Basics of Power Systems: Review of Phasorscs620/PhasorReview...2 Back to Phasors 3 Phasor...

Review ofPhasors

Prof S. A.Soman

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Working withComplexExponential

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Basics of Power Systems: Review of Phasors

Prof S. A. Soman

Visiting FacultyDepartment of Electrical and Computer Engineering

Virginia [email protected]

Department of Electrical EngineeringIIT-Bombay

August 29, 2012

Prof S. A. Soman (IIT-Bombay) Review of Phasors August 29, 2012 1 / 41

Review ofPhasors

Prof S. A.Soman

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Examples

1 Sinusoidal Nature of Voltage and Currents

2 Back to Phasors

3 Phasor Computation

4 Behaviour of Circuit Elements

5 Working with Complex Exponential

6 Examples


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Examples

Flux Linking and EMF Generation

Flux linked to coil is given by

φ =

∫

~B · d~s = BA cos θ

Suppose coil rotates around itsaxis at speed ω

θ(t) = ωt + θ0

⇒φ(t) = BA cos (ωt + φ)

As per Faraday’s law

e = −Ndφ

dt[generator convention]

= +NBAω sin (ωt + θ0)

W

A


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Examples

Flux Linking and EMF Generation

Flux linked to coil is given by

φ =

∫

~B · d~s = BA cos θ

Suppose coil rotates around itsaxis at speed ω

θ(t) = ωt + θ0

⇒φ(t) = BA cos (ωt + φ)

As per Faraday’s law

e = −Ndφ

dt[generator convention]

= +NBAω sin (ωt + θ0)

W

A

0

0

v

φ

ωt

ωt


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Examples

Convention

Generator

+

−

R V

I

e+

Motor

e

I

v+

−+

e − v = 0

⇒v = e = −Ndφ

dt

v + e = 0

⇒v = −e = Ndφ

dt=

dλ

dt


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Example of Motor Convention

V

+

−

e

v =dλ

dt= L

di

dt[∵ λ = Li ]


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Back to Phasors

Voltage generated by generator at its terminal is sinusoidalin nature.

Whether it is sine or cosine depends upon where origin ort = 0 is placed on the sinusoid.

If there are 3 coils displaced in space by 120, we wouldget 3-phase voltage,

va(t) = Vmsin(ωt + φ)

vb(t) = Vmsin

(

ωt + φ − 2π

3

)

vc(t) = Vmsin

(

ωt + φ − 4π

3

)

= Vmsin

(

ωt + φ +2π

3

)


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Prof S. A.Soman

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Recap

AC generators produces sinusoidal voltages.

The generated voltages in multi-phase system e.g.3-phase, 5-phase, etc. have a symmetry about them.


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Addition of Sinusoids at Same Frequency

Let,

v1(t) = Vm1 sin(ωt + φ1)

= Vm1 sin ωt cos φ1 + Vm1 cos ωt sin φ1

v2(t) = Vm2 sin(ωt + φ2)

= Vm2 sin ωt cos φ2 + Vm2 cos ωt sin φ2

Then,

v1(t) + v2(t) = [Vm1 cos φ1 + Vm2 cos φ2] × sin ωt

+ [Vm1 sin φ1 + Vm2 sin φ2] × cos ωt

= A sin ωt + B cos ωt

=p

A2 + B2

»

A√A2 + B2

sin ωt +B√

A2 + B2cos ωt

–

=p

A2 + B2 [cos θ sin ωt + sin θ cos ωt]

= Vm sin(ωt + θ)

tan θ =Vm1 sin φ1 + Vm2 sin φ2

Vm1 cos φ1 + Vm2 cos φ2

V 2m = (Vm1 cos φ1 + Vm2 cos φ2)

2 + (Vm2 sin φ1 + Vm2 sin φ2)2

= V 2m1 + V 2

m2 + 2Vm1Vm2 cos (φ1 − φ2)


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Geometrical Visualization of Vector Addition

φ1Vm sin1

φ2Vm sin2

φ1Vm cos1 φ2Vm cos2

Vm

2Vm

φ1φ

φ21Vm

Can be inductively extended to addition of multiple sinusoids.


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Examples

Root Mean Square Value

Xm

Xrms = Xm/√

2

Let, x(t) = Xm sin(ωt + φ); ω = 2πT

. Then,

Xrms =1

T

√

∫ t+T

t

x(τ)2dτ


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What is a Phasor?

Sinusoids at a given frequency (x(t) = Xm sin (ωt + φ))can be represented by vectors.

