Electrical Sciences

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Dr. A. Amalin Prince

BITS - Pilani K K Birla Goa Campus
Department of Electrical and Electronics Engineering
Lecture – 12,13,14,15
AC Analysis: Time- and Frequency-domain analysis, Important Power Concepts

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Some History: AC Power Become
Standard
A Victory for AC Power:
The 1891 International Electro-Technical Exhibition
Lauffen to Frankfurt 1891 The beginning of modern
electric power in the world

The three phase generator installed at Lauffen am

Neckar, Germany. All historic photos provided by
the Historisches Museum, Frankfurt, Germany

Village of Lauffen in 1891 with the Neckar River in

the foreground

Cologne become the first German city to adopt ac system

The Helios display. Over theEEE

6 month
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period 3
1,200,000 visitors attended the exposition
Sinusoids

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Phasors

 Convenient form
 Phasor is a complex number that represents the amplitude
and phase of a sinusoid
 Charles Steinmetz in 1893

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Phasor: Frequency Domain Representation

V is thus the phasor representation of the

sinusoid v(t).
In other words, a phasor is a complex
representation of the magnitude
and phase of a sinusoid.

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Phasor: Frequency Domain Representation

When function Ae j (t  ) is depicted in the complex

plane, the vector formed by its imaginary and real parts
rotates around the origin. Its magnitude is A, and it The sum of phasors as
completes one cycle every 2π/ω seconds. θ is the angle
addition of rotating
it forms with the real axis at t = n•2π/ω, for integerEEE F111 7
values of n. vectors
Time Domain vs Phasor

 The differences between v(t) and V should be emphasized:

 v(t) is the instantaneous or time-domain representation, while V is the
frequency or phasor-domain representation.
 v(t) is time dependent, while V is not. (This fact is often forgotten by
students.)
 v(t) is always real with no complex term, while V is generally complex.
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Practice Problems
Evaluate these complex numbers:

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Practice Problems
Transform these sinusoids to phasors :

Find the sinusoids represented by these phasors:

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Phasor Relationships For Circuit Elements

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Phasor Relationships for Resistor

i (t )  I m cos(t   )  Re(Ie jt )

v(t )  i (t ) R  RI m cos(t   )
V  RI m  =RI

Phasor voltage and current of a

resistor are in phase
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Time Domain Frequency Domain
Phasor Relationship for Resistor

Frequency Domain v(t )  i (t ) R  RI m cos(t   )

V  RI m  =RI

Time Domain Voltage and current of a resistor

are in phase

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Phasor Relationships for Inductor
di d
v(t )  L  L I m cos(t   )   LI m sin(t   )   LI m cos(t    90)
dt dt
V   LI m (  90)= LI m e j e j 90  j LI

Phasor current of an inductor

LAGS the voltage by 90 degrees.
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Time Domain Frequency Domain
Phasor Relationships for Inductor

Frequency Domain
Phasor current of an inductor
LAGS the voltage by 90 degrees.

Time Domain

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Phasor Relationships for Capacitor

dv d
i (t )  C  C Vm cos(t   )  CVm sin(t   )  CVm cos(t    90)
dt dt
I
I  CVm (  90)=CVm e j e j 90  jCV V=
j C

Phasor current of a capacitor LEADS

Time Domain Frequency Domain
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Phasor Relationships for Capacitor

Phasor current of a capacitor

Frequency Domain LEADS the voltage by 90 degrees.

Time Domain

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Phasor Relationships for Circuit Elements

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Phasor Relationships for Circuit Elements

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Impedance and Admittance
 The Impedance Z of a circuit is the ratio of phasor voltage V to the phasor
current I.

V
Z or V =ZI
I
 The Admitance Y of a circuit is the reciprocal of impedance measured in
Simens (S).
I 1
Y
 or I =YV
V Z
 Impedances and Admitances of passive elements.
Element Impedance Admitance
1
R Z=R Y=
R
1
L Z  j L Y=
j L
1
C Z= Y  jC
j C
Impedance as a Function of Frequency
 The Impedance Z of a circuit is a function of the frequency.
Element Impedance Admitance
1
L Z  j L Y=
j L
1
C Z= Y  jC
j C
 Inductor is SHORT CIRCUIT at DC and OPEN CIRCUIT at high frequencies.
Capacitor is OPEN CIRCUIT at DC and SHORT CIRCUIT at high frequencies.
Z L  j L
ZL  0   0 (Short at DC)
Z L      (Open as   )
 0
1
ZC =
j C
Z C     0 (Open at DC)
ZC  0    (Open as   )
Impedance of Joint Elements
 The Impedance Z represents the opposition of the circuit to the flow of
sinusoidal current.

