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Fall 2008 Physics 231 Lecture 10-1 Chapter 30 Inductance

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Chapter 30 Inductance. Magnetic Effects. As we have seen previously, changes in the magnetic flux due to one circuit can effect what goes on in other circuits The changing magnetic flux induces an emf in the second circuit. Mutual Inductance. Suppose that we have two coils, - PowerPoint PPT Presentation

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Page 1: Chapter 30 Inductance

Fall 2008Physics 231 Lecture 10-1

Chapter 30Inductance

Page 2: Chapter 30 Inductance

Fall 2008Physics 231 Lecture 10-2

Magnetic Effects

As we have seen previously, changes in the magnetic flux due to one circuit can effect what goes on in other circuits

The changing magnetic flux induces an emf in the second circuit

Page 3: Chapter 30 Inductance

Fall 2008Physics 231 Lecture 10-3

Suppose that we have two coils,Coil 1 with N1 turns and Coil 2 with N2 turns

dt

dN B2

22

Coil 1 has a current i1 which produces a magnetic flux, , going through one turn of Coil 2If i1 changes, then the flux changes and an emf is induced in Coil 2 which is given by

Mutual Inductance

Page 4: Chapter 30 Inductance

Fall 2008Physics 231 Lecture 10-4

Mutual InductanceThe flux through the second coil is proportional to the current in the first coil

12122 iMN B

where M21 is called the mutual inductance

dt

diM

dt

dN B 1

212

2 or

dt

diM 1

212

Taking the time derivative of this we get

Page 5: Chapter 30 Inductance

Fall 2008Physics 231 Lecture 10-5

If we were to start with the second coil having a varying current, we would end up with a similar equation with an M12

We would find that MMM 1221

The two mutual inductances are the same because the mutual inductance is a geometrical property of the arrangement of the two coils

To measure the value of the mutual inductance you can use either

dt

dIM 1

2 ordt

dIM 2

1

Mutual Inductance

Page 6: Chapter 30 Inductance

Fall 2008Physics 231 Lecture 10-6

Units of Inductance

2Amp

J1

Amp

secV1 Henry 1

Page 7: Chapter 30 Inductance

Fall 2008Physics 231 Lecture 10-7

Self InductanceSuppose that we have a coil having N turns carrying a current I

That means that there is a magnetic flux through the coil

This flux can also be written as being proportional to the current

ILN B

with L being the self inductance having the same units as the mutual inductance

Page 8: Chapter 30 Inductance

Fall 2008Physics 231 Lecture 10-8

If the current changes, then the magnetic flux through the coil will also change, giving rise to an induced emf in the coil

This induced emf will be such as to oppose the change in the current with its value given by

dt

dIL

If the current I is increasing, then

If the current I is decreasing, then

Self Inductance

Page 9: Chapter 30 Inductance

Fall 2008Physics 231 Lecture 10-9

There are circuit elements that behave in this manner and they are called inductors and they are used to oppose any change in the current in the circuit

As to how they actually affect a circuit’s behavior will be discussed shortly

Self Inductance

Page 10: Chapter 30 Inductance

Fall 2008Physics 231 Lecture 10-10

What Haven’t We Talked AboutThere is one topic that we have not mentioned with

respect to magnetic fields

Just as with the electric field, the magnetic field has energy stored in it

We will derive the general relation from a special case

Page 11: Chapter 30 Inductance

Fall 2008Physics 231 Lecture 10-11

Magnetic Field EnergyWhen a current is being established in a circuit, work has to be done

If the current is i at a given instant and its rate of change is given by di/dt then the power being supplied by the external source is given by

dt

diiLiVP L

The energy supplied is given by PdtdU

The total energy stored in the inductor is then

2

02

1ILdiiLU

I

Page 12: Chapter 30 Inductance

Fall 2008Physics 231 Lecture 10-12

This energy that is stored in the magnetic field is available to act as source of emf in case the current starts to decrease

We will just present the result for the energy density of the magnetic field

0

2

2

1

B

uB

This can then be compared to the energy density of an electric field

202

1EuE

Magnetic Field Energy

Page 13: Chapter 30 Inductance

Fall 2008Physics 231 Lecture 10-13

R-L CircuitWe are given the following circuit

and we then close S1 andleave S2 open

It will take some finite amount of time for the circuit to reach its maximum current which is given by RI

Kirchoff’s Law for potential drops still holds

Page 14: Chapter 30 Inductance

Fall 2008Physics 231 Lecture 10-14

Suppose that at some time t the current is i

The voltage drop across the resistor is given by

RiVab The magnitude of the voltage drop across the inductor is given by

dtdiLVbc

The sense of this voltage drop is that point b is at a higher potential than point c so that it adds in as a negative quantity

0dt

diLRi

R-L Circuit

Page 15: Chapter 30 Inductance

Fall 2008Physics 231 Lecture 10-15

We take this last equation and solve for di/dt

iL

R

Ldt

di

Notice that at t = 0 when I = 0 we have that Ldt

di

initial

Also that when the current is no longer changing, di/dt = 0, that the current is given by R

I

as expected

But what about the behavior between t = 0 and t =

R-L Circuit

Page 16: Chapter 30 Inductance

Fall 2008Physics 231 Lecture 10-16

R-L CircuitWe rearrange the original equation and then integrate

dtL

R

Ri

di

/

ti

dtL

R

Ri

di

00

'/'

'

