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ENE 104
Electric Circuit Theory
Lecture 10:
AC Power Circuit Analysis
http://webstaff.kmutt.ac.th/~dejwoot.kha/
(ENE) Mon, 19 Mar 2012 / (EIE) Wed, 28 Mar 2012
Week #11 : Dejwoot KHAWPARISUTH
Objectives : Ch11
Week #11
Page 2
ENE 104
• the instantaneous power
• the average power
• the rms value of a time-varying waveform
• complex power: average and reactive power
• the power factor, how to improve.
AC Circuit Power Analysis
Instantaneous Power:
Week #11
Page 3
ENE 104 AC Circuit Power Analysis
𝑝 𝑡 = 𝑣 𝑡 𝑖(𝑡)
the device is a resistor:
𝑝 𝑡 = 𝑖2 𝑡 𝑅 =𝑣2 𝑡
𝑅
the device is entirely inductive:
𝑝 𝑡 = 𝐿𝑖 𝑡𝑑𝑖 𝑡
𝑑𝑡=
1
𝐿𝑣(𝑡) 𝑣 𝑡′ 𝑑
𝑡
−∞𝑡′
the device is entirely capacitive:
𝑝 𝑡 = 𝐶𝑣 𝑡𝑑𝑣 𝑡
𝑑𝑡=
1
𝐶𝑖(𝑡) 𝑖 𝑡′ 𝑑
𝑡
−∞𝑡′
Instantaneous Power:
Week #11
Page 4
ENE 104 AC Circuit Power Analysis
Consider the series RL circuit,
𝑖 𝑡 =𝑉0𝑅
1 − 𝑒−𝑅𝑡 𝐿 𝑢(𝑡)
The total power delivered by the source
𝑝 𝑡 = 𝑣 𝑡 𝑖 𝑡 =𝑉0
2
𝑅1 − 𝑒−𝑅𝑡 𝐿 𝑢(𝑡)
Instantaneous Power:
Week #11
Page 5
ENE 104 AC Circuit Power Analysis
Consider the series RL circuit,
𝑖 𝑡 =𝑉0𝑅
1 − 𝑒−𝑅𝑡 𝐿 𝑢(𝑡)
The power delivered to the resistor is
𝑝𝑅 𝑡 = 𝑖2 𝑡 𝑅 =𝑉0
2
𝑅(1 − 𝑒−𝑅𝑡 𝐿 )2𝑢(𝑡)
Instantaneous Power:
Week #11
Page 6
ENE 104 AC Circuit Power Analysis
Consider the series RL circuit,
𝑖 𝑡 =𝑉0𝑅
1 − 𝑒−𝑅𝑡 𝐿 𝑢(𝑡)
to determine the power absorbed by the inductor, first
𝑣𝐿(𝑡) = 𝐿𝑑𝑖(𝑡)
𝑑𝑡
= 𝑉0𝑒−𝑅𝑡 𝐿 𝑢 𝑡 +
𝐿𝑉0
𝑅1 − 𝑒−𝑅𝑡 𝐿 𝑑𝑢(𝑡)
𝑑𝑡
= 𝑉0𝑒−𝑅𝑡 𝐿 𝑢 𝑡
Then,
𝑝𝐿(𝑡) = 𝑣𝐿 𝑡 𝑖(𝑡) =𝑉0
2
𝑅𝑒−𝑅𝑡 𝐿 (1 − 𝑒−𝑅𝑡 𝐿 )2𝑢(𝑡)
Instantaneous Power:
Week #11
Page 7
ENE 104 AC Circuit Power Analysis
Consider the series RL circuit,
𝑝 𝑡 = 𝑝𝑅 𝑡 + 𝑝𝐿(𝑡)
Instantaneous Power:
Week #11
Page 8
ENE 104 AC Circuit Power Analysis
Power Due to Sinusoidal Excitation:
the response is Where
)cos()( tIti m
222 LR
VI m
m
R
L 1tan
Practice: 11.1
Week #11
Page 9
ENE 104 AC Circuit Power Analysis
A current source of 12cos(2000t) A., a 200-ohm
resistor, and a 0.2-H inductor are in parallel.
