Lecture in AC Ckts

ELECTRIC CIRCUITS 2 AC Circuits

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Prepared by: Joana Joy N. Abonalla and Pal Maleter Domingo


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BASIC CONCEPTS GENERATION OF AC VOLTAGE AC can be developed in a coil of wire in one of the three ways,

1. Changing the flux through the coil 2. By moving the coil through a magnetic field so that the flux cutting results 3. By altering the direction of the flux with respect to the coil.

In the first of these, the voltage is said to be an induced emf and in accordance with Faraday’s law. Its

magnitude at any instant of time is given by equation,

e=N𝑑𝜙

𝑑𝑡 x 10-8 Volts

Where: N = number of turns in the coil

𝑑𝜙

𝑑𝑡= rate at which the flux, in maxwells, changes through the coil.

Note: By this method of developing an emf, there is no physical motion of coil or magnet; the

current through the exciting coil that is responsible for the magnetism is altered to change the flux through the coil in which the voltage is induced.

By the second or third, There is actual physical motion of the coil of magnet, and in altered position of

coil or magnet flux through the coil changes.

A voltage developed in either of these ways is said to be generated emf and is given by the equation,

e=Blv x 10-8 Volts

where: B = flux density in lines per sq.in l = length of wire moved relatively in the flux v = velocity of the wire, in/sec, with respect to the flux


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How AC is generated


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Sine Wave or Sinusoidal Wave CYCLE

A complete change in value and direction of an alternating quantity. A cycle of alternating voltage and current completes 360 electrical degrees. There are two alternations per cycle.

FREQUENCY (f)

The number of cycles per second expressed in hertz (Hz)

f = PN

120

where: f = frequency or hertz or cps

P = no. of poles N = Speed, rpm

PERIOD (T) The time it takes to complete one cycle.

T = 1

f

WAVELENGTH (λ)

The length of one complete cycle or wave or the distance traveled by the wave in one cycle.

λ = velocity,𝑣

frequency, f


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Note: 1. For electromagnetic waves,

𝒗𝐯𝐚𝐜𝐮𝐮𝐦 = 𝟏𝟖𝟔,𝟎𝟎𝟎 𝒎𝒊

𝐬𝐞𝐜 𝒐𝒓 𝟑 𝒙 𝟏𝟎𝟏𝟎

𝒄𝒎

𝐬𝐞𝐜

2. For sound waves,

𝒗𝐚𝐢𝐫 = 𝟏,𝟏𝟑𝟎 𝐟𝐭

𝐬𝐞𝐜

INSTANTANEOUS VALUE The value of alternating quantity at any instant.

MAXIMUM VALUE The value attained by an alternating quantity during positive or negative half cycle. This called peak value or amplitude of alternating quantity.

AVERAGE VALUE The average of all the instantaneous value of half (either positive or negative) cycle of alternating

quantity.

For sine waves,

Vave =2Vm

π= 0.636 Vm

Iave =2Im

π= 0.636 Im

For other waves,

Vave =Area

Base=

1

T 𝑣 t dt

T

0

Iave =Area

Base=

1

T 𝑖 t dt

T

0

where: Base = Period (T) Area = dA = 𝑣 t dt


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From the above figure, v(t) = Vm sin ωt volts where: v(t) = instantaneous value of voltage, volts Vm = maximum value of voltage, volts ω = angular frequency (2πf), rad/sec t = time, seconds ωt = radian Similarly, if the above voltage wave is a current wave, then the equation is

i(t) = Im sin ωt Amperes

where: i(t) = instantaneous value of current, amperes Im = maximum value of current, amperes Note: ωt whose unit is radian can be replaced by θ expressed in degrees.

Points in the figure:

1. v(t) = Vm sin (ωt+α1) + slope (zero point)

2. v(t) = Vm cos (ωt+α2) + slope (max. point)

3. v(t) = -Vm sin (ωt+α3) - slope (zero point)

4. v(t) = -Vm cos (ωt+α3) - slope (max. point)

EFFECTIVE VALUE The value of the alternating quantity which when applied to a given circuit for a given time produces

the same expenditure of energy as when dc is applied to the same circuit for the same interval of time.

Also called “root-mean-square” (rms) value. For sine waves,

Vrms or V =Vm

2= 0.707 Vm

Irms or I =Im

2= 0.707 Im


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For other waves,

Vrms2 =

1

T 𝑣2 t dt

T

0

Irms2 =

1

T 𝑖2 t dt

T

0

Form Factor = Effective value

Average value

Note: For sine wave, FF = 1.11

Crest or Peak Factor = Maximum value

Effective value

Note: For sine wave, Crest Factor = 1.414

One of the most important excitation is the sinusoidal forcing function. In EE, sinusoidal functions are extremely important for numbers of reasons. The carrier signals generated for communication purposes are sinusoids, and of course, it is the dominant signal in the electric power industry. Almost every useful signal in EE can be resolved into sinusoidal components. The more general sinusoidal expression is given by,

v(t) = Vm sin (ωt+α)

where: Φ = α + β And can be expressed as; v1 leads v2 by Φ v2 lags v1 by Φ


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Thus far we have considered sine functions rather than cosine functions in defining sinusoids. It does not matter which form we use since,

A + B = 90 complementary angles In trigonometry,

sin A = cos (90 – A) = cos B so;

sin ωt = cos ωt−𝜋

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If we will rewrite v1; Example: v1(t) = Vm sin (ωt+α) 1. sin 30 = cos 60 but; 0.5 = 0.5 A = Then; 2. v1= 4 sin (2t+30) v1(t) = Vm sin A v2= 6 sin (2t-12)

= Vm cos 𝜋

2– (ωt + α) so;

v1(t) = -Vm cos (𝛚𝐭+ 𝛂)–𝝅

𝟐 30- (-12) = 42°

PHASE DIFFERENCE From the above figure: where: v(t) = Vm sin ωt α1= leading or ahead v1(t) = Vm sin (ωt+α1) α2= lagging or behind v2(t) = Vm sin (ωt+α2) α = phase angle or angle of displacement


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REPRESENTATION OF PHASORS Phasors are rotating vectors. It has magnitude and direction. To specify a vector quantity, a magnitude and direction must be given. These two quantities can be merged into a single expression to facilitate the use of this vector in equations. A possible notation is:

A = A ∠θ

Where: |A| = magnitude θ = phase angle

= measured in counterclockwise direction with respect to the positive x – axis. It is used to specify direction.

This vector is shown, Represents, A = AX + jAY rectangular form of a vector Note: Two quantities cannot be combined to form a single number.

j = −1 j = operator which is irreducible that would keep j2 = -1 the two quantities separate from each other. j3 = -j j4 = 1 Relationship, Polar to Rectangular Form AX = |A| cosθ = the real component

Ay = |A| sinθ = the quadrature component

A

AX

AY


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Z1

Z2

Z = Z1 +

Z2

Z1

-Z2 Z = Z1 -

Z2

Z2

Rectangular to Polar Form

|A| = AX2 + AY

2 θ = tan−1 AY

AX

From the Mclaurin’s series expansion of sine, cosine and exponential functions that:

ejθ = cosθ+ j sinθ = 1∠θ

where:

j = −1 Note:

ej = ∠ and the Phasor A ∠θ is read as “A cisθ”, where cis stands for cosine,imaginary,sine. i = j in vector analysis

OPERATIONS OF PHASORS ADDITION AND SUBTRACTION OF PHASORS Given: Z1 = X1 + j Y1

Z2 = X2 + j Y2

Thus; Za = Z1 + Z2 = (X1+ X2) + j (Y1+ Y2) ZS = Z1 - Z2 = (X1- X2) + j (Y1- Y2) MULTIPLICATION OF PHASORS

1. Rectangular Form: Two phases expressed in rectangular forms can be multiplied using the distribution rule. Given: Z1 = X1 + j Y1

Z2 = X2 + j Y2

Then; Z1 Z2 = (X1 + j Y1)( X2 + j Y2)

