Physics for Scientists and Engineers II, Summer Semester 2009 1 Lecture 19: July 8 th 2009 Physics for Scientists and Engineers II

Physics for Scientists and Engineers II , Summer Semester 2009

1

Lecture 19: July 8th 2009

Physics for Scientists and Engineers II


2

RLC Series Circuits –solved using Kirchhoff’s loop

tVv sinmax

0

0 :rule loop sKirchhoff'

2

2

C

q

dt

qdL

dt

dqRΔv

C

q

dt

dILRIΔv

Cv

i

CLR

LvRv


3


0sin2

2

max C

q

dt

qdL

dt

dqRtΔV

tAtqtAtqtAtq cos)(sin)(cos)(

:solution Try this2

0coscossinsin

:DE intoit Plug

2max t

C

AtLAtRAtV

sinsincoscoscos

sincoscossinsin

:identities theseUse

ttt

ttt

0sinsincoscossincoscossinsin 2max

ttLAC

AttRAtV


4


R

XX

RC

LL

CR

ωt

LAC

ARAt

LAC

ARAVt

ttLAC

AttRAtV

CL

1

tan0cos1

sin

term)cos (from

0cossincos

0sincossin

0sinsincoscossincoscossinsin

2

2max

2max


5


22

222

222

222

22

22

max

maxmax

max

2max

1

tan1cos

coscos

sincos

circuit) theof impedance"" thecalled is (Zsintancos

0sin1

cos

0sincos

term)in (from

CL

CLCL

XXRZ

XXRR

XXRZ

RR

Z

RRZ

RRZI

V

A

V

CLR

A

V

LAC

ARAV

ωts


6


tVv sinmax

Cv

i

CLR

LvRv

R

XX

XXRZ

VZI

tII

tVv

CL

CL

1

22

maxmax

max

max

tan

where

sin

sin


7

RLC Series Circuits - book’s method

tVv sinmax

t

t

sinIi :assume sLet'

series).in are they (because elementscircuit threeallfor phase) and (amplitude

same theiscurrent The ).determined be to(also I amplitudean has and

much) how determined be (to voltage with thephase ofout iscurrent The

sinVv

max

max

max

Cv

i

CLR

LvRv


8

RLC Series Circuits - book’s method

tVv sinmax

current) behind 90 lags C across (voltagecos2

sinXI

current) of ahead 90 is L across (voltage cos2

sinXI

current) with phasein is Rin (voltage sin sinI

Cmax

Lmax

max

tVtv

tVtv

tVtRv

CC

LL

RR

Cv

i

CLR

LvRv


9

Phasor Diagram

maxI

RV

LV

CVThe voltages across the components are out of phaseas shown in the phasor diagram.They need to be added as vectors to get the total voltage.


10

Phasor Diagram

maxI

RV

LVCV

The voltages across L and C can simply be subtracedfrom each other (180 degrees out of phase).

CL VV


11

Phasor Diagram

maxI RV

CL VV

maxV

22max

2maxmax

2max

22max

CL

CL

CLR

XXRI

XIXIRI

VVVV

R

XX

R

XX

RI

XIXI

XXRI

VZ CL

CL1CL

max

CmaxLmax

R

CL

22

max

max

tanV

VVtan

:diagramphasor thefrom angle phase thedetermine alsocan We

e)resistenac toanalogousquantity (a Zimpedance""an definecan We


12

Power in an AC Circuit

ttVItVtIviP sinsinsinsin maxmaxmaxmax

sincoscossinsin :Use ttt

source) by the deliveredpower ousinstantane theis (This

sincossincossin maxmax2

maxmax ttVItVIP

sincossincossin

:average) time(takePower Average

maxmax2

maxmax ttVItVIPavg

2

1sin2 t 02sin

2

1cossin ttt

coscos22

cos2

1 maxmaxmaxmax rmsrmsavg VI

VIVIP


13

Power in an AC Circuit

cosrmsrmsavg VIP “Power factor”

