find the equivalent inductance of the circuit
shown in Fig. 6.31.
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Example 6.11
• Solution:
Series : 20H, 12H, 10H
 42H
7  42
 6H
Parallel :
7  42
 Leq  4  6  8  18H
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Practice Problem 6.11
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Example 6.12
• Find the circuit in Fig. 6.33, i (t )  4(2  e
If i2 (0)  1 mA, find : (a) i (0)
(b) v(t ), v1 (t ), and v2 (t ); (c) i1 (t ) and i2 (t )
10t
)mA.
1
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Example 6.12
Solution:
10 t
(a ) i(t )  4(2  e )mA  i(0)  4(2  1)  4mA.
 i1 (0)  i (0)  i2 (0)  4  (1)  5mA
(b) The equivalent inductance is
Leq  2  4 || 12  2  3  5H
di
10 t
10 t
 v(t )  Leq  5(4)(1)(10)e mV  200e mV
dt
di
10 t
10 t
v1 (t )  2  2(4)(10)e mV  80e mV
dt
10 t
 v2 (t )  v(t )  v1(t )  120e mV
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Example 6.12
1 t
(c) i  0 v(t ) dt  i (0) 
L
1 t
120 t 10t
i1 (t )  0 v2 dt  i1 (0) 
e
dt

5
mA

4
4 0
10 t t
 3e
 5 mA  3e 10t  3  5  8  3e 10t mA
0
1 t
120 t 10t
i2 (t )  0 v2 dt  i2 (0) 
e dt  1mA

0
12
12
10 t t
 e
 1mA  e 10t  1  1  e 10t mA
0
Note that i1 (t )  i2 (t )  i (t )
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