= X1X2 + j X1Y2 + j X2Y1 + j2 Y1 Y2


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Hence;

Z1 Z2 = (X1X2 - Y1Y2) + j (X1Y2 + X2Y1)

2. Polar Form: Given: Z1 = Z1 ∠θ1

Z2 = Z2 ∠θ2 Then; Z1Z2 = Z1 ∠θ1 Z2 ∠θ2 = Z1 cosθ1 + j Z1 sinθ1 Z2 cosθ2 + j Z2 sinθ2 = Z1 Z2 cosθ1cosθ2 − sinθ1sinθ2 + j cosθ1sinθ2 + sinθ1cosθ2 = Z1 Z2 cos θ1 + θ2 − j sin θ1 + θ2 Thus; 𝐙𝟏𝐙𝟐 = 𝐙𝟏 𝐙𝟐 ∠ 𝛉𝟏 + 𝛉𝟐


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DIVISION OF PHASORS

1. Rectangular Form: Given: Thus;

Z1 = X1 + j Y1 𝒁𝟏

𝒁𝟐=

𝐗𝟏𝐗𝟐 + 𝐘𝟏𝐘𝟐 + 𝐣 (𝐗𝟐𝐘𝟏−𝐗𝟏𝐘𝟐)

𝐗𝟐 𝟐 + 𝐘𝟐 𝟐

Z2 = X2 + j Y2

Then;

𝑍1

𝑍2=

X1 + j Y1

X2 + j Y2

X2− j Y2

X2− j Y2

2. Polar Form: Given: Z1 = Z1 ∠θ1

Z2 = Z2 ∠θ2 Then;

𝑍1

𝑍2=

Z1 ∠θ1

Z2 ∠θ2

= Z1 cosθ1+j sinθ1

Z2 cosθ2+jsinθ2

cosθ2− j sinθ2

cosθ2− j sinθ2

= Z1

Z2

cosθ1cosθ2+sinθ1sinθ2− j cosθ1sinθ2+ j cosθ2sinθ1

cos2θ2 + sin2θ2

𝑍1

𝑍2= Z1

Z2 cos θ1 − θ2 + j sin θ1 − θ2

Thus;

𝒁𝟏𝒁𝟐

= 𝐙𝟏

𝐙𝟐 ∠ 𝛉𝟏 − 𝛉𝟐


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THE FUNCTIONAL TRANSFORMATION Ac voltage and current circuits, expressed as functions of time, must undergo a functional transformation in order to convert them into Phasors.

FUNCTIONAL TRANSFORMATION converts the original function into a new function. The two functions are related to each other in a definite manner. In addition, the original

function should be obtainable from the new function by an inverse transformation process.

Why do we perform functional transformation? The good reason is to avoid the use of differential equations. With the use of Phasors, we can write and solve circuit equation using phasor algebra. In addition, circuit elements represented by phasors can be dealt with in the same manner as circuit in a dc circuit.

In general, ac voltage v is transformed in the manner. 𝑣 t = Vm sin (ωt + θ) to V = Vrms ∠θ It also requires a conversion from peak value to rms values. This enable us to use circuit equations developed for DC circuits with little or no modification. Phasor Solutions are applicable only for single frequency or sinusoidal forms only. It can only give the steady state condition. The phasor angles are defined at the time, the “snap-shot” is taken. Sinusoidal waveforms will maintain the same relative phase positions even as time advances. SUM OF TWO VOLTAGES EXPRESSED AS FUNCTIONS IN TIME Given: v1(t) = Vm1 sin (ωt+β) v2(t) = Vm2 sin (ωt+α) Then;

V = v1(t) + v2(t) = Vm1 sin (ωt+β) + Vm2 sin (ωt+α) = (Vm1 sinωt cosβ + Vm1 cosωt sinβ ) + (Vm2 sinωt cosα + Vm2 cosωt sinα )

= sinωt (Vm1cosβ + Vm2 cosα) + cosωt (Vm1sinβ + Vm2 sinα)


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Let: A = Vm1cosβ + Vm2 cosα B = Vm1sinβ + Vm2 sinα

C = A2 + B2 From which, v1+ v2 = A sinωt + B cosωt The sum of sine and cosine wave.

SUM OF THE SINE WAVE AND COSINE WAVE

The sum of sine and cosine wave of the same frequency is another sinusoid of that frequency.

A cosωt + B sinωt = A2 + B2 A

A2 + B2 cosωt +

B

A2 + B2 sinωt

Consider;

A cosωt + B sinωt = A2 + B2 cosωt cosθ + sinωt sinθ

A cosωt + B sinωt = A2 + B2 cos ωt− θ Where by:

θ = tan−1 B

A


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v = Ri

+

-

i

R v = RI

+

-

I

R

Resistive elementcarrying sinusoidalcurrent or timedomain.

Frequency domainequivalent circuit ofa resistor

v(t)

t

v

i

COMPLETE RESPONSE It is the sum of a natural and forced response

1. The Natural Response

It is obtained from the dead ciruit and therefore is independent of the sources or excitations.

2. The Forced Response (if) It depends directly on the type of excitation applied to the circuit.

*In case of dc source, the forced response is a dc steady state response and so on.*

FREQUENCY DOMAIN ANALYSIS First we must establish the relationship between the current and phasor voltage at terminals of the passive circuit elements. Relationship for RESISTOR

i = Im cos (ωt + θi ) v = Vm cos (ωt+ θv) Waveform Diagram: since, v = Ri Then,

Vm ej(ωt+ θv) = R Im ej(ωt+ θi)

By dividing both sides by ej ωt ,

Vm ej θv = R Im ej θi V = RI A plot shows that the voltage is and

current at the terminals are “IN PHASE”.


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PURE RESISTANCE

where: V = effective or RMS voltage I = effective or RMS current R = effective or ac resistance For sinusoidal voltage supply,

V =Vm

2= 0.707 Vm

I =Im

2= 0.707 Im

∴ Vm = Im R


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v(t)

t

v

i

90°

v =XLi

+

-

i

L

+

-

I

Inductive elementcarrying sinusoidalcurrent or timedomain.

Frequency domainequivalent circuit ofa inductor.

V =XLI XL

Relationship for INDUCTOR

i = Im cos (ωt + θi )

v L= L di

dt Waveform Diagram:

since,

v L= L di

dt

Then,

di

dt =

d Im cos ωt + θi

dt

di

dt = −ωIm sin ωt + θi

Now; 𝑉 = −ωLIm sin ωt + θi

𝑉 = −ωLIm cos ωt + θi − 90 Voltage leads the Current by 90° (ELI)

The phasor representation; or Current lags Voltage by 90°

𝑉 = −ωLIm ej( θi−90)

= −ωLIm ejθie−j90 We can also express the phase shift in

= jωLIm ejθi seconds. A phase shift of 90° corresponds to ¼

𝑽 = 𝐣𝛚𝐋𝐈 of period; hence the voltage leads the current by T/4

or 1/4f seconds.

Note: e−j90 = cos 90− j sin90 = −j But;

V = XLI

= ωL∠90 Im∠θi

𝐕 = 𝛚𝐋𝐈𝐦 ∠𝟗𝟎+ 𝛉𝐢

Or

𝐕 = 𝛚𝐋𝐈𝐦𝐜𝐨𝐬 (𝛚𝐭+ 𝛉𝐢 + 𝟗𝟎)


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PURE INDUCTANCE

where: f= frequency ; L = inductance in Henry XL = 2πfL = ωL Ω = inductive reactance For sinusoidal voltage supply,

V =Vm

2 ; I =

Im

2

∴ Vm = Im XL


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v =XCi

+

-

i

C

+

-

I

Capacitive elementcarrying sinusoidalcurrent or timedomain.

Frequency domainequivalent circuit ofa capacitor.

