![Page 1: INC 111 Basic Circuit Analysis Week 12 Complex Power Complex Frequency](https://reader035.vdocuments.mx/reader035/viewer/2022062408/56649e5e5503460f94b57433/html5/thumbnails/1.jpg)
INC 111 Basic Circuit Analysis
Week 12
Complex Power
Complex Frequency
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Example
AC
+
- 50cos(2t)
i(t)
2.5H5Ω
3Ω
2H
AC
50sin(2t)
-
+
Find i(t)
AC
+
- 50∟90
i(t)
j55
3
j4
AC
-
+
50∟0
ω = 2
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AC
+
- 50∟90
i(t)
j55
3
j4
i1(t)Use superposition
2.2335.65.283.5
5)43(
5)43(5
j
jj
jjZ total
8.6687.72.2335.6
90501
totalZ
VI
i(t) can be found from current divider
3.8515.4)43(5
51
I
jj
jI
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i(t)
j55
3
j4
AC
-
+
50∟0
i2(t)2.6874.625.65.2
5)43(
5)43(5
j
j
jjZ total
2.6842.72.6874.6
0502
totalZ
VI
i(t) can be found from current divider
3.8515.4)43(5
52
I
jI
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3.8530.83.8515.43.8515.4 totalI
)488.12sin(30.8)(
)3.852sin(30.8)(
tti
tti
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Average Power in AC circuits
R
tvRtitp
22 )(
)()(
In AC circuits, voltage and current values oscillate.This makes the power (instantaneous power) oscillate as well.
However, electric power is best represented as one value.Therefore, we will use an average power.
Average power can be computed by integration ofinstantaneous power in a periodic signal divided by time.
From instantaneous power
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)sin()( tAtv Let v(t) in the form Change variable of integration to θ
sinAv We got Then find the instantaneous power
R
A
R
vp
222 sin integrate from 0 to 2π (1 period)
R
A
R
A
dA
dR
A
dR
AP
24
2sin
2
1
2
2
2cos1
2sin
2
sin2
1
22
0
2
2
0
22
0
22
2
0
22
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Compare with power from DC voltage source
DC
AC
+
-
Asin(ωt+Ф) R
i(t)
AC
+
-A R
i(t)
R
AP
2
2
R
AP
2
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Root Mean Square Value (RMS)
In DC circuits,R
VRIP
22
In AC, we define Vrms and Irms for convenience to calculate power.Vrms and Irms are defined such that,
R
VRIP rms
rms
22 Note: Vrms and Irms are constant all the time
For sine wave,2
_A
valuerms
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V (volts)
t (sec)
311V
V peak (Vp) = 311 VV peak-to-peak (Vp-p) = 622VV rms = 220V
3 ways to tell voltage
0
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Reactive Power
Capacitors and inductors have average power = 0 because they have voltage and current with 90 degree phase difference.
)sin()( tAtv )90sin()( tBti
Change variable of integration to θ
sinAv cos)90sin( BBi
cossinABvip
Then integrate from 0 to 2π (1 period)
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0)0(sin)2(sin
4
24
)(sinsin2
cossin2
1
22
2
0
2
0
2
0
sin
AB
AB
dAB
dABP
Capacitors and Inductors do not have average power although there are voltage and current.
Therefore, reactive power (Q) is defined
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Complex Power
Power can be divided into two parts: real and imaginary
Complex power S = P + jQ
P = real power Q = Reactive power
Inductor has no real power P =0But it has complex power, computed by V, I that are perpendicularto each other.
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sin2
sin
cos2
cos
2
pprmsrms
pprmsrms
pprmsrms
IVIVQ
IVIVP
IVIVS
jQPS
Complex power = S is a vector
2
2
prms
prms
II
VV
where
for sine waves
S
P
Q
θ
θ is the angle difference of (V-I)or (S-P)
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Phasor Diagram of an inductor
v
i
Power = (vi cosθ)/2 = 0
Phasor Diagram of a resistor
v
i
Power = (vi cosθ)/2 = vi/2
Note: No power consumed in inductorsi lags v
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Phasor Diagram of a capacitor
v
i
Power = (vi cosθ)/2 = 0
Phasor Diagram of a resistor
v
i
Power = (vi cosθ)/2 = vi/2
Note: No power consumed in capacitorsi leads v
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Power Conservation Theorem
“The sum of real power and reactive power in a circuit is equal to zero”
This is because power cannot disappear. It can only change form.
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Example
AC
+
-
5sin(3t+π/3)
i(t)
3H
2Ω
+ vR(t) -
+vL(t)
-
Find i(t), vL(t)
AC
+
-
5∟60
i(t)
j9
2
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V
IVR
VL47.1754.0 I
53.7288.4 LV
47.1708.1 RV
Phasor Diagram
1. Resistor consumes power WRIP rms 292.02
254.0 22
2. Inductor consumes no real power P = 0 but it has reactive power
VARVIQ rmsrms 318.12
88.454.0
θ = 77.47
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3. Voltage source supplies power
318.1))47.17(60sin(2
54.05sin
2sin
292.0))47.17(60cos(2
54.05cos
2cos
pprmsrms
pprmsrms
IVIVQ
IVIVP
0
318.1292.0
318.10
0292.0
SLR
S
L
R
PPP
jP
jP
jP
Conservation of energy holds
Resistor 2 ohm
Inductor 3H
Voltage source
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Power Factor
cosFactorPower rmsrmsIV
P
S
P
Power factor of a source is the ratio of real power to the complex power
S
P
Q
θ = 77.47
Complex Power Diagram
Consider the voltage source,
The voltage source supplies 0.292W real power and 1.318VAR reactive power.
Power factor = 0.217
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Complex Frequency
ste is a fundamental waveform of electrical engineering
where s can be a complex number in a rectangular form
tjetetjte
eeeeettt
tjttjttjst
sincos)sin(cos
)(
js
It is composed of real and imaginary parts.
The theory of phasor can be extended to supportste signal
Consider only real part
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tjte tj sincos AC
+
-
sinωt
tjetee tttj sincos)( AC
+
-
eσtsinωt
Define js
s is called “complex frequency”
If σ = 0
The voltage source will be in the format that we already know.
If σ is not 0, we get a new form of signal.
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Summary of Procedures
• Change voltage/current sources in to phasor form
• Change R, L, C value into phasor form
R L C R sL 1/sC
• Use DC circuit analysis techniques normally, but the value of voltage, current, and resistance can be complex numbers
• Change back to the time-domain form if the problem asks.
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Example
AC
+
-20e-2tcos(4t+π/2)
i(t) 1H
5Ω 0.1f
ic(t)
Find i(t), ic(t)
AC
+
-20∟90
i(t) s
5 1/0.1s42 j
js
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AC
+
-20∟90
i(t) s
5 1/0.1s
ic(t)
105
50105
105
50)
10||5(
2
s
ss
ss
ssZ total
We then substitute s = -2+j4 and got
1.1435.25.12105
50105 2
j
s
ssZ total
1.5381.1435.2
9020
totalZ
VI
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1.538 I
)9268.04cos(8)(
)1.534cos(8)(2
2
teti
tetit
t
ic(t) can be computed from current divider
535.26944.8
1.538565.26118.11.538)5.01(
1.538105
51.538
/105
5
j
s
s
sIC
)4631.04cos(944.8)(
)535.264cos(944.8)(2
2
teti
tetit
C
tC
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Complex Frequency Characteristics
te t sin
00
01
1
1