4-ac circuit analysis
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17 August 2005 Engineer M S Ayubi 1
ELECTRIC CIRCUITS
THEORY 1
These lecture slideshave beencompiled by
MohammedSalahUdDin Ayubi.
LECTURE 4 contd...AC Circuit Analysis
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8) AC Circuit AnalysisSinusoidal voltages and currents
8.1) Description of Sinusoidal Signal
3 parameters
Magnitude Vm
Phase
Frequency = 2 f
Ac circuit analysis is conducted at one frequency at a
time. Only two variables left, magnitude and phase.
If more than one frequency source is present. Use the
principle of superposition
)cos()( += tVtv m
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8.2 Advantages of Sinusoidal Signals
Sinusoidal functions yield further sinusoidalfunctions when integrated or differentiated
)cos()( += tVtv m
)sin()(
+= tV
dttv m
)sin()(
+= tVdt
tdvm
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Signal v(t) and integral (represented by iv(t)) and
differential (represented by dv(t))
0 0.5 1 1.5 2 2.5 3 3.5 460
48
36
24
12
0
12
24
36
48
6010V rms ac signal at 0.5 Hz
Time in seconds
Voltageinvolts
44.429
44.429
v t n( )dv t n( )
iv t n( )
40 t n
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8.3 Representation of Amplitude and Phase of a Signal on the
Complex Plane
+ j
+ real- real
- j
t
Vm
= 0
= 90 or /2
= -90 or - /2
= 180 or
imaginary
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+ j
+
real- real
- j
tV
m
= 0
= 90 or /2
= -90 or - /2
= 180 or
15 10 5 0 5 10 1515
13.5
12
10.5
9
7.5
64.5
3
1.5
010V rms ac signal at 0.5 Hz
voltage in volts
angularfrequencytimestim
einradians
0
12.566
tn
14.14214.142 v real t( ) n
0 5 10 1515
12
9
6
3
0
3
6
9
12
1510V rms ac signal at 0.5 Hz
angular frequency times time in radians
Voltageinvolts
14.142
14.142
v imag t n( )
12.5660 t( )n
)sin(
isaxis)(imaginaryjon thephasor
rot
atingtheofprojectionThe
tVv mimag =
)cos(
isaxisrealon thephasor
rotatingtheofprojectionThe
tVv mreal =
( ) ( )
caseparticularIn this
5.02cos102cos ttfVv m ==
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8.4 Relationship between rms and peak
and i and j Electrical engineers generally use a slightly different
nomenclature to mathematicians for complexnotation
Firstly we use j, rather than i, for the square root of 1
(i gets confused with current)
Secondly, we normally use rms values rather than peakvalues to describe the amplitude. For sine waves the peakis always 1.414 times the rms. (rms values rather than peakare used to describe voltages and currents)
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8.5 Representation of d/dt by j We know that a signal represented by the differentiation
of a sinusoidal voltage or current is the equivalent of theoriginal signal Multiplied by
Advanced in phase by 90 degrees or /2
Now multiplying by j gives a phase shift of +90 degrees
Thus the operation of d/dt on a sinusoidal signal is likemultiplying by j
(in the complex plane)
)cos()( += tVtv m
)2
cos()sin()(
++=+= tVtVdt
tdvmm
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8.6 Impedance of resistors, inductors and capacitors
Impedance is a term used to describe therelationship between sinusoidal voltage and
current for
a single passive component
a group of passive components
(passive components are R, L and C. They
are passive because they are linear and
have no gain)
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Impedance is given the symbol Z
Note v and i are complex quantities here. (Some books use bold type to
distinguish between complex and absolute values of sinusoidal
voltages and currents or between complex and dc values)
Zv
i
v = i Z
v = i R
iLjdtdiLv == i
Cji
Cjv
thus
vCjdt
dvCi
==
==
1
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Thus
C
jZ
LjZ
RZ
C
L
R
=
=
=
We have two components to the impedance.One has no j term and is referred to as real
The other is all j terms and is referred to as imaginary
The same is true for voltage and current.
All complex values can be expressed as a magnitude and phase angle
Or as a complex quantity.
