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1
Homework for Ch.29 Alternating Current Circuits
• 19, 23, 25, 31, 39
2
29 Overview
• why & how to use rms values
• determine impedance of L & C
• why & how: phase relationships in ac circuits
3
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
7
capacitors: VC ~ Q
)cos( tCQ o
)sin(1
)sin()cos( tC
tCtCdt
dQ
dt
dI o
oo
current “leads” voltage
8
impedance Z = “ac R” ZI
LL
IL oo
Z :
RR
IR oo Z :
ωCZ
CIC oo
1
1 :
9
Example: 55mH Inductor, r = 0, connected to household 120VAC (60 hertz).
)377cos(19.8 tI
AL
I oo 19.8
)1055)(377(
1703
10
Example: 10F capacitor: connected to household 120VAC (60 hertz).
)377cos(0064.0 tI
AC
I oo 0064.0
)1010)(377(1
170
1 6
11
exponential notation
sincos iei
used to replace cosine or sine dependence
1
12
i
i
a
b
ebaiba i
1
22
tan
12
exp derivatives
xixdt
d
xixdt
d
exx tio
22
2
)(
)(
13
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:
14
ac LR lab
• measure: voltages
• calculate: L & phase angle
15
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
16
Trig Calculations
2
cos2
cos2coscosBABA
BA
)8cos(54.5
)8cos()8cos(6)4cos(3)cos(3
:
t
ttt
Ex
17
Phasor Calculation
)cos()cos( 21 tt
phase
22
221 )sin()cos(
18
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
20
Exercise• Use trig identity & phasor method to show
that
• has amplitude 5.66 and phase 45°.)2cos(4)cos(4 tt
21
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
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
24
Transformer
25
AC Power
RIRIRIP avgavgavgavg222 )()()(
2212 )( peakavg II
average
26
AC Power
RIRIP peakavgavg2
212 )(
peakpeakavgrms IIII 707.0)( 212
2212
peakrms II
RIRIP rmspeakavg22
21 )(
27
Example I(t)
= 0.577 Io
28
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?
30
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)
31
Alternating Current Generators
)sin()( tt peak
m = NBAcos.
32
Generators
m = NBAcos: ( = t + when rotating )
emf = -dm/dt = -NBA(-sin(t + ))
emf = NBAsin(t + )
(emf)peak = NBA.
)cos()sin()( 2 ttt peakpeak
33
)cos()sin()( 2 ttt peakpeak
)cos(/)( tRtI peak
AC Generator applied to Resistor