Transcript L11_Inductance_orig

Coils sharing the same magnetic flux, BA
These can be used
as “transformers”.
(See next slide.)
In air
With iron yoke
Transformers
Faraday’s Law of Induction tells us the
relationship between the magnitudes of the
voltage at the terminals and the rate of change
of magnetic flux through each coil:
d 2
d 1
V2  N 2
dt
dt
But the flux through each coil is the same:
V1  N1
1   2
d1 d 2

dt
dt
V1 V2

N1 N 2
So the output (secondary) voltage can be raised or
lowered compared to the input (primary) voltage, by
the ratio of turns. A transformer that raises (lowers)
voltage is called a “step-up” (“step-down”) transformer.
Transformers lose very little power:
P  V1I1  V2 I 2
N2
V2 
V1
N1
V1
N1
I 2  I1 
I1
V2
N2
So as the voltage is increased, the current is lowered by the same factor.
A.C. power networks
Transmission lines:
155,000-765,000 V
Local substation:
7,200 V
House:
110/220 V
For a given length of
power line, P=RI2. If
the power line voltage
is 550,000 V, the
current is reduced by a
factor of 5000
compared to that at 110
V, and the power loss is
reduced by a factor of
25 million.
Local
substation
High voltage
transmission lines
Transmission
substation
Power plant
Inductance of a solenoid
As we know, when a solenoid has a current passing
through it, there is an approximately uniform magnetic
field passing through its core. In the circuit at right,
the battery is a source of emf, and the current is
determined only by the internal resistance of the
battery and any resistance in the wire (via I=V/R).
But what would happen if we had a variable emf, and
could increase the current? Lenz’s Law tells us that
the solenoid would create a magnetic field trying to
oppose the increase in B, and Faraday’s Law tells us
that the solenoid itself would develop an emf opposite
in direction to the battery (reducing the total emf).
 0 N 2 A  dI
d solen
d
d N 

emf  V  
  N BA    N  0 IA   
dt
dt
dt 
l

 l  dt
The last quantity in parentheses is the “inductance” of a solenoid, “L”:
Lsolen 
0 N 2 A
l
General:
dI
V  L
dt
L unit:
“Henry”
1H 1
V s
A
A solenoid is an example of an “inductor” (L)
This is a new circuit element with the following symbol:
What does an inductor “do” in a circuit:
(1) Like a resistor or a capacitor, it can have a voltage drop or rise across its
terminals. A resistor develops a voltage difference V, when a current flows
through it. But an inductor develops a voltage difference V when the current
through it changes.
(2) A capacitor can store charge, but an inductor stores magnetic field.
(3) An inductor opposes rapid changes of the current passing through it.
Inductors can be made out of loops, solenoids, toroids, or any other shape that
creates a magnetic field threading its own loops.
The inductance of any such device can be greatly increased by filling its
magnetic volume with ferromagnetic material.
Comparing behavior of circuit devices: R vs L
Energy stored in an inductor
For all types of inductor, we have the basic equation:
V  L
dI
dt
As an inductor is charged, the power into the inductor is the rate at which
energy is being deposited in the device:
dI dU
P  VI  LI

dt
dt
So during each time interval dt, we have simply: dU  LIdI
We can integrate dU to find the total potential energy stored in the
inductor as it is energized, from I = 0 to the final current:
UB
I
1
U   dU   LIdI  LI 2  U B
0
0
2
This is magnetic potential energy stored in the magnetic field
inside the inductor when it has reached the final current. Note
the similarities with the corresponding capacitor derivation, both
in the steps taken, and in the form of the final result.
Energy density of a magnetic field
Just as we used a capacitor to find the energy density of an electric field,
we can use a solenoid to find the energy density of a magnetic field:
L
0 N 2 A
l
 0 n 2 Al
and
B  0 nI
Putting these into the expression for the stored magnetic energy:
2


1 2 1
B
B2
2
 
 Al 
U B  LI  (  0 n Al )
2
2
2 0
 0n 
But Al is the solenoid volume. We divide by it to find the energy density in
the magnetic field:
B2
uB 
2 0
This result also applies to magnetic fields in free space. We can
combine it with the electric field energy density to find the expression for
the total field energy density in any given region:
2

1
B
2

utotal  u E  u B    0 E 
2
0 
Increasing the inductance
The inductance of any inductor may be increased by filling its
volume—where the magnetic field is stored—with a ferromagnetic
material. The very large increase in the magnetic field caused by
the magnetization factor of the material increases the induced emf
(the voltage across the terminals) by the same factor.
The RL circuit: de-energizing an inductor
This circuit has a resistor in series with an
inductor which has been energized by t = 0. We
want to find the current as a function of time, as
the resistor dissipates energy and the inductor
de-energizes.
The sum of voltage drops around the loop = 0:
VR  VL
RI   L
dI
dt
(Minus sign since L is de-energizing.)
Now separate the equation and integrate each side:
dI
R
R
t
  dt
ln( I )  ln( I 0 )  ln  I    t  
I
L
 I0 
L

I
Now exponentiate:
Giving:
I  I 0e

I0
e
t

 I 0e

t

R
 t
L
with
 
L
R
Recall RC circuit!
The RL circuit: energizing an inductor
If we start instead with a de-energized inductor,
and close switch S1, the battery will energize the
inductor through the resistor, R. The analysis is
similar to that on the previous slide.
The sum of voltage drops around the loop = 0:
Vbatt  VR  VL  Vbatt  RI  L
dI
0
dt
After more mathematical thrashing (see text):
t

Vbatt 
1  e 
I
R 




with
 
L
R
The LC Oscillator Circuit
We’ve seen that in RC and RL circuits, the current rises
or falls exponentially with time. What happens if we
connect an inductor and capacitor in series—with nothing
else in the circuit—and then energize one of them?
The sum of voltage drops around the loop = 0:  L
dI Q
 0
dt C
2
dI
d
dQ
d
Q


Since current is the flow of charge:
 
 2
dt dt  dt  dt
Putting this into the loop equation we get:
d 2Q Q
d 2Q
1
or
L 2  0

Q0
C
dt
dt 2 LC
This equation says that the charge Q is oscillating with time, as we now show:
d 2Q
dQ
2



Q0 cos(t )
Try: Q  Q0 cos(t )
 Q0 sin( t )
2
dt
dt
Put this Q, and its second derivative,
into the loop equation, and cancel!
  2 Q0 cos(t ) 
1
Q0 cos(t )  0
LC

1
LC
The LC oscillator: “watching the oscillations”
The RLC circuit: oscillator circuit with damping
1. Charge the capacitor with the switch, as
shown.
2. Move the switch to position a and the LC
circuit will start to oscillate.
3. The resistor will dissipate energy as the
current flows, damping the oscillations.