The Ups and Downs of Circuits The End is Near! Quiz – Nov 18 th – Material since last quiz....

The Ups and Downs of The Ups and Downs of CircuitsCircuits

The End is Near!

• Quiz – Nov 18th – Material since last quiz. (Induction)

• Exam #3 – Nov 23rd – WEDNESDAY• LAST CLASS – December 2nd • FINAL EXAM – 12/5 10:00-12:50 Room

MAP 359• Grades by end of week. Hopefully

Maybe.

A circular region in the xy plane is penetrated by a uniform magnetic field in the positive direction of the z axis. The field's magnitude B (in teslas) increases with time t (in seconds) according to B = at, where a is a constant. The magnitude E of the electric field set up by that increase in the magnetic field is given in the Figure as a function of the distance r from the center of the region. Find a.[0.030] T/s

VG

r

For the next problem, recall that

R

L

eR

Ei LRt

constant time

)1( /i

For the circuit of Figure 30-19, assume that = 11.0 V, R = 6.00 , and L = 5.50 H. The battery is connected at time t = 0.

(a) How much energy is delivered by the battery during the first 2.00 s?[23.9] J (b) How much of this energy is stored in the magnetic field of the inductor? [7.27] J(c) How much of this energy is dissipated in the resistor?[16.7] J

6

5.5H

Let’s put an inductor and a capacitor in the SAME circuit.

At t=0, the charged capacitor is connected to the inductor. What would you expect to happen??

Current would begin to flow….

High

Low

Low

High

202

1Eu Energy Density in Capacitor

Energy Flows from Capacitor to the Inductor’s Magnetic Field

Energy Flow

202

1EuC 2

02

1BuL

Energy

LC Circuit

High

Low

Low

High

LC

idt

id

dt

idLi

C

dt

diL

C

q

EquationLoop

1

0

01

0

2

22

2

2

2

)sin()cos(

)cos()sin(

1

0

2

22

2

tBtAdt

di

tBtAiLC

idt

id

When t=0,

i=0so

B=0When t=0, voltageacross the inductor= Q0/C

LC

Q

LCLC

Q

LC

QA

ALC

Q

dt

diL

000

0

/

][

0At t

)sin()cos(

)cos()sin(

1

0

2

22

2

tBtAdt

di

tBtAiLC

idt

id

The Math Solution:

)1

()( 0 tLC

SinLC

Qti

Energy

2

22

2

1

:2

1

2

1

:

LiE

InductorC

QCVE

Capacitor

L

C

Inductor

tSinQC

tSinLQE

tLC

SinLC

QLE

tLC

SinLC

Qti

C

C

220

2220

2

0

0

2

1

2

1

)1

(2

1

)1

()(

The Capacitor

)(2

1

)(1

2

1)(

2

1

2

1

)(

:Eq Diff

:Capacitor For the

)()1

()(

220

22022

2220

422

02

00

tCosQC

tCosQCL

CLtCosQCLCv

tCosQLvdt

diL

C

Q

From

CvQ

tSinQtLC

SinLC

Qti

Add ‘em Up …

Constant2

1 2220 tCostSinQ

CEtotal

Add Resistance

Actual RLC:

)'(2

Energy Total

)2/('

)'(

0

/

0

20

22

2/0

2

2

tCoseC

QU

LR

tCoseQq

C

q

dt

dqR

dt

qdL

LRttotal

LRt

New Feature of Circuits with L and C These circuits can produce oscillations in the currents and

voltages Without a resistance, the oscillations would continue in an

un-driven circuit. With resistance, the current will eventually die out.

The frequency of the oscillator is shifted slightly from its “natural frequency”

The total energy sloshing around the circuit decreases exponentially

There is ALWAYS resistance in a real circuit!

