Transcript Chapter-6-update

L
6.1 The Inductor
i
Is a passive element ( can not generate energy)
 v 
However the inductor is capable of storing energy in the magnetic field which can be released
Represented graphically as a coiled wire
Symbolized by the letter L
Measured in henrys (H)
The voltage drop across the inductor the inductor and the current in the direction
of voltage drop are related as follows
di
v=L
dt
Example 6.1
 0
i =  5t
10te

i
v


100 mH
t<0
t>0
(b) At what instant of time is the current maximum ?
 0
i =  5t
10te
0.736
0.2
Since the maximum will occur were the derivative of the current is zero
di
=0
dt
di
= 10(  5te 5t  e 5t )
dt
= 10e5t (1  5t) = 0
As
(1  5t)  0
t
1
s  0.2 s
5
t<0
t>0
(c) Find the voltage across the terminals of the inductor ?

v

i
 0
i =  5t
10te
100 mH
t<0
t>0

di
v=L
dt
=
(0.1)10(  5te
=
0
V
5t
e
5t
t<0
(d) Sketch the voltage across the terminals
of the inductor ?
)
=
e
5t
(1  5t) V
t>0
(e) Are the voltage and the current at a maximum at the same time ?
Since the voltage across the inductor is
proportional to di/dt not i

100 mH
v

i
the voltage maximum is not at the same time as
current maximum

 0
i =  5t
10te
0.736
t<0
t>0
0.2
0

v =  5t
e (1  5t)
t<0
t>0
(f) At what instant of time does the voltage change polarity ?

100 mH
v

i
At t = 0.2 s which corresponds to the
moment when di/dt is passing through
zero changing sign

 0
i =  5t
10te
0.736
t<0
t>0
0.2
0

v =  5t
e (1  5t)
t<0
t>0
(g) Is there ever an instantaneous change in voltage across the inductor ? If so at what time ?
Yes, At t = 0 s

100 mH
v

i
Note that the voltage can change instantaneously
across the terminals of an inductor

 0
i =  5t
10te
0.736
t<0
t>0
0.2
0

v =  5t
e (1  5t)
t<0
t>0
Current in an Inductor in Terms of the Voltage Across the Inductor
L
 v 
Since
di
v=L
dt
Multiplying both side by the differential
i
di 
v dt = L 
dt

 dt 
di
v dt = L  dt
 dt 
vdt = Ldi
Integrating both side with respect to the differentials
i(t )
L

i(t 0 )
t

dx = v d 
t0
di
OR
and
dt
dt
Ldi = vdt
we have
Note we used x and  the variables of integration, were
become limits on the integrals
( x and  become dummy variables )
t

L  i(t ) - i(t 0 )  = v d 
t0
t

1
i(t ) =
v d   i(t 0 )
L
t0
i
and
t
Example 6.2 The voltage pulse applied to the 100mH inductor. Assume
v(t)
(a) Sketch the voltage v(t) ?
i
i = 0 for
 0
v (t ) =  10t
20te
t<0
t>0
t ≤0
v(t)
i(t)
 0
v (t ) =  10t
20te
t<0
t>0
(b) Find the inductor current i(t) ?
1
i(t) =
0.1
t

20 e 10 d   0
= 2 1  10te 10t  e 10t  A t > 0
0
(c) Sketch the current i(t) ?
Note the inductor current i(t) approaches a constant vlue of 2 A as t increases.
Power and Energy in the Inductor
t
di
v=L
dt
L

1
i(t ) =
v d   i(t 0 )
L
t0
 v 
i
di
p = vi =  L
 dt
 i = Li di

dt

dw
=
dt
Integrating both side with respect to the differentials
w

0
i

dx = L y dy
0
1 2
w = Li
2
dw = Li di
dw
and di were w=
0
if
i= 0
L
The Inductor
i
 v 
t
di
v=L
dt

1
i(t ) =
v d   i(t 0 )
L
1
w = Li 2
2
di
p = Li
dt
t0
i
L
The Capacitor
 v 
dv
i =C
dt
1
v(t ) =
C
t
 i d  v (t
t0
0
)
1
w = Cv 2
2
p = Cv
dv
dt
L
 v 
i
L
 v 
i
di
v=L
dt
dv
i =C
dt
t

1
i(t ) =
v d   i(t 0 )
L
t0
1
w = Li 2
2
di
p = Li
dt
v(t ) =
1
C
t

i d   v (t 0 )
t0
1 2
w = Cv
2
dv
p = Cv
dt
Inductors in Series
1 1
1
  
Lequ  L1 L 2
1 


Ln 
Therefore inductor combine like resistor
5.1.1 Capacitors in Series
5.1.1 Capacitors in Parallel
Therefore capacitor combine like conductor
Example 5.1
A voltage source is applied to a 5-F capacitor as
shown.
Sketch the capacitor current and the stored energy as a
function of time.
dv t 
i t   C
dt
dv t 
5
dt
1
2
w t   C v t 
2
1
  5v s2 t 
2
Example 5.2
A current source is applied to a 5-F capacitor.
Sketch the capacitor voltage as a function of time.
t
1
v t    i s  d
5 
The capacitor voltage is related to
area Under current source.
For example area at t = 1 s is
Area of Triangle is (1)(10)=10
V(1s)=10/5= 2V.
For example area at t=3 s is
10+10+5 = 25 hence
V(3s)=25/5= 5V.
Example 5.3
A current source is applied to a 5-H inductor as shown. Sketch
the voltage across the inductor versus time.
di t 
v t   L
dt
1 2
w t   Li t 
2
Example 5.4
A voltage source is applied to a 5-H inductor as
shown. Sketch the inductor current versus time.
1
i t  
L
t
 v  d 

1
i t  
L
t
 v  d 
