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Chapter 6. Capacitance and
inductance
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Contents
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1. Capacitors
2. Inductors
3. Capacitor and inductor combinations
4. RC operational amplifier circuits
5. Application examples
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Usage of inductors and capacitors
Power supply board
inductor
capacitor
PC motherboard
Cell phone
inductor
capacitor
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1. Capacitors
• Capacitance is defined to be the ratio of charge to voltage difference.
• Used to store charges
Q
• Used to store electrostatic energy
C
V
q
Q0
V 0
VS
VS
E
VS
V  VS
If the voltage difference between the
terminals of the capacitor is equal to
the supply voltage, net flow of
charges becomes zero.
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Generation of charges : battery
Electrons(-) are absorbed.
(+) charges are generated
Electrons(-) are generated.
(+) charges are absorbed.
2NH 4  2e   2 NH3  H 2
Zn  Zn 2  2e 
Electrons are generated via
electro-chemical reaction.
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Usage of capacitors
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• Used to store charges
• Used to store electrostatic energy
• Slow down voltage variation
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Type of capacitors
t
1
 (t )   i (t ) dt
C0
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Frequently used formulas on capacitors
Capacitance : C 
Current :
Voltage :
Power :
Energy :
i
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q

dq
d
C
dt
dt
1 t
1 t0
1 t
   i ( )d   i ( )d   i ( )d
C 
C 
C t0
1 t
  (t0 )   i ( )d
C t0
p(t )   (t )i (t )  C (t )
t
d (t ) d  1

  C 2 
dt
dt  2

d
  d
W (t )    ( )i ( )d  

t
1
1
2
C

d


C [ (t )]2


2
2

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iυ relation of capacitors
i
0
t

0
t

1 t
i ( )d



C
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Example 6.2
The voltage across a 5-μF capacitor has the waveform shown in Fig. 6.4a. Determine
the current waveform.
iC
d
dt
24

0  t  6ms
 6  103 t
  24
 (t )  
t  96 6  t  8ms
3
2

10

0
t  8ms


i (t )
 20mA 0  t  6ms

i (t )   60mA 6  t  8ms
 0
t  8ms

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Properties of capacitors
RS
i

VS
C
C

• Capacitor voltage cannot change instantaneously due to finite current supply.

 (t0   )   (t0   )
t0
i  C
t
d

dt
• In steady state, capacitor behaves as if open circuited.
RS
VS
i
iC
d DC
0
dt
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Example 6.3
Determine the energy stored in the electric field of the capacitor in Example 6.2 at
t=6 ms.
1
W (t )  C [ (t )]2  1440 [  J ]
2
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Example 6.4
The current in an initially uncharged 4μF capacitor is shown in Fig. 6.5a. Let us
derive the waveforms for the voltage, power, and energy and compute the energy
stored in the electric field of the capacitor at t=2 ms.
 16
 2  103 t 0  t  2ms

i (t )    8
2  t  4ms

0
t  4ms

1 t
3
2
 4 0 8  10 xdx  1000t 0  t  2ms
 1 t
 (t )    ( 8)dx  4  2t  8 2  t  4ms
0
4
0
t  4ms


 8t 3
0  t  2ms

p(t )  i (t ) (t )  16t  64 2  t  4ms
 0
t  4ms


2t 4
t

W (t )   p( x )dx  8t 2  64  109 t  128  1012
0

0

0  t  2ms
2  t  4ms
t  4ms
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2. Inductors
 L (t )  L
di
dt
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Two important laws on magnetic field
Current
B-field
Current generates magnetic field
(Biot-Savart Law)
Current
Vinduced
Time-varying magnetic field generates
induced electric field that opposes the
variation. (Faraday’s law)
V
B-field
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Biot-Savart Law
B  
C
ˆ
Idr  R
, R  r  r
4 R 2
Faraday’s Law
Vind  
d
dt
   B  da  Li
S
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Self induced voltage
=
Vind  
d
di
L
dt
dt
• The induced voltage is generated such that it opposes the applied magnetic flux.
• The inductor cannot distinguish where the applied magnetic flux comes from.
• If the magnetic flux is due to the coil itself, it is called that the induced voltage
is generated by self-inductance.
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Frequently used formulas on inductors
N
i
Inductance :
L
Voltage :
L
Current :
Power :
Energy :
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di
dt
1 t
1 t0
1 t
i    ( )d    ( )d    ( )d
L 
L 
L t0
1 t
 i (t0 )    ( )d
L t0
d 1 2
 di (t ) 
p(t )   (t )i (t )   L
i
(
t
)

Li 


dt  2
 dt 

t
d
  d
W (t )    ( )i ( )d  

t
1
1 2
2
 Li d  L [i (t )]
2
2

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Properties of inductors
RS
i

