A sinusoidal current source (independent or dependent) produces a currentThat varies sinusoidally with time

t

t T

2 2

Tt 12 f

OR t

We define the Root Mean Square value of v(t) or rms as

Take the square oot R

calculte the ean M

quare (t)vS

The Root Mean Square value of

Expand using trigonometric identity

1T ms 1 100 0 f HZT

2 2000 /f rad s

20(0) cos( )i 10 co20 s( )

1

2010cos 60o

( ) cos(20002 60 ) A0 oi t t

20

2rmsI 14.44 A

9.2 The Sinusoidal Response

In this chapter, we develop a technique for calculating the response directly without solving the differential equation

2 H

3 ( )i t

( ) Rv t

( ) Lv t

( ) cos(3 410 )0SoV t t

c 40os(3 )10 odiL Ri tdt KVL

( ) cos(31.58 1 56 )3 . oi t t

( ) cos(33 ).1 31.56oRv t t

( ) cos(39 ).5 58.43o

Lv t t

Solution for i(t) should be a sinusoidal of frequency 3

We notice that only the amplitudeand phase change

( )SdiL Ri V tdt

2 H

3 ( )i t

( ) Rv t

( ) Lv t

( )SV t

Time Domain Complex Domain

Deferential Equation Algebraic Equation

cos( ) sin( )j je

cos( ) { }je sin( ) { }je

9.3 The phasor

The phasor is a complex number that carries the amplitude and phase angle information of a sinusoidal function

The phasor concept is rooted in Euler’s identity

Euler’s identity relates the complex exponential function to the trigonometric function

We can think of the cosine function as the real part of the complex exponential and the sine function as the imaginary part

Because we are going to use the cosine function on analyzing the sinusoidal steady-state we can apply

cos( ) { }je

cos( ) sin( )j je cos( ) { }je sin( ) { }je

cos( )mv V t ( ){ }j tmV e { }j t j

mV e e

{ }j j tmv V e e

jmV e cos( ){ } mV tP

Phasor Transform

Were the notation cos( ){ } mV tP

Is read “ the phasor transform of cos( )mV t

We can move the coefficient Vm inside

jmV e The quantity is a complex number define to be the phasor

that carries the amplitude and phase angleof a given sinusoidal function

=V

Summation Property of Phasor

1 2where , , are sinusoida lnv v v

)can be shown(

Since

Next we derive y using phsor method

cos() ( )m ii t tI

cos( )m iR tI

(( )) Rv it t

cos( )m iR tI

The VI Relationship for a Resistor

Let the current through the resistor be a sinusoidal given as

(( )) Rv it t

voltage phase

( ) cos( )m iR tt Iv

Is also sinusoidal with

amplitude mmV RI And phase iv

The sinusoidal voltage and current in a resistor are in phase

R

(t)i

( )v t

cos() ( )m ii t tI

cos( ( )) m iR tv t I

Now let us see the pharos domain representation or pharos transform of the current and voltage

i

mjI e I mI i

RV I

i

mjRI e V mRI i

m

V

v

Which is Ohm’s law on the phasor ( or complex ) domain

R

I

V

R

(t)i

( )v t

cos() ( )m ii t tI Phasor Transform

cos( ( )) m iR tv t I Phasor Transform R= I

RV IR

I

V

The voltage and the current are in phase

Real

Imaginary

I

V

iv

mV

mI

i

The VI Relationship for an Inductor

(( )) dLv ittt

d

cos() ( )m ii t tI Let the current through the resistor be a sinusoidal given as

(( )) dLv ittt

d sin( )m itIL

The sinusoidal voltage and current in an inductor are out of phase by 90o

The voltage lead the current by 90o or the current lagging the voltage by 90o

You can express the voltage leading the current by T/4 or 1/4f seconds were T is the period and f is the frequency