Complex voltages and currents are called phasors
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8.7 Phasors, Reference Phasor and CIVIL
A phasor is a term given to a voltage or current which has a real and complex part. Represented as one of the following forms
( ) ( )
partimaginarytheofsignthebetoof
signtherequiresfunctioncostheUsingNote
coscosarccos
sincos
1
22
=
=
=
+=
=
+=
=
+=
m
real
m
real
m
real
imagrealm
m
mm
j
m
imagreal
v
v
v
va
v
v
vvv
wherevv
vjvv
evv
vjvv
+real
-real
- j
vm
= 0
= 90 or /2
= -90 or - /2
= 180 or
In circuits with several voltages and currents
We need to define we need to take
a current or voltage as a reference (i.e. let =0 for that variable)
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Taking the same voltage applied across a parallel combination of aresistor, a capacitor and an inductor
v ZR
iRZC
iCZL
iL
Plotting the voltage and three current phasors on the complex plane
produces a phasor diagram. Voltage is common to all so we take thatas the reference phasor in this case
+ j
+ real- real
- j
= 0
= 90 or /2
= -90 or - /2
= 180 or
C
jZ
LjZ
RZ
C
L
R
=
=
=
Z
vi =
v
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Taking the same voltage applied across a parallel combination of aresistor, a capacitor and an inductor
v ZR
iRZC
iCZL
iL
Plotting the voltage and three current phasors on the complex plane
produces a phasor diagram. Voltage is common to all so we take thatas the reference phasor in this case
+ j
+ real- real
- j
= 0
= 90 or /2
= -90 or - /2
= 180 or
C
jZ
LjZ
RZ
C
L
R
=
=
=
Z
vi =
viR
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Taking the same voltage applied across a parallel combination of aresistor, a capacitor and an inductor
v ZR
iRZC
iCZL
iL
Plotting the voltage and three current phasors on the complex plane
produces a phasor diagram. Voltage is common to all so we take thatas the reference phasor in this case
+ j
+ real- real
- j
= 0
= 90 or /2
= -90 or - /2
= 180 or
C
jZ
LjZ
RZ
C
L
R
=
=
=
Z
vi =
viR
iC
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Taking the same voltage applied across a parallel combination of aresistor, a capacitor and an inductor
v ZR
iRZC
iCZL
iL
Plotting the voltage and three current phasors on the complex plane
produces a phasor diagram. Voltage is common to all so we take thatas the reference phasor in this case
+ j
+ real- real
- j
= 0
= 90 or /2
= -90 or - /2
= 180 or
C
jZ
LjZ
RZ
C
L
R
=
=
=
Z
vi =
viR
iC
iL
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8.8 Phasor Addition and Complex Arithmetic to Find the
Combined Current for the Previous Circuit
Find i
v ZR
iRZC
iCZL
iL
i
+=
++=
++=
++=
LCj
Rvi
LjCjR
v
ZZZ
vi
iiii
LCR
LCR
11
1
1
11111
- real
j
By complex algebra and arithmetic
+ real
- j
v
iL
By phasor
addition
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8.8 Phasor Addition and Complex Arithmetic to Find the
Combined Current for the Previous Circuit
Find i
v ZR
iRZC
iCZL
iL
i
+=
++=
++=
++=
LCj
Rvi
LjCjR
v
ZZZ
vi
iiii
LCR
LCR
11
1
1
11111
j
By complex algebra and arithmetic
+ real- real
- j
v
iRiL
By phasor
addition
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8.8 Phasor Addition and Complex Arithmetic to Find the
Combined Current for the Previous Circuit
Find i
v ZR
iRZC
iCZL
iL
i
+=
++=
++=
++=
LCj
Rvi
LjCjR
v
ZZZ
vi
iiii
LCR
LCR
11
1
1
11111
j
By complex algebra and arithmetic
+ real- real
- j
v
iR
iC
iL
By phasor
addition
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8.8 Phasor Addition and Complex Arithmetic to Find the
Combined Current for the Previous Circuit
Find i
v ZR
iRZC
iCZL
iL
i
+=
++=
++=
++=
LCj
Rvi
LjCjR
v
ZZZ
vi
iiii
LCR
LCR
11
1
1
11111
+ real- real
- j
v
iR
iC
iL
j
i i
By complex algebra and arithmetic
By phasor
addition
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8.9 General Circuit Solution
We can solve ac networks with the same tools andmethods used for dc networks
Must use complex representation of voltages, currentsand impedances