VS
L
L

• Inductor current cannot change instantaneously due to finite current supply.
i (t0   )  i (t0   )
i
t0
  L
t
di

dt
• In steady state, inductor behaves as if short circuited.
RS
VS
i
L
diDC
0
dt
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Example 6.5
Find the total energy stored in the circuit of Fig. 6.8a.
VC1  9
 3
VC1
0
6
9
 3VC1  27  54  2VC1  5VC1  81  0
VC1  16.2 [V ], VC2  16.2 
I L1 
1
( 2  103 )( 1.2) 2  1.44 [mJ ]
2
1
WL2  ( 4  103 )(1.8) 2  6.48 [mJ ]
2
1
WC1  ( 20  106 )(16.2) 2  2.62 [mJ ]
2
1
WC2  (50  106 )(10.8) 2  2.92 [mJ ]
2
9  VC1
6
6
 10.8 [V ]
9
 1.2 [ A], I L2 
VC1
9
 1.8 [ A]
WL1 
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Example 6.6
The current in a 10-mH inductor has the waveform shown in Fig. 6.9a. Determine the
voltage waveform.

20  103
t

2  103

 20  103
i (t )  
t  40  103
3
 2  10
0


0  t  2ms
2  t  4ms
t  4ms
3

3 20  10
0  t  2ms
 (10  10 ) 2  103  100 [mV ]

3

3 20  10
 (t )   (10  10 )
 100 [mV ] 2  t  4ms
3
2

10

0
t  4ms


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Example 6.7
The current in a 2-mH inductor is
i (t )  2 sin(377t ) [ A]
Determine the voltage across the inductor and the energy stored in the inductor.
 L (t )  L
WL (t ) 
di
d [2 sin(377t )]
 ( 2  103 )
 1.508 cos(377t ) [V ]
dt
dt
1
1
L[i (t )]2  ( 2  103 )[2 sin(377t )]2  0.004 sin 2 (377t ) [ J ]
2
2
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Example 6.8
The voltage across a 200-mH inductor is given by the expression
(1  3t )e 3t [mV ] t  0
 (t )  
0
t0

Let us derive the waveforms for the current, energy, and power.
 103 t
3 x
 3t

(
1

3
x
)
e
dx

5
te
[mA] t  0

i   200 0

0
t0
5t (1  3t )e 6t [ W ] t  0
p(t )   (t )i (t )  
0
t0

2.5t 2e 6t [ J ] t  0
1
2
W (t )  L[i (t )]  
2
0
t0

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Capacitor and inductor specifications
Standard tolerance
values are ; 5%, ;
10%, and ; 20%.
Tolerances are
typically 5% or
10% of the
specified value.
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Example 6.10
The capacitor in Fig. 6.11a is a 100-nF capacitor with a tolerance of 20%. If the
voltage waveform is as shown in Fig. 6.11b, let us graph the current waveform for the
minimum and maximum capacitor values.
iC
d
dt
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6.3 Capacitor and Inductor Combinations
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=
N
1
1
1
1
1
 


CS i 1 Ci C1 C2
CN
=
N
CP   Ci  C1  C2  C3    C N
i 1
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=
N
LS   Li  L1  L2  L3    LN
i 1
=
N
1
1
1 1
1
1
     
LP i 1 Li L1 L2 L3
LN
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6.4 RC Operational Amplifier Circuits
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Op-amp differentiator
iC
C1
d
  
(1    )  o
 i
dt
R2
      0, i  0
o   R2C1
d1 (t )
dt
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Op-amp integrator
1   
R1
1
R1
 C2
 C2
o  
d
(o    )  i
dt
do (t )
dt
1
R1C2

t

      0, i  0
1 ( x )dx  
1 t
1 t

(
x
)
dx


(
0
)


1 ( x )dx
1
o


0
0
R1C2
R1C2
o (0)  0
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Example 6.17
The waveform in Fig. 6.26a is applied at the input of the differentiator circuit shown
in Fig. 6.25a. If R2=1 kΩ and C1=2 μF, determine the waveform at the output of the
op-amp.
o   R2C1
d1 (t )
d (t )  4 [V ] 0  t  5 ms
 ( 2)103 1  
dt
dt
 4 [V ] 5  t  10 ms
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Example 6.18
If the integrator shown in Fig. 6.25b has the parameters R1=5 kΩ and C2=0.2μF,
determine the waveform at the op-amp output if the input waveform is given as in Fig.
6.27a and the capacitor is initially discharged.
 103 ( 20)103 t  20t 0  t  0.1 [ s ]
1 t
o  
1 ( x )dx  
R1C2 0
20t
t  0.1 [ s ]

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