The solution for this is tLReR

i /1

Which looks like

Page 17: Chapter 30 Inductance

Fall 2008Physics 231 Lecture 10-17

As we had with the R-C Circuit, there is a time constant associated with R-L Circuits

R

L

Initially the power supplied by the emf goes into dissipative heating in the resistor and energy stored in the magnetic field

dt

diiLRii 2

After a long time has elapsed, the energy supplied by the emf goes strictly into dissipative heating in the resistor

R-L Circuit

Page 18: Chapter 30 Inductance

Fall 2008Physics 231 Lecture 10-18

We now quickly open S1 and close S2

The current does not immediatelygo to zero

The inductor will try to keep the current, in the same direction, at its initial value to maintain the magnetic flux through it

R-L Circuit

Page 19: Chapter 30 Inductance

Fall 2008Physics 231 Lecture 10-19

R-L CircuitApplying Kirchoff’s Law to the bottom loop we get

where I0 is the current at t = 0

0dt

diLiR

Rearranging this we have dtL

R

i

di

and then integrating this

t

I

dtL

R

i

di

0

0

''

'

0

tLReII /0

Page 20: Chapter 30 Inductance

Fall 2008Physics 231 Lecture 10-20

This is a decaying exponentialwhich looks like

The energy that was stored in the inductor will be dissipated in the resistor

R-L Circuit

Page 21: Chapter 30 Inductance

Fall 2008Physics 231 Lecture 10-21

L-C CircuitSuppose that we are now given a fully charged capacitor and an inductor that are hooked together in a circuit

Since the capacitor is fully charged there is a potential difference across it given by Vc = Q / C

The capacitor will begin to discharge as soon as the switch is closed

Page 22: Chapter 30 Inductance

Fall 2008Physics 231 Lecture 10-22

We apply Kirchoff’s Law to this circuit

0C

q

dt

diL

Remembering thatdt

dqi

We then have that 2

2

dt

qd

dt

di

The circuit equation then becomes 01

2

2

qLCdt

qd

L-C Circuit

Page 23: Chapter 30 Inductance

Fall 2008Physics 231 Lecture 10-23

This equation is the same as that for the Simple Harmonic Oscillator and the solution will be similar

)cos(0 tQq

The system oscillates with angular frequency

The current is given by )sin(0 tQdt

dqi

LC

1

is a phase angle determined from initial conditions

L-C Circuit

Page 24: Chapter 30 Inductance

Fall 2008Physics 231 Lecture 10-24

Both the charge on the capacitor and the current in the circuit are oscillatory

For an ideal situation, this circuit will oscillate forever

The maximum charge and the maximum current occur seconds apart

L-C Circuit

Page 25: Chapter 30 Inductance

Fall 2008Physics 231 Lecture 10-25

L-C Circuit

Page 26: Chapter 30 Inductance

Fall 2008Physics 231 Lecture 10-26

Just as both the charge on the capacitor and the current through the inductor oscillate with time, so does the energy that is contained in the electric field of the capacitor and the magnetic field of the inductor

Even though the energy content of the electric and magnetic fields are varying with time, the sum of the two at any given time is a constant

BETotal UUU

L-C Circuit

Page 27: Chapter 30 Inductance

Fall 2008Physics 231 Lecture 10-27

Instead of just having an L-C circuit with no resistance, what happens when there is a resistance R in the circuit

Again let us start with the capacitor fully charged with a charge Q0 on it

The switch is now closed

L-R-C Circuit

Page 28: Chapter 30 Inductance

Fall 2008Physics 231 Lecture 10-28

L-R-C CircuitThe circuit now looks like

The capacitor will start to discharge and a current will start to flow

We apply Kirchoff’s Law to this circuit and get

0C

q

dt

diLiR

And remembering thatdt

dqi we get

01

2

2

qLCdt

dq

L

R

dt

qd

Page 29: Chapter 30 Inductance

Fall 2008Physics 231 Lecture 10-29

The solution to this second order differential equation is similar to that of the damped harmonic oscillator

The are three different solutions

Underdamped

Critically Damped

Overdamped

Which solution we have is dependent upon the relative values of R2 and 4L/C

L-R-C Circuit

Page 30: Chapter 30 Inductance

Fall 2008Physics 231 Lecture 10-30

L-R-C CircuitUnderdamped:

C

LR

42

The solution to the second differential equation is then

tL

R

LCeQq

tL

R

2

22

04

1cos

This solution looks like The system still oscillates but with decreasing amplitude, which is represented by the decaying exponential

This decaying amplitude is often referred to as the envelope

Page 31: Chapter 30 Inductance

Fall 2008Physics 231 Lecture 10-31

L-R-C CircuitCritically Damped:

C

LR

42

Here the solution is given by

tL

R

etL

RQq

2

0 21

This solution looks like

This is the situation when the system most quickly reachesq = 0

Page 32: Chapter 30 Inductance

Fall 2008Physics 231 Lecture 10-32

L-R-C Circuit

Overdamped:C

LR

42

Here the solution has the form

ttt

LR

eLR

eLR

eQ

q ''20

'21

'21

2

withLCL

R 1

4'

2

2

This solution looks like

Page 33: Chapter 30 Inductance

Fall 2008Physics 231 Lecture 10-33

L-R-C Circuit

The solutions that have been developed for this L-R-C circuit are only good for the initial conditions at t = 0 that q = Q0 and that i = 0