Has two degree of freedoms viz-a-viz, amplitude Xm or rms

value Xrms =Xm√

2and angle φ.

This complex number or vector representation is called aphasor.Can be represented in polar form as well.

Time Domain ⇔ Phasor Domain

Need to define one reference phasor for convenience.


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Examples

What is a Phasor? cntd. . .

1 Let x(t) = 1 × cos(ωt) be reference signal. Now,represent phasors for the following:

a 10 cos (ωt + 30)b 10 cos (ωt − 30)

c5√2

sin (ωt + 30)

2 Let, x1(t) = Xm1 sin (ω1t + φ1) andx2(t) = Xm2 sin (ω2t + φ2). If ω1 6= ω2, can we representthem by phasors?


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Examples

Phasor representation of KCL and KVL

i (t)1

i (t)2

i (t)N

i (t)3

1V (t)2V (t)

3V (t)

4V (t)

N∑

i=1

Imicos (ωt + φi ) = 0

⇒N

∑

i=1

Imi= 0

N∑

i=1

Vmicos (ωt + φi ) = 0

⇒N

∑

i=1

Vmi= 0


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Examples

Instantaneous Power in Single Phase Circuit

Instantaneous power at a time t is given by

p(t) = v(t) × i(t)

With v(t) = Vm sin (ωt) and i(t) = Im sin (ωt − φ)

p(t) = VmIm sin ωt sin(ωt − φ)

=VmIm

22 sin ωt sin(ωt − φ)

=Vm√

2

Im√2[cos φ − cos(2ωt − φ)]

= Vrms Irms [cos φ − cos(2ωt − φ)]


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Instantaneous Power in Single Phase Circuit

Instantaneous power at a time t is given by

p(t) = v(t) × i(t)

With v(t) = Vm sin (ωt) and i(t) = Im sin (ωt − φ)

p(t) = VmIm sin ωt sin(ωt − φ)

=VmIm

22 sin ωt sin(ωt − φ)

=Vm√

2

Im√2[cos φ − cos(2ωt − φ)]

= Vrms Irms [cos φ − cos(2ωt − φ)]

0.25

0.50

0.75

−0.25

T2p(t)

t

Pavg


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Examples

Observation from Instantaneous Power Expression

We have

p(t) = Vrms Irms [cos φ − cos(2ωt − φ)]

Pulsating with twice the frequency, i.e. 2ω

We,therefore, calculate average power over a cycle, termedas real or active power

P =1

T0

∫ t+T0

t

p(τ)dτ = Vrms Irms cos φ

Unit is watt (W)


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Examples

Power Factor

We define S = Vrms Irms as the apparent power associatedwith the current.

The ratio PS

is in a sense efficiency of power transfer. It iscalled power factor.

Hence, the power factor is given by cos φ ∈ [−1, 1].

Current lags voltage, we say lagging power factor; cosφ ispositive.

Current leads voltage, we say it as leading power factor;cos φ is negative.

For example, power factor 0.5 lagging or 0.5 leading.


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Examples

Reactive Power

I cos

II sin

V

Active power is a product of voltage and projection ofcurrent phasor on voltage phasor

Similarly, product of voltage phasor with quadraturecomponent of current (VI sinφ) can be computed

Termed as reactive powerSignificant in power system computation


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Notion of Complex Power

Consider,

~S = VI ∗

= Vrms∠0[Irms∠ − φ]∗

= Vrms Irms∠φ

= Vrms Irms cos φ + jVrms Irms sinφ

⇒ ~S = P + jQ

|~S |2 = P2 + Q2

cos φ and sinφ are dimensionless. Hence, power factor hasno unit.

Q = VI sinφ has a unit of VA. However, it is written asVAR. It indicates VA reactive.

In power systems, MW and MVAR are commonly usedunits.


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Examples

Power Triangle at Different Power Factors

Leading

Q

P

S

Lagging

Q

P

S

Unity

P

S


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Prof S. A.Soman

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Examples

Review Questions I

Prove that R.M.S value of a sinusoid is 1√2

of its peak

value.

Let a voltage source, v(t) = Vm sin(ωt + φ), deliver powerto a resistive circuit (resistance R) as shown in figurebelow.

R

i(t)

v(t)

Calculate the instantaneous current and power.Calculate the average power.Draw the phasor diagram depicting voltage and currentphasor.Draw the power triangle (P + jQ).