V =XCI XC

v(t)

t

i

v

90°

Relationship for CAPACITOR

iL= C dv

dt

v= Vm cos (ωt + θv ) Phasor Diagram:

then,

V= 1

jωC I

Equation demonstrates that the equivalent Circuit for the capacitor in the phasor domain is,

Current leads Voltage by 90° (ICE)

or Voltage lags Current by 90°

Note: 𝑋𝑐 =1

jωC

The voltage across the terminals of a capacitor behind the current by exactly 90°. But; V = XCI

= 1∠−90

ωC Im∠θi

𝐕 =𝐈𝐦𝛚𝐂

∠𝛉𝐢 − 𝟗𝟎

Or

𝐕 =𝐈𝐦𝛚𝐂

𝐜𝐨𝐬 (𝛚𝐭+ 𝛉𝐢 − 𝟗𝟎)


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PURE CAPACITANCE

where:

XC = 1

2πfC=

1

𝜔𝐶 = capacitive reactance in ohms

C = capacitance in farad (f) For sinusoidal voltage supply,

V =Vm

2 ; I =

Im

2

∴ Vm = Im XC


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OWERS AND POWER TRIANGLE

1. Real or True or Active or Average Power

P = VI cosθ watts

P =Vm Im

2 cosθ watts

𝐏 = 𝐈𝟐𝐑 𝐰𝐚𝐭𝐭𝐬

2. Reactive Power

Q = VI sinθ volt− amp− reactive (vars)

Q =Vm Im

2 sinθ vars

3. Apparent Power

S = VI va

S =Vm Im

2 va

where: θ = power factor (pf) angle or phase angle pf = power factor = cos θ = P/S f= reactive factor = sin θ = Q/S tan θ = Q/P


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ENERGY EXPENDED:

1. 𝐖 = 𝐏𝐭 𝐢𝐧 𝐣𝐨𝐮𝐥𝐞𝐬 𝐨𝐫 𝐰𝐚𝐭𝐭 − 𝐬𝐞𝐜 2. ∴ For pure L, since θ = 90°

P = VI cos 90° = 0 watts Q = VI sin 90° vars = VI

∴ S = Q

𝐖𝐋 =𝐈𝟐𝐋

𝟐 𝐢𝐧 𝐣𝐨𝐮𝐥𝐞𝐬 𝐨𝐫 𝐰𝐚𝐭𝐭 − 𝐬𝐞𝐜

3. ∴ For pure C, since θ = 90°

P = VI cos 90° = 0 watts Q = VI sin 90° vars = VI

∴ S = Q

𝐖𝐂 =𝐕𝟐𝐂

𝟐 𝐢𝐧 𝐣𝐨𝐮𝐥𝐞𝐬 𝐨𝐫 𝐰𝐚𝐭𝐭 − 𝐬𝐞𝐜

IMPEDANCE, ADMITTANCE, CONDUCTANCE AND SUSCEPTANCE

Impedance is defined as the opposition of circuit to flow of alternatine current.

It is denoted by Z and its unit is ohms.

Admittance is the reciprocal of impedance.

It is the overall ability of an electric circuit to pass alternating current

The letter symbol for admittance is Y

where Y= 1/Z = G ± jB in siemens or mhos.

Conductance is the measure of the ability of resistance to pass electric current.

Reciprocal of resistance.

The letter symbol for conductance is G. where G = 1/R.

The unit of conductance is the siemens or mho.

Similarly, the reciprocal of reactance is susceptance.

It is the ability of inductance or capacitance to pass alternating current.

The letter symbol for susceptance is B.

where B = 1/X.


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R and L in Series

∴ V = VR2 + VL

2

V = I R2 + XL2

𝐕 = 𝐈𝐙 𝐯𝐨𝐥𝐭𝐬 where:

Z = impedance in ohms For R-L series circuit, it can be observed from the

phasor diagram that the current lags behind the

applied voltage by an angle Φ. From the voltage

triangle, we can write,

tan θ = VL

VR =

XL

R, and

pf = cos θ = VR

V =

R

Z

rf = sin θ = VL

V =

XL

Z

If all the sides of the voltage are divided by current, we get Impedance Triangle where:

Z = impedance =V

I

R = resistance = VR

I

XL = inductive reactance = VL

I

From this impedance triangle, we can see that the x-component of impedance Z is R, and it is given by

𝐑 = 𝐙 𝐜𝐨𝐬 𝛉 and y-component of impedance is XL and is given by

𝐗𝐋 = 𝐙 𝐬𝐢𝐧 𝛉 In rectangular form, the impedance is denoted as,

𝐙 = 𝐑+ 𝐣𝐗𝐋 while in polar form, it is denoted as,


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𝐙 = 𝐙 ∠ 𝛉 where:

𝐙 = R2 + XL2

𝛉 = tan−1 XL

R

Solving for powers of R-L Circuits:

P = VI pf = I 2R =VR

2

R watts

Q = VI rf = I 2XL =VL

2

XL vars

S = VI = I 2Z =V2

Z va

R and C in Series

∴ V = VR2 + VC

2

V = I R2 + XC2

For R-C series circuit, it can be observed from the

phasor diagram that the current lags behind the

applied voltage by an angle Φ. From the voltage

triangle, we can write,

tan θ = VC

VR =

XC

R, and

pf = cos θ = VR

V =

R

Z

rf = sin θ = VC

V =

XC

Z

If all the sides of the voltage are divided by current, we get Impedance Triangle where:

Z = impedance =V

I

R = resistance = VR

I

XC = capacitive reactance = VC

I

From this impedance triangle, we can see that the x-component of impedance Z is R, and it is given by


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𝐑 = 𝐙 𝐜𝐨𝐬 𝛉 and y-component of impedance is Xc and is given by

𝐗𝐂 = 𝐙 𝐬𝐢𝐧 𝛉 In rectangular form, the impedance is denoted as,

𝐙 = 𝐑− 𝐣𝐗𝐂 while in polar form, it is denoted as,

𝐙 = 𝐙 ∠ 𝛉

where:

𝐙 = R2 + XC2; 𝛉 = tan−1

XL

R

Solving for powers of R-L Circuits:


2

R watts

Q = VI rf = I 2XC =VC

2

XC vars

S = VI = I 2Z =V2

Z va

R LC in Series

V = VR2 + VL − VC

2

V = I R2 + XL − XC 2

𝐕 = 𝐈𝐙 𝐯𝐨𝐥𝐭𝐬

where: Z = impedance in ohms

Z = R2 + XL − XC 2

pf = cos θ = VR

V =

R

Z

rf = sin θ = VL−VC

V =

XL−XC

Z

XC = 1

2πfC=

1

𝜔𝐶 Ω tan θ =

VL−VC

VR =

XL−XC

R

XL = 2πfL = ωL Ω

Note:

Transpose XL and XC if XC >XL


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2

R watts

Q = VI rf = I 2 XL − XC =VL

2

XL−

VC2

XC vars

S = VI = I 2Z =V2

Z va

Also,

S = P2 + Q2 RESONANCE IN SERIES RLC CIRCUIT We know that both XL and XC are the functions of frequency f. When f is varied both XL and XC also get varied. At a particular frequency, XL becomes equal to XC. Such a condition when XL = XC for a certain frequency is called series resonance. At resonance, the reactive part in the impedance of RLC series circuit is zero. The frequency at which the resonance occurs is called resonant frequency denoted as ωr, rad/sec or fr, Hz. CHARACTERISTICS OF SERIES RESONANCE

In a series resonance, the voltage applied is constant and frequency is variable. Hence, following parameters o series RLC circuit get affected due to change in frequency:

1. XL; As XL=2πfL. As frequency is changed from 0 to ∞, XL increases linearly and graph of XL against f is straight line passing through origin.

2. XC; As XC=1/2πfC. As frequency is changed from 0 to ∞, XC reduces and graph of XC against f is rectangular hyperbola. Mathematically, sign of XC is opposite to XL hence graph of XL versus f is shown in the first quadrant while XC versus f is shown in the third quadrant.

