Review ofPhasors
Prof S. A.Soman
SinusoidalNature
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Behaviour ofCircuitElements
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
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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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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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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
Prof S. A. Soman (IIT-Bombay) Review of Phasors August 29, 2012 3 / 41
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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
Prof S. A. Soman (IIT-Bombay) Review of Phasors August 29, 2012 4 / 41
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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
)
Prof S. A. Soman (IIT-Bombay) Review of Phasors August 29, 2012 6 / 41
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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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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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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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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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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)]
Prof S. A. Soman (IIT-Bombay) Review of Phasors August 29, 2012 14 / 41
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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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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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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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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.
Prof S. A. Soman (IIT-Bombay) Review of Phasors August 29, 2012 18 / 41
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Power Triangle at Different Power Factors
Leading
Q
P
S
Lagging
Q
P
S
Unity
P
S
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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).
Prof S. A. Soman (IIT-Bombay) Review of Phasors August 29, 2012 20 / 41
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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?
Prof S. A. Soman (IIT-Bombay) Review of Phasors August 29, 2012 22 / 41
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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
Prof S. A. Soman (IIT-Bombay) Review of Phasors August 29, 2012 23 / 41
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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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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
Prof S. A. Soman (IIT-Bombay) Review of Phasors August 29, 2012 25 / 41
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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
Prof S. A. Soman (IIT-Bombay) Review of Phasors August 29, 2012 27 / 41
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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
Prof S. A. Soman (IIT-Bombay) Review of Phasors August 29, 2012 28 / 41
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Behaviour of Circuit ElementsCapacitor cntd. . .
2
4
6
−2
−4
i(t)v(t)p(t)
+ × + = + − × + = −− × − = ++ × − = −
Prof S. A. Soman (IIT-Bombay) Review of Phasors August 29, 2012 29 / 41
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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
Prof S. A. Soman (IIT-Bombay) Review of Phasors August 29, 2012 30 / 41
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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
Prof S. A. Soman (IIT-Bombay) Review of Phasors August 29, 2012 31 / 41
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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.
Prof S. A. Soman (IIT-Bombay) Review of Phasors August 29, 2012 32 / 41
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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
Prof S. A. Soman (IIT-Bombay) Review of Phasors August 29, 2012 33 / 41
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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)
Prof S. A. Soman (IIT-Bombay) Review of Phasors August 29, 2012 36 / 41
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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)
Prof S. A. Soman (IIT-Bombay) Review of Phasors August 29, 2012 36 / 41
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Geometrical representation of ejwt
1
e jω T4
e jω T2
e jω 3T4
ωtωt + 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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Geometrical representation of ejwt
1
e jω T4
e jω T2
e jω 3T4
ωtωt + 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)
Prof S. A. Soman (IIT-Bombay) Review of Phasors August 29, 2012 36 / 41
Review ofPhasors
Prof S. A.Soman
SinusoidalNature
Back toPhasors
PhasorComputation
Behaviour ofCircuitElements
Working withComplexExponential
Examples
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)
Prof S. A. Soman (IIT-Bombay) Review of Phasors August 29, 2012 36 / 41
Review ofPhasors
Prof S. A.Soman
SinusoidalNature
Back toPhasors
PhasorComputation
Behaviour ofCircuitElements
Working withComplexExponential
Examples
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)
Prof S. A. Soman (IIT-Bombay) Review of Phasors August 29, 2012 36 / 41
Review ofPhasors
Prof S. A.Soman
SinusoidalNature
Back toPhasors
PhasorComputation
Behaviour ofCircuitElements
Working withComplexExponential
Examples
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)
Prof S. A. Soman (IIT-Bombay) Review of Phasors August 29, 2012 36 / 41
Review ofPhasors
Prof S. A.Soman
SinusoidalNature
Back toPhasors
PhasorComputation
Behaviour ofCircuitElements
Working withComplexExponential
Examples
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)
Prof S. A. Soman (IIT-Bombay) Review of Phasors August 29, 2012 36 / 41
Review ofPhasors
Prof S. A.Soman
SinusoidalNature
Back toPhasors
PhasorComputation
Behaviour ofCircuitElements
Working withComplexExponential
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.
Prof S. A. Soman (IIT-Bombay) Review of Phasors August 29, 2012 37 / 41
Review ofPhasors
Prof S. A.Soman
SinusoidalNature
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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.
Prof S. A. Soman (IIT-Bombay) Review of Phasors August 29, 2012 38 / 41
Review ofPhasors
Prof S. A.Soman
SinusoidalNature
Back toPhasors
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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
?
Prof S. A. Soman (IIT-Bombay) Review of Phasors August 29, 2012 39 / 41
Review ofPhasors
Prof S. A.Soman
SinusoidalNature
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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]
Prof S. A. Soman (IIT-Bombay) Review of Phasors August 29, 2012 40 / 41
Review ofPhasors
Prof S. A.Soman
SinusoidalNature
Back toPhasors
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Thank You
Prof S. A. Soman (IIT-Bombay) Review of Phasors August 29, 2012 41 / 41
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