“Electric Circuits Analysis ||”

Electric Circuits Analysis || Reference Books. Fundamentals of Electric Circuits. By: Charles K. Alexander Matthew N. O. Sadiku Any other book of Electric Circuits

Historical Contribution by Scientists Heinrich Rudorf Hertz (1857-1894), demonstrated that the Electromagnetic Waves obeys the same principle as light.

Heinrich Rudorf Hertz work is confirmed by the Clerk Maxwell’s.

Heinrich Rudorf Hertz was born in Germany. Hertz did his doctorate under the well-known and prominent physicist Hermann von Helmholtz.

Hertz successfully generated and detected Electromagnetic waves.

It is due to Hertz contribution that Electromagnetic waves paved the way for practical use of such waves in Radio, Television and other Communication Systems.

The unit of the frequency is given by his name Hertz Introduction of Sinusoid Signal

DC sources were the main sources of providing electric power un-till the late 1800s.

AC ( Alternating Current ) is introduced in the beginning of 1900 century.

The biggest advantage of the AC is that is more efficient to transmit over long distances.

Sinusoid is a signal that has the form of the sine or cosine function.

Sinusoid Current is referred as AC, the AC current reverses at regular time interval and has alternative ( +ve ) and ( -ve ) values.

AC circuits are those which are driven by sinusoid Current or sinusoidal voltages

Why Sinusoid ?

Nature it self Characteristically sinusoid. Examples are, Motion of pendulum Vibration of String Water waves or the ripples on the ocean surface The biggest advantage of the Sinusoid is that it is easy to generate and transmit

Sinusoid signal is in the form of voltage generated throughout the world, then supplies to homes, factories, laboratories, universities and so on.

Sinusoid signals are easy to handle mathematically, the derivatives and integrals of the sinusoid are them selves sinusoid.

Sinusoid Mathematics Sinusoid Voltage is define as, Vm = the amplitude of the sinusoid

tVtv m ωsin)( =w = the angular frequency in radian/s

Time Period= Sinusoid Signal repeats itself every T seconds, the T is called period of the sinusoid. In sinusoid the repeat it selves every T seconds, so this can be replaced by, t+T

ωπ2

=T

)()(sin)(

sin)()2sin()(

)2(sin)(

)(sin)(

TtvtvtVtv

tVTtvtVTtv

tVTtv

TtVTtv

m

m

m

m

m

+==

=++=+

+=+

+=+

ωω

πωωπω

ω Periodic Function

A Periodic function is define as, the function that satisfies,

)()( nTtftf +=

For all t and for all integers n. Frequency, Time Period

Periodic T of the periodic function is the time of the one complete cycle, this can be define in other words as number of the cycles per seconds.

The reciprocal of the number of the cycles per second is known as frequency of sinusoid

Tf 1

=

f is in hertz (Hz) General Expression for Sinusoid

The general Expression for the sinusoid signal is given as

)sin()( φω += tVtv m

• The amplitude is Vm • The phase is • The angular frequency is

φω

Examine the two Sinusoids

The two sinusoid can be examine as,

and

tVv m ωsin1 =

)sin(2 φω += tVv m Leading and Lagging

The figure shows that leads by by 2v 1v φ

2v1v

This can be said as lags by by

If =0 then and are said to be in phase.

If 0 then and are said to be out of phase.

It is not necessary that and have the same apmlitude.

Important Trigonometry Identities

A sinusoid can be expressed in either sine or cosine form. When comparing two sinusoids, it is expedient to express both as either sine or cosine with positive amplitude. This can be done by using following identities

By looking at the identities, it can be seen that

Sinusoids Addition

The addition of two sinusoids and , A and B are the magnitudes, the resultant sinusoid in cosine form is given as following.

