Transcript Inductance

Chapter 31B - Transient
Currents and Inductance
A PowerPoint Presentation by
Paul E. Tippens, Professor of Physics
Southern Polytechnic State 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 a coil connected to resistance R and
voltage V. When switch is closed, the rising
current I increases flux, producing an internal
back emf in the coil. Open switch reverses 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 DI/Dt.
Di
E  L ;
Dt
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 Di/ Dt
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?
Di/ Dt = 2 A/s
4 mV
R
Di
E  L ;
Dt
(0.004 V)
L
2 A/s
E
L
Di / Dt
L = 2.00 mH
Note: We are following the practice of using
lower case i for transient or changing current
and upper case I for steady current.
Calculating the Inductance
Recall two ways of finding E:
D
E  N
Dt
Di
E  L
Dt
Increasing Di/ Dt
R
Setting these terms equal gives:
D
Di
N
L
Dt
Dt
Thus, the inductance L
can be found 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
0 NI
and  = BA
R
Inductance L
Combining the last
two equations gives:

0 NIA
L
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:
0 N A (4 x 10
L

2
l
-7 Tm
A
2
2
)(100) (0.002 m )
0.300 m
L = 8.38 x 10-5 H
A
R
Note: L does NOT depend
on current, but on physical
parameters of the coil.
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
L = 8.38 x 10-5 H
A
R
Di
E  L
Dt
(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 Di/Dt, we have:
R
Di
P  Ei  Li
Dt
Since the power P = Work/t, Work = P Dt. Also
the average value of Li is Li/2 during rise to the
final current I. Thus, the total energy stored is:
Di
EL ;
Dt
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

2
B
u
20
B
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 )
u = 0.268 J/m3
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:
Di
V – E = iR
EL
Dt
Di
V  L  iR
Dt
V
S1
S2
R
i
L
E
Initially, Di/Dt is large, making the back emf large
and the current i small. The current rises to its
maximum value I when rate of change is 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 t:
L
t 
R
i
I
0.63 I
Current
Rise
t
Time, t
In an inductor, the current will rise to 63% of its
maximum value in one time constant t = L/R.
The R-L Decay
Now suppose we close S2
after energy is in inductor:
Di
EL
E = iR
Dt
For current L Di  iR
Dt
decay in L:
V
S1
S2
R
i
L
E
Initially, Di/Dt is large and the emf E driving the
current is at its maximum value I. The current
decays to zero when the emf plays 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 t:
L
t 
R
Current
Decay
0.37 I
t
Time, t
In an inductor, the current will decay to 37%
of its maximum value in one time constant t.
Example 5: The circuit below has a 40-mH
inductor connected to a 5-W resistor and a
16-V battery. What is the time constant and
what is the current after one time constant?
L 0.040 H
t 
R
5W
16 V
5W
L = 0.04 H
After time t:
i = 0.63(V/R)
R
Time constant: t = 8 ms
V
i  (1  e ( R / L )t )
R
 16V 
i  0.63 

 5W 
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, Q/C is small, making the back emf small
and the current i is a maximum I. As the charge
Q builds, the current decays to zero when Eb = V.
Rise of Charge
Qmax
q
Q
t = 0, Q = 0,
V   iR
0.63 I
I = V/R
C
t =  , i = 0, Qm = C V
Q  CV (1  e
 t / RC
)
The time constant t:
t  RC
Capacitor
Increase in
Charge
t
Time, t
In a capacitor, the charge
Q will rise to 63% of its
maximum value in one
time constant t.
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 t:
t  RC
Current
Decay
0.37 I
t
Time, t
As charge Q increases
The current will decay to 37% of its maximum
value in one time constant t; the charge rises.
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, Q is large and the emf E driving the
current is at its maximum value I. The current
decays to zero when the emf plays out.
i
Current Decay
V  t / RC
i
e
R
I
i
Capacitor
t  RC
At t = 0, I = V/R
At t = , I = 0
As the current decays,
the charge also decays:
Current
Decay
0.37 I
t
Q  CVe
Time, t
 t / RC
In a discharging capacitor, both current and
charge decay to 37% of their maximum values
in one time constant t = RC.
Example 6: The circuit below has a 4-F
capacitor connected to a 3-W resistor and a
12-V battery. The switch is opened. What is
the current after one time constant t?
12 V
3W
C = 4 F
After time t:
i = 0.63(V/R)
t = RC = (3 W)(4 F)
R
Time constant: t = 12 s
V
i  (1  e  t / RC )
R
 12V 
i  0.63 

 3W 
i = 2.52 A
Summary
Di
E  L ;
Dt
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
t 
R
I
i
Inductor
Current
Rise
0.63I
t
Time, t
In an inductor, the current will rise to 63% of its
maximum value in one time constant t = L/R.
The initial current is zero due to fast-changing
current in coil. Eventually, induced emf becomes
zero, resulting in the maximum current V/R.
Summary (Cont.)
V ( R / L )t
i e
R
The initial current,
I = V/R, decays to
zero as emf in coil
dissipates.
I
i
Inductor
Current
Decay
0.37I
t
Time, t
The current will decay to 37% of its maximum
value in one time constant t = L/R.
Summary (Cont.)
When charging a capacitor the charge rises
to 63% of its maximum while the current
decreases to 37% of its maximum value.
Qmax
q
Capacitor
I
Increase in
Charge
0.63 I
t
Capacitor
Current
Decay
0.37 I
Time, t
Q  CV (1  et / RC )
i
t  RC
t
Time, t
V  t / RC
i e
R
CONCLUSION: Chapter 31B
Transient Current - Inductance