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Chapter 6Chapter 6
Capacitors and InductorsCapacitors and Inductors
電路學電路學 (( 一一 ))
2
Capacitors and Inductors Capacitors and Inductors Chapter 6Chapter 6
6.1 Capacitors6.2 Series and Parallel Capacitors6.3 Inductors6.4 Series and Parallel Inductors
3
6.1 Capacitors (1)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).
4
6.1 Capacitors (2)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).
• 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
vCq d
AC
and
F (10–6)
5
6.1 Capacitors (3)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
td
vdCi )(
10
0
tvtdiC
vt
t and
6
6.1 Capacitors (4)6.1 Capacitors (4)• The energy, w, stored in
the capacitor is
• A capacitor is – an open circuit to dc (dv/dt = 0). – its voltage cannot change abruptly.– The ideal capacitor does not dissipate energy.
2
2
1vCw
7
6.1 Capacitors (5)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)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
9
6.1 Capacitors (7)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)6.1 Capacitors (8)Example 4Obtain the energy stored in each capacitor in the figure under dc conditions.
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6.1 Capacitors (9)6.1 Capacitors (9)Example 5Under dc condition, find the energy stored in the capacitors in the figure.
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6.2 Series and Parallel 6.2 Series and Parallel Capacitors (1)Capacitors (1)
• The equivalent capacitance of N parallel-connected capacitors is the sum of the individual capacitances.
Neq CCCC ...21
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6.2 Series and Parallel6.2 Series and Parallel Capacitors (2) Capacitors (2)
• The equivalent capacitance of N series-connected capacitors is the reciprocal of the sum of the reciprocals of the individual capacitances.
Neq CCCC
1...
111
21
14
6.2 Series and Parallel 6.2 Series and Parallel Capacitors (3)Capacitors (3)
Example 6
Find the equivalent capacitance seen at the terminals of the circuit in the circuit shown below:
Answer:
Ceq = 40F
15
6.2 Series and Parallel 6.2 Series and Parallel Capacitors (4)Capacitors (4)
Example 7Find 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 6.2 Series and Parallel Capacitors (5)Capacitors (5)
Example 8Find the voltage across each of the capacitors in the circuit shown below:
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6.3 Inductors (1)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)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).
• The unit of inductors is Henry (H), mH (10–3) and H (10–6).
td
idLv
l
ANL
2
and
19
6.3 Inductors (3)6.3 Inductors (3)• The current-voltage relationship of an inductor:
• The power stored by an inductor:
)()(1
00
titdtvL
it
t
2
2
1iLw
• An inductor acts like a short circuit to dc (di/dt = 0)• The current through an inductor cannot change
abruptly.
20
6.3 Inductors (4)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.
21
6.3 Inductors (5)6.3 Inductors (5)Example 10The 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.
22
6.3 Inductors (6)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 = 2 A
vC = 10 V
wL = 4 J
wC = 50 J
23
6.3 Inductors (7)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.
24
6.4 Series and Parallel6.4 Series and Parallel Inductors (1) Inductors (1)
• The equivalent inductance of series-connected inductors is the sum of the individual inductances.
Neq LLLL ...21
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6.4 Series and Parallel 6.4 Series and Parallel Inductors (2)Inductors (2)
• The equivalent capacitance of parallel inductors is the reciprocal of the sum of the reciprocals of the individual inductances.
Neq LLLL
1...
111
21
26
6.4 Series and Parallel6.4 Series and Parallel Inductors (3) Inductors (3)
Example 13
Calculate the equivalent inductance for the inductive ladder network in the circuit shown below:
Answer:
Leq = 25mH
27
6.4 Series and Parallel6.4 Series and Parallel InductorsInductors (4) (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)
28
6.4 Series and Parallel 6.4 Series and Parallel Inductors (5) Inductors (5)
• Current and voltage relationship for R, L, C
+
+
+
29
at t = 0−, inductance acts as a short circuit
21 s
1 2
(0 )R
i iR R
12 s L
1 2
(0 ) (0 )R
i i iR R
at t = 0+, iL cannot change instantaneously.
1L s 2
1 2
(0 ) (0 )R
i i iR R
1(0 ) 0i
+v2
−
+vL−
L
2 2 2
(0 ) 0
(0 ) (0 )
v
v i R
2 2 2
L 2
(0 ) (0 )
(0 ) (0 )
v i R
v v
30
2c s
1 2
(0 )R
v vR R
at t = 0−, capacitor acts as a open circuit
at t = 0+, vc cannot change instantaneously.
2c c s
1 2
(0 ) (0 )R
v v vR R
ic
i2
c
s2
1 2
(0 ) 0
(0 )
i
vi
R R
c2
2
c 2
(0 )(0 )
(0 ) (0 )
vi
R
i i
31
at t = 0−,
iL(0−) = 10/5 = 2 A vc(0−) = iL(0−) 3 = 6 V
at t = 0+,
iL(0+) = iL(0−) = 2 A vc(0+) = vc(0−) = 6 V
ic(0−) = 0vL(0−) = 0
ic(0+) = −iL(0+) = −2 A vL(0+) = vc(0+) − iL(0+) 3 = 0
ic +vL
−
32
Example Find iL(0+), vc(0+), dvc(0+)/dt, and diL(0+)/dt
iL(0+) = iL(0−) = 0 vc(0+) = vc(0−) = −2 V
L L(0 ) (0 )2 A/s
di v
dt L
c c(0 ) (0 )12 V/s
dv i
dt C
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6.5 Applications (1)6.5 Applications (1)
Integrator An op amp circuit whose output is proportional to the integral of the input signal
0
1( )
t
o iv v t dtRC
34
Example 15If 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.
6.5 Applications (2)6.5 Applications (2)
35
6.5 Applications (3)6.5 Applications (3)
Differentiator An op amp circuit whose output is proportional to the rate of change of the input signal
io
dvv RC
dt
36
Example 16Sketch the output voltage for the circuit, given the input voltage in the figure. Take vo = 0 at t = 0.
6.5 Applications (4)6.5 Applications (4)
vi(V)
37
6.5 Applications (5)6.5 Applications (5)
Analog Computer
38
2
2 2 10sin4 with (0) 4, '(0) 1o oo
d v dvv t v v
dt dt
6.5 Applications (6)6.5 Applications (6)
2
2 10sin4 2o oo
d v dvt v
dt dt
'
010sin4 2 (0)
to o
o o
dv dvt v dt v
dt dt
39
6.5 Applications (7)6.5 Applications (7)
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6.5 Applications (8)6.5 Applications (8)