physics for scientists and engineers ii, summer semester 2009 1 lecture 18: july 6 th 2009 physics...

Physics for Scientists and Engineers II , Summer Semester 2009

1

Lecture 18: July 6th 2009

Physics for Scientists and Engineers II


2

Energy loss in the circuit

LRt

LRt

LRt

etRI

edt

dILILI

dt

d

edt

dQ

C

Q

C

Q

dt

d

)(dt

dU

resistor) (in theheat tosystemin energy of Loss

2

1

dt

dU :Inductor

2

1

dt

dU :Capacitor

:components in theenergy of change of Rate

2R

2L

2c


3

Energy loss in the circuit

tttI

t

ddd

d

sineQcose2L

RQ)(

coseQQ(t)

2LRt-

max2L

Rt-max

2LRt-

max

teC

Q

C

tQt L

Rt2

2max

2

c cos22

1 U:Capacitorin Energy

tt

L

Rtt

L

RQLe

tILt

ddd

dddL

Rt sincos

2sincos

42

1

2

1 U:Inductorin Energy

2222

22max

2L

dtttL

Rtt

L

RQt

dttIRtt

ddd

dddL

Rt

0

2222

22maxR

2R

sincos2

sincos4

ReU

dU :resistorin heat toconvertedEnergy


4

Example from book: Chapter 32, Problem 18

00.4

V10 00.4

00.8

H00.1

r. thereafte timeoffunction

a asswitch in thecurrent theandinductor in thecurrent Find

0.at t closed is time)long afor open (which wasswitch The


5

Use Kirchhoff’s Rules

00.4

V10 00.4

00.8

H00.11 2

1I2I

3I

)3(000.100.800.4 :2 Loop

)2(000.400.4-10V :1 Loop

)1( :RuleJunction

223

31

321

dt

dIHII

II

III


6

Solving the system of equations

)3(000.100.800.4 :2 Loop

)2(000.400.4-10V :1 Loop

)1( :RuleJunction

223

31

321

dt

dIHII

II

III

)4(000.8I00.4I-10V

000.4I00.4)I(I-10V :(2) into )1(

32

332

)4(000.20.20I-10V

000.200.16I00.4I-10V :(3) 2 )4(

22

222

dt

dIH

dt

dIH


7


)4(000.20.20I-10V 22

dt

dIH

22 I0.20I20.0-10V x:onSubstituti ddx

tH

x

x

t

exx

tHx

xdt

Hx

dx

dtHx

dx

dt

dxHx

0.10

0

0 0

0.10ln0.10

0.1000.20

00.2

0

tH

tH ee

0.100.10

2

2

10V0I20.0-10VI20.0-10V

I20.0-10V x:againon Substituti


8


tH

tH

tH

eAe

e

0.100.10

2

0.10

2

15.020.0

10V-10VI

10VI20.0-10V

tH

tH

tH

AA

VV

I10.0-

10.0-

3

3

10.0-

32

e-125.025.100.8

e-100.20.10

000.8I00.4e-10.5A-10V

)4(000.8I00.4I-10V

Eq.(4) intoResult


9


t

HeA0.10

2 15.0I

t

HAAI10.0-

3 e-125.025.1

tH

tH

tH

AA

AAAIII

10.0-

10.0-10.0-

321

e-125.025.1

e-125.025.1e-15.0


10


t

HeA0.10

2 15.0I

t

HAAI10.0-

3 e-125.025.1

t

HAAI10.0-

1 e-125.025.1

A5.0I2

AI 5.11

AI 0.13

For long times t:

00.4

V10 00.4

00.8 67.2

67.6totalR

AV

I 5.167.6

101


11

AC (“alternating current”) Sources

R

tVv sinmax

msf

T 7.161

srad377 60Hzf : USin they electricit Household

Rv

-2

0

vmaxV

T


12

Resistors in AC Circuits

R

tVv sinmax

tItR

V

R

ΔviRi-Δv RR sinsin0 :rule loop sKirchhoff' max

max

Rv

Ri

tRIRiΔv RR sin :resistor across drop Voltage max

resistorrough Current th

-1.3

0

Current

Voltage

maxI

maxV


13

Resistors in AC Circuits

tIiR sinmax

tVΔvR sinmax

-1.3

0

Current

Voltage

maxI

maxV

Applied voltage and resulting current reach their peaks at the same time They are “in phase”.


