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電路學(一)
Chapter 6
Capacitors and Inductors
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Capacitors and Inductors
Chapter 6
6.1
6.2
6.3
6.4
Capacitors
Series and Parallel Capacitors
Inductors
Series and Parallel Inductors
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6.1 Capacitors (1)
• A capacitor is a passive element designed
to store energy in its electric field.
• A capacitor consists of two conducting plates
separated by an insulator (or dielectric).
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6.1 Capacitors (2)
• Capacitance C is the ratio of the charge q on one
plate of a capacitor to the voltage difference v
between the two plates, measured in farads (F).
qC v
and
C
 A
d
• Where  is the permittivity of the dielectric material
between the plates, A is the surface area of each plate,
d is the distance between the plates.
• Unit: F, pF (10–12), nF (10–9), and F (10–6)
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6.1 Capacitors (3)
• If i is flowing into the +ve
terminal of C
– Charging => i is +ve
– Discharging => i is –ve
• The current-voltage relationship of capacitor
according to above convention is
dv
iC
dt
and
1
v
C

t
t0
i d t  v(t0 )
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6.1 Capacitors (4)
• The energy, w, stored in
the capacitor is
1
w  C v2
2
• A capacitor is
– an open circuit to dc (dv/dt = 0).
– its voltage cannot change abruptly.
– The ideal capacitor does not dissipate energy.
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6.1 Capacitors (5)
Example 1
(a) Calculate the charge stored on a 3-pF capacitor with 20 V
across it.
(b) Find the energy stored in the capacitor.
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6.1 Capacitors (6)
Example 2
The current through a 100-F capacitor is
i(t) = 50 sin(120 t) mA.
Calculate the voltage across it at t =1 ms and
t = 5 ms.
Take v(0) =0.
Answer:
v(1ms) = 93.14mV
v(5ms) = 1.7361V
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6.1 Capacitors (7)
Example 3
An initially uncharged 1-mF capacitor has the
current shown below across it.
Calculate the voltage across it at t = 2 ms and
t = 5 ms.
Answer:
v(2ms) = 100 mV
v(5ms) = 500 mV
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6.1 Capacitors (8)
Example 4
Obtain the energy stored in each capacitor in the figure under
dc conditions.
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6.1 Capacitors (9)
Example 5
Under dc condition, find the energy stored in the capacitors in
the figure.
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6.2 Series and Parallel
Capacitors (1)
• The equivalent capacitance of N parallelconnected capacitors is the sum of the individual
capacitances.
C eq  C1  C 2  ...  C N
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6.2 Series and Parallel
Capacitors (2)
• The equivalent capacitance of N series-connected
capacitors is the reciprocal of the sum of the
reciprocals of the individual capacitances.
1
1
1
1


 ... 
Ceq C1 C2
CN
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6.2 Series and Parallel
Capacitors (3)
Example 6
Find the equivalent capacitance seen at the
terminals of the circuit in the circuit shown below:
Answer:
Ceq = 40F
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6.2 Series and Parallel
Capacitors (4)
Example 7
Find the voltage across each of the capacitors in
the circuit shown below:
Answer:
v1 = 15 V
v2 = 10 V
v3 = 5 V
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6.2 Series and Parallel
Capacitors (5)
Example 8
Find the voltage across each of the capacitors in
the circuit shown below:
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6.3 Inductors (1)
• An inductor is a passive element designed
to store energy in its magnetic field.
• An inductor consists of a coil of conducting wire.
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6.3 Inductors (2)
• Inductance is the property whereby an inductor
exhibits opposition to the change of current
flowing through it, measured in henrys (H).
di
vL
dt
and
N2  A
L
l
• The unit of inductors is Henry (H), mH (10–3)
and H (10–6).
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6.3 Inductors (3)
• The current-voltage relationship of an inductor:
1
i
L
t
 v (t ) d t  i (t )
t0
0
• The power stored by an inductor:
1
w  L i2
2
• An inductor acts like a short circuit to dc (di/dt = 0)
• The current through an inductor cannot change
abruptly.
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6.3 Inductors (4)
Example 9
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.
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6.3 Inductors (5)
Example 10
The terminal voltage of a 2-H
inductor is
v = 10(1-t) V
Find the current flowing through it at
t = 4 s and the energy stored in it
within 0 < t < 4 s. Assume i(0) = 2 A.
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6.3 Inductors (6)
Example 11 Consider the circuit, under dc
conditions, find:
(a) i, vC, and iL
(b) the energy stored in the capacitor and inductor.
Answer:
i=2A
vC = 10 V
wL = 4 J
wC = 50 J
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6.3 Inductors (7)
Example 12
Determine vc, iL, and the energy stored in the
capacitor and inductor in the circuit of circuit shown
below under dc conditions.
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6.4 Series and Parallel
Inductors (1)
• The equivalent inductance of series-connected
inductors is the sum of the individual inductances.
Leq  L1  L2  ...  LN
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6.4 Series and Parallel
Inductors (2)
• The equivalent capacitance of parallel inductors
is the reciprocal of the sum of the reciprocals of
the individual inductances.
1
1
1
1


