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Homework for Ch.29 Alternating Current Circuits

• 19, 23, 25, 31, 39

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29 Overview

• why & how to use rms values

• determine impedance of L & C

• why & how: phase relationships in ac circuits

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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

4

basic circuits with: )cos( to

5

resistors: VR ~ I

)cos()cos(

tIR

t

RI o

o

6

inductors: VL ~ dI/dt

)cos()cos(

tLL

t

dt

dI oo

)sin()sin(

)cos( tL

t

Ldtt

LI ooo

voltage “leads” current

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capacitors: VC ~ Q

)cos( tCQ o

)sin(1

)sin()cos( tC

tCtCdt

dQ

dt

dI o

oo

current “leads” voltage

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impedance Z = “ac R” ZI

LL

IL oo

Z :

RR

IR oo Z :

ωCZ

CIC oo

1

1 :

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

)377cos(19.8 tI

AL

I oo 19.8

)1055)(377(

1703

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

)377cos(0064.0 tI

AC

I oo 0064.0

)1010)(377(1

170

1 6

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exponential notation

sincos iei

used to replace cosine or sine dependence

1

12

i

i

a

b

ebaiba i

1

22

tan

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exp derivatives

xixdt

d

xixdt

d

exx tio

22

2

)(

)(

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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:

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ac LR lab

• measure: voltages

• calculate: L & phase angle

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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

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Trig Calculations

2

cos2

cos2coscosBABA

BA

)8cos(54.5

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

:

t

ttt

Ex

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Phasor Calculation

)cos()cos( 21 tt

phase

22

221 )sin()cos(

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Phasor Calculation

phase

22

221 )sin()cos(

)cos(

)sin(tan

21

21

phase

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

54.5121.2121.5

)45sin(3)45cos(33

22

22

5.22)45cos(33

)45sin(3tan 1phase

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Exercise• Use trig identity & phasor method to show

that

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

21

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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

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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

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Transformer

25

AC Power

RIRIRIP avgavgavgavg222 )()()(

2212 )( peakavg II

average

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AC Power

RIRIP peakavgavg2

212 )(

peakpeakavgrms IIII 707.0)( 212

2212

peakrms II

RIRIP rmspeakavg22

21 )(

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Example I(t)

= 0.577 Io

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

29

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

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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.

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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