3. Total Reactance 4. Impedance 5. Current, I 6. power factor, cos θ

At f = fr the value of XL = XC at this frequency. As X = XL - XC, the graph of X against f is shown in the figure below. For f < fr the value of XC >XL and net reactance is capacitive while for f > fr the value of XC < XL and net reactance is inductive.


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Note:

𝑍 = 𝑅 + 𝑗𝑋 = 𝑅 + 𝑗 𝑋𝐿 − 𝑋𝐶 but at f = fr the value of XL = XC and X = 0. Hence, the net impedance Z = R which is purely resistive. So the impedance is minimum and current is maximum at series resonance.

Now power factor cos θ = R/Z and at f=fr as Z= R, the power factor is unity and at its maximum at series resonance. For f < fr it is leading in nature while for f > fr it is lagging in nature. EXPRESSION FOR RESONANT FREQUENCY Let fr be the resonant frequency in Hz at which,

XL = XC …. Series resonance

∴ 2πfrL =1

2πfrC

∴ fr 2 =

1

4π2LC

∴ fr =1

2π LC 𝐻𝑧

∴ ωr =1

LC 𝑟𝑎𝑑/𝑠𝑒𝑐


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BANDWIDTH OF SERIES R-L-C CIRCUIT At series resonance, I is maximum and Z is minimum. Now, power consumed in a circuit is proportional to the square of the current as P=I2R. So at series resonance as current is maximum, power is also at its maximum, Pm. The figure below shows the graph of current and power against frequency.

It can be observed that at two frequencies f1 and f2 the power is half of its maximum value. These frequencies are called half power frequencies.

The difference between the half power frequencies f1 and f2 at which power is half of its maximum is called bandwidth of series RLC circuit.

∴ 𝑩𝑾 = 𝒇𝟐 − 𝒇𝟏 In the bandwidth, the power is more than half the maximum value. The bandwidth decides selectivity. The selectivity is defined as the ratio of the resonant frequency to the bandwidth.

∴ 𝑺𝒆𝒍𝒆𝒄𝒕𝒊𝒗𝒊𝒕𝒚 = 𝒇𝒓𝑩𝑾

= 𝒇𝟐 − 𝒇𝟏𝑩𝑾

Thus, if the bandwidth is more, the selectivity of the circuit is less. Out of the two half power frequencies, the frequency f2 is called upper cut-off frequency while frequency f1 is called lower cut-off frequency.


29


EXPRESSION FOR LOWER AND UPPER CUT-OFF FREQUENCIES The current in a series RLC circuit is given by the equation,

𝑰 = 𝑉

𝑍 but Z = R + j XL − XC

𝑰 = 𝑉

R2 + XL − XC

2

= 𝑉

R2 + ωL −1

ωC

2

At resonance, 𝑰𝒎 = 𝑉

𝑅 (maximum value)

and 𝑃𝑚 = 𝐼𝑚2𝑅

At half power point,

𝑷 =𝑷𝒎𝟐

=𝑰𝒎

𝟐

𝟐𝑹 =

𝑰𝒎

𝟐 𝟐

𝑹

∴ 𝐼 = 𝐼𝑚

2 at half power frequency

Equating;

𝑰 = 𝑉

R2 + ωL −1

ωC

2

𝑉

2R=

𝑉

R2 + ωL −1

ωC

2

2R = R2 + ωL−1

ωC

2

2R2 = R2 + ωL−1

ωC

2


30


R2 = ωL−1

ωC

2

∴ ± 𝐑 = 𝛚𝐋 −𝟏

𝛚𝐂

From the equation above, we can find two values of half power frequencies which are ω1 and ω2 corresponding to f1 and f2.

∴ + 𝐑 = 𝛚𝟐𝐋 −𝟏

𝛚𝟐𝐂

and

∴ −𝐑 = 𝛚𝟏𝐋 −𝟏

𝛚𝟏𝐂

Adding equations;

ω1L−1

ω1C+ ω2L−

1

ω2C = −R + + R

ω1 +ω2 L− 1

ω1+

1

ω2

1

C= 0

ω1 +ω2 L = 1

ω1+

1

ω2

1

C

ω1 +ω2 L = ω1 +ω2

ω1ω2

1

C

𝛚𝟏𝛚𝟐 =𝟏

𝐋𝐂

But: 𝛚𝐫 =𝟏

𝐋𝐂

Then; ω1ω2 = ωr

2

f1f2 = fr 2


31


The equation shows that the resonant frequency is the geometric mean of the two half power frequencies.

𝐟𝐫 = 𝐟𝟏𝐟𝟐 QUALITY FACTOR The quality factor of the R-L-C series circuit is the voltage magnification in the circuit at resonance.

𝑽𝒐𝒍𝒕𝒂𝒈𝒆 𝒎𝒂𝒈𝒏𝒊𝒇𝒊𝒄𝒂𝒕𝒊𝒐𝒏 = 𝑽𝒐𝒍𝒕𝒂𝒈𝒆 𝒂𝒄𝒓𝒐𝒔𝒔 𝑳 𝒐𝒓 𝑪

𝑺𝒖𝒑𝒑𝒍𝒚 𝒗𝒐𝒍𝒕𝒂𝒈𝒆

Now, 𝐕𝐋 = 𝐈𝐦𝐗𝐋 = 𝐈𝐦𝛚𝐫𝐋 at resonance

And at resonance, 𝐼𝑚 = 𝑉

𝑅 and VL =

𝑉ωr L

𝑅

Therefore;

𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝑚𝑎𝑔𝑛𝑖𝑓𝑖𝑐𝑎𝑡𝑖𝑜𝑛 =

𝑉ωrL𝑅𝑉

=ωrL

𝑅

This is nothing but Quality Factor, Q.

𝑄 =ωrL

𝑅 but ωr =

1

LC

𝑄 =1

𝑅

L

𝐶 while BW =

R

2πL

𝑄 ∙ 𝐵𝑊 =1

𝑅

L

𝐶 ∙

R

2πL=

1

2π LC= fr

Hence; 𝑸 =𝐟𝐫

𝐁𝐖


32


SIGNIFICANCE OF QUALITY FACTOR CAN BE STATED AS,

1. It indicates the selectivity or sharpness of the tuning of a series circuit. 2. It gives the correct indication of the selectivity of such series RLC circuit which is used in many radio

circuits. Note: At the resonant frequency, the impedance is minimum and hence the circuit is known as acceptor

circuit at resonance. AC PARALLEL CIRCUITS A parallel circuit is one which has two or more impedances connected in parallel across the supply voltage. Each impedance may be a separate series circuit. Each impedance is called branch of the parallel circuit.

The circuit shown is a parallel circuit consisting of three impedances connected in parallel across an ac supply of V volts.

Note: The voltage across each impedance is the same as the voltage supply. Thus,

𝑽𝑻 = 𝑽𝟏 = 𝑽𝟐 = 𝑽𝟑 = 𝑽 The current taken by each impedance is different.

𝐼𝑇 = 𝐼1 + 𝐼2 + 𝐼3

𝑉𝑇𝑍𝑇

=𝑉1

𝑍1+𝑉2

𝑍2+𝑉3

𝑍3


33


𝑉

𝑍𝑇=𝑉

𝑍1+𝑉

𝑍2+𝑉

𝑍3

Therefore;

𝟏

𝒁𝑻=𝟏

𝒁𝟏+𝟏

𝒁𝟐+𝟏

𝒁𝟑

where ZT is called equivalent impedance. This result is applicable for “n” such impedance connected in parallel. CURRENT DIVIDER PRINCIPLE Just like in DC circuit, the process of solving branch currents is the same. The only difference is that we’re no longer dealing with scalar quantity “R” but we’re considering an impedance “Z” which is a vector quantity having magnitude and direction. If there are two impedance connected in parallel and if IT is the total current, then current division rule can be applied to find the individual branch currents.