 ... 
Leq L1 L2
LN
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6.4 Series and Parallel
Inductors (3)
Example 13
Calculate the equivalent inductance for the
inductive ladder network in the circuit
shown below:
Answer:
Leq = 25mH
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6.4 Series and Parallel
Inductors (4)
Example 14
For the circuit, i= 4(2 – e-10t) mA. If i2(0) = -1 mA,
find: (a) i1(0); (b) v(t), v1(t), and v2(t); (c) i1(t)
and i2(t)
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6.4 Series and Parallel
Inductors (5)
• Current and voltage relationship for R, L, C
+
+
+
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+vL−
+
v2
−
at t = 0−, inductance acts as a short circuit

R2
v
(0
)0
R

L


1
i1 (0 ) 
is i2 (0 ) 
is  iL (0 )
R1  R2
R1  R2
v2 (0 )  i2 (0 ) R2
at t = 0+, iL cannot change instantaneously.
R1

iL (0 ) 
is  i2 (0 )
R1  R2
i1 (0 )  0
v2 (0 )  i2 (0 ) R2
vL (0 )  v2 (0 )
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ic
i2
at t = 0−, capacitor acts as a open circuit
R2
vc (0 ) 
vs
R1  R2

ic (0 )  0
vs
i2 (0 ) 
R1  R2

at t = 0+, vc cannot change instantaneously.

v
(0
)

c
R2
i2 (0 ) 


vc (0 )  vc (0 ) 
vs
R2
R1  R2
ic (0 )  i2 (0 )
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ic
+
vL
−
at t = 0−,
iL(0−) = 10/5 = 2 A
vc(0−) = iL(0−) 3 = 6 V
ic(0−) = 0
vL(0−) = 0
at t = 0+,
iL(0+) = iL(0−) = 2 A
vc(0+) = vc(0−) = 6 V
ic(0+) = −iL(0+) = −2 A
vL(0+) = vc(0+) − iL(0+) 3 = 0
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Example Find iL(0+), vc(0+), dvc(0+)/dt, and diL(0+)/dt
diL (0 ) vL (0 )

 2 A/s
dt
L
iL(0+) = iL(0−) = 0
vc(0+) = vc(0−) = −2 V
dvc (0 ) ic (0 )

 12 V/s
dt
C
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6.5 Applications (1)
Integrator An op amp circuit
whose output is proportional
to the integral of the input
signal
1 t
vo  
vi (t )dt

RC 0
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6.5 Applications (2)
Example 15
If v1 = 10 cos 2t mV and v2 = 0.5t mV, find vo in
the op amp circuit. Assume the voltage across the
capacitor is initially zero.
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6.5 Applications (3)
Differentiator An op amp circuit whose output is
proportional to the rate of change of the input signal
dvi
vo   RC
dt
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6.5 Applications (4)
Example 16
Sketch the output voltage for the circuit, given the
input voltage in the figure. Take vo = 0 at t = 0.
vi(V)
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6.5 Applications (5)
Analog Computer
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6.5 Applications (6)
d2 vo
dvo
2
 vo  10 sin 4t with v(0)  4, v '(0)  1
2
dt
dt
d2 vo
dvo

10
sin
4
t

2
 vo
2
dt
dt
t
dvo
dvo

    10 sin 4t  2
 vo  dt  vo' (0)
0
dt
dt


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6.5 Applications (7)
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6.5 Applications (8)
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