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Chapter 2AC Circuits
1
2.1 - The sinusoidal function2.2 - The capacitor and the inductance2.3 - Phasors2.4 - Impedance and admitance2.5 - Mutual inductance2.6 - Instantaneous power in AC systems.
2.7 - Characteristic powers P, Q y S.2.8 - Power factor correction.
TEMA 2. AC CIRCUITS
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Chapter 2AC Circuits
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Lecture 3
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2.1. The sinusoidal function
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Chapter 2AC Circuits
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The sinusoidal functionSinusoidal waveform: )()( tsenXtx
M
Parameters:
-XM= amplitude o maximum value (peak value)
- T =period [s]
- = angular frequency [rad/s]:
-f= frequency [Hz]:
-t= phase angle [rad]
- = initial phase lag [rad] (usually written in )
Periodic function: )()( txTtx
fT
22
Tf
1
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Chapter 2AC Circuits
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Leading/lagging the phase
To compare a sinusoidal function withanother of the same frequency todetermine the phase difference should
be expressed:
- With positive amplitudes.- In sinus and cosinus functions
Considering 2 sinusoidal functions:
)()(
)()(
22
11
tsenXtx
tsenXtx
M
M
It is said thatx1(t) leadsx2(t) rads, oralso thatx2(t) lagsx1(t) rads.
If both functions are in phase
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Chapter 2AC Circuits
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2.2. The capacitor and the inductance
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Capacitor
A capacitor consists of two conductingsurfaces separated by a dielectric material andstores energy as electric charge.
The capacity Cis measured in
Farads [F]q
CV
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Chapter 2AC Circuits
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Voltage [V]:
Current [A]:
Power [W]:
Energy [J]:
Capacitor Equations
t
dxxiC
tv )(1)(
dt
tdvtCvtitvtp
)()()()()(
The capacitor stores energy as electric charge
21( ) ( ) ( )2
t
Cw t p t C v t
( )dv
i t Cdt
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The capacitor leads 90 the sinusoidal waveform of
the current with respect to the voltage waveform.In DC current the capacitor behaves as an opencircuit.
Current in a Capacitor
( ) sin ; ( )
( ) cos sin 2
dvv t V t i t C
dt
i t V C t V C t
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Combination of capacitors
C1
v1+ -
C2
v2+ -
Cn
vn+ -
i(t)
v(t)+ -
Other parts of the circuit
n
n
i iS CCCCC11111
211
n
n
i
iP CCCCC
211
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Chapter 2AC Circuits
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InductorsAn inductor is a coiled conductor (usually
around a ferromagnetic core) and has theability to store energy in the form ofmagnetic
flux.
The inductanceL is measured in
Henrys [H]
LI
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The inductor lags 90 the sinusoidal waveform of the
current with respect to the voltage waveform.In DC current the inductor behaves as a short circuit
Current in an Inductor
1( ) sin ; ( )
( ) cos sin 2
v t V t i t v dt L
V V
i t t t L L
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Combination of inductors
L1
v1+ -
L2
v2+ -
Ln
vn+ -
i(t)
v(t)+ -
Other parts of the circuit
n
n
i
iS LLLLL
211 n
n
i iP LLLLL11111
211
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Capacitor-Inductor dualityCapacitor Inductor
dttditLitp )()()(
dt
tdiLtv
)()(
2)(
2
1)( tiLtwL
Cvq
t
dxxiC
tv )(1
)(
dttdvtCvtp )()()(
2)(
2
1)( tvCtwC
dt
tdvCti
)()(
t
dxxvL
ti )(1
)(
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2.3. Phasors
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Phasors
Time domain (sinusoidal
functions)
Frequency domain
(phasors)
tAcos A A
tAsen 90A A
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Phasor for a resistor
Voltage and current in phase
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Phasors for an inductor
Voltage leads 90 the current
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Phasors for a capacitor
Current leads 90 the voltage
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Summary of phasors for R, L y C
Element RatioV/I
PhasorRatio
Phase
Capacitor dt
tdvCti
)()(
90
vVC
Cj
VIIleads 90
V
Inductordt
tdiLtv
)()(
90
iIL
Lj
IV
Vleads 90 I
Resistor )()( tRitv 0
IR
RIV
Vand I
are in phase
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Example
E7.6 The current in a 0.05 H inductor is equal toI = 4
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Ejemplo
E7.7 The current in a 150 F capacitor is I = 3.6
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2.4. Impedance and admitance
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Impedance It can be understood as the relationship
(magnitude and phase) between the voltage andcurrent phasor in an electrical circuit.
Ohm, [], is the magnitude of the impedance
The impedance value is, normally, a complex
number but not a phasor.
