Physics 1502 lecture 23 today s agenda
This presentation is the property of its rightful owner.
Sponsored Links
1 / 32

Physics 1502: Lecture 23 Today’s Agenda PowerPoint PPT Presentation


  • 106 Views
  • Uploaded on
  • Presentation posted in: General

Announcements: RL - RV - RLC circuits Homework 06: due next Wednesday … Maxwell’s equations / AC current. Physics 1502: Lecture 23 Today’s Agenda. X X X X X X X X X. Induction. Self-Inductance, RL Circuits. long solenoid. Energy and energy density. In series (like resistors). a.

Download Presentation

Physics 1502: Lecture 23 Today’s Agenda

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript


Physics 1502 lecture 23 today s agenda

Announcements:

RL - RV - RLC circuits

Homework 06: due next Wednesday …

Maxwell’s equations / AC current

Physics 1502: Lecture 23Today’s Agenda


Induction

X X X X X X X X X

Induction

Self-Inductance, RL Circuits

long solenoid

Energy and energy density


Inductors in series and parallel

In series (like resistors)

a

a

a

a

L1

L2

Leq

Leq

L1

L2

b

b

b

b

And finally:

And:

Inductors in Series and Parallel

  • In parallel (like resistors)


Summary of e m

J. C. Maxwell (~1860) summarized all of the work on electric and magnetic fields into four equations, all of which you now know.

However, he realized that the equations of electricity & magnetism as then known (and now known by you) have an inconsistency related to the conservation of charge!

Gauss’ Law

Faraday’s Law

Gauss’ Law

For Magnetism

Ampere’s Law

Summary of E&M

I don’t expect you to see that these equations are inconsistent with conservation of charge, but you should see a lack of symmetry here!


Ampere s law is the culprit

Gauss’ Law:

!

Ampere’s Law is the Culprit!

  • Symmetry: both E and B obey the same kind of equation (the difference is that magnetic charge does not exist!)

  • Ampere’s Law and Faraday’s Law:

  • If Ampere’s Law were correct, the right hand side of Faraday’s Law should be equal to zero -- since no magnetic current.

  • Therefore(?), maybe there is a problem with Ampere’s Law.

  • In fact, Maxwell proposes a modification of Ampere’s Law by adding another term (the “displacement” current) to the right hand side of the equation! ie


Displacement current

Remember:

Iin

Iout

changing electric

flux

Displacement current

FE


Maxwell s displacement current

Can we understand why this “displacement current” has the form it does?

circuit

Maxwell’s Displacement Current

  • Consider applying Ampere’s Law to the current shown in the diagram.

    • If the surface is chosen as 1, 2 or 4, the enclosed current = I

    • If the surface is chosen as 3, the enclosed current = 0! (ie there is no current between the plates of the capacitor)

Big Idea: The Electric field between the plates changes in time.

“displacement current” ID = e0 (dfE/dt)= the real current I in the wire.


Maxwell s equations

Maxwell’s Equations

  • These equations describe all of Electricity and Magnetism.

  • They are consistent with modern ideas such as relativity.

  • They even describe light


Physics 1502 lecture 23 today s agenda

R

C

L

e

~

w

AC Current


Power production an application of faraday s law

S

S

S

N

N

N

End View

Side View

Power ProductionAn Application of Faraday’s Law

  • You all know that we produce power from many sources. For example, hydroelectric power is somehow connected to the release of water from a dam. How does that work?

  • Let’s start by applying Faraday’s Law to the following configuration:


Power production an application of faraday s law1

Power ProductionAn Application of Faraday’s Law

  • Apply Faraday’s Law


Power production an application of faraday s law2

Water

Power ProductionAn Application of Faraday’s Law

  • A design schematic


Rlc circuits

Add resistance to circuit:

R

C

L

e

RLC - Circuits

  • Qualitatively: Oscillations created by L and C are damped (energy dissipation!)

  • Compare to damped oscillations in classical mechanics:


Ac circuits intro

Answer:Yes, if we can supply energy at the rate the resistor dissipates it! How? A sinusoidally varying emf (AC generator) will sustain sinusoidal current oscillations!

R

+

+

C

L

-

-

  • Last time we discovered that a LC circuit was a natural oscillator.

AC Circuits - Intro

  • However, any real attempt to construct a LC circuit must account for the resistance of the inductor. This resistance will cause oscillations to damp out.

  • Question:Is there any way to modify the circuit above to sustain the oscillations without damping?


Ac circuits series lcr

We could solve this equation in the same manner we did for the LCR damped circuit. Rather than slog through the algebra, we will take a different approach which uses phasors.

R

C

L

e

~

AC CircuitsSeries LCR

  • Statement of problem:

  • Given e = emsinwt , find i(t).