Review ofPhasors

Prof S. A.Soman

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Review Questions II

If current lags voltage phasor which if by 30. Would youdesignate it as Irms∠ − 30 or Irms∠ + 30? Justify?

Draw circuits to correspond to be following 4 quadrants ofP − Q diagram as shown in figure below.

P +veQ +ve

P +veQ −ve

P −veQ +ve

P −veQ −ve

P

Q


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Review Questions III

If, The first quadrant represents a circuit whichconsumes active and reactive power e.g., aninductive current.

Then, The forth quadrant represents a current whichconsumes active power but generates reactivepower e.g., a capacitive current.

What do IInd and III ndquadrant represent?


Review ofPhasors

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Examples

Behaviour of Circuit ElementsResistor

R

i(t)

v(t) I

V

Let,v(t) = Vm sin(ωt + φ)

Then,

i(t) =Vm

Rsin(ωt + φ)

p(t) = v(t) × i(t) =V 2

m√2

sin2(ωt + φ)

Pav =1

T

Z t+T0

t

p(τ)dτ = Vrms Irms

cos φ = 1


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Behaviour of Circuit ElementsResistor Cntd. . .

2

4

6

8

10

−2

−4

Pavg

i(t)v(t)p(t)


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Examples

Behaviour of Circuit ElementsInductor

i(t)

v(t) L

LV

LI

Following equation relates inductor voltage and current,

v(t) = Ldi

dt

Let, i(t) = Im sin(ωt + φ)

⇒ v(t) = ωLIm cos(ωt + φ)

⇒ p(t) = Vm sin (ωt + φ) × Im cos (ωt + φ) [Vm = ωLIm]

=VmIm

2sin (2ωt + 2φ)

⇒ Pav =1

T0

Z T0

0p(τ)dτ = 0

Additionally, S = P + jQ = VI⋆ = 0 + jVrms Irms


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Behaviour of Circuit ElementsInductor cntd. . .

2

4

6

−2

−4

i(t)v(t)p(t)

+ × + = + + × − = −− × − = +− × + = −


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Behaviour of Circuit ElementsInductor cntd. . .

Following can be said about power associated with an inductor

Power across an inductor is pulsating at double thefrequency of voltage source

In a quarter cycle, the inductor absorbs energy and storesit as magnetic energy (when its current is increasing)

In the next quarter cycle, it returns the same to the source

This behavior in steady state proceeds add infinum


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Behaviour of Circuit ElementsCapacitor

i(t)

v(t) C

Ic

Vc

Following equation relations capacitor voltage and current

ic (t) = Cdvc

dt

Let, vc(t) = Vm sin(ωt + φ)

⇒ ic (t) = ωCVm cos(ωt + φ)

⇒ p(t) = VmIm sin (ωt + φ) cos (ωt + φ) [Im = ωCVm]

= Vrms Irms sin (2ωt + 2φ)

⇒ Pav =1

T0

Z T0

0p(τ)dτ = 0

Additionally, S = P + jQ = VI⋆ = 0 − jVrms Irms


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Behaviour of Circuit ElementsCapacitor cntd. . .

2

4

6

−2

−4

i(t)v(t)p(t)

+ × + = + − × + = −− × − = ++ × − = −


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Examples

Behaviour of Circuit ElementsCapacitor cntd. . .

Following can be said about power associated with a capacitor

Power across a capacitor is pulsating at double thefrequency of voltage source

In a quarter cycle, the capacitor absorbs energy and storesas electrostatic energy (when its voltage is increasing)

In the next quarter cycle, it returns the same to the source

This behavior in steady state proceeds add infinum


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Behaviour of Circuit ElementsComments on Inductor and Capacitor

It is often said that inductor consumes and capacitorgenerates reactive power

Average power consumed is zero

Voltage and current are 90 out of phase; in case ofinductor voltage leads the current while voltages lagscurrent in capacitor


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Examples

Euler’s Identity

ex can be represented by following infinite series

ex = 1 +

x

1!+

x2

2!+

x3

3!+ · · ·

=

∞X

r=0

x r

r !

If we permit complex exponential, then with x = jθ, we have

ejθ =

∞X

r=0

j rθr

r !

=∞

X

r=0

j2rθ2r

(2r)!+

∞X

r=0

j2r+1θ2r+1

(2r + 1)!

=∞

X

r=0

`

j2´r

θ2r

(2r)!+

∞X

r=0

j`

j2´r

θ2r+1

(2r + 1)!