L

(t)i

( )v t

cos() ( )m ii t tI

si( n) ( )m iv t tIL

Now we rewrite the sin function as a cosine function (remember the phasor is defined in terms of a cosine function)

ocos( 90 )( ) m iv t tIL

The pharos representation or transform of the current and voltage

i

mjI e I mI i

o( 90 ) ijmL eI V

o 90 ij jm

jL e eI

i

mjLIj e

cos() ( )m ii t tI

ocos( 90 )( ) m iv t tIL

But since 901

ojj e 1 o90

Therefore i

mjIj L e V 90

o ij jmL e eI ( 90 )

oijmeIL mIL ( 90 )o

i

m

V

v

m mIV L and 90i ov

L

(t)i

( )v t

j L

V

I

j LV I

m mIV L and 90i ov

The voltage lead the current by 90o or the current lagging the voltage by 90o

Real

Imaginary

I

i

V

vmV

mI

The VI Relationship for a Capacitor

C

(t)i

( )v t

(( )) dCi vttt

d

cos(( ) )vmv t tV Let the voltage across the capacitor be a sinusoidal given as

(( )) dCi vttt

d sin( ) vm tVC

The sinusoidal voltage and current in an inductor are out of phase by 90o

The voltage lag the current by 90o or the current leading the voltage by 90o

The VI Relationship for a Capacitor

C

(t)i

( )v t

cos(( ) )vmv t tV sin( )( ) vmi C tt V

The pharos representation or transform of the voltage and current

cos(( ) )vmv t tV v

mjV e V mV v

sin( )( ) vmi C tt V cos ( 90 )vmoC tV o( 90 ) vj

mL eV I

o 90 v

mj

j jC e eV

I v

mjVj C e j C V

j CV I

90 1

o

i

mj

je

e Cj

I

( 90 )

oijme

CI

mC

I ( 90 )o

i

m

mIV

C and 90i ov

m

V

v

m

mIV

C and 90i ov

j CV I

1 j C

V

I

The voltage lag the current by 90o or the current lead the voltage by 90o

Real

Imaginary

I

i

V

v

mV

mI

R

I

V

j L

V

I

RV I

Real

Imaginary

I

V

iv

mV

mI

Real

Imaginary

I

i

V

vmV

mI

j LV I

1 j C

V

I

Real

Imaginary

I

i

V

v

mV

mI

j CV I

cos() ( )m ii t tI

si( n) ( )m iv t tIL

cos() ( )m ii t tI

cos( ( )) m iR tv t I

R

(t)i

( )v t

C

(t)i

( )v t

cos(( ) )vmv t tV

sin( )( ) vmi C tt V

Time-Domain Phasor ( Complex or Frequency) Domain

(( )) Rv it t

(( )) dLv ittt

d

RV I

j LV I

j CV I

(( )) dCi vttt

d

L

(t)i

( )v t

Impedance and Reactance

The relation between the voltage and current on the phasor domain (complex or frequency) for the three elements R, L, and C we have

RV I

j C

V I j LV I

When we compare the relation between the voltage and current , we note that they are all of form:

ZV I Which the state that the phasor voltage is some complex constant ( Z ) times the phasor current

This resemble ( شبه ) Ohm’s law were the complex constant ( Z ) is called “Impedance” (أعاقه )

Recall on Ohm’s law previously defined , the proportionality content R was real and called “Resistant” (مقاومه )

Z VISolving for ( Z ) we have

The Impedance of a resistor is

j C 1 I

Z R R

1Z

C j C

Z L j LThe Impedance of an indictor is

The Impedance of a capacitor is

In all cases the impedance is measured in Ohm’s

The reactance of a resistor is X 0R

1X C C

X L LThe reactance of an inductor is

The reactance of a capacitor is

The imaginary part of the impedance is called “reactance”

We note the “reactance” is associated with energy storage elements like the inductor and capacitor

The Impedance of a resistor is Z R R

1Z

C j C

Z L j LThe Impedance of an indictor is

The Impedance of a capacitor is

In all cases the impedance is measured in Ohm’s

RV I j LV I j C

1V I

Z VI

Impedance

Note that the impedance in general (exception is the resistor) is a function of frequency