𝑽𝟏 = 𝑽𝑻 𝐼1𝑍1 = 𝐼𝑇𝑍𝑇

𝐼1 = 𝐼𝑇 𝑍𝑇

𝑍1

𝐼1 = 𝐼𝑇

𝑍1𝑍2𝑍1+𝑍2

𝑍1

Hence;

𝑰𝟏 = 𝑰𝑻 𝒁𝟐

𝒁𝟏+𝒁𝟐 and 𝑰𝟐 = 𝑰𝑻

𝒁𝟏

𝒁𝟏+𝒁𝟐

Z1 Z2

I1 I2IT


34


CONCEPT OF ADMITTANCE Just as the conductance, the reciprocal of resistance, prove to be useful quantity in the analysis of resistive circuits, so does the reciprocal of the impedance offer some convenience in the sinusoidal steady state analysis of a general RLC circuit.

Admittance is defined as the reciprocal of the impedance. It is denoted as Y and is measured in unit siemens or mhos. Now, the current equation of the three impedance comnnected in parallel is,

𝐼𝑇 = 𝐼1 + 𝐼2 + 𝐼3

𝑉𝑇 1

𝑍𝑇 = 𝑉1

1

𝑍1 + 𝑉2

1

𝑍2 + 𝑉3

1

𝑍3

𝑉𝑇𝑌𝑇 = 𝑉1𝑌1 + 𝑉2𝑌2 + 𝑉3𝑌3

𝑉𝑌𝑇 = 𝑉𝑌1 + 𝑉𝑌2 + 𝑉𝑌3 Hence,

𝒀𝑻 = 𝒀𝟏 + 𝒀𝟐 + 𝒀𝟑 where Y is the admittance of the total circuit. The three impedances connected in parallel can be replaced by an equivalent circuit, where three admittance are connected in series,

Admittance is also defined as the ratio of phasor current to phasor voltage. Thus,

𝒀 = 𝑰

𝑽=𝟏

𝒁

and thus, 𝒀 = 𝑮+ 𝒋𝑩


35


where G is the real part of admittance called conductance and B is the imaginary part of the called susceptance.

The equations should be crutinized carefully. To obtain the relation of Y and Z, we may rationalize it. making,

𝐺 + 𝑗𝐵 =1

𝑍

=1

𝑅 + 𝑗𝑋∙𝑅 − 𝑗𝑋

𝑅 − 𝑗𝑋

𝑮+ 𝒋𝑩 =𝑹− 𝒋𝑿

𝑹𝟐 + 𝑿𝟐

Separating the real and imaginary part of the admittance makes,

𝑮+ 𝒋𝑩 =𝑹

𝑹𝟐 + 𝑿𝟐− 𝒋

𝑿

𝑹𝟐 + 𝑿𝟐

From the above equation, we can realize that,

𝑮 =𝑹

𝑹𝟐+𝑿𝟐 and 𝑩 =

𝑿

𝑹𝟐+𝑿𝟐

Therefore, it does not state that the real part of the admittance is equal to the reciprocal of the real part of the impedance.

𝑮 ≠𝟏

𝑹 except in th purely resistive case (X=0)

Similarly that the imaginary part of the admittance is equal to the reciprocal of the imaginary part of the impedance.

𝑩 ≠𝟏

𝑿 except in th purely reactive case (R=0)

Note: Admittance, Conductance and Susceptance are all measured in siemens (S) or mho. REMEMBER : The term “IMMITANCE” is the combination of the words IMPEDANCE and ADMITTANCE.


36


Note also: If

G1 B1Z1 Z2

Y1

G2 B2

Y2

R and L in Parallel where: Z = impedance in ohms

Z =RXL

R2+XL2

pf = cos θ = IR

IT =

Z

R

rf = sin θ = IL

IT =

Z

XL

tan θ = IL

IR =

R

XL

P = VI pf = I 2R =V2

R watts

Q = VI rf = I 2XL =V2

XL vars

R

X

G B


37


S = VI = IT2Z =

V2

Z va

∴ IT = IR2 + IL

2 = V R2+XL

2

RXL

IT =V

Z Amperes

R and C in Parallel where: Z = impedance in ohms

Z =RXC

R2+XC2

pf = cos θ = IR

IT =

Z

R

rf = sin θ = IC

IT =

Z

XC

tan θ = IC

IR =

R

XC

P = VI pf = IR 2R =V2

R watts

Q = VI rf = IC 2XC =V2

XC vars

S = VI = IT2Z =

V2

Z

∴ IT = IR2 + IC

2 = V R2+XC

2

RXC

IT =V

Z Amperes


38


R LC in Parallel

∴ IT = IR2 + IL − IC

2 = V XL

2XC2+R2 XL−XC

2

RXL XC

IT =V

Z Amperes

where:

Z = impedance in ohms

Z =RXL XC

XL2XC

2+R2 XL−XC 2

pf = cos θ = IR

IT =

Z

R

XC = 1

2πfC=

1

𝜔𝐶 Ω rf = sin θ =

IL−IC

IT =

Z XL−XC

XL XC

XL = 2πfL = ωL Ω tan θ = IL−IC

IR =

R XL−XC

XL XC

IR =V

R; IL =

V

XL; IC =

V

XC Amperes


39


RESONANCE IN PARALLEL RLC CIRCUIT

Similar to a series a.c. circuit, there can be resonance in parallel ac circuit. When the power factor of a

parallel ac circuit is unity, that is when the voltage and current is in phase at a particular frequency, then the

parallel circuit is said to be at resonance. The frequency at which the parallel resonance occurs is called

resonant frequency denoted as fr. Unlike a series circuit, however, the total current “feeding” all branches is

the minimum because algebraic sum of quadrature current component is zero.

R L C

IRIT IL IC

VT

At resonance,

IL − IC = 0

IL = IC

V

XL=

V

XC

1

XL=

1

XC

𝐁𝐋 = 𝐁𝐂

and,

IT = IR2 + IL − IC

2

0

IT = IR

VT

Z=

VT

R

𝐙 = 𝐑 maximum


40


∴ 𝐈𝐓 =𝐕𝐓

𝐙

maximum

minimum

The term ANTIRESONANCE is sometimes used for unity power factor parallel circuits to distinguish it

from a similar condition I series circuits.

CHARACTERISTICS OF PARALLEL RESONANCE

Consider a practical parallel circuit used for the parallel resonance as showm in the figure.

The one branch consists of resistance R in series with

inductor L. So it is series RL circuit with impedance ZL. The other

branch is pure capacitive with a capacitor C. Both the branches are

connected in parallel across a variable frequency constant voltage

source.

The current drawn by inductive load is IL while the current

drawn to capacitive load is IC.

IL =V

ZL where ZL = R + jXL

And

IC =V

XC where XC =

1

2πfC

The current IL lags the voltage V by an angle ΦL which is

decided by R and XL while the current IC leads the voltage V by 90°. The

total current I is the phasor addition of IL and IC.

For the parallel resonance V and I must be in phase. To achieve

this unity pf condition,


41


I = IL cosϕL and IC = IL sinϕL

From the impedance triangle of RL series circuit, we can write,

tanϕL = XL

𝑅 ; cosϕL =

𝑅

ZL ; sinϕL =

XL

ZL

As the frequency is increased, inductive reactance XL increases due to which ZL = R2 + XL2 also

increases. Hence power factor cosϕL decreases and reactive factor sinϕL increases. As ZL increases, the

current IL also decreases.

At resonance f = fr and IL cosϕL is at its minimum. Thus, at resonance, current is minimum while the

total impedance of the circuit is at maximum. As admittance is reciprocal of impedance, as frequency is

changed, admittance decreases and is minimum at resonance. The three curves are characteristics of parallel

circuit shown below.

At resonance,

IC = IL sinϕL

V

XC=

V

ZL∙

XL

ZL

V

XC=

VXL

ZL2

ZL2 = XLXC

R2 + XL2 = XLXC

R2 + 2πfrL 2 = 2πfrL ∙1

2πfr C as f = fr

R2 + 2πfrL 2 = 𝐿

𝐶


42


2πfrL 2 = 𝐿

𝐶− R2

2πfrL 2 = 1

𝐿𝐶−

R2

L2

fr = 1

2π

1

𝐿𝐶−

R2

L2

Thus if R is very small compared to L and C, R2

L2 ≪1

𝐿𝐶

∴ fr = 1

2π 𝐿𝐶

This is the same as that for series resonance.