Ziv
M
M
iM
vM ZI
V
I
V
I
VZ
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Impedance In rectangular notation:
R= Real part (resistor):
X = Imaginary (reactance):
jXR Z
ZZR cos
ZZX sen
2 2
1
:
: tanZ
agnitude Z R X
XPhase
R
X > 0: Inductive reactance
X < 0: Capacitive reactance
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Impedance of the elements of a circuit
Element Impedance
CapacitorC
j
C
jX
CjC
90
11cZ
Inductor LjLjXLj L 90LZ
Resistor 0 RRRZ
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Series Impedance
ZS
Z1
Zn
Z2
n21iS ZZZZZ
n
i 1
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Paralel impedance
n21iP ZZZZZ
11111
1
n
i
ZPZ1 ZnZ2
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Admitance In the inverse value of the impedance, hence:
The admitance is measured in Siemens, [S].
As the impedance is a complex number (not a
phasor), the same can be said for the admitance.
Yvi
M
M
vM
iM YV
I
V
I
V
I
ZY
1
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Admitance In rectangular notation:
G = Real Part (conductance):
B = Imaginary Part (susceptance):
jBG Y
YYG cosG
BFase
BGYMdulo
Y
1
22
tan:
:
YYB sen
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Admitance
22 XR
XB
Important: The conductance is NOT the inverse value of the resistance:
G 1/R The susceptance is NOT the inverse value of the reactance:
B 1/X
For finding the inverse value multiply and divide for theconjugated value
22
XR
RG
jXRjBG
1
jXR
jXR
jXRjXR
111
Z
22
XR
jXR
Y
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Admitance of the elements
Element Admitance
Capacitor90
1
1 CCj
Cj
cY
InductorL
j
LLj
90
11LY
Resistor1
(only in a resistive case)GR
RY
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Series admitance
YS
Y1
Yn
Y2
n21iS YYYYY
11111
1
n
i
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Paralel admitance
YPY1 YnY2
n21iP YYYYY
n
i 1
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AC Ci it
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ExampleE7.11 Draw the diagram of the phasors that show all the
currents and voltages involved in this circuit:
Solucin:
AC Ci it
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ExampleSolved exercice 7.16 Find Vo in the following circuits,
using: nodes analysis, mesh analysis, the Thvenintheorem and the Norton theorem.
Solution:
Vo = 4
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2.5. Mutual inductance
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Mutual inductance The mutual inductance appears when 2 or more inductors
a so close that they share a common magnatic flux.
Considering two inductors:
dt
idL
dt
idMtv
dt
idM
dt
idLtv
22
12
2111
)(
)(
Inductancia mutua
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2212
2111
IIV
IIV
LjMj
MjLj
41
dt
idL
dt
idMtv
dt
idM
dt
idLtv
22
12
2111
)(
)(
i1(t)+
v1(t) L1
i2(t)+
v2(t)L2
M
Mutual impedance
Time Domain Frequency Domain
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Calculus method
Step 1. A dependent voltage source is connected in serieswith the inductor. Its value should be :jMIj
Step 2. The signs + and are assigned to the terminals ofthe dependent source considering the sense of the currentwith respect the point, that is showing the inductive
coupling at the inductorj
Step 3. Make the culculus considering the circuit found at
the second step.
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Step 2. Dependent voltage source
i1(t)
?
v1(t)
L1
i2(t)+
v2(t)
L2
jMI2 jMI1?
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Calculus method
Step 1. A dependent voltage source is connected in serieswith the inductor. Its value should be :jMIj
Step 2. The signs + and are assigned to the terminals ofthe dependent source considering the sense of the currentwith respect the point, that is showing the inductive
coupling at the inductorj
Step 3. Make the culculus considering the circuit found at
the second step.
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Step 2. Sense of the dependent voltagesource
Convention:
A current entering to the point is a positive current.
Hence, it is analyzed ifij is getting in or out of the inductorsLjpoint:
- Ifij gets in the point, a positive sign + should be assigned at thedependent sources (in series with the inductor i) corresponding
with the point of the inductori.- Ifij gets out the point, a positive sign - should be assigned at thedependent sources (in series with the inductor i) correspondingwith the point of the inductor i.
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Step 2. Sense of the dependent voltagesource
i1(t)
+
-
v1(t)L1
i2(t)
+
v2(t)L2
+
-jMI
2 jMI1
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Step 2. Examples
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Step 2. Examples
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Calculus method
Step 1. A dependent voltage source is connected in serieswith the inductor. Its value should be :jMIj
Step 2. The signs + and are assigned to the terminals ofthe dependent source considering the sense of the currentwith respect the point, that is showing the inductive
coupling at the inductorj
Step 3. Make the culculus considering the circuit found at
the second step.
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p
50
Step 3. Solve the circuit
The circuite resulting from step 2 (with the dependantsources) is solved by any of the methods presented in thecourse.