tttttttt

ωωωωωωωω

sin)90cos(,cos)90sin(cos)180cos(,sin)180sin(

m=±±=±−=±−=±

1v vφ 2

1v v≠φ1v2

v2

sin(BABABABABABA

sinsincoscos),sincoscossin)

m=±=± ±

cos(

tA ωcos tB ωsin

)cos(sincos θωωω −=+ tCtBtA

Here,

AB1tan −=θ

, 22 BAC +=

Sinusoids Addition Example The example shows the addition of two sinusoid,

add and , tωcos3 tωsin4−

3)4(tan 1−=θ

)1.53cos(5

sin4cos3

+

−

t

tt

ω

ωω5)4()3( 22 =+=C

Exercise Problems. Add the following sinusoids,

tttt

tttttt

tttt

tttt

tt

ωωωωωωωωωωωω

ωωωωωω

ωω

sin4,cos5,10sin,cos6,9

sin2,cos5,8sin8,cos7,7sin3,cos4,6

sin3,cos2,5sin5,cos6,4

sin5,cos7,3sin3,cos4,2

sin2,cos5,1

−−−−−

−−−−

−−−−−

−−−

−

Problem 2 Find the amplitude, phase , period and frequency of the sinusoid ? )1050cos(12)( += ttv

• The amplitude Vm is 12 • The phase is 10 • The angular frequency is = 50 rad/s

φω

• The period T is 0.1257 s • The frequency f is 7.958 Hz

Problem 2.1 Find the amplitude, phase , period and frequency of the sinusoid ?

)604sin(5)( 0−= ttv π

• The amplitude Vm is 5 • The phase is -60 • The angular frequency is = 12.57 rad/s

φω

• The period T is 0.5 s • The frequency f is 2 Hz

Exercise Problems. Find the amplitude, phase, period and frequency of the sinusoids.

)1510cos(12)()256sin(10)(

)403sin(16)()105cos(4)(

)309cos(8)()2533sin(10)(

0

0

0

0

0

0

−=

−−=

+=

−=

−=

+=

ttvttv

ttvttv

ttvttv

π

π

π Problem 2.2 Calculation of phase angle and leading sinusoid ?

The phase can be calculate in three ways, the two ways are by using trigonometric identities, while the third one is by using graphical method.

Method 1: Let ,

)10sin(12),50cos(10 21 −=+−= tvtv ωω )230cos(10

),130cos(10)18050cos(10)50cos(10

1

1

1

+=−=

−+=+−=

tvortv

ttv

ωω

ωω

Problem 2.2 Calculation of phase angle and leading sinusoid ? Cont’

)100cos(12)9010cos(12)10sin(12

2

2

−=−−=−=

tvttv

ωωω

)30130cos(122 +−= tv ωWe can write V2 as, v

)30130cos(12)130cos(10

2

1

+−=−=

tt

ωω

v

This shows that the phase difference between V1 and V2 is 30. Problem 2.2 Calculation of phase angle and leading sinusoid ? Cont’ Method 2: We can express V1 in sine form,

)10sin(12,

)3010sin(10)40sin(10)9050sin(10)50cos(10

2

1

1

−=

−−=−=−+=+−=

t

v

tttt

ω

ωωω ω

v but

vComparing V2 and V1 , it shows that V1 lags V2 by 30. Problem 2.2 Calculation of phase angle and leading sinusoid ? Cont’ Method 3:

)10sin(12),50cos(10 21 −=+−= tvtv ω ωFrom the above it can be express V1 as , with phase shift of

tωcos10−

+50. with the phase shift of -10

“PHASORS” A Phasor is complex number that represent the amplitude and phase of a sinusoid.

The advantage of the phasors that, sinusoids are easily expressed in terms of phasors, that are more convenient to work with than sine and cosine function.

The idea of solving AC circuits using phasors was first introduced by Charles Steinmetz in 1893

Phasors provides a simple means of analyzing linear circuits excited by sinusoidal sources, solution of such circuits would be difficult otherwise.

Mathematics of Phasors

A complex number z can be written in rectangular form as jyxz +=

where x is real part of z, y is imaginary part of z. ,1−=j

The complex form z can be written in polar or exponential

form as, φφ jrerz =∠=

r is magnitude of z, and is the phase of z.

Mathematics of Phasors (cont’) Representation of Phasors.

jyxz +=o Rectangular Form

o Polar form φ∠= rz

o Exponential Form φjrez =

Where, , So, Mathematics operations on Phasors

when dealing with phasors, addition and subtraction of complex numbers are better perform in rectangular form.

Multiplication and division are better done in polar form.