At = 0 (DC), we have the following

Z L j L (0)j L 0 short

1Z

C j C(

1 0) j C

open

Time Domain

Phasor (Complex) Domain

9.5 Kirchhoff’s Laws in the Frequency Domain ( Phasor or Complex Domain)

Consider the following circuit

1 1 1( ) cos( )v t V t

2 2 2( ) cos( )v t V t

4 4 4( ) cos( )v t V t

3 3 3( ) cos( )v t V t

KVL 21 3 4( ) 0) ( )( ( )vv t v tt tv

3 3 4 41 1 2 2 cos cos(cos( )co ( s( ) ) 0 ) V V tV ttt V

Using Euler Identity we have 42 313 421{ } { } { } { } 0j j tj j j j jtt j tV e e V eVV e ee ee

Which can be written as 42 313 421{ } 0j j t j jj j t tj jt VVV e Vee ee ee e

Factoring j te 321 4

431 2{( ) } 0j tjj j jV eV ee V eeV

1

V

2

V

3

V

4

V

2( ) v t

4( ) v t

3 ( )

+

v t

L

C

1( ) v t

1R

2R

21 43+ + + = 0V VV V

KVL on the phasor domain

11 1= jV e V

22 2= jV e V

33 3= jV e V

44 4= jV e V

Phasor Transformation

PhasorCan not be zero

So in general n1 2+ + + = 0V V V

Kirchhoff’s Current Law

A similar derivation applies to a set of sinusoidal current summing at a node

n1 2+ + + = 0I I I

1 2( ) ( ) ( ) 0ni t i t i t

1 1 1( ) cos( )i t I t 2 2 2( ) cos( )i t I t ( ) cos( )n n ni t I t

11 1= jI e I 2

2 2= jI e I = nn n

jI e IPhasor Transformation

KCL

KCL on the phasor domain

9.6 Series, Parallel, Simplifications

and Ohm’s law in the phosor domain

Example 9.6 for the circuit shown below the source voltage is sinusoidal

)a (Construct the frequency-domain (phasor, complex) equivalent circuit?

o( ) 750cos(5000 30 )sv t t

LZ j L (5000)(32 X 10 )j 160 j The Impedance of the indictor is

1Z

C j CThe Impedance of the capacitor is 6

1 (5000) (5 X 10 )j

40 j

o30=750sjeVThe source voltage pahsor transformation or equivalent

o=750 30

)b (Calculte the steady state current i(t)?

ab

sIZ

V

To Calculate the phasor current Io30750

90 160 40

jej j

o3075090 120

jej

o

o

750 30

150 53.13

o 5 23.13 A

o( ) 5cos(5000 23.13 ) Ai t t

o( ) 750cos(5000 30 )sv t t

and Ohm’s law in the phosor domain

Example 9.7 Combining Impedances in series and in Parallel

)a (Construct the frequency-domain (phasor, complex) equivalent circuit?

)b (Find the steady state expressions for v,i1, i2, and i3? ?

( ) 8cos(200,000 ) Asi t t

)a(

Delta-to Wye Transformations

to Y

Y to

Example 9.8

9.7 Source Transformations and Thevenin-Norton Equivalent Circuits

Source Transformations

Thevenin-Norton Equivalent Circuits

Example 9.9

Example 9.10

Source Transformation

Since then

Next we find the Thevenin Impedance

KVL

Thevenin Impedance

TT T

T

Find interms of then form the ratio V

I VI

Ta bFind and interms of I I V

12 60

9.8 The Node-Voltage Method

Example 9.11

KCL at node 1

KCL at node 2 Since

)1(

)2(

Two Equations and Two Unknown , solving

To Check the work

9.9 The Mesh-Current Method

Example 9.12

KVL at mesh 1

KVL at mesh 2 Since

Two Equations and Two Unknown , solving

)1(

)2(

9.12 The Phasor Diagram