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Title: Electrical Engineering principles and Applications : Chapter 3 Inductance And Capacitance
Description: Detailed Lecture notes with illustrations and lots of solved examples
Description: Detailed Lecture notes with illustrations and lots of solved examples
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Electrical Engineering
Principles & Applications
Chapter 3- Inductance and
Capacitance
Slide 1
Inductance and Capacitance
1
...
Parallel and series connections of capacitors or
inductors
3
...
Plot i(t) and q(t)
q = Cv
Slide 8
dv
i = C
dt
Determining Voltage for a Capacitance Given
Current
Example:
The current through a 0
...
5 sin(104t)
...
5 ×10−4[1 − cos104 t ]
Slide 9
v (t ) = q (t ) / C = 500[1 − cos 10 4 t ]
Capacitor as Energy Storage Device
+
iC
Instantaneous power
pC (t) = vC (t)iC (t)
vC
−
dv c
(t )
dt
dv
pC (t ) = CvC (t ) c
dt
iC (t ) = C
W
Energy is the integral of power
t2
wC (t 2 , t1 ) = ∫ pC ( x )dx
1 t
q (t )
vC (t ) = ∫ iC ( x ) dx = C
C −∞
C
pC (t ) =
t1
d 1 2
pC ( t ) = C vC (t )
dt 2
pC ( t ) =
1
dq
qC (t ) C (t )
C
dt
1 d 1 2
qc (t )
C dt 2
If t1 is minus infinity we talk about “energy stored at time t 2”
1 2
1 2
wC (t 2 , t1 ) = Cv C ( t2 ) − Cv C (t1 )
2
2
1 2
1 2
wC ( t 2 , t1 ) = qC ( t2 ) − qC ( t1 )
C
C
If both limits are infinity then we talk about the “total energy stored”
1
q 2 (t )
2
Slide 10
w (t ) =
2
Cv
(t ) =
2C
Example
The voltage v(t) is applied to a 10 µF capacitance
...
1000 t 0 < t < 1
v (t ) = 1000 1 < t < 3
500 ( 5 − t ) 3 < t < 5
p (t ) = v(t )i (t )
Slide 11
10×10−3 A 0 < t <1
dv(t)
i(t) = C
= 0A 1 < t < 3
dt
− 5×10−3 A 3 < 3 < 5
1
w(t ) = Cv2 (t)
2
Connections of Capacitors
Parallel Connection
i = C eq
Series Connection
dv
dt
v =
1
C
eq
t
∫ i (t )dt
t
0
To get a high voltage than the source, charge capacitors in
parallel then connect them in series
...
85×10
Fm
Construction of Practical Capacitors
• Parallel- plate capacitors are too large
– Rolled-type capacitors are used
• A dielectric with high dielectric constant is used
• Dielectric becomes a conductor if the voltage is high è
capacitors have a maximum voltage rating
Slide 14
Inductance
A time varying flux
causes a voltage to
appear at the
terminals of the device
Inductors store energy in the
form of magnetic field
Slide 15
Inductance
A time varying magnetic flux
induces a voltage
dφ
vL =
dt
Induction law
For a linear inductor the flux is
proportional to the current
φ = Li
1
i (t ) =
L
t
di
v (t ) = L
dt
∫ v (t )dt
Current through an inductor has
to be continuous
+ i (t 0 )
t0
1
Energy: w (t ) =
Li
2
2
(t )
L is the inductance with units of
henries (H) [volt seconds per ampere]
Slide 16
If i is constant
èv=0
DC or steady state behavior
An inductor in steady state acts as a
SHORT CIRCUIT
Example
The current through a 5-H inductance is given
di
v (t ) = L
dt
p(t) =v(t)i(t)
Slide 17
w (t ) =
1
Li 2 (t )
2
Example
1
i (t ) =
L
Slide 18
t
∫ v (t )dt
t0
+ i (t 0
)
Connections of Inductors
di
v=L
dt
=
Slide 19
1
L
∫ v (t )dt
Connections of Inductors
Slide 20
Title: Electrical Engineering principles and Applications : Chapter 3 Inductance And Capacitance
Description: Detailed Lecture notes with illustrations and lots of solved examples
Description: Detailed Lecture notes with illustrations and lots of solved examples