Assume steady-state conditions exist. At t = 1ms.,
find the power being absorbed by the
(a) resistor
(b) inductor
(c) sinusoidal source
Practice: 11.1
Week #11
Page 10
ENE 104 AC Circuit Power Analysis
Average Power:
Week #11
Page 11
ENE 104 AC Circuit Power Analysis
For Periodic Waveforms:
2
1
)(1
12
t
t
dttptt
P
)()( Ttftf
Tt
t
dttpT
P1
1
)(1
1
Tt
t
x
x
x
dttpT
P )(1
Average Power:
Week #11
Page 12
ENE 104 AC Circuit Power Analysis
For Periodic Waveforms:
2
2
)(1
nT
nT
dttpnT
P
,...3,2,1)(1
ndttpnT
P
nTt
t
x
x
2
2
)(1
lim
nT
nT
dttpnT
Pn
Example:
Week #11
Page 13
ENE 104 AC Circuit Power Analysis
find the average power delivered to a resistor R
by the periodic sawtooth current waveform
TtTTtT
Iti m 2),()(
TttT
Iti m 0,)(
R
tvRtititvtp
)()()()()(
22
Example:
Week #11
Page 14
ENE 104 AC Circuit Power Analysis
TtTTtT
Iti m 2),()(
TttT
Iti m 0,)(
TtRtIT
tp m 0,1
)( 22
2
RIdttT
RI
TP m
T
m 2
0
2
2
2
3
11
TtTTtRIT
tp m 2,)(1
)( 22
2
Average Power:
Week #11
Page 15
ENE 104 AC Circuit Power Analysis
In the Sinusoidal Steady State
the instantaneous power is
Or
By inspection:
)cos()( tIti m
)cos()( tVtv m
)cos()cos()( ttIVtp mm
)cos(2
1 mmIVP
)2cos(2
1)cos(
2
1)( tIVIVtp mmmm
Example:
Week #11
Page 16
ENE 104 AC Circuit Power Analysis
Given the time-domain voltage V.,
find both the average power and an expression for the
instantaneous power that result when the corresponding
phasor voltage V = 4 0o V. is applied across an
impedance Z = 2 60o ohm.
Sol:
6/cos4 tv
Example:
Week #11
Page 17
ENE 104 AC Circuit Power Analysis
W
IVP mm
2)60cos()2)(4(2
1
)cos(2
1
V. V = 4 0o V.
Z = 2 60o ohm.
Sol:
The phasor current is V/Z = 2 -60o
So the average power is:
6/cos4 tv
606/cos2 ti
Example:
Week #11
Page 18
ENE 104 AC Circuit Power Analysis
)603
cos(42)606
cos()6
cos(8
)2cos(2
1)cos(
2
1
)cos()cos()(
ttt
tIVIV
ttIVtp
mmmm
mm
the instantaneous power is:
6/cos4 tv 606/cos2 ti
Example:
Week #11
Page 19
ENE 104 AC Circuit Power Analysis
)603
cos(42)( t
tp
6/cos4 tv 606/cos2 ti
Practice: 11.2
Week #11
Page 20
ENE 104 AC Circuit Power Analysis
Given the phasor voltage V = V. across
an impedance Z = 16.26 19.3o ohm., obtain an
expression for the instantaneous power, and compute
the average power if = 50 rad/s.
Sol:
452115
)cos()cos()( ttIVtp mm
)2cos(2
1)cos(
2
1)( tIVIVtp mmmm
Practice: 11.2
Week #11
Page 21
ENE 104 AC Circuit Power Analysis
Average Power:
Week #11
Page 22
ENE 104 AC Circuit Power Analysis
Absorbed by an Ideal Resistor:
Or
Absorbed by Purely Reactive Elements:
mmmmR IVIVP2
1)0cos(
2
1
R
VRIP m
mR22
1 22
0xP
Example:
Week #11
Page 23
ENE 104 AC Circuit Power Analysis
Find the average power being delivered to an
impedance ZL = 8-j11 ohm. by a current I = 5 20o A.
