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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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2

Lecture 3


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2.1. The sinusoidal function


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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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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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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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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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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

Ch t 2AC Circuits


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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

Ch t 2AC Circuits


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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

Ch t 2AC Circuits


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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





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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Chapter 2

49

Calculus method





the second step.



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p

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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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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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