Note: The net susceptance of the whole circuit is zero at resonance.

A paralle circuits can exhibit properties of resonance only when one or more branches contain an

excess of inductive reactance with respect to other branches that have an excess of capacitive reactance.

In this case of two branches connected in parallel, one path being series RL and the other a series RC. When

the two branch parallel cirucit is in resonance, the quadrature component of currents IC and IL are equal.

Thus,


43


Z1 Z2

IL ICIT

VT = V1 = V2

At resonance, IC = IL

V1BL = V2BC

BL = BC

XL

R2 + XL2 =

XC

R2 + XC2

ωL

R2 + ωL 2=

1ωC

R2 + 1ωC

2

ωL

R2 +ω2L2=

ωC

R2ω2C2 + 1

R2ω2C2L + L = R2C +ω2L2C

R2ω2C2L−ω2L2C = R2C− L

ω2LC R2C− L = R2C− L

ω2LC = 1

ω2 = 1

LC

ω = 1

LC

2πfr = 1

LC

∴ fr = 1

2π 𝐿𝐶


44


DYNAMIC IMPEDANCE AT RESONANCE

The impedance offered by the parallel circuit at resonance is called dynamic impedance denoted as

ZD. This is maximum at resonance. As current drawn at resonance is minimum, the parallel circuit at resonance

is called rejector circuit. This is indicates that it rejects the unwanted frequencies and hence it is used as filters

in radio receiver.

FromIC = IL sinϕL , we have seen that,

ZL2 =

L

C

while

I = IL cosϕL = V

ZL∙

R

ZL=

VR

ZL2

I = VR

ZL2 =

VR

LC

= V

LRC

∴ 𝐈 = 𝐕

𝐙𝐃

where

𝐙𝐃 =𝐋

𝐑𝐂 is called the Dynamic Impedance


45


QUALITY FACTOR OF PARALLEL CIRCUIT

The parallel circuit is used to magnify the current and hence known as current resonance circuit.

The Quality Factor of the parallel circuit is defined as the current magnification in the circuit at

resonance.

The current magnification is defined as,

𝑪𝒖𝒓𝒓𝒆𝒏𝒕 𝒎𝒂𝒈𝒏𝒊𝒇𝒊𝒄𝒂𝒕𝒊𝒐𝒏 = 𝑪𝒖𝒓𝒓𝒆𝒏𝒕 𝒊𝒏 𝒕𝒉𝒆 𝒊𝒏𝒅𝒖𝒄𝒕𝒊𝒗𝒆 𝒃𝒓𝒂𝒏𝒄𝒉

𝑪𝒖𝒓𝒓𝒆𝒏𝒕 𝒊𝒏 𝑺𝒖𝒑𝒑𝒍𝒚 𝒂𝒕 𝒓𝒆𝒔𝒐𝒏𝒂𝒏𝒄𝒆= 𝑰𝑳𝑰

=

𝑉𝑍𝐿𝑉𝑍𝐷

= 𝑍𝐷𝑍𝐿

=

LRC

LC

= 1

R

L

C

as

𝑍𝐿 = 𝑿𝑳𝑿𝑪 = L

C

This is nothing but the Quality Factor at resonance.

𝑸 = 𝟏

𝐑 𝐋

𝐂


46


POWER IN COMPLEX FORM

1. Voltage Conjugate Method 2. Current Conjugate Method

S = Current x conjugate of voltage S = Voltage x conjugate of Current

S = IV∗ S = VI∗

S = P ± jQ S = P∓ jQ

S = P2 + Q2∠± θ va S = P2 + Q2∠∓ θ va

Note: (used in Series) Note: (used in Parallel)

+j = Q is capacitive (pf is leading) -j = Q is capacitive (pf is leading)

-j = Q is inductive (pf is lagging) +j = Q is inductive (pf is lagging)

Where:

S = Apparent power, va P = Real or true or average power, watts Q = Reactive power, vars I = current, amperes V = voltage, volts V∗ = conjugate of voltage I∗ = conjugate of current θ = power factor angle


47


POWER FACTOR

Power Factor is defined as a factor by which the apparent power must be multiplied in order to obtain

the true power.

It is the ratio of true power to apparent power

𝐏𝐨𝐰𝐞𝐫 𝐅𝐚𝐜𝐭𝐨𝐫 = 𝐓𝐫𝐮𝐞 𝐩𝐨𝐰𝐞𝐫

𝐀𝐩𝐩𝐚𝐫𝐞𝐧𝐭 𝐩𝐨𝐰𝐞𝐫= 𝐕 𝐈 𝐜𝐨𝐬 𝛉

𝐕 𝐈= 𝐜𝐨𝐬𝛉

The numerical value of cosine of the phase angle between the applied voltage and the current drawn

from the supply voltage gives the power factor. It cannot be greater than 1.

It is also defined as the ratio of resistance to the impedance.

𝐜𝐨𝐬 𝛉 =𝐑

𝐙

Note:

The nature of power factor is always determined by the position of current with respect to the

voltage.

If the current lags voltage, power factor is said to be lagging. If current leads voltage, power factor

is said to be leading.

So, for pure inductance, the power factor is cos (90). Therefore, it means that the power factor is

zero lagging. For pure capacitance, the power factor is cos (90). Therefore, it means that the power factor is

zero leading. For purely resistive circuit, voltage and current are in phase. Therefore the theta θ = 0. The

power factor is cos (0) = 1. Such circuit is called unity power factor.

Note: θ = is the angle between supply voltage and current.

Nature of power factor always tells the position of the currrent with respect to the voltage.


48


DISADVANTAGES OF LOW POWER FACTOR

The power factor plays an importance role in a.c. circuits since power consumed depends upon this factor.

𝐏 = 𝐕𝐈 𝐜𝐨𝐬 𝛉

∴ 𝐈 =𝐏

𝐕𝐜𝐨𝐬 𝛉

It is clear from above that for fixed power and voltage, the load current is inversely proportional to the power factor. Lower the power factor, higher is the load current and vice-versa. A power factor less than unity results in the following disadvantages:

(i) Large kVA rating of equipment.

The electrical machinery (e.g., alternators, transformers, switchgear) is always rated in *kVA.

Now, kVA = kW cos θ

It is clear that kVA rating of the equipment is inversely proportional to power factor. The smaller the power factor, the larger is the kVA rating. Therefore, at low power factor, the kVA rating of the equipment has to be made more, making the equipment larger and expensive.

(ii) Greater conductor size.

To transmit or distribute a fixed amount of power at constant voltage, the conductor will have to carry more current at low power factor. This necessitates large conductor size. For example, take the case of a single phase a.c. motor having an input of 10 kW on full load, the terminal voltage being 250 V. At unity p.f., the input full load current would be 10,000/250 = 40 A. At 0·8 p.f; the kVA input would be 10/0·8 = 12·5 and the current input 12,500/250 = 50 A. If the motor is worked at a low power factor of 0·8, the cross-sectional area of the supply cables and motor conductors would have to be based upon a current of 50 A instead of 40 A which would be required at unity power factor.

(iii) Large copper losses.

The large current at low power factor causes more I2R losses in all the elements of the supply system. This results in poor efficiency.

(iv) Poor voltage regulation.

The large current at low lagging power factor causes greater voltage drops in alternators, transformers, transmission lines and distributors. This results in the decreased voltage available at the


49


supply end, thus impairing the performance of utilisation devices. In order to keep the receiving end voltage within permissible limits, extra equipment (i.e., voltage regulators) is required.

(v) Reduced handling capacity of system.

The lagging power factor reduces the handling capacity of all the elements of the system. It is because the reactive component of current prevents the full utilisation of installed capacity.

