1 homework for ch.29 alternating current circuits 19, 23, 25, 31, 39
TRANSCRIPT
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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
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basic circuits with: )cos( to
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resistors: VR ~ I
)cos()cos(
tIR
t
RI o
o
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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
19
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
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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