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p
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ExampleE8.2 Find the currents I1 e I2 and the voltage Vo in the
following circuit:
Solution:
I1 = 4.29
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Lecture 4
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2.6. AC Instantaneous power
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Instantaneous power
AC voltage and currents waveforms:v(t) = VMcos(t+v)
i(t) =IMcos(t+i)
The instantaenous power are defined as:
p(t) = v(t) i(t) = VMIMcos(t+v) cos(t+i)
p(t) = VMIM[cos(v-i) + cos(2t+v +i)]
Constant Double frequency sinusoidalterm
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Purely resistive circuit
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Purely inductive circuit
VI=LI2
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Inductive circuit (45)
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Average PowerThe average power results from integrating the
instantaneous power over one period and dividing thisvalue by the period:
ivMM
iMvM
Tt
t
Tt
t
-IV=
dt+tItVT
=dtp(t)T
=P
cos2
1
coscos110
0
0
0
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RMS valuesThe RMS value of the voltage or the current in AC is
a constant value (DC) that would provide the sameaverage power in a resistor R during a period T.
21 ( )o
o
t T
rms
t
I i t dtT
RMSmeans root-mean-square
2 2( )
( )
1 ( )o
o
t T
rms AC DC t
RI R i t dtT
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RMS valuesRMS values of a sinusoidal waveform:
2
( ) cos( )
1( ) ... 2
o
o
M i
t T
rms
t
i t I t
I i t dtT
Using the RMS values the expression for the average
power can be rewritten as:1
cos( ) ... cos( )2 M v i rms rms v i
P V I V I
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2.7. Characteristic powers P, Q y S
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Active Power P
Purely resistive circuit :
Purely reactive circuit : cos 0 ; 90rms rms v i v iP = V I - -
The active power, P, is the average value of the
instantaneous power and it is measured in [W]
2
cos ; 0rmsrms rms v i rms rms rms v iV
P = V I - V I RI -R
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Reactive Power Q
Purely resistive circuit :
Purely reactive circuit :
sin 0 ; 0rms rms v i v iQ = V I - -
The reactive power, Q, represents the maximum
energy storage speed of a reactive element, The activepower is measured in [var]
2
sin ; 90rmsrms rms v i rms rms rms v iV
Q = V I - V I XI -X
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Aparent Power S
Generic circuit :
rms rmsS = V I
The aparent power, S, represents the maximum
active power that could be consumed by animpedance Z=RjX or developed by an AC soure.The units are [VA]
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Complex PowerThe complex power is a mathematical tool that is
defined as the product of the voltage phasor by theconjugated value of the current phasor:
*
rms rmsS IV
*rms rms rms rms rms
rms rms rms rms
V I V I
V I cos V I sin
rms v i v i
v i v ij
P j Q
S V I
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Complex Power (S)
rms rms
rms rms
rms rms
V I [VA]
ngulo entre e[W]Re( ) V I cos
Im( ) V I sin [var]
Lv i Z
v i
v i
S
P
Q
S
S V I
S
S
cosrms rms v iP V I
sinrms rms v iQ V I
rms rmsS V I S
v i
tan v iQ
P
22 QPS
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2.8. Power factor correction
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Power factor
We defined:Activepower: [W]
Aparentpower : [VA]
Thepower factor is a relationship between P and S:
cos( )rms rms v iP V I
rms rmsS V I
pf ... cos( )v iP
S
The phase angle of the previous expresion is the sameangle of the load impedance
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PF correction Among the total power delivered by the electrical
companies only the active power is consumed, so that itis "interesting" having a unitary power factor at the loadPF = P/S = P/P = 1
But almost all loads have an inductive behaviour andtherefore placing in parallel a capacitor bank is necessaryin order to compensate the reactive power and obtain a pfclose to one.
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PF compensation
2cQC
V
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ExampleE9.13 Calculate the value of the capacitor bank needed to
change to 0.95 inductive power factor in the followingsystem:
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Steps for the correction of the pf
Step 1. Find QorigconsideringPL y origor fporig
Step 2. Fing Qnue considering the desired pfnue (known)
Step 3. Find the reactive power of the capacitor bank
Step 4. Determine the value of the capacitors at the bank
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Step 1. Find Qorig
Original phase between the voltage and the current:
arccos(fdp ) arccos(0,7) 45
Original reactive potencia :
tan(45 ) 102.020,4 var
orig orig
orig orig Q P
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Step 2. Find Qnue
nue
The active power keeps equal:
New phase between voltage and current:arccos(fdp ) arccos(0,95) 18,19
New reactive power:tan(18,19 ) 32.865,4 var
nue orig
nue
nue nue
P P
Q P
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Steps for the correction of the pf
Step 1. Find QorigconsideringPL y origor fporig
Step 2. Fing Qnue considering the desired pfnue (known)
Step 3. Find the reactive power of the capacitor bank
Step 4. Determine the value of the capacitors at the bank
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Paso 3. Find Qcap
Power at the capacitor bank:
62.152 varcap nue orig Q Q Q
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Paso 4. Find C
2
22
1 VCCj
V
Z
V
Q ccap
Capacity of the capacitor bank to be placed in paralel with theload:
2
cQ
C V
Hence:
42
6.2157,96 10 796
2 60480capQ F F
f 2
V = Voltage at thecapacitor bank
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ExampleE9.10 (modified)
An industrial load consumes 100 kW with a pf of 0.707inductive (lagging). The line voltage at the load is 480
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