Causes of Low Power Factor Low power factor is undesirable from economic point of view. Normally, the power factor of the whole load on the supply system in lower than 0·8. The following are the causes of low power factor:

(i) Most of the a.c. motors are of induction type (1Φand 3Φ induction motors) which have low lagging power factor. These motors work at a power factor which is extremely small on light load (0·2 to 0·3) and rises to 0·8 or 0·9 at full load.

(ii) Arc lamps, electric discharge lamps and industrial heating furnaces operate at low lagging power

factor. (iii) The load on the power system is varying; being high during morning and evening and low at other

times. During low load period, supply voltage is increased which increases the magnetisation current. This results in the decreased power factor.


50


POWER FACTOR CORRECTION

The electrical energy is almost exclusively generated, transmitted and distributed in the form of

alternating current. Therefore, the question of power factor immediately comes into picture. Most domestic

loads (such as washing machines, air conditioners, and refrigerators) and industrial loads (such as induction

motors) are inductive and operate at a low lagging power factor. The low power factor is highly undesirable as

it causes an increase in current, resulting in additional losses of active power in all the elements of power

system from power station generator down to the utilisation devices. In order to ensure most favourable

conditions for a supply system from engineering and economical standpoint, it is important to have power

factor as close to unity as possible.

In order to improve the power factor, some device taking leading power should be connected in

parallel with the load. One of such devices can be a capacitor. The capacitor draws a leading current and partly

or completely neutralises the lagging reactive component of load current. This raises the power factor of the

load. The process of increasing the power factor without altering the voltage or current to the original load is

known as power factor correction.

Since most loads are inductive, as shown in Fig. (a), a load’s power factor is improved or corrected by

deliberately installing a capacitor in parallel with the load, as shown in Fig. (b). The effect of adding the

capacitor can be illustrated using either the power triangle or the phasor diagram of the currents involved.

Figure 11.28 shows the latter, where it is assumed that the circuit in Fig. (a) has a power factor of cos θ1, while

the one in Fig. (b) has a power factor of cos θ2. It is evident from Fig. that adding the capacitor has caused the

phase angle between the supplied voltage and current to reduce from θ1 to θ2, thereby increasing the power

factor. We also notice from the magnitudes of the vectors in Fig. that with the same supplied voltage, the

circuit in Fig. (a) draws larger current IL than the current I drawn by the circuit in Fig. (b). Power companies

charge more for larger currents, because they result in increased power losses (by a squared factor, since

P=I2R). Therefore, it is beneficial to both the power company and the consumer that every effort is made to

minimize current level or keep the power factor as close to unity as possible. By choosing a suitable size for

the capacitor, the current can be made to be completely in phase with the voltage, implying unity power

factor.


51


(a) (b)

We can look at the power factor correction from another perspective. Consider the power triangle,

We can look at the power factor correction from another perspective. Consider the power triangle in Figure in the left. If the original inductive load has apparent power S1, then

P = S1 cos θ1, Q1 = S1 sin θ1 = P tan θ1 If we desire to increase the power factor from cos θ1 to cos θ2 without altering the real power (i.e., P = S2 cos θ2), then the new reactive power is Q2 = P tan θ2

The reduction in the reactive power is caused by the shunt capacitor, that is,

QC = Q1 − Q2 = P (tan θ1 − tan θ2)

But from equation QC = V2rms/XC = ωCV2

rms. The value of the required shunt capacitance C is determined as

𝐂 = 𝐐𝐜

𝛚𝐕𝐫𝐦𝐬𝟐

=𝐏 (𝐭𝐚𝐧 𝛉𝟏 − 𝐭𝐚𝐧 𝛉𝟐)

𝛚𝐕𝐫𝐦𝐬𝟐


52


Note that the real power P dissipated by the load is not affected by the power factor correction because the average power due to the capacitance is zero. Although the most common situation in practice is that of an inductive load, it is also possible that the load is capacitive, that is, the load is operating at a leading power factor. In this case, an inductor should be connected across the load for power factor correction. The required shunt inductance L can be calculated from

𝐐𝐋 = 𝐕𝐫𝐦𝐬𝟐

𝐗𝐋=𝐕𝐫𝐦𝐬𝟐

𝛚𝐋

∴ 𝐋 = 𝐕𝐫𝐦𝐬𝟐

𝛚𝐐𝐋

where QL = Q1 − Q2, the difference between the new and old reactive powers. E X

Importance of Power Factor Improvement The improvement of power factor is very important for both consumers and generating stations as discussed below : (i) For consumers.

A consumer* has to pay electricity charges for his maximum demand in kVA plus the units consumed. If the consumer imporves the power factor, then there is a reduction† in his maximum kVA demand and consequently there will be annual saving due to maximum demand charges. Although power factor improvement involves extra annual expenditure on account of p.f. correction equipment, yet improvement of p.f. to a proper value results in the net annual saving for the consumer. (ii) For generating stations.

A generating station is as much concerned with power factor improvement as the consumer. The generators in a power station are rated in kVA but the useful output depends upon kW output. As station output is kW = kVA × cos φ, therefore, number of units supplied by it depends upon the power factor. The greater the power factor of the generating station, the higher is the kWh it delivers to the system. This leads to the conclusion that improved power factor increases the earning capacity of the power station.


53



54


v(t)

t

v

i

90°

Inductance Circuit

𝒆 = 𝑬𝒎 𝐬𝐢𝐧𝝎𝒕 = 𝑳 𝒅𝒊

𝒅𝒕

𝑑𝑖 =𝐸𝑚𝐿

sin𝜔𝑡 𝑑𝑡

𝑑𝑖 = 𝐸𝑚𝐿

sin𝜔𝑡 𝑑𝑡

𝑖 =𝐸𝑚𝐿 sin𝜔𝑡 𝑑𝑡

u=ωt du= ω dt nf= 1/ω

Then,

𝑖 =𝐸𝑚𝜔𝐿

− cos𝜔𝑡

𝑖 = −𝐸𝑚𝜔𝐿

cos𝜔𝑡

but;

cos𝜔𝑡 = sin(90−𝜔𝑡)

So;


sin(90−𝜔𝑡)


− sin(𝜔𝑡 − 90)


sin(𝜔𝑡 − 90)

note:

90 =𝜋

2; 𝐼𝑚 =

𝐸𝑚𝜔𝐿

Hence,

𝒊 = 𝑰𝒎 𝐬𝐢𝐧 𝝎𝒕 −𝝅

𝟐

This means;

Voltage leads the Current by 90° (ELI)

or Current lags the Voltage by 90°

when:

𝜔𝑡 = 0; 𝑖 = − 𝐸𝑚𝜔𝐿

= −𝐼𝑚

𝜔𝑡 =𝜋

2; 𝑖 = 0

𝜔𝑡 = 𝜋; 𝑖 = 𝐸𝑚𝜔𝐿

= 𝐼𝑚

𝜔𝑡 =3𝜋

2; 𝑖 = 0

Now,

𝐼𝑚 =𝐸𝑚𝜔𝐿

𝝎𝑳 =𝐸𝑚𝐼𝑚

=

𝐸 2

𝐼 2

= 𝑿𝑳

Then;

𝑿𝑳 = 𝝎𝑳 =𝑬

𝑰


55


Considering 𝜔𝑡 = 𝜋

Then,


sin 𝜔𝑡 −𝜋

2

𝐼𝑚 =𝐸𝑚𝜔𝐿

sin 𝜋 −𝜋

2

𝑋𝐿 =𝐸𝑚𝐼𝑚

sin𝜋

2

Hence;

𝑿𝑳 =𝑬

𝑰 ∠𝟗𝟎 = 𝒋𝝎𝑳

Power of Inductor;

𝒑 = 𝒗𝒊

𝑝 = 𝐸𝑚 sin𝜔𝑡 − 𝐼𝑚 cos𝜔𝑡

𝑝 = − 𝐸𝑚 𝐼𝑚 sin𝜔𝑡 cos𝜔𝑡

note: 2sinAcosA = sin 2A 2sin 𝜔𝑡 cos 𝜔𝑡 = sin 2 𝜔𝑡 but: 𝜃 = 𝜔𝑡

then;

