Download Transients and inductance

Chapter 31B - Transient
Currents and Inductance
AA PowerPoint
PowerPoint Presentation
Presentation by
by
Paul
Paul E.
E. Tippens,
Tippens, Professor
Professor of
of Physics
Physics
Southern
Southern Polytechnic
Polytechnic State
State University
University
©
2007
Objectives: After completing this
module, you should be able to:
• Define and calculate inductance in
terms of a changing current.
• Calculate the energy stored in an
inductor and find the energy density.
• Discuss and solve problems involving
the rise and decay of current in
capacitors and inductors.
Self-Inductance
Consider
Consider aa coil
coil connected
connected to
to resistance
resistance RR and
and
voltage
voltage VV.. When
When switch
switch isis closed,
closed, the
the rising
rising
current
current II increases
increases flux,
flux, producing
producing an
an internal
internal
back
back emf
emf in
in the
the coil.
coil. Open
Open switch
switch reverses
reverses emf.
emf.
Increasing I
R
Lenz’s Law:
The back emf
(red arrow) 
must oppose
change in flux:
Decreasing I
R
Inductance
The back emf E induced in a coil is proportional
to the rate of change of the current I/t.
i
E  L ;
t
L  inductance
An inductance of one henry
(H) means that current
changing at the rate of one
ampere per second will induce
a back emf of one volt.
Increasing i/ t
R
1V
1 H
1 A/s
Example 1: A coil having 20 turns has an
induced emf of 4 mV when the current is
changing at the rate of 2 A/s. What is the
inductance?
i/ t = 2 A/s
4 mV
R
i
E L ;
t
(0.004 V)
L
2 A/s
E
L
i / t
2.00 mH
mH
LL == 2.00
Note: We
We are
are following
following the
the practice
practice of
of using
using
Note:
lower case
case ii for
for transient
transient or
or changing
changing current
current
lower
and upper
upper case
case II for
for steady
steady current.
current.
and
Calculating the Inductance
Recall two ways of finding E:

E  N
t
i
E  L
t
Increasing i/ t
R
Setting these terms equal gives:

i
N
L
t
t
Thus,
Thus, the
the inductance
inductance LL
can
can be
be found
found from:
from:
Inductance L
N
L
I
Inductance of a Solenoid
Solenoid
B
l
The B-field created by a
current I for length l is:
B
R
Inductance L
Combining the last
two equations gives:

 0 NI

0 NIA

L
and  = BA
N
L
I
0 N 2 A

Example 2: A solenoid of area 0.002 m2 and
length 30 cm, has 100 turns. If the current
increases from 0 to 2 A in 0.1 s, what is the
inductance of the solenoid?
First we find the inductance of the solenoid:
L
0 N A (4 x 10
2


l
-7 Tm
A
2
2
)(100) (0.002 m )
0.300 m
8.38 xx 10
10-5-5 HH
LL == 8.38
A
R
Note: LL does
does NOT
NOT depend
depend
Note:
on current,
current, but
but on
on physical
physical
on
parameters of
of the
the coil.
coil.
parameters
Example 2 (Cont.): If the current in the 83.8H solenoid increased from 0 to 2 A in 0.1 s,
what is the induced emf?
l
-5 H
-5
L
=
8.38
x
10
L = 8.38 x 10 H
A
R
i
E  L
t
(8.38 x 10-5 H)(2 A - 0)
E
0.100 s
E  1.68 mV
Energy Stored in an Inductor
At an instant when the current
is changing at i/t, we have:
R
i
P  Ei  Li
t
Since the power P = Work/t, Work = P t. Also
the average value of Li is Li/2 during rise to the
final current I. Thus, the total energy stored is:
i
EL ;
t
Potential energy
stored in inductor:
U  12 Li 2
Example 3: What is the potential energy
stored in a 0.3 H inductor if the current rises
from 0 to a final value of 2 A?
U  12 Li 2
L = 0.3 H
R
I=2A
U  12 (0.3 H)(2 A)2  0.600 J
U = 0.600 J
This energy is equal to the work done in
reaching the final current I; it is returned
when the current decreases to zero.
Energy Density (Optional)
The energy density u is the
energy U per unit volume V
l
A
R
L
0 N 2 A

