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7/29/2019 Revision of Phasors
http://slidepdf.com/reader/full/revision-of-phasors 1/22
Revision of Phasors
• A phasor is a complex number
• A phasor is a way of writing down a sinusoidal signal
a
jb X
+ - x of complex numbers
A complex number can be
In Cartesian form: a + jb
In Polar form: X
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a
jb X
22 ba X
a
btan 1
θ
θcos X a θ sin X b
a+ jb X
To + and - complex numbers
Complex numbers should be in Cartesian form
Add/subtract real and imaginary parts separately
Eg. What is 3 - j7 + 431 ?
+ 431 = 3.43 + j2
3 - j7 + 3.43 + j2 = 6.43 – j5
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To x and of complex numbers
If in Cartesian form:
Multiply ad bc jdbac jd c jba Divide:
22
d c
ad bc jbd ac
jd c jd c
jd c jba
jd c
jba
2222
d c
ad bc j
d c
bd ac
jd c
jba
If in polar form:
1θ X 2θY 21 θθ XY x =
212
1θθ
θ
θ
Y
X
Y
X Divide:
Multiply
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Representation of a sinusoidal signal
In a linear circuit, all signals at the same frequency = 2 f
But signals will have different magnitude and have different
phase displacements to each other
Eg. The following are two signals in a circuit
10B
80
A
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10B
80
A
If we say that A goes through zero at t =0
Then A is
And B is
t sinω10 010
80ω4 t sin 804
The "magnitude" becomes the magnitude of the
complex number
The phase displacement becomes the angle of the
complex number
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10B
80
A
Actually, the magnitude of the phasor (complex number)is, by convention, the RMS value of the signal
(peak/root 2).
The phasor for A is thus
The phasor for B is thus or
(the angle can be either degrees or radians)
41832 .. or
0077.
80832.
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A
B
10
The phasors can be put on an Argand diagram:
0077.
80832.
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10
A B
60
Question 1
Go from "waveform on a scope" to "phasor" to "argand
diagram". E.g. you see B on a scope
Write down its phasor and put it on the Argand diagram
A is 0077. B is 12010
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Adding sinusoidal currents or voltages etc.
You can now add sinusoidal signals together. You must add
them as phasors (complex numbers). NEVER JUST ADDTHEIR MAGNITUDES.
Consider two AC currents and , both of peak 10A, flowing
in the circuit below:
Ammeter 1
Ammeter 2
Ammeter 3
?
2 I ~
1 I ~
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101 I ̂ Ammeter 1
Ammeter 2
Ammeter 3
?
102 I ˆ
Q3 If ALL ammeters are set to "AC", what do
Ammeters 1 and 2 read?
Answer:
What does Ammeter 3 read?
Answer:
Q2. If ALL the ammeters are set to "DC", what will they
read? Answer: All 0A
Both 7.07A
Don’t know. Not told phase of sinusoids
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Question 4
Let and be as in the example above: 1 I ~
2 I ~
10
80
00771 . I ~ 8083.2~2 I
What will Ammeter 3 read? Answer:
I 1 = 7.07 + j0
I 2= 2.83{cos(-80)+jsin(-80)} = 0.49 - j2.78
I 1+ I 2 = 7.56-j2.78 = 8.05
-20. It will read 8.05A
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2 I ~
1 I ~
10
Question 5
Let1 I
~2 I
~and be as follows:
What will Ammeter 3 read? Answer : 0 A !!
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R, C and L in AC Circuits
In steady state AC circuits, R, L and C all have an
IMPEDANCE.
This complex for inductors and capacitors:
Resistor (purely real)Inductor (purely imaginary)
Capacitor (purely imaginary)
You will normally be given the values of R , L, C and = 2 f , where f = 50Hz (60Hz in the US).
The quantity =2f is called the angular frequency of the
AC in radians per second.
R X R L j X L ω
C j / X C ω1 C / j ω
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Why should L go to jL in AC circuits?
Consider Faraday's Law for a coil:dt
t di L Rt it v
)()()(
Let the current be t sin I ˆ )t ( i ω
dt
t sin I ̂ d Lt sin I ̂ R )t ( v
ωω
t cos I ̂ Lt sin I ̂ R )t ( v ωωω
90ωωω t sin I ̂ Lt sin I ̂ R )t ( v
j I ~
L j I ~
RV ~ 0ω01
I ~
L j I ~
RV ~
ω I ~
X I ~
RV ~
L
change to phasors:
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Worked Example R1
A 50Hz voltage of 220V rms is applied to
a series combination of R = 10 and L = 38mH.
(a) What is the rms value of the current flowing?
(b) What is the rms voltage across R and L?
(a) Impedance of R = 10
Impedance of L =
Combined series Impedance
v
R
L
= 10 + j12 L jX R Z jXL
R
Z
Impedance Phasors:
2 x 50 x 0.038 = 12
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I ~
X RV ~
L 1210
0220
j X R
V ~
I ~
L
2.501.142.500
6.15
220
2.506.15
0220~ I
v
R
L
iNote that a reactive impedance (with
+j component), causes the current
to lag the voltage (i.e. current has a
-j component)
A useful mnemonic is C I V I L:
In a Capacitive circuit I leads V ;
whereas V leads I in an inductive circuit (L)
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LV L
RV (b) What is the rms voltage across R and L?
I ~ L j I ~ RV ~V ~V ~ L R ω
2.501412.501.1410~~
I RV R
Voltage and Current Phasors:
V
-j
I
IR
-50.2°
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LV L
RV
2.501699012.501.1412~~
j I L jV L
8.39169
~ LV
-j
V
Voltage and Current
Phasors:
jIXLI
IR
Note that although
This is because VR and VL are sinusoids and NOT in phase
L R V ~V ~V ~ , 220 141 + 169 !!
Note: 141
2
+ 169
2
= 220
2
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Ammeter 1
Ammeter 2
Ammeter 3
?
Worked example R2
1 I ~
2 I ~
t sin )t ( i ω31 02
31 I ~
t cos )t ( i ω42 902
4
2
I
~
Let
Let
t cost sin )t ( i )t ( i ω4ω321
θω
21
t sin X )t ( i )t ( iResult will be sinusoid. Must find
magnitude and phase
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t cost sin )t ( i )t ( i ω4ω321
θω
21
t sin X )t ( i )t ( it cos sin X cost sin X )t ( i )t ( i ωθθω21
3θ cos X 4θ sin X Hence: and
222222 43θθ sin X cos X 5X 252 , X
3
4
θ
θ cos X
sin X
3
4θ
1 tan
Not a good way of adding sinusoids!!
53ω521 t sin )t ( i )t ( i A / I ~
253
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Much better to use Phasors:
02
3
1
I ~
902
4
2 I ~
3
4
2
5 13
tan I ~
And the answer is obtained almost immediately
02
31 I ~
902
42 I
~
Have and
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Revision of Phasors
• A phasor is a complex number
• A phasor is a way of writing down a sinusoidal signal
• Animations on WebCT
a
jb X
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