Addition

Subtraction

Multiplication

Division

Reciprocal

Square root

xyyxr 2 tan, −=+= φ φφ sin,cos ryrx ==12

)sin(cos φφφ jrrjyxz +=∠=+=

)()( 212121 yyjxxzz +++=+

)()( 212121 yyjxxzz −+−=−

)( 212121 φφ ∠+∠= rrzz)( 21

2

1

2

1 φφ ∠−∠=rr

zz

)(11 φ−∠=rz

)2( φ∠=z r

Phasors Representation

The idea of Phasor representation is based on the Euler’s Identity.

)Im(sin)Re(cos

φ

φ

φ

φj

j

ee

=

= given as, φφφ sincos je j ±=± Given a sinusoid, )cos()( φω += tVtv m

φ

φω

φ

ω

ωφ

φω

∠==

=

=

=+= +

mj

m

tj

tjjm

tjmm

VeVVwhere

Vetvthus

eeVtv

eVtVtv

,)Re()(

,)Re()(

)Re()cos()( )( Phasors Representation

V is the phasor representation of sinusoid of v(t). A phasor is a complex representation of magnitude and phase of sinusoid.

Phasor has magnitude and phase “ direction”, it behaves as vector.

Phasors , and , Graphical representation of phasor is known as a Phasor.

=VV φ∠m θ−∠= II m

Phasors Representation

Sinusoid-Phasor Transformation

Time-Domain Representation Phasor-domain Representation

ω

ωθθφφ

jV

VjIIVV

m

m

m

m

90

90

−∠∠

−∠∠

∫

++++

vdtdtdv

tItItVtV

m

m

m

m

)sin()cos()sin()cos(

θωθωφωφω

Problems. Evaluate these complex number.

( ){ }( ) ( )( ){ }

( ) ({ }

Taking square root.

Problems 2 Evaluate these complex number

( )( )

31.160565.048.12208.2666.3773.14

)53)(42()43(3010

,tan,

2214966.11

)53)(42()43()566.8(

)53)(42()43(3010

)43)(42()43()30sin(30cos10

)53)(42()43(3010

)53)(42()43(3010),2(

*

221

*

*

*

−∠=∠−∠

=−+

−+−∠

+=⎟⎠⎞

⎜⎝⎛=

+−−

=++

−+−=

−+−+−∠

++−+−+−

=−+

−+−∠−+

−+−∠

−

jjj

yxrxybut

jj

jjjj

jjj

jjjj

jjj

jjj

θ

)

( ) ( )

81.1291.6)30205040(

63.2503.4364.20tantan

72.4764.20028.43,

64.2003.43)302050401032.1764.30708.25)30205040

1032.1730sin30cos203064.30708.2550sin50cos4050

,)30205040();1

21

11

2222

21

∠=−∠+∠

=⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛=

=+=+=

+=−∠+∠−++=−∠+∠−=−+−=−∠

+=+=∠

−∠+∠

−−

xy

yxr

jj

jjj

θ

(solution4020((

where

Exercise Problems Evaluate these complex number

sors.

[ ]

sors.

he phasor form of v is,

[ ]

Problems 2 Transformation of sinusoids to Phaansformation of sinusoids to Pha

.

.

TThe phasor form of v is,

301043

403510),2(

605)41)(25(),1( *

∠++−

∠++

∠−+−+

jj

jj

)14030cos(4)905030cos(4)5030(4

∠==+=

+=++=+−=

,)5030sin(4),1( +−=

1404)14030cos(4 Vtv

ttt

wheretv

ttttttt

ωωωω

vv

tωωωω

sin)90cos(,cossincos)180cos(,sin)180sin(

m=±±)90( =±−=± ± = −

Problems 2.1 ansformation of sinusoids to Phasors.

he phasor form of I is,

roblems 2.2 oids to Phasors

ransformation of sinusoids to Phasors.