Sol:
.100852
1
2
1 22 WRIP mR
Practice: 11.3
Week #11
Page 24
ENE 104 AC Circuit Power Analysis
Calculate the average power delivered to the
impedance 6∠25°Ω by the current 𝐈 = 2 + 𝑗5 A.
Practice: 11.4
Week #11
Page 25
ENE 104 AC Circuit Power Analysis
Compute the average power delivered to each of the
passive elements.
Sol:
Practice: 11.4
Week #11
Page 26
ENE 104 AC Circuit Power Analysis
For the circuit in the figure below, compute the
average power delivered to each of the passive
elements. Verify your answer by computing the
power delivered by the two sources.
Practice: 11.4
Week #11
Page 27
ENE 104 AC Circuit Power Analysis
Practice: 11.4
Week #11
Page 28
ENE 104 AC Circuit Power Analysis
Maximum Power Transfer:
Week #11
Page 29
ENE 104 AC Circuit Power Analysis
An independent voltage source in series with an
impedance Zth delivers a maximum average power
to that load impedance ZL which is the conjugate of
Zth, or ZL=Z*th
Example 11.5:
Week #11
Page 30
ENE 104 AC Circuit Power Analysis
If we are assured that the voltage source is delivering
maximum average power to the unknown impedance,
what is its value?
Sol:
3500*
? jZZ th
Practice: 11.5
Week #11
Page 31
ENE 104 AC Circuit Power Analysis
If the 30-mH inductor of Example 11.5 is
replaced with a 10-𝜇𝐹 capacitor, what is the
value of the inductive component of the
unknown impedance 𝐙? if it is known that 𝐙? is
absorbing maximum power?
Practice: 11.5
Week #11
Page 32
ENE 104 AC Circuit Power Analysis
Average Power:
Week #11
Page 33
ENE 104 AC Circuit Power Analysis
For Nonperiodic Function: for example
the average power delivered to a 1 ohm resistor:
ttti sinsin)(
Watt
dtttttP
12
1
2
1
)sinsin2sin(sin1
lim2
2
22
Average Power:
Week #11
Page 34
ENE 104 AC Circuit Power Analysis
For
the average power delivered to a resistor R:
tItItIti NmNmm cos...coscos)( 2211
RIIIP mNmm
22
2
2
1 ...2
1
Practice: 11.6
Week #11
Page 35
ENE 104 AC Circuit Power Analysis
A voltage source 𝑣𝑠 is connected across a 4-Ω
resistor. Find the average power absorbed by
the resistor if 𝑣𝑠 equals (a) 8𝑠𝑖𝑛200𝑡 (b)
8𝑠𝑖𝑛200𝑡 − 6cos(200𝑡 − 45°) V. (c) 8𝑠𝑖𝑛200𝑡 −4𝑠𝑖𝑛100𝑡 V. (d) 8𝑠𝑖𝑛200𝑡 − 6 cos 200𝑡 − 45° −5𝑠𝑖𝑛100𝑡 + 4 V.
Practice: 11.6
Week #11
Page 36
ENE 104 AC Circuit Power Analysis
Effective Values:
Week #11
Page 37
ENE 104 AC Circuit Power Analysis
If the resistor received the
same average power in part
(a) and (b), then the effective
value of i(t) is equal to Ieff,
and the effective value of v(t)
is equal to Veff
Effective Values:
Week #11
Page 38
ENE 104 AC Circuit Power Analysis
of a Periodic Waveform:
the average power delivered
to the resistor,
the power delivered by the
direct current is:
RIP eff
2
TT
dtiT
Rdti
TP
0
2
0
21
Effective Values:
Week #11
Page 39
ENE 104 AC Circuit Power Analysis
of a Periodic Waveform:
we get
note: the effective value is
often called the root-mean-
square value, or simply the
rms value.
