1 29 Overview why & how to use rms values determine impedance of L & C why & how: phase...

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1

29 Overview

• why & how to use rms values

• determine impedance of L & C

• why & how: phase relationships in ac circuits

2

sinusoidal current “ac”

• I ~ sine, cosine variation with time:(I = Io cos(wt + phi))

• w = 2pf, e.g. US grid uses 60 cycles/sec, w = 2p(60) = 377 rad/s

-15

-10

-5

0

5

10

15

-20 -15 -10 -5 0 5 10 15 20

3

basic circuits with: )cos( to

4

resistors: VR ~ I

)cos()cos(

tIR

t

RI o

o

5

inductors: VL ~ dI/dt

)cos()cos(

tLL

t

dt

dI oo

)sin()sin(

)cos( tL

t

Ldtt

LI ooo

voltage “leads” current

6

capacitors: VC ~ Q

)cos( tCQ o

)sin(1

)sin()cos( tC

tCtCdt

dQ

dt

dI o

oo

current “leads” voltage

7

impedance Z = “ac R” ZI

LL

IL oo

Z :

RR

IR oo Z :

ωCZ

CIC oo

1

1 :

8

Example: 55mH Inductor, r = 0, connected to household 120VAC (60 hertz).

)377cos(19.8 tI

AL

I oo 19.8

)1055)(377(

1703

9

Example: 10F capacitor: connected to household 120VAC (60 hertz).

)377cos(0064.0 tI

AC

I oo 0064.0

)1010)(377(1

170

1 6

10

Example I(t)

= 0.577 Io

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Summary

• sine dependent I has I rms = 0.707 Io

• other rms values from direct calculation

• phase relations: R: phi = 0L: voltage on inductor leads I. C: I to capacitor leads voltage.

• impedance & resonance in RLC circuit

12

exponential notation

sincos iei

used to replace cosine or sine dependence

1

12

i

i

a

b

ebaiba i

1

22

tan

13

exp derivatives

xixdt

d

xixdt

d

exx tio

22

2

)(

)(

14

RLC exp application: tioec

dt

dxb

dt

xda

2

2

CcRbLaQx 1 , , ,

2 cbia

ex

tio

R

LX

CX

R

XX CL1-tan

b

ca

ecab

e

dt

dxi

tio

1-

222222tan

)(

From dx/dt = I, Z and phase are:

15

ac LR lab

• measure: voltages

• calculate: L & phase angle

16

Student Data (L ~ 1mH, f ~ 10,000Hz)

15ohm 60ohm 100ohm

V 6.7 6.3 6.5

V-ind 6.6 4.8 3.9

V-R 1.0 4.3 5.4

angle 79 50 36 ))((2cos

222

R

indR

VV

VVV

17

Trig Calculations

2

cos2

cos2coscosBABA

BA

)8cos(54.5

)8cos()8cos(6)4cos(3)cos(3

:

t

ttt

Ex

18

Phasor Calculation

)cos()cos( 21 tt

phase

22

221 )sin()cos(

19

Phasor Calculation

phase

22

221 )sin()cos(

)cos(

)sin(tan

21

21

phase

20

phasor )4cos(3)cos(3: ttEx

54.5121.2121.5

)45sin(3)45cos(33

22

22

5.22)45cos(33

)45sin(3tan 1phase

21

Exercise• Use trig identity & phasor method to show

that

• has amplitude 5.66 and phase 45°.)2cos(4)cos(4 tt

22

Resonance in an RLC Circuit• min. Z: when XL = XC

• result: large currents

• application: radio tuner

• hi power at tuned freq.

• low power at other f’s

• Ex. calc LC for f = 10,000

23

Transformer

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AC Power

RIRIRIP avgavgavgavg222 )()()(

2212 )( peakavg II

average

25

AC Power

RIRIP peakavgavg2

212 )(

peakpeakavgrms IIII 707.0)( 212

2212

peakrms II

RIRIP rmspeakavg22

21 )(

26

An I(t) current source continuously repeats the following pattern: {1 seconds @ 3 ampere, 1 second @ 0 ampere} Calculate average, rms I.

27

If a sinusoidal generator has a maximum voltage of 170V, what is the root-mean-square voltage of the generator?

28

R settingActual R

10 ohm 30 ohm 60 ohm 100 ohm

Vapp(V)

Vind(V)

VR(V)

Table 2: Calculated Data

cosf

f(degrees)

VL = Vsinf

Vr = Vcosf - VR

r = RVr/VR

L = RVL/(wVR)

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Alternating Current Generators

)sin()( tt peak

m = NBAcos.

30

Generators

m = NBAcos: ( = t + when rotating )

emf = -dm/dt = -NBA(-sin(t + ))

emf = NBAsin(t + )

(emf)peak = NBA.

)cos()sin()( 2 ttt peakpeak

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)cos()sin()( 2 ttt peakpeak

)cos(/)( tRtI peak

AC Generator applied to Resistor