𝑝 = − 𝐸𝑚 𝐼𝑚sin 2𝜔𝑡

2

but;

𝐸𝐼 = 𝐸𝑚

2 𝐼𝑚

2 =

𝐸𝑚 𝐼𝑚2

Hence;

𝒑 = − 𝑬𝑰 𝐬𝐢𝐧𝟐𝝎𝒕

or

𝑷𝒂𝒗𝒆𝑳 = 𝑽𝑰 =𝟏

𝑻 𝑝

𝑇

0

𝑑𝑡

For Energy Stored in Inductor;

W(joules) = pt (watt-sec)

Now,

𝑊 = 𝑝𝑡

𝑑𝑊 = 𝑝𝑑𝑡

𝑑𝑊 = − 𝐸𝐼 sin 2𝜔𝑡 𝜋

𝜋2

𝑑𝑡

= − 𝐸𝐼 sin 2𝜔𝑡𝜋

𝜋2

𝑑𝑡

Note: u=2ωt du= 2ω dt nf= 1/2ω

𝜃 = 𝜔𝑡

Then;

𝑊 = − 𝐸𝐼

2𝜔 −𝑐𝑜𝑠 2𝜃 𝜋

2

𝜋

= − 𝐸𝐼

2𝜔 −cos 2𝜋 − − cos 2

𝜋

2

= − 𝐸𝐼

2𝜔 −1 +−1

=𝐸𝐼

𝜔= 𝜔𝐿𝐼 𝐼

𝜔

𝑊 = 𝐿𝐼2 = 𝐿 𝑖

2

2


56


v(t)

t

i

v

90°

Hence;

𝑾 = 𝑳𝒊𝟐

𝟐

or

𝒘(𝒕)𝒂𝒗𝒆𝑳 = 𝑝𝑡

0

𝑑𝑡

Capacitive Circuit

𝒆 = 𝑬𝒎 𝐬𝐢𝐧𝝎𝒕 =𝟏

𝑪 𝒊𝒕

𝟎

𝒅𝒕

note;

𝑉 =𝑄

𝐶 𝑜𝑟 𝑣 =

𝑞

𝐶

and

𝑞 = 𝑖𝑡

0

𝑑𝑡

Then;

𝑒 =𝑞

𝐶

𝐸𝑚 sin𝜔𝑡 =𝑞

𝐶

𝑞 = 𝐶𝐸𝑚 sin𝜔𝑡

𝑑𝑞

𝑑𝑡= 𝐶𝐸𝑚

𝑑 sin𝜔𝑡

𝑑𝑡

note:

𝑖 =𝑑𝑞

𝑑𝑡;𝑑 sin 𝑢

𝑑𝑡= cos𝑢

𝑑𝑢

𝑑𝑡

So;

= 𝐶𝐸𝑚 cos𝜔𝑡𝑑 𝜔𝑡

𝑑𝑡

= 𝜔𝐶𝐸𝑚 cos𝜔𝑡𝑑 𝑡

𝑑𝑡

𝑖 = 𝜔𝐶𝐸𝑚 cos𝜔𝑡

but;

± cos𝜔𝑡 = sin(90∓𝜔𝑡)

So;

𝑖 = 𝜔𝐶𝐸𝑚 sin 90 +𝜔𝑡

𝑖 = 𝜔𝐶𝐸𝑚 sin 𝜔𝑡 + 90

note:

90 =𝜋

2; 𝐼𝑚 =

𝐸𝑚𝜔𝐿

Hence,

𝒊 = 𝑰𝒎 𝐬𝐢𝐧 𝝎𝒕 +𝝅

𝟐

This means;

Current leads the Voltage by 90° (ICE)

Voltage lags the Current by 90°

when:


57


𝜔𝑡 = 0; 𝑖 = − 𝐸𝑚𝜔𝐿

= −𝐼𝑚

𝜔𝑡 =𝜋

2; 𝑖 = 0

𝜔𝑡 = 𝜋; 𝑖 = 𝐸𝑚𝜔𝐿

= 𝐼𝑚

𝜔𝑡 =3𝜋

2; 𝑖 = 0

Now,

𝐼𝑚 = 𝜔𝐶𝐸𝑚∠90 = 𝑗𝜔𝐶𝐸𝑚

𝟏

𝒋𝝎𝑪=𝐸𝑚𝐼𝑚

=

𝐸 2

𝐼 2

= 𝑿𝑪

Then;

𝑿𝑪 =𝟏

𝒋𝝎𝑪=𝑬

𝑰

Considering 𝜔𝑡 = 𝜋

Then,

𝑖 = 𝜔𝐶𝐸𝑚 sin 𝜔𝑡 +𝜋

2

𝐼𝑚 = 𝜔𝐶𝐸𝑚 sin 𝜋 +𝜋

2

𝑋𝐶 = 𝜔𝐶𝐸𝑚 sin3𝜋

2

Hence;

𝑿𝑪 =𝑬

𝑰 ∠ − 𝟗𝟎 = −𝒋

𝟏

𝝎𝑪

Power of capacitor;

𝒑 = 𝒗𝒊

𝑝 = 𝐸𝑚 sin𝜔𝑡 𝐼𝑚 cos𝜔𝑡

𝑝 = 𝐸𝑚 𝐼𝑚 sin𝜔𝑡 cos𝜔𝑡

note: 2sinAcosA = sin 2A 2sin 𝜔𝑡 cos 𝜔𝑡 = sin 2 𝜔𝑡 but: 𝜃 = 𝜔𝑡

then;

𝑝 = 𝐸𝑚 𝐼𝑚sin 2𝜔𝑡

2

but;

𝐸𝐼 = 𝐸𝑚

2 𝐼𝑚

2 =

𝐸𝑚 𝐼𝑚2

Hence;

𝒑 = 𝑬𝑰𝐬𝐢𝐧 𝟐𝝎𝒕

or

𝑷𝒂𝒗𝒆𝑪 = 𝑽𝑰 =𝟏

𝑻 𝑝

𝑇

0

𝑑𝑡

For Energy Stored in Capacitor;

W(joules) = pt (watt-sec)

Now,

𝑊 = 𝑝𝑡

𝑑𝑊 = 𝑝𝑑𝑡

𝑑𝑊 = 𝐸𝐼 sin 2𝜔𝑡

𝜋2

0

𝑑𝑡

= 𝐸𝐼 sin 2𝜔𝑡

𝜋2

0

𝑑𝑡

Note: u=2ωt


58


du= 2ω dt nf= 1/2ω

𝜃 = 𝜔𝑡

Then;

𝑊 = 𝐸𝐼

2𝜔 −𝑐𝑜𝑠 2𝜃 0

𝜋2

= 𝐸𝐼

2𝜔 − cos 2

𝜋

2 − −cos 2(0)

= 𝐸𝐼

2𝜔 1 + 1

=𝐸𝐼

𝜔= 𝐸 𝜔𝐶𝐸

𝜔

𝑊 = 𝐶𝐸2 = 𝐶 𝑣

2

2

Hence;

𝑾 = 𝑪𝒗𝟐

𝟐

or

𝒘(𝒕)𝒂𝒗𝒆𝑳 = 𝑝𝑡

0

𝑑𝑡

SUMMARY

INDUCTOR CAPACITOR

𝑣 𝐿 = L di

dt 𝑖 =

1

𝐶 𝑖 𝑑𝑡 + 𝑣 (𝑡0)

𝑡

0

𝑖 = 1

𝐿 𝑣 𝑑𝑡 + 𝑖 (𝑡0)

𝑡

0

𝑣 𝐶 = C d𝑣

dt

𝑝 = 𝑣 𝐿i = Li di

dt 𝑝 = 𝑣 𝐶i = Cv

d𝑖

dt

𝑊𝐿 = 1

2Li2 𝑊𝐶 =

1

2C𝑣2

Lecture in AC Ckts

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Transcript of Lecture in AC Ckts