V
Z   R  jX  +
I Z I
=Resistance + j  Reactance V
-
= Z 

X
Z  R X 2 2
  tan 1

R
R  Z cos X  Z sin 

 The Reactance is Inductive if X is positive and it is Capacitive if X is negative.

Practice Problem
Determine v(t) and i(t)
Admittance of Joint Elements
 The Admittance Y represents the admittance of the circuit to the flow of
sinusoidal current.
The admittance is measured in Siemens (s)
+
Y I
V
-

1 I
Y    G  jB
Z V
 Conductance + j  Suseptance= Y 

1 R  jX R  jX
Y  G  jB   2
R  jX R  jX R  X 2
R X
G 2 B 2
R X 2
R  X2
Sinusoidal Steady State Analysis

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Steps to Analyze AC Circuits
 Transform the circuit to the Phasor Domain.

 Solve the problem using circuit techniques listed below

 Krichhoff’s Law
 Current and Voltage division
 Nodal Analysis
 Mesh Analysis
 Superposition
 Source transformation
 Thevenin or Norton Equivalents

 Transform the resulting circuit back to time domain.

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Practice Problem

Determine the input impedance of the circuit at

ω=10 rad/sec

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Practice Problem
Calculate the vo in the circuit given

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Practice Problem: Nodal Analysis
Find ix in the circuit using nodal analysis

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Take-home Problem: Nodal Analysis
Compute V1 and V2 in the circuit

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Practice Problem: Mesh Analysis
Determine current Io in the circuit using mesh analysis

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Practice Problem: Thevenin’s Theorem
Find the Thevenin equivalent of the circuit
as seen from terminals a-b.

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Practice Problem: Norton’s Theorem
Obtain current Io using Norton’s theorem

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AC Power Analysis

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Instantaneous AC Power
 Instantaneous Power p(t) is the power at any instant of time.

Assume a sinusoidal voltage with phase v , v(t )  Vm cos(t  v )

Assume a sinusoidal current with phase i , i (t )  I m cos(t  i )
p(t )  v(t )i(t )
1 1
p(t )  v(t )i (t )  Vm I m cos(v  i )  Vm I m cos(2t  v  i )
2 2
p(t )  CONSTANT POWER+SINUSOIDAL POWER (frequency 2 )

1 1
p(t )  v(t )i (t )  Vm I m cos(v  i )  Vm I m cos(2t  v  i )
2 2

 The instantaneous power is composed of two parts.

• A constant part.

• The part which is a function of time.

Instantaneous AC Power
 Instantenous Power p(t) is the power at any instant of time.
v(t )  Vm cos(t   v ) i (t )  I m cos(t  i )
1 1
p(t )  v(t )i (t )  Vm I m cos( v  i )  Vm I m cos(2t  v  i )
2 2
Instantaneous and Average Power
 The instantaneous power p(t) is composed of a constant part (DC) and a time
dependent part having frequency 2ω.
p (t )  v(t )i (t )
v(t )  Vm cos(t   v ) i (t )  I m cos(t  i )
1 1
p (t )  Vm I m cos( v  i )  Vm I m cos(2t   v  i )
2 2

Instantenous Power p(t)

Average Power
P  12 Vm I m cos( v  i )
Instantaneous and Average Power

p(t )  12 Vm I m cos(v i )  12 Vm I m cos(2t  v  i )  p1 (t )  p2 (t )

Average Power
The average power P is the average of the instantaneous power over one period .

p(t )  v(t )i (t ) Instantaneous Power

1 T
P   p(t )dt Average Power
T 0
v(t )  Vm cos(t  v ) i (t )  I m cos(t  i )
1 T 1 T1 1 T1
P   p(t )dt   2 Vm I m cos( v   i )dt   2 Vm I m cos(2t   v   i )dt
T 0 T 0 T 0
1 T 1 T
P  Vm I m cos(v  i )  dt  2 Vm I m  cos(2t  v  i )dt
1
2
1
T 0 T 0
= 12 Vm I m cos(v  i )  0 (Integral of a Sinusoidal=0)

P  12 Vm I m cos(v  i )
1
P  Re  VI   Vm I m cos(v  i )
1
2


2
Average Power
The average power P, is the average of the instantaneous power over one period .

P  12 Vm I m cos(v  i )
1
P  Re  VI   Vm I m cos(v  i )
1
2


2
 A resistor has (θv-θi)=0º so the average power becomes:

PR  Vm I m  I m R  I R
1 1 2 1 2
2 2 2

1. P is not time dependent.

2. When θv = θi , it is a purely resistive load case.
3. When θv– θi = ±90o, it is a purely reactive load case.
4. P = 0 means that the circuit absorbs no average power.
Instantaneous and Average Power
Calculate the instantaneous power and average power absorbed by a passive
linear network if:

v(t )  80 cos (10 t  20)

i (t )  15 sin (10 t  60)
1 1
p (t )  Vm I m cos( v  i )  Vm I m cos(2t   v  i )
2 2
=385.7  600cos(20t  10) W

