Download Chapter 30: Inductance

Chapter 30
1
Mutual Inductance
Consider a changing current in coil 1
We know that
B1=m0i1N1
And if i1 is changing with time,
dB1/dt=m0N1 d(i1)/dt
But a changing B-field across coil 2
will initiate an EMF2 such that
EMF2=-N2A2 dB1/dt
Since dB1/dt is proportional to di1/dt
then the
EMF2 
Where M is the mutual
inductance which is based on the
sizes of the coils, and the number
of turns
di1
dt
di1
EMF2   M
dt
2
Mutual-mutual Inductance
But it could be that the changes are
happening in coil 2. Then
EMF1 
di2
dt
di2
EMF1   M
dt
It turns out that this value of M is
identical to the previously
discussed M so
EMF1   M
di2
dt
and
EMF2   M
di1
dt
where M 
N 2  B 2 N1 B1

i1
i2
3
My favorite unit—the henry
The Henry (H) is the unit of inductance
 Equivalent to:



1H=1 Wb/A = 1 V*s/A = 1 W*s = 1 J/A2
H is large unit; typically we use small
units such as mH and mH.
4
Self Inductance



But a coil of wire with a
changing current can
produce an EMF within
itself.
This EMF will oppose
whatever is causing the
changing current
So a coil of wire takes
on a special name called
the inductor
5
Inductor


Definition of inductance
is the magnetic flux per
current (L)
For an N-turn solenoid,
L is

L= N/I

N turns= (n turns/length)*(l length)

The near center solution
of inductance depends
only on geometry
Electrical symbol
N  nlBA
where B  m 0in
N  m 0 n 2liA
N
If L 
i
L  m 0 n 2lA
Near center
L
 m0 n 2 A
l
6
Inductor
N
If L 
 Li  N
i
and if
d
di
EMF   N
 L
dt
dt
Electrical symbol
Again, the EMF acts to oppose the change in current
i (increasing)
i (decreasing)
High potential
VL
acts like a
Low potential
Low potential
VL
acts like a
High potential
7
RL Circuits
Initially, S is
open so at t=0,
i=0 in the
resistor, and the
current through
the inductor is
0.
 Recall that
i=dq/dt

R
A
S
B
V
L
8
Switch to A
Initially, the inductor acts
against the changing current
but after a long time, it behaves
like a wire
Rt


V
L
i  1  e  and
R

S
H
i
V
di
 iR  L  V  0
dt
di
V  iR  L
dt
Ansatz
R
A
B
L
L
di V  RtL
 e
dt L
9
Voltage across the resistor and
inductor
Potential across resistor, VR
VR  iR 
Rt
Rt
 
 

V
1  e L  R  V 1  e L 



R 



R
A
S
Potential across
capacitor, VC
B
V
L
di
V  RtL
VL  L  L e
dt
L
VL  Ve

Rt
L
At t=0, VL=V and VR=0
At t=∞, VL=0 and VR=V
10
L/R—Another time constant
L/R is called the “time constant” of the
circuit
 L/R has units of time (seconds) and
represents the time it takes for the
current in the circuit to reach 63% of its
maximum value
 When L/R=t, then the exponent is -1 or
e-1
 tL=L/R

11
Switch to B

The current is at a
steady-state value of
i0 at t=0
di
 iR  L  0
dt
dq
di
R
 L
dt
dt
i (t )  i0 e

V
R
A
S
B
L
Rt
L
12
Energy Considerations
di
Rate at which energy is supplied from battery
V L
dt
and P  Vi
Rate at which energy is stored in the
magnetic field of the inductor
di
Vi  Li
dt
Energy of the magnetic field, UB
dU B
di
P
 Li
dt
dt
1 2
1 2
U B  Li
(recall KE  mv )
2
2
13
Energy Density, u

Consider a solenoid of area A and length, l
UB
U
 B
volume A  l
1 2
Li
L
2
u
but
 m0 n 2 A
Al
l
1
u  m 0 n 2i 2 but B  m 0in
2
0E2
B2
uB 
and
uE 
2m0
2
u
Energy stored at any
point in a magnetic field
Energy stored at any
point in a magnetic
field
14
L-C Oscillator – The Heart of
Everything
di q
 0
dt C
dq
di d dq d 2 q
i
 
 2
dt
dt dt dt dt
d 2q q
d 2q 1
L 2  0 2 
q0
dt
C
dt
LC
Ansatz : q  Q cost   
dq
 Q sin t     i
dt
d 2q
  2Q cost   
2
dt
so
1
  2Q cost    
Q cost     0
LC
1
1
2 
 0  2 
LC
LC
L
L
C
If the capacitor has a
total charge, Q
15
Perpetual Motion?
16
Starting Points
Charge q
Current i
t
The phase angle, , will determine when
the maximum occurs w.r.t t=0
The curves above show what happens if
the current is 0 at t=0
17
Energy considerations



A quick and dirty way to
solve for i at any time t
in terms of Q & q
At t=0, the total energy
in the circuit is the
energy stored in the
capacitor, Q2/2C
At time t, the energy is
shared between the
capacitor and inductor


(q2/2C)+(1/2 Li2)
Q2/2C= (q2/2C)+(1/2 Li2)
i
1
Q2  q2
LC
18
Oscillators is oscillators is oscillators
19
Give me an “R”!
Consider adding a resistor, R to the
circuit
The resistor dissipates the energy. For
example, consider a child on a swing.
His/her father pushes the child and
gets the child swinging. In a perfect
system, the child will continue swinging
forever.
The resistor provides the same action
as if the child let their feet drag on the
ground. The amplitude of the child’s
swing becomes smaller and smaller
until the child stops.
The current in the LRC circuit oscillates
with smaller and smaller amplitudes
until there is no more current
20
Mathematically
di q
  iR  0
dt C
dq
di d 2 q
i
  2
dt
dt dt
d 2q
dq q
L 2 R
 0
dt
dt C
Ansatz
L
q (t )  Qe
where

Rt
2L
If R is small,
underdamped
When
oscillation
stops due to
R, critically
damped
Very large
values of R,
overdamped
cos t   
R2
    2
4L
2
21
Why didn’t I use a voltage source?

The practical applications of the LC, LR,
and LRC circuits depend on using a
sinusoidally varying voltage source:

An AC voltage source
22