maxI RV

CL VV

maxV

rms

RR

V

V

V

V

2

cosmax

RIRI

IV

VVIP rms

rmsrms

rms

Rrmsrmsavg

2

2

2

2

RIRIV rmsR 2:resistor For the max

up). heats(resistor resistor in theenergy internal

toconverted is source by the deliveredpower Average:means2RIP rmsavg


14

Implications of power factor

cosrmsrmsavg VIP “Power factor”

rmsrmsavg VIP 1)0cos(cosresistor Only

:1 Example

inductor) pure a power toany deliver not does average,on source,power AC(an

00)90cos(cosinductor Only

:2 Example

avgP

capacitor) pure a power toany deliver not does average,on source,power AC(an

00)90cos(coscapacitor Only

:3 Example

avgP


15

Resonance in a Series RLC Circuit – the current

circuit. theoffrequency resonance thecalled is

phase.in are voltageapplied

theandcurrent and valuemaximum a hasCurrent :1

For

1at0C

1-LX

tan:Also

10

C

1-L

for maximum a has anddependent frequency is

1

0

0

0L

0

0

22

22

LC

LCRR

X

LC

I

CLR

V

XXR

V

Z

VI

C

rms

rms

CL

rmsrmsrms


16

0

0 20 40 60 80

R=10 Ohm

R=20 Ohm

R=30 Ohm

Resonance

R=0 Ohm

Resonance in a Series RLC Circuit – the Current

)( srad

0

rmsI

VV

mFC

HL

rms 10

0.1

0.1

22 1

CLR

V

Z

VI rmsrmsrms


17

Resonance in a Series RLC Circuit – power

0

220

2222

22

22222

22

2222

22

2222

22

22

2

22

2

2

22

when maximum a has

1

11

1

avg

rmsrms

rmsrms

rms

CL

rmsrmsrmsavg

P

LR

RV

CLLR

RV

CLR

RV

CLR

RV

CLR

RV

XXR

RVR

Z

VRIP


18

0

0 10 20 30 40 50 60

R=10 Ohm

R=20 Ohm

R=30 Ohm

Resonance

Resonance in a Series RLC Circuit – power

0 )( srad

avgP

220

2222

22

LR

RVP rmsavg R

LQ 00 :factorQuality

Describes sharpness of resonance.Q is larger for smaller R.

maximum) halfat width (full


19

AC Transformer – a simple design

currents.eddy reduces Lamination

them).contains and field magnetic (increases

coreiron laminatedSoft

Primary Winding (input)

1v

S

LR2v

B

1N2N

Secondary Winding (output)


20

AC Transformer - Voltage Transformation

ngeach windigh Flux throu :

law) s(Faraday'11

B

B

dt

dNv

1v

S

LR2v

1I B

1N2N

law) s(Faraday'12 dt

dNv B

11

22 v

N

Nv


21

Step-up versus Step-down Transformers

former"down trans-Step":

er" transformup-Step":

12

12

11

22

NN

NN

vN

Nv


22

AC Transformer – Connecting the load

1v LR 2v

1I B

1N2N

2I

222L :Rin dissipatedPower VIP rms values

LLLeqLeq

RN

NR

VNN

VR

V

VR

R

V

R

V

VIVI2

2

12

11

2

21

22

21

22

21

2211

:rmation)in transfo losspower (noer transformIdeal

side.primary from ed when view,resistance

load of resistance equivalent:R eq

matching" impedance"

for used becan r Transforme


23

Transmission line economics

TR resistance has lineon Transmissi

consumer at

resistance Load :R Lgenerator AC

LR

TR

circuit Equivalent


24

Transmission line economics

LR

TR

power)(1KW cleaner vacuumrunsConsumer

consumerlostconsumerL

TconsumerLT

L

consumerLTgenerator

TL

consumerTlost

L

consumerLCconsumer

PPPR

RPRR

R

PRRIP

RR

PRIP

R

PIRIΔVIP

2

2

22

Documents

Physics for Scientists and Engineers II, Summer Semester 2009 1 Lecture 19: July 8 th 2009 Physics for Scientists and Engineers II