Must choose one current or voltage as reference phasorand relate all others to it in terms of the phase angle
Evaluation of power needs care (rms values help)
Evaluation of stored energy needs care
8 10 I l d i
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8.10 Instantaneous power, real power and reactive power
0 1 2 3 42
1
0
1
22
2
v t n( )
i t n( )
P t n( )
40 t n
0 1 2 3 42
1
0
1
22
2
v t n( )
i t n( )
P t n( )
40 t n
0 1 2 3 42
0
2
43
2
v t n( )
i t n( )
P t n( )
40 t n
Instantaneous power waveforms for a voltage of 2V peak and a current of 1.5A peak
Flowing separately in a resistor, a capacitor and an inductor
Resistor case
Average power
Pav =0.5Vm*ImPav =vrms *irms
Inductor case
Pav = 0
Capacitor case Pav = 0
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Zvi
v = i Z Consider a series connected resistor and inductor
( ) ( ) ( )
( )( ) ( )
( ) ( ) ( )( )
( )
( )
sin
Q.symbol
given theisThis.componentsreactiveinstorageenergyforeresponsiblproductcurrent
voltagetheofcomponentthedefineshpower whicreactivetheispartimaginarytheAnd
cos
powertheaspartrealonly thetakeWe
sincos
Thus
referenceasvoltagethetakeweifand
cosandiwhere
1
0
0
2222
22
=
=
===
=
+=+==
+
=
+
=
+==
+=
ivQ
ivP
jiveiveeiviv
evv
LR
Ra
LR
veii
LR
LjR
LjRLjR
LjR
LjRv
Zvi
LjRZ
jjj
j
j
8 11 Solutions to Problems with Power Evaluation When
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8.11 Solutions to Problems with Power Evaluation When
Neither Voltage or Current Phasor is the Reference Phasor
( ) ( )
( ) ( )( )
( )
( )
( ) ( )
( )
0)(tryQforsigncorrectthegivesthissince
sinhavemustBut we
functioncostheofsignthechangenotdoesangletheofsignthesince
cosorcos
beshouldanswercorrectThe
sin
cos
powertheaspartrealonly thetakeWesincos
now
isphasorscurrentandvoltageebetween thangleThe
referenceasvoltagethenot takedoweifand
=
=
=
+=
+=
++=
===
=
=
=
+
v
vi
ivvi
iv
iv
iviv
jj
vi
j
j
ivQ
ivivP
ivQ
ivP
jiv
eiveeivivP
evv
eii
ivij
v
v
i
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17 August 2005 Engineer M S Ayubi 26
Solutions to Problems with Power Evaluation When
Neither Voltage or Current Phasor is the Reference Phasor
( ) ( )
( ) ( )( )
( )
( )vi
iv
iviv
jjj
ivQ
ivQ
ivP
jiv
eiveeiviv
iv
iviv
=
=
=
==
sin
ofpartimaginarytheispowerreactiveThus
onlyvoltagetheofconjugatethemust takewe
Q,powerreactivefor thesigncorrectwant theweIf
cos
powertheaspartrealonly thetakeWe
sincos
product.in thecurrenttheofconjugateor thevoltagethe
ofconjugatethetakecircuit wein thepowertheevaluatetoThus
*
*
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17 August 2005 Engineer M S Ayubi 27
8.12 Power Dissipated in Resistors With Complex
Currents and Voltages
In dc circuits the power dissipated in resistors was
R
vP
RiP
2
2
=
=
In ac circuits with complex phasor notation the
power dissipated in resistors is
R
vP
RiP
2
2
=
=
Rv i v = i R
8 13 P F i P S
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8.13 Power Factor in Power Systems The angle between the voltage and current ( in our previous
calculation) in a power network is called the power factor angle and isusually given the symbol .
The term cos( ) is called the power factor
Most power networks in factories and shops etc. have a lagging powerfactor. That is the load is inductive and resistive and is negative.
The charge for industrial electricity is based on the power consumed
and the maximum VA product (Maximum demand).
This is because
The cables and transformers required to supply a load are rated accordingto their maximum current (power lost in device conductors is dependenton current squared).
The bigger the rated VA the greater the cost of the equipment required tosupply it
Thus it is necessary to charge for maximum demand because otherwiseyou could have a high current demand with virtually no power and norevenue to cover your investment in expensive power supply equipment.
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