inc 112 basic circuit analysis week 12 complex power complex frequency
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INC 112 Basic Circuit Analysis
Week 12
Complex Power
Complex Frequency
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
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
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
3.8530.83.8515.43.8515.4 totalI
)488.12sin(30.8)(
)3.852sin(30.8)(
tti
tti
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.
)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
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
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
V (volts)
t (sec)
311V
V peak (Vp) = 311 VV peak-to-peak (Vp-p) = 622VV rms = 220V
3 ways to tell voltage
0
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)
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
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.
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
V
IVR
VL47.1754.0 I
53.7288.4 LV
47.1708.1 RV
Phasor Diagram
Resistor consumes power WRIP rms 292.02
254.0 22
Inductor consumes no real power P = 0 but it has reactive power
VARVIQ rmsrms 318.12
88.454.0
θ = 77.47
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.
Definition: Power factor = cos θ
Power factor = 0.217
cosFactorPower rmsrmsIV
P
S
P
Complex Frequency
ste is a fundamental waveform of electrical engineering
What if s is a complex number?
tjetetjte
eeeeettt
tjttjttjst
sincos)sin(cos
)(
js
Let s be a complex number composed of real and imaginary parts.
AC
+
-
5cos(3t+π/3) 3H
2Ω
AC
+
-
5∟60 j9
2
)sin()cos()( tjte tj
PhasorFrequency domain
Time domain Euler’s Identity
sincos je j
Complex Frequency
tjte tj sincos AC
+
-
sinωt
tjetee tttj sincos)( AC
+
-
eσtsinωt
Define js
s is called “complex frequency”
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.
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
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
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
Complex Frequency Characteristics
te t sin
00
01
1
1