; U  LI ; V  A
1
2
2
Substitution gives u = U/V :
 0 N A  2
U 
I ;
  
2
1
2
 0 N 2 AI 2 



2
U 

u 
V
A
u
0 N 2 I 2
2
2
Energy Density (Continued)
Energy
u
density:
l
A
R
2
2
Recall formula for B-field:
B
0 NI

2

0  NI  0 B 
u
 2

 
2   
2  0 
2
0 N 2 I 2

NI B

0

2
B
u
2 0
Example 4: The final steady current in a
solenoid of 40 turns and length 20 cm is 5 A.
What is the energy density?
B
0 NI

(4 x 10-7 )(40)(5 A)

0.200 m
B = 1.26 mT
2
A
R
-3
2
B
(1.26 x 10 T)

u
-7 Tm
2 0 2(4 x 10 A )
uu == 0.268
0.268 J/m
J/m33
l
Energy density is
important for the
study of electromagnetic waves.
The R-L Circuit
An inductor L and resistor
R are connected in series
and switch 1 is closed:
i
V – E = iR
EL
t
i
V  L  iR
t
V
S1
S2
R
i
L
E
Initially,
i/t isis large,
Initially, i/t
large, making
making the
the back
back emf
emf large
large
and
small. The
The current
current rises
rises to
to its
its
and the
the current
current ii small.
maximum
maximum value
value II when
when rate
rate of
of change
change isis zero.
zero.
The Rise of Current in L
V
 ( R / L )t
)
i  (1  e
R
At t = 0, I = 0
At t = , I = V/R
The time constant 
L
 
R
i
I
0.63 I
Current
Rise

Time, t
In
In an
an inductor,
inductor, the
the current
current will
will rise
rise to
to 63%
63% of
of its
its
maximum
.
== L/R
L/R.
maximum value
value in
in one
one time
time constant
constant 
The R-L Decay
Now suppose we close S2
after energy is in inductor:
i
EL
E = iR
t
For current L i  iR
t
decay in L:
V
S1
S2
R
i
L
E
Initially,
i/t isis large
Initially, i/t
large and
and the
the emf
emf EE driving
driving the
the
current
current isis at
at its
its maximum
maximum value
value II.. The
The current
current
decays
decays to
to zero
zero when
when the
the emf
emf plays
plays out.
out.
The Decay of Current in L
V  ( R / L )t
i e
R
i
I
At t = 0, i = V/R
At t = , i = 0
The time constant 
L
 
R
Current
Decay
0.37 I

Time, t
In
In an
an inductor,
inductor, the
the current
current will
will decay
decay to
to 37%
37%
of

of its
its maximum
maximum value
value in
in one
one time
time constant
constant 
Example 5: The circuit below has a 40-mH
inductor connected to a 5- resistor and a
16-V battery. What is the time constant and
what is the current after one time constant?
L 0.040 H
 
R
5
16 V
5
L = 0.04 H
After time 
i = 0.63(V/R)
R
Time
Time constant:
constant:  =
= 88 ms
ms
V
i  (1  e  ( R / L )t )
R
 16V 
i  0.63 

 5 
i = 2.02 A
The R-C Circuit
Close S1. Then as charge
Q builds on capacitor C, a
back emf E results:
Q
V – E = iR
E
C
Q
V   iR
C
V
S1
S2
C
R
i
E
Initially,
Initially, Q/C
Q/C isis small,
small, making
making the
the back
back emf
emf small
small
and
. As
and the
the current
current ii isis aa maximum
maximum II.
As the
the charge
charge
QQ builds,
= V.
V.
builds, the
the current
current decays
decays to
to zero
zero when
when EEbb =
Rise of Charge
Qmax
q
t = 0, Q = 0,
Q
V   iR
0.63 I
I = V/R
C
t =  , i =  , Qm = C V
Q  CV (1  e
 t / RC
)
The time constant 
  RC
Capacitor
Increase in
Charge