P= 385.7 W is the average power flow

Practice Problem
Calculate the average power absorbed by each of the five elements in the circuit
given.
Practice Problem
Maximum Average Power Transfer

 Finding the maximum average power which can be transferred from a linear circuit to a
Load connected.

a) Circuit with a load b) Thevenin Equivalent circuit

• Represent the circuit to the left of the load by its Thevenin equiv.
• Load ZL represents any element that is absorbing the power generated by the circuit.
• Find the load ZL that will absorb the Maximum Average Power from the circuit to which it is
connected.
Maximum Average Power Transfer
Condition
• Write the expression for average power associated with ZL: P(ZL).
ZTh = RTh + jXTh ZL = RL + jXL

2 RL
VTh VTh 1 2 VTh
I  P I RL  2
ZTh  Z L ( RTh  jX Th )  ( RL  jX L ) 2 ( RTh  RL ) 2  ( X Th  X L ) 2
Ajust R L and X L to get maximum P
VTh RL ( X Th  X L )
2
P

X L ( R  R ) 2  ( X  X ) 2  2
 Th L Th L 
P VTh ( RTh  RL )  ( X Th  X L )  2 RL ( RTh  RL ) 
2 2 2


RL 2 2
2 ( RTh  RL )  ( X Th  X L ) 
2

P P
 0  X L   X Th 0  RL  RTh 2  ( X Th  X L ) 2  RTh
X L RL
Z L  RL  jX L  RTh  jX Th  ZTh
Maximum Average Power Transfer
Condition

• Therefore: ZL = RTh - XTh = Z*Th will generate the maximum power transfer.
• Maximum power Pmax
2 2
I L RL VTh
Pmax  
2 8 RTh

 For Maximum average power transfer to a load impedance ZL we must choose ZL as the
complex conjugate of the Thevenin impedance ZTh.

Z L  RL  jX L  RTh  jX Th  Z Th
2
VTh
Pmax 
8 RTh
Practice Problem
Calculate the load impedance for maximum power transfer and the maximum
average power.
Practice Problem
Maximum Average Power for Resistive
Load
 When the load is PURELY RESISTIVE, the condition for maximum power transfer is:

XL  0  RL  RTh 2  ( X Th  X L )2  RTh 2  X Th 2  ZTh

 Now the maximum power can not be obtained from the Pmax formula given
before.
 Maximum power can be calculated by finding the power of RL when XL=0.

● RESISTIVE
LOAD

●
Maximum Average Power for Resistive
Load
Calculate the resistive load needed for maximum power transfer and the
maximum average power.
Maximum Average Power for Resistive
Load

RL

 Notice the way that the maximum power is calculated using the Thevenin
Equivalent circuit.
Effective or RMS Value
 The EFFECTIVE Value or the Root Mean Square (RMS) value of a periodic current
is the DC value that delivers the same average power to a resistor as the periodic
current.

a) AC circuit b) DC circuit

1 T R T
P   i (t ) Rdt   i (t ) 2 dt  I eff 2 R  I Rms 2 R
2

T 0 T 0
1 T 1 T
I eff  I Rms   Veff  VRms  
2 2
i (t ) dt v (t ) dt
T 0 T 0
Effective or RMS Value of a Sinusoidal
 The Root Mean Square (RMS) value of a sinusoidal voltage or current is equal
to the maximum value divided by square root of 2.

1 T 2 Im2 T 1 Im
   tdt   (1  cos 2 t )dt 
2
I Rms I m cos
T 0 T 0 2 2

P  12 Vm I m cos(v i )  VRms I Rms cos(v i )

 The average power for resistive loads using the (RMS) value is:

2
V
PR  I Rms 2 R  Rms
R
Practice Problem
 Find the RMS value of the current waveform. Calculate the average power if the
current is applied to a 9  resistor.

4t
8-4t T 2
 4t 0  t 1
i (t )  
8  4t 1  t  2

  i dt   
1 T 2 1 1 2
0  1 
2 2 2
I (4t ) dt (8 4t ) dt
2  
rms
T 0

16  1 2  1  t 3
 2  16
I rms    
t dt  (4  4t  t ) dt 
2 2 2
2
2
I rms  8    4t  2t   1  
2
0 1  3  3  3
16
I rms   2.309A  16 
3 PI 2
rms R   (9)  48W
 3
Take-home Problem
Take-home Problem
Determine the rms value of the current waveform in Figure. If the current is passed through a
2Ω resistor, find the average power absorbed by the resistor.

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