RIdtiT
Reff
T2
0
2
T
eff dtiT
I0
21
Effective Values:
Week #11
Page 40
ENE 104 AC Circuit Power Analysis
of a sinusoidal Waveform:
which has a period
to obtain the effective value
)cos()( tIti m
T
eff dtiT
I0
21
2T
2
0 21
21
0
22
)]22cos([2
)(cos1
dttI
dttIT
I
m
T
meff
Effective Values:
Week #11
Page 41
ENE 104 AC Circuit Power Analysis
of a sinusoidal Waveform
)cos()( tIti m
T
eff dtiT
I0
21
2
][4
)]22cos([2
)(cos1
2
2
0
0 21
21
0
22
meff
m
m
T
meff
II
tI
dttI
dttIT
I
Effective Values:
Week #11
Page 42
ENE 104 AC Circuit Power Analysis
Use of RMS Values to
Compute Average Power
R
VRIRIP
eff
effm
2
22
2
1
2
meff
II
)cos()cos(2
1 effeffmm IVIVP
Effective Values:
Week #11
Page 43
ENE 104 AC Circuit Power Analysis
with Multiple-Frequency
Circuits
RIIIP Neffeffeff
22
2
2
1 ...
2
meff
II
22
2
2
1 ... Neffeffeffeff IIII
Effective Values: Example
Week #11
Page 44
ENE 104 AC Circuit Power Analysis
Ex22 Page 384: find the effective value of:
T
eff dtiT
I0
21
Practice: 11.7
Week #11
Page 45
ENE 104 AC Circuit Power Analysis
Calculate the effective value of each of the
periodic voltages: (a) 6𝑐𝑜𝑠25𝑡; (b) 6𝑐𝑜𝑠25𝑡 +4sin(25𝑡 + 30°) V. (c) 6𝑐𝑜𝑠25𝑡 + 5𝑐𝑜𝑠2(25𝑡) V.
(d) 6𝑐𝑜𝑠25𝑡 + 5𝑠𝑖𝑛30𝑡 + 4 V.
Practice: 11.7
Week #11
Page 46
ENE 104 AC Circuit Power Analysis
Practice: 11.7
Week #11
Page 47
ENE 104 AC Circuit Power Analysis
Practice: 11.7
Week #11
Page 48
ENE 104 AC Circuit Power Analysis
Apparent power and Power factor:
Week #11
Page 49
ENE 104 AC Circuit Power Analysis
The sinusoidal voltage
is applied to the network and the resultant
sinusoidal current is
the average power delivered to the network
the apparent power:
the power factor:
)cos( tVv m
)cos( tIi m
effeff IV
)cos()cos(2
1 effeffmm IVIVP
effeff IV
PPF
Example:
Week #11
Page 50
ENE 104 AC Circuit Power Analysis
Calculate values for the average power delivered
to each of the two loads, the apparent power
supplied by the source, and the power factor of
the combined load.
Example:
Week #11
Page 51
ENE 104 AC Circuit Power Analysis
Calculate values
for the average
power delivered
to each of the two
loads, the
apparent power
supplied by the
source, and the
power factor of
the combined
load.
)cos( angIangVIVP effeff
effeff IV
effeff IV
PPF
Example:
Week #11
Page 52
ENE 104 AC Circuit Power Analysis
.432
)]1.53(0cos[)12()60(
)cos(
W
angIangVIVP effeffS
rmsAjj
IS .13.5312)51()12(
060
Example:
Week #11
Page 53
ENE 104 AC Circuit Power Analysis
.720)12()60( VAIV effeff
the apparent power
supplied by the source:
the power factor of the
combined load.
6.0)12()60(
432
effeff IV
PPF
2)12( 2 P
1)12( 2 P
Practice: 11.8
Week #11
Page 54
ENE 104 AC Circuit Power Analysis
For the circuit of the figure below, determine the
power factor of the combined loads if 𝑍𝐿 = 10Ω.
Practice: 11.8
Week #11
Page 55
ENE 104 AC Circuit Power Analysis
Complex Power:
Week #11
Page 56
ENE 104 AC Circuit Power Analysis
the average power, express as,
the complex power,
)cos( effeff IVP
*)( ReReRe effeff
j
eff
j
eff
j
effeff IVeIeVeIVP
)sin( effeff IVQ
jQPeIVIVS j
effeffeffeff )(*
Power Measurement:
Week #11
Page 57
ENE 104 AC Circuit Power Analysis
The complex power
delivered to several
interconnected loads is
the sum of the complex
power delivered to
each of the individual
loads.