Types of Current

Direct Current Create New forms of life

Alternating Current Let there be light

-1.5

-1

-0.5

0

0.5

1

1.5

0 1 2 3 4 5 6 7 8 9 10

Time

Vo

lts

Alternating

emf

Sinusoidal

DC

Sinusoidal Stuff

)sin( tAemf

“Angle”

Phase Angle

Same Frequencywith

PHASE SHIFT

Different Frequencies

Note – Power is delivered to our homes as an oscillating source (AC)

Producing AC Generator

x x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x x

The Real World

The Flux:

tAR

emfi

tBAemf

t

BA

bulb

sin

sin

cos

AB

OUTPUT

)sin(0 tVVemf

WHAT IS AVERAGE VALUE OF THE EMF ??

Average value of anything:

Area under the curve = area under in the average box

T

T

dttfT

h

dttfTh

0

0

)(1

)(

T

h

Average Value

T

dttVT

V0

)(1

0sin1

0

0 T

dttVT

V

For AC:

So …

Average value of current will be zero. Power is proportional to i2R and is ONLY

dissipated in the resistor, The average value of i2 is NOT zero because

it is always POSITIVE

Average Value

0)(1

0

T

dttVT

V

2VVrms

RMS

2

2)(

2

2)

2(

2

1

)2

(1

0

02

0

20

0

20

0

20

220

VV

VdSin

VV

tT

dtT

SinT

TVV

dttT

SinT

VtSinVV

rms

rms

T

rms

T

rms

Usually Written as:

2

2

rmspeak

peakrms

VV

VV

Example: What Is the RMS AVERAGE of the power delivered to the resistor in the circuit:

E

R

~

Power

tR

VRt

R

VRitP

tR

V

R

Vi

tVV

22

0

2

02

0

0

sin)sin()(

)sin(

)sin(

More Power - Details

R

VVV

RR

VP

R

VdSin

R

VP

tdtSinR

VP

dttSinTR

VP

tSinR

VtSin

R

VP

rms

T

T

200

20

20

2

0

22

0

0

22

0

0

22

0

22

022

0

22

1

2

1

2

1)(

2

1

)(1

2

)(1

Resistive Circuit

We apply an AC voltage to the circuit. Ohm’s Law Applies

Con

sid

er

this

cir

cuit

CURRENT ANDVOLTAGE IN PHASE

R

emfi

iRe

Alternating Current Circuits

is the angular frequency (angular speed) [radians per second].

Sometimes instead of we use the frequency f [cycles per second]

Frequency f [cycles per second, or Hertz (Hz)] f

V = VP sin (t -v ) I = IP sin (t -I )

An “AC” circuit is one in which the driving voltage andhence the current are sinusoidal in time.

v

V(t)

t

Vp

-Vp

v

V(t)

t

Vp

-Vp

V = VP sin (wt - v )Phase Term

Vp and Ip are the peak current and voltage. We also use the

“root-mean-square” values: Vrms = Vp / and Irms=Ip /

v and I are called phase differences (these determine whenV and I are zero). Usually we’re free to set v=0 (but not I).

2 2

Alternating Current Circuits

V = VP sin (t -v ) I = IP sin (t -I )

v

V(t)

t

Vp

-Vp

Vrms

I/

I(t)

t

Ip

-Ip

Irms

Example: household voltage

In the U.S., standard wiring supplies 120 V at 60 Hz. Write this in sinusoidal form, assuming V(t)=0 at t=0.



This 120 V is the RMS amplitude: so Vp=Vrms = 170 V.2



This 120 V is the RMS amplitude: so Vp=Vrms = 170 V.This 60 Hz is the frequency f: so =2f=377 s -1.

2



This 120 V is the RMS amplitude: so Vp=Vrms = 170 V.This 60 Hz is the frequency f: so =2f=377 s -1.

So V(t) = 170 sin(377t + v).Choose v=0 so that V(t)=0 at t=0: V(t) = 170 sin(377t).