406),40

)4050cos(6

−∠=−

Tr

− T 50cos(6=

PTransformation of sinus The phasor form of V is,

=

Ii

ti

2207

)2202cos(7)402cos(7)180402cos(7)40

∠=

+=+−=++=+

V

ttvtv

solutionv

tttttttt

ωωωω

t

)402cos(7 +−= t:

ω

Problems 2.3 T

2cos(7−= t

ωωωsin)90cos(,cos)90sin(

cos)180cos(,sin)180sin(m=±±=±−==± − ±

,solution tω cos(4)1010sin(4 =+= ti

Problems 2.4 Find the sum of Sinusoids

hehe is change from sine to cosine form

hasor can be written as,

804),8010cos(4)8010cos(4)1010sin(4

)901010

)1010sin(4

−∠=−=−=+=

−+

+=

Itittit

ti

ttttttt

ωωω

T can be written as, T P The addition of Transforming into time do

main,

ωωωωsin)90cos(,cos

cos)180cos(,sin)180sin(m=±±=

± = − ± = −)90sin( ±

)20sin(5)2()30cos(4)1(

2

1

−=+=

titi

ωω

)(1 ti 304),30cos(4)( 11 = ∠=+ Itti ω)(2 ti

}

)97.56(

97.56218.3

)110sin()110cos(5)30sin()30cos(4110

)110cos(5))9020cos(5)20sin(5)

21

2

2

2

−∠

−+−++−∠

−=(

{ } {

cos(218.3) −=

698.2.1 −754698.471.12.3 −−+= 464

5304 +∠=+

1105 −∠=

(

==

==

−−=−=

i

II

jIjjI

I

I

ttittti

ωω ω

2i+1ii =II

jj

t tω

Problems ind the sum of S

2.5 F inusoids.

he sinusoid is change from sine to cosine.

sing the phasor approach, determine the current in a circuit escribe by integrodifferential equation.

)45cos(20)30sin(10

2

1

−=+= −

tvtv

ωω

T

Problems 2.6 Ud

{ )45cos(20 +−= }j

jttt

)45sin()45

66.85)120cos(10)9030cos(10)30sin(10

2

1

1

−

+−=+=v +−=

ωω = ω + +

v

tv cos(202 −= ωv

j14.1414.141 −=v

)94.30cos(66.10)(

65.10,94.30tan

48.514.9)14.1466.8()14.145(

221

21

−=

=+==⎟⎠⎞

⎜⎝⎛=

−=−++−=

+=

−

ttv

yxrxy

jvj

vv

ω

φ

)(ti

vv

Transfer domain from time to phasor.

sing the phasor approach, determine the current in a circuit escribe by integrodifferential equation.

Representation of voltage or current in the phasor domain involving passive elements R,L and C in the circuit, the requirement is that transform the voltage-current relationship from time domain to phasor domain.

The current through resistor R is,

Now convert to time domain.

Practice Problems Ud

Phasor Relationship for Circuit Element

∫

∫

+=++

−=++

)104sin(2045),(

)305cos(201052),

tvdtvdvb

tvdtvdtdva(

dt

)2.1432cos(642.4)(

2.143642.42.6277.10

75501047550

7550)644(2,

)752cos(5038

+=

∠=−∠

∠=

−∠

=

∠=−−=

+=−+ ∫

tt

jI

jjhere

j

tdtdiidti

ωω

)(tv

4

7550384 ∠=−+ IjII ω

WI

i

by applying Ohm’s l aw, the voltage across it is given by,

the phasor form of the voltage is,

hasor Relationship Current-Voltage for RVI rel

VI relationship for R in Phasor domain.

the current I is represented in phasor form as,

Pesistor

ationship for R in time domain.

for the resistor the current and voltage are in phase.

Phasor Relationship Current-Voltage for Inductor. VI relationship for L in time domain.

VI relationship for L in Phasor domain

The voltage and current are 90 out of Phase, the current lags the voltage by 90.

Phasor Relationship Current-Voltage for Capacitor.

VI relationship for L in time domain.

VI relationship for L in Phasor domain

The voltage and current are 90 out of Phase, the current Leads the voltage by 90.