  • Everything else will follow.

  • Procedure: start with loop equation?


E r circuit

Before introducing phasors, per se, begin by considering simple circuits with one element (R,C, or L) in addition to the driving emf.

R

i

R

e

~

m

m / R

Note: this is always, always, true… always.

0

0

m

m / R

t

0

t

0

x

eR Circuit

  • Begin with R: Loop eqn gives:

Þ

Voltage across R in phase with current through R


Lecture 23 act 1a

Consider a simple AC circuit with a purely resistive load. For this circuit the voltage source is e = 10V sin (2p50(Hz)t) and R = 5 W. What is the average current in the circuit?

R

e

~

A) 10 A

B) 5 A

C) 2 A

D) √2 A

E) 0 A

Lecture 23, ACT 1a


Chapter 28 act 1b

Consider a simple AC circuit with a purely resistive load. For this circuit the voltage source is e = 10V sin (2p50(Hz)t) and R = 5 W. What is the average power in the circuit?

R

e

~

A) 0 W

B) 20 W

C) 10 W

D) 10 √2 W

Chapter 28, ACT 1b


Rms values

Average values for I,V are not that helpful (they are zero).

Thus we introduce the idea of the Root of the Mean Squared.

In general,

RMS Values

So Average Power is,


E c circuit

Voltage across C lags current through C by one-quarter cycle (90°).

C

e

~

Cm

m

Is this always true?

YES

0

0

m

Cm

0

t

t

0

x

  • Now consider C: Loop eqn gives:

eC Circuit

Þ

Þ


Lecture 23 act 2

A circuit consisting of capacitor C and voltage source e is constructed as shown. The graph shows the voltage presented to the capacitor as a function of time.

Which of the following graphs best represents the time dependence of the current i in the circuit?

e

t

i

i

i

(c)

(a)

(b)

t

t

t

Lecture 23 , ACT 2


E l circuit

Voltage across L leadscurrent through L by one-quarter cycle (90°).

L

e

~

m

m L

0

0

Yes, yes, but how to remember?

m

m L

0

t

t

0

x

x

  • Now consider L: Loop eqn gives:

eL Circuit

Þ

Þ


Phasors

A phasor is a vector whose magnitude is the maximum value of a quantity (eg V or I) and which rotates counterclockwise in a 2-d plane with angular velocity w. Recall uniform circular motion:

y

y

  • R: V in phase with i

Þ

Phasors

  • C: V lags i by 90°

Þ

  • L: V leads i by 90°

Þ

The projections of r (on the vertical y axis) execute sinusoidal oscillation.

w

x


Phasors for l c r

Suppose:

i

w

i

0

wt

t

w

i

i

0

wt

w

i

i

wt

0

ß

Phasors for L,C,R


Lecture 23 act 3

A series LCR circuit driven by emf e = e0sinwt produces a current i=imsin(wt-f). The phasor diagram for the current at t=0 is shown to the right.

At which of the following times is VC, the magnitude of the voltage across the capacitor, a maximum?

t=0

f

i

i

t=0

t=tc

t=tb

i

i

(c)

(a)

(b)

Lecture 23, ACT 3


Series lcr ac circuit

Here all unknowns, (im,f) , must be found from the loop eqn; the initial conditions have been taken care of by taking the emf to be: e = em sinwt.

R

C

L

e

~

  • Assume a solution of the form:

Series LCRAC Circuit

  • Consider the circuit shown here: the loop equation gives:

  • To solve this problem graphically, first write down expressions for the voltages across R,C, and L and then plot the appropriate phasor diagram.


Phasors lcr

From these equations, we can draw the phasor diagram to the right.

  • Given:

R

C

L

e

~

Þ

w

Phasors: LCR

  • Assume:

Þ

  • This picture corresponds to a snapshot at t=0. The projections of these phasors along the vertical axis are the actual values of the voltages at the given time.


Phasors lcr1

w

R

C

L

e

~

The unknowns (im,f) can now be solved for graphically since the vector sum of the voltages VL + VC + VR must sum to the driving emf e.

Phasors: LCR

  • The phasor diagram has been relabeled in terms of the reactances defined from:

Ohms ->


Phasors lcr2

Phasors:LCR

Þ

ß


Phasors tips

y

x

imXL

em

f

imR

Z

| XL-XC|

| 

imXC

R

“Full Phasor Diagram”

“ Impedance Triangle”

Phasors:Tips

  • This phasor diagram was drawn as a snapshot of time t=0 with the voltages being given as the projections along the y-axis.

  • Sometimes, in working problems, it is easier to draw the diagram at a time when the current is along the x-axis (when i=0).

From this diagram, we can also create a triangle which allows us to calculate the impedance Z:


  • Login