2

Resistors in AC Circuits

ER

~EMF (and also voltage across resistor): V = VP sin (t)Hence by Ohm’s law, I=V/R:

I = (VP /R) sin(t) = IP sin(t) (with IP=VP/R)

V and I“In-phase”

V

t

I

This looks like IP=VP/R for a resistor (except for the phase change). So we call Xc = 1/(C) the Capacitive Reactance

Capacitors in AC Circuits

E

~C Start from: q = C V [V=Vpsin(t)]

Take derivative: dq/dt = C dV/dtSo I = C dV/dt = C VP cos (t)

I = C VP sin (t + /2)

The reactance is sort of like resistance in that IP=VP/Xc. Also, the current leads the voltage by 90o (phase difference).

V

t

I

V and I “out of phase” by 90º. I leads V by 90º.

I Leads V???What the **(&@ does that mean??

I

V

Current reaches it’s maximum at

an earlier time than the voltage!

1

2

I = C VP sin (t +/2)

Capacitor Example

E

~

CA 100 nF capacitor isconnected to an AC supply of peak voltage 170V and frequency 60 Hz.

What is the peak current?What is the phase of the current?

MX

f

C 65.2C

1

1077.3C

rad/sec 77.360227

Also, the current leads the voltage by 90o (phase difference).

Again this looks like IP=VP/R for aresistor (except for the phase change).

So we call XL = L the Inductive Reactance

Inductors in AC Circuits

LV = VP sin (t)Loop law: V +VL= 0 where VL = -L dI/dtHence: dI/dt = (VP/L) sin(t).Integrate: I = - (VP / L cos (t)

or I = [VP /(L)] sin (t - /2)

~

Here the current lags the voltage by 90o.

V

t

I

V and I “out of phase” by 90º. I lags V by 90º.

Phasor Diagrams

Vp

Ipt

Resistor

A phasor is an arrow whose length represents the amplitude ofan AC voltage or current.

The phasor rotates counterclockwise about the origin with the angular frequency of the AC quantity.

Phasor diagrams are useful in solving complex AC circuits.The “y component” is the actual voltage or current.

A phasor is an arrow whose length represents the amplitude ofan AC voltage or current.

The phasor rotates counterclockwise about the origin with the angular frequency of the AC quantity.

Phasor diagrams are useful in solving complex AC circuits.The “y component” is the actual voltage or current.

Phasor Diagrams

Vp

Ipt

Vp

Ip

t

Resistor Capacitor

A phasor is an arrow whose length represents the amplitude ofan AC voltage or current.The phasor rotates counterclockwise about the origin with the angular frequency of the AC quantity.Phasor diagrams are useful in solving complex AC circuits.The “y component” is the actual voltage or current.


Phasor Diagrams

Vp

Ipt

Vp

Ip

t

Vp Ip

t

Resistor Capacitor Inductor



+ +i

+++

+

i

i

i

i

i

LC Circuit

time

U UB UE 12LI2 1

2q2

C

dU

dt d

dt(1

2LI2 1

2

q2

C)0

LIdI

dtq

C

dq

dt0 L(

dq

dt)d2q

dt2 q

C

dq

dt

Ld2q

dt 2

1

Cq 0

Analyzing the L-C Circuit

Total energy in the circuit:

Differentiate : N o change in energy

U UB UE 12LI2 1

2q2

C

dU

dt d

dt(1

2LI2 1

2

q2

C)0

LIdI

dtq

C

dq

dt0 L(

dq

dt)d2q

dt2 q

C

dq

dt

Ld2q

dt 2

1

Cq 0

Analyzing the L-C Circuit

Total energy in the circuit:

Differentiate : N o change in energy

The charge sloshes back andforth with frequency = (LC)-1/2

The charge sloshes back andforth with frequency = (LC)-1/2

tqq

qdt

qd

p

cos

022

2

The Ups and Downs of Circuits The End is Near! Quiz – Nov 18 th – Material since last quiz....

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