Physics 1502: Lecture 23Today’s Agenda

• Announcements:– RL - RV - RLC circuits

• Homework 06: due next Wednesday …Homework 06: due next Wednesday …

• Maxwell’s equations / AC current

InductionSelf-Inductance, RL Circuits

X X X X X X X X X

RI

ε

a

b

L

I

long solenoid

Energy and energy density

Inductors in Series and Parallel• In series (like resistors) a

b

L2

L1

a

b

Leq

And:

a

b

L2L1

a

b

Leq

• In parallel (like resistors)

And finally:

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!

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!

Gauss’ Law

Gauss’ LawFor Magnetism

Faraday’s Law

Ampere’s Law

Ampere’s Law is the Culprit!• Gauss’ Law:

• 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:

E

Iin Iout

changing electric flux

Maxwell’s Displacement Current

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

• 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 = ε0 (dE/dt) = the real current I in the wire.

circuit

Maxwell’s Equations• These equations describe all of Electricity and

Magnetism.

• They are consistent with modern ideas such as relativity.

• They even describe light

LC

ε

R

φ

φ

φ

imR

imL

imC

εm

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:

N

S

N

S

Side View

N

S

End View

Power ProductionAn Application of Faraday’s Law

• Apply Faraday’s Law

Power ProductionAn Application of Faraday’s Law

• A design schematic

Water

RLC - Circuits

• Add resistance to circuit:LC

ε

R

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

• Compare to damped oscillations in classical mechanics:

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?

• 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!

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

LC+ +

- -

R

AC CircuitsSeries LCR• Statement of problem:

Given ε = εmsint , find i(t).

Everything else will follow.

• Procedure: start with loop equation?

• 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.

LC

ε

R

εR Circuit• Before introducing phasors, per se, begin by considering

simple circuits with one element (R,C, or L) in addition to the driving emf.

• Begin with R: Loop eqn gives:

Voltage across R in phase with current through R

εiR

R

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

0

V R

tx

εm

εm

0

0 t

iRεm / R

εm / R

0

Lecture 23, ACT 1a

• Consider a simple AC circuit with a purely resistive load. For this circuit the voltage source is ε = 10V sin (250(Hz)t) and R = 5 . What is the average current in the circuit?

ε

R

A) 10 A B) 5 A C) 2 A D) √2 A E) 0 A

Chapter 28, ACT 1b• Consider a simple AC circuit with a

purely resistive load. For this circuit the voltage source is ε = 10V sin (250(Hz)t) and R = 5 . What is the average power in the circuit?

ε

R

A) 0 W B) 20 W C) 10 W D) 10 √2 W

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,

So Average Power is,

εC Circuit• Now consider C: Loop eqn gives:

ε

C iC

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

Is this always true?

YES

0

V R

tx

εm

εm

0

t

0

iC

0

Cεm

Cεm

Lecture 23 , ACT 2• A circuit consisting of capacitor C and voltage

source ε 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?

(a) (b) (c)i

t

i

t t

i

ε

t

εC

εL Circuit• Now consider L: Loop eqn gives:

Voltage across L leads current through L by one-quarter cycle (90).

ε

LiL

Yes, yes, but how to remember?

0

V R

tx

εm

εm

0

t

iL

x

εm L

εm L 0

0

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 . Recall uniform circular motion:

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

• R: V in phase with i

• C: V lags i by 90

• L: V leads i by 90

x

y y

Phasors for L,C,Ri

t

V

R

i

t

VL

i

t

VC

t

iV

R

0

VC

0

i

i

0

VL

2

Suppose:

Lecture 23, ACT 3• A series LCR circuit driven by emf ε = ε0sint

produces a current i=imsin(t-). 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?

LC

∼ε

R

i

t=0

(a) (b) (c)i

t=0

i

t=tb

i

t=tc

Series LCR

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

• Here all unknowns, (im,) , must be found from the loop eqn; the initial conditions have been taken care of by taking the emf to be: εεm sint.

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

LC

ε

R

• Assume a solution of the form:

Phasors: LCR

• Assume:

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

φ

φ

φ

imR

imL

imC

εm

LC

ε

R

• 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.

• Given:

Phasors: LCR

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

φ

φ

φ

imR

εm

imXC

imXL

LC

ε

R

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

Phasors:LCR

imR

εm

φ

im(XL-X C)

φφ

φ

imR

εm

imXC

imXL

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.

φφ

φ

imR

εm

imXC

imXLy

x

imR

imXL

imXC

εm

“Full Phasor Diagram”

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

• 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).

“ Impedance Triangle”

Z

|

R

| XL-XC |