14

Phasor Diagram

-1.3

0

Current

Voltage

maxI

maxV

In phasor diagram: Vectors rotate counterclockwise with frequency . The projection of the rotating vectors onto the vertical axis yields the instantaneous values of voltage and current.

RR vi ,

maxI

maxV

t


15

Average current, rms current, average power

Current

-4

00 iR

Current Squared

-4

0

2max

2R 2

1 i I

i2R

Ri

2Ri

maxmax

maxmax

2

2

max2max

22

707.02

707.02

22

1

:Rin dissipatedpower average The

VV

V

II

I

RIP

RI

RI

Ri RiP

rms

rms

rms

RR


16

Measuring AC voltage and AC current with volt/current meters.

02

V indicate

will voltmeter themode, AC theusing measuredproperly If

present. is voltage

no that believe you to leading zero, be willread valueThe

:mode DCin voltmetera using voltageACan measureyou If

:Careful

.I / V read meter will themode, AC""In

.I / V read meter will themode, DC""In

:AC and DCfor settings separate have meterscurrent Volt /

maxrms

rmsrms

averageaverage

V


17

-1.3

0

Current

Voltage

Inductors in an AC Circuits

L

tVv sinmax

2sin

cossin

sin

sin0 :rule loop sKirchhoff'

max

maxmax

max

max

tL

Vi

tL

Vdtt

L

Vi

dttL

Vdi

dt

diLtV

dt

diL-Δv

L

L

L

LL

Lv

Li

maxI

maxV


18


-1.3

0

Current

Voltage

LL vi ,

maxI

maxV

t

For sinusoidal voltages:The phase of the current through the inductor lags behind the voltageacross the inductor by 90 degrees (one quarter cycle in time).

Lv

Li


19


ω)). toaccording (reacts ωon

depends that form a hasXfact that thefrom comes reactance term(the

.reactance" inductive" called is LωXquantity The

e).(resistanc of units has

resistor. afor to thisCompare

L

L

maxmax

maxmax

L

R

VI

L

VI

0IlimXlim

small). is(current large very is X voltages,changingrapidly For very

Ilim0Xlim

large). is(current small very is X voltages,changingslowly For very

:implies LωX

maxω

Lω

L

max0ω

L0ω

L

L

LX

V

L

VI maxmax

max


20

-1.3

0

Current

Voltage

Capacitors in an AC Circuits

C

tVv sinmax

2sin

cos

sin

sin0 :rule loop sKirchhoff'

max

max

max

max

tVCi

tVCdt

dqi

dttVCqC

qtV

C

q-Δv

C

C

Cv

Ci

maxI

maxV


21

-1.3

0

Current

Voltage


CC vi ,

maxI

maxV

t

For sinusoidal voltages:The phase of the current to the capacitor leads the voltageacross the capacitor by 90 degrees (one quarter cycle in time).

CvCi


22


.reactance" capacitive" called is 1

Xquantity The

e).(resistanc of units has1

resistor. afor to thisCompare

C

maxmaxmaxmax

C

C

R

VIVCI

maxω

Cω

C

max0ω

C0ω

C

C

Ilim0Xlim

large). is(current small very is X voltages,changingrapidly For very

0IlimXlim

small). is(current large very is X voltages,changingslowly For very

:implies 1

X C

CX

VVCI max

maxmax

physics for scientists and engineers ii, summer semester 2009 1 lecture 18: july 6 th 2009 physics...

Documents

summer semester

circuit slide

ac current

ac circuits

average power slide

measuring ac voltage

average current

system of equations