=∞

X

r=0

(−1)rθ2r

(2r)!+ j

∞X

r=0

(−1)rθ2r+1

(2r + 1)!⇒ e

jθ = cos θ + j sin θ

The above is a well known Euler’s identity which plays a very important

role in analysis of steady state behavior of circuits.


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Euler’s Identity cntd. . .

It follows that

e−jθ = cos θ − j sin θ

Hence, cos θ =e jθ + e−jθ

2

sin θ =e jθ − e−jθ

2j

If we replace θ with (ωt + φ) we get,

|A| cos(ωt + φ) =Ae jφe jωt + Ae−jφe−jωt

2

⇒ |A| cos(ωt + φ) =Ae jωt + A⋆e−jωt

2

‖ly, |A| sin(ωt + φ) =

Ae jωt − A⋆e−jωt

2j

where,

A = |A|e jφ

= |A| cos φ + j |A| sin φ

= Areal + jAimag


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Examples

Euler’s Identity cntd. . .

Verify following

Multiplicatione jθ × e jα = e j(θ+α)

Differentiationde jθ

dθ= je jθ


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Choice of Reference Wave

Can be either sinωt or cos ωt

Just a matter of convenience

In steady state analysis, sinusoids recur from −∞ to ∞ intime

Reference is decided by that point in time axis which isdesignated to t = 0


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Geometrical representation of ejwt

1

e jω T4

e jω T2

e jω 3T4

ωt

ωt + φ

cos ωt is the projection of the rotating e jωt vector onX-axis

sinωt is the project of rotating unit e jωt vector onquadrature of Y-axis

If the initial position of the unit vector is at position φ, wewill generate cos(ωt + φ) and sin(ωt + φ)


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1

e jω T4

e jω T2

e jω 3T4

ωtωt + φ





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Examples


1

e jω T4

e jω T2

e jω 3T4

ωt

ωt + φ





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Examples

Series R-L circuit I

Claim

Consider a simple series R-L circuit described by Ri + L didt

= v(t). Let,

v(t) = Ae jωt be the forcing function. The steady state response, i(t), is given byBe jωt .

We can verify the claim as follows:

RBe jωt + jωLBe jωt = Ae jωt

⇒(R + jωL)Be jωt = Ae jωt

Hence, B =A

R + jωL

=|A|e jφ

|Z |e jθ

=|A||Z | e

j (φ − θ)

A is the voltage phasor.R + jωL is called the impedance of thecircuit (Z).B is called the current phasor.


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Series R-L circuit II

Thus, the response to input Ae jωt is given by AZ

e jωt i.e.

i(t) =|A||Z | e

j(ωt+φ)−θ

=|A||Z | [cos(ωt + φ − θ) + j sin(ωt + φ − θ)]

Due to linear nature of the circuit, the real part of A is the response to|A| cos(ωt + φ) and imaginary part of it is response to |A| sin(ωt + φ).

Thus if, v(t) = Vm cos(ωt + φ)

Then, i(t) =Vm

|Z | cos(ωt + φ − θ)

and if, v(t) = Vm sin(ωt + φ)

Then, i(t) =Vm

|Z | sin(ωt + φ − θ)

V = A = Vme jφ; I = B =Vm

Ze jφ

⇒ V = Z I

Voltage and current phasors satisfiesOhm’s law (with complex quantities).Notice that unit of impedance is ohms.


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Series R-L-C Circuit

ClaimFor a series R-L-C circuit as shown in figure described by

Ri + Ldi

dt+

1

C

Z T

−∞

i(τ)dτ = v(t)

the phasor response is given by

»

R + jωL − 1

jωC

–

I = V

Hence, complex impedance is given by

Z = R + j(ωL − 1

ωC)

What happens when ωL = 1ωC

?


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Parallel R-L-C Circuit

ClaimConsider a parallel R-L-C network. The governing network equation is given by:

i(t) =v(t)

R+

1

L

Z t

∞

v(τ)dτ + Cdv

dt

The phasor response is given by

I =

»

1

R+

1

jωL+ jωC

–

V

Further , with Y defined such that I = Y V , impedance of the circuit is given by

Y =1

Z=

1

R+

1

jωL+ jωC =

1

R+ j(ωC − 1

ωL)

Admittance Inverse of impedance. Its unit is called mho or siemens and is

denoted either by1

Ωor simply .

What happens if ωL = 1ωC

? [Parallel Resonance]


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Basics of Power Systems: Review of Phasorscs620/PhasorReview...2 Back to Phasors 3 Phasor...

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