S = VI* = V(I1+I2)
* = V(I1*+I2
*) = VI1*+VI2
*
Example:
Week #11
Page 58
ENE 104 AC Circuit Power Analysis
An industrial consumer
is operating a 50-kW
induction motor at a
lagging PF of 0.8. The
source voltage is 230
Vrms. In order to obtain
lower electrical rates,
the customer wishes to
raise the PF to 0.95
lagging. Specify a
suitable solution.
Example:
Week #11
Page 59
ENE 104 AC Circuit Power Analysis
A purely reactive load must be added to the
system, in parallel, since the supply voltage must
not change. The complex power supplied to the motor must
have a real part of 50-kW and an angle of
cos-1(0.8)=36.9o
Hence, S1
kVAj 5.3750
8.0
9.3650
Example:
Week #11
Page 60
ENE 104 AC Circuit Power Analysis
S1 =
To achieve a PF of 0.95, the total complex power
must become:
S
Thus the corrective load is
S2 =
kVAj 5.37508.0
9.3650
kVAj 07.21
kVAj 43.1650
)95.0(cos95.0
50 1
Example:
Week #11
Page 61
ENE 104 AC Circuit Power Analysis
S2 =
We select a phase angle of 0o for the voltage
source, and the current drawn by Z2 is
Or
Therefore,
.6.91230
210702*
2 Ajj
V
SI
kVAj 07.21
.6.912 AjI
51.26.91
230
2
2 jjI
VZ
Practice: 11.9
Week #11
Page 62
ENE 104 AC Circuit Power Analysis
Find the complex power
absorbed by the
(a) 1-ohm R
(b) -j10 C
(c) 5+j10 Z
(d) source
Practice: 11.9
Week #11
Page 63
ENE 104 AC Circuit Power Analysis
Practice: 11.9
Week #11
Page 64
ENE 104 AC Circuit Power Analysis
Practice: 11.9
Week #11
Page 65
ENE 104 AC Circuit Power Analysis
Practice: 11.10
Week #11
Page 66
ENE 104 AC Circuit Power Analysis
A 440-Vrms source supplies power to a load
𝑍𝐿 = 10 + 𝑗2Ω through a transmission line
having a total resistance of 1.5Ω. Find (a) the
average and apparent power supplied to the
load; (b) the average and apparent power lost in
the transmission line; (c) the average and
apparent power supplied by the source; (d) the
power factor at which the source operates.
Practice: 11.10
Week #11
Page 67
ENE 104 AC Circuit Power Analysis
Example: Final 2/47
Week #11
Page 68
ENE 104 AC Circuit Power Analysis
จากวงจรตามรูป ใหห้า 1) the average power ท่ีจ่ายโดย แหล่งจ่าย (source) 2) ค่า power factor ท่ีแหล่งจ่าย (source) 3) ขนาดของ ตวัเกบ็ประจุ (capacitor) ท่ีเม่ือน าไปต่อ
ขนานกบั แหล่งจ่าย ท าใหค่้า power factor = 1 4 ohm
j16 ohm120 Vrms 60Hz
12 ohm
Example: Final 2/47
Week #11
Page 69
ENE 104 AC Circuit Power Analysis
Example: Final 2/47
Week #11
Page 70
ENE 104 AC Circuit Power Analysis
4 ohm
j16 ohm120 Vrms 60Hz
12 ohm
Example: Final 2/47
Week #11
Page 71
ENE 104 AC Circuit Power Analysis
Ex: Page 72
ENE 104
1 µF
7.5 ΩVg
50 µH
5 Ω
Find the average power, the reactive power,
and the apparent power supplied by the voltage source in the circuit if vg = 50 cos 105t V.
Ex: Page 73
ENE 104
1 µF
7.5 ΩVg
50 µH
5 Ω
Summary:
Week #11
Page 74
ENE 104 AC Circuit Power Analysis
Hw:
Week #11
Page 75
ENE 104 AC Circuit Power Analysis
Reference: Page 76
ENE 104
W.H. Hayt, Jr., J.E. Kemmerly, S.M. Durbin, Engineering Circuit Analysis, Sixth Edition.
Copyright ©2002 McGraw-Hill. All rights reserved.
Circuit Analysis and Electrical Engineering
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