Summary of Voltage – Current Relationship. Elements

Time Domain

Frequency Domain

R

L V Problems 3.

s applied to a 0.1H inductor,

iRv =

The voltage iFind the steady state current through the inductor. Solution: For the inductor, rad/sec.

dtdiLv =

dtdiCi =

IRV =

LIjV = ω

CjIVω

=

)4560cos(12 += tv

60, == ωω whereLIjV

4512)4560cos(12= tv + = ∠

Converting to time domain,

roblems 3.1

e applied tur

,

PThe voltag

rent through the capacitor. o a 50uF

capacitor, Find the cSolution: For the capac tor rad/sec.

i

C

onverting to time domain,

A

I

AI

jjVI

)4560cos(2

,452

452

61.060

−=

−∠=

−∠===

×Lj45124512

tti )(

LjV 4512 ∠

906∠

∠=

××∠

==ω

ω

6 )30100cos(= tv −

100, =ωrej

I=

ωwhe

CV

)30100cos(6= tv −

)60100cos(30)

603090

105010033

6

+=

∠=∠××−∠=××−∠=

×××−−

−

tti

mAIjI

jI

Cjv

ω306, −∠

I= =

105306105306

(

Impedance The impedance Z of a circuit is the ratio of the phasor voltage V to the phasor current I measured in ohms Ώ.

voltage-current relationship for passive

e equat he ratio of the

nce The impedance represent the opposition, which the circuit exhibit to the flow of sinusoidal current.

The impedance is the ratio of two phasor it is not a phasor. The impedance for inductor and for capacitor is

r ( for DC sources),

s uit.

We obtained theelements as,

, , IRV = LIjV

Thes ion can be written in terms of tphasor voltage to the phasor corrent.

From the above expression we obtain the Ohm’s law in phasor form for the above type of elements as,

OR

Z is a frequency dependant quantity known as impedance, measured in ohms.

Impeda

Two cases fo

The inductor act like short circuit , where as capacitor actlike open circ

The two cases for ( for high frequencies)

ω= CjωV 1=

RI

V= Lj

Iω=

VCjI

Vω1

=

IVZ = ZIV =

LjZL = ω0=ω

CC ω

∞=LZ 0jZ −=

=CZ

∞=ω∞=LZ 0CZ =

The inductor act like open circuit , where as capacitor act short circuit at high frequency.

like

Impedance

when impedance is inductive X is positiv pacitive X is negative.

The imped ve or lagging since current lags voltage.

Impedance

Impedance is a complex quantity, the impedance may be expressed in rectangular form as,

R is Real part of Z is the resistance, X is the Imaginary part of of Z which is reactance. X may be negative or positive,

e, when impedance is caance is said to be inducti

jXZ R +=

The impedance is said to be capacitive or leading since current leads voltage.

mpedance

The impedance may be expressed in the rectangular form as,

The impedance may be expressed in polar form as,

Wher

A iprocal of impedance, admittance

ittance can be write as,

jXRZ +=

jXRZ = −I

θ∠=+= ZjXRZ

θ∠= ZZ Z θ∠=+= ZjXR

e,

dmittance The admittance Y is the recmeasured in siemens (S). The admittance Y of an element (or a circuit) is the ratio ofthe phasor current through it to phasor voltage

As a complex quantity adm

2X+2RZ =RX1tan−=θ

VI

ZY ==

1

jBGY = +

G is the real part of admittance called conductance, B is the imaginary part of the admi susceptance.

y

m

lement Admittance

PrFin

ttance called B

rationalization,

I pedances and admittances of Passive Elements E Impedance R Z=R L C

oblem 4 d and in the circuit shown in Figure?

jXRjBG

+=+

1

22

22

XRRXR

RG

+

+=

B −=

2

1R

jBG 2 XRjXR

jXRjXR

jX +−

=−−

•+

=+

RY 1=

LjY

ω1

=LjZ ω=

CjZ

ω1

=CjY = ω

)( ti )(tv

From the voltage source

ss ca

shou rrent leads the voltage by 90.

The impedance is, Hence the current is, The voltage acro pacitor is,

It ld be noted that the Cu

VtAtti

AIVV

VV

jCjZ C

)43.634)57.264cos(789.1)(

57.26789.143.6347.4

43.6347.4904.0

57.26789.11.04

−+=

∠=−∠=

−∠=∠∠

=

××

tv cos(47.4)( =

IIV 57.26789.1 ∠===

ω

4,4cos10 =ωt010,4cos10 ==v Ss Vt ∠

AjI

jjZ

VI

Z

RjXRZ

S

57.26789.18.06.15.25

)5.25.25

010

11

1

22

∠=+=++

=−∠

=

+=+=Cjω

jjj

5.254.0

51.04

5 Ω−=×

+=××

+=

5(10=

Problem 4 Find and in the circuit shown in Figure?

)(ti )(tv