Time, t
In
In aa capacitor,
capacitor, the
the charge
charge
QQ will
will rise
rise to
to 63%
63% of
of its
its
maximum
maximum value
value in
in one
one
time
time constant
constant 
Of course, as charge rises, the current i will decay.
The Decay of Current in C
V t / RC
i e
R
i
Capacitor
I
At t = 0, i = V/R
At t = , i = 0
The time constant 
  RC
Current
Decay
0.37 I
Time, t
As charge Q increases

The
The current
current will
will decay
decay to
to 37%
37% of
of its
its maximum
maximum
value
the
the charge
charge rises.
rises.
value in
in one
one time
time constant
constant 
The R-C Discharge
Now suppose we close S2
and allow C to discharge:
Q
E
E = iR
C
Q
For current
 iR
C
decay in L:
V
S1
S2
C
R
E
Initially,
Initially, QQ isis large
large and
and the
the emf
emf EE driving
driving the
the
current
current isis at
at its
its maximum
maximum value
value II.. The
The current
current
decays
decays to
to zero
zero when
when the
the emf
emf plays
plays out.
out.
i
Current Decay
V  t / RC
i
e
R
I
i
Capacitor
  RC
At t = 0, I = V/R
At t = , I = 0
As the current decays,
the charge also decays:
Current
Decay
0.37 I

Q  CVe
Time, t
 t / RC
In
In aa discharging
discharging capacitor,
capacitor, both
both current
current and
and
charge
charge decay
decay to
to 37%
37% of
of their
their maximum
maximum values
values
in
in one
one time
time constant
constant 
== RC.
RC.
Example 6: The circuit below has a 4-F
capacitor connected to a 3- resistor and a
12-V battery. The switch is opened. What is
the current after one time constant ?
12 V
3
C = 4 F
After time 
i = 0.63(V/R)
 = RC = (3 )(4 F)
R
Time
Time constant:
constant:  =
= 12
12 s
s
V
i  (1  e  t / RC )
R
 12V 
i  0.63 

 3 
i = 2.52 A
Summary
i
E  L ;
t
L
l
L  inductance
0 N A
2

Potential Energy
Energy Density:
A
N
L
I
R
U  Li
1
2
2
2
B
u
2 0
Summary
V
 ( R / L )t
i  (1  e
)
R
L
 
R
I
i
Inductor
Current
Rise
0.63I

Time, t
In
In an
an inductor,
inductor, the
the current
current will
will rise
rise to
to 63%
63% of
of its
its
maximum
.
== L/R
L/R.
maximum value
value in
in one
one time
time constant
constant 
The
The initial
initial current
current isis zero
zero due
due to
to fast-changing
fast-changing
current
current in
in coil.
coil. Eventually,
Eventually, induced
induced emf
emf becomes
becomes
zero,
zero, resulting
resulting in
in the
the maximum
maximum current
current V/R.
V/R.
Summary (Cont.)
V ( R / L )t
i e
R
The
The initial
initial current,
current,
II =
, decays
= V/R
V/R,
decays to
to
zero
zero as
as emf
emf in
in coil
coil
dissipates.
dissipates.
I
i
Inductor
Current
Decay
0.37I

Time, t
The
The current
current will
will decay
decay to
to 37%
37% of
of its
its maximum
maximum
value
== L/R.
L/R.
value in
in one
one time
time constant
constant 
Summary (Cont.)
When
When charging
charging aa capacitor
capacitor the
the charge
charge rises
rises
to
to 63%
63% of
of its
its maximum
maximum while
while the
the current
current
decreases
decreases to
to 37%
37% of
of its
its maximum
maximum value.
value.
Qmax
q
Capacitor
I
i
Capacitor
Current
Decay
Increase in
0.37 I
Charge
0.63 I

Time, t
Q  CV (1  e  t / RC )
  RC

Time, t
V t / RC
i e
R
CONCLUSION: Chapter 31B
Transient Current - Inductance