Electric Circuits Analysis

“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 9 Heinrich Rudorf Hertz (1857-1894), demonstrated that the Electromagnetic Waves obeys the same principle as light. 9 Heinrich Rudorf Hertz work is confirmed by the Clerk Maxwell’s. 9 Heinrich Rudorf Hertz was born in Germany. 9 Hertz did his doctorate under the well-known and prominent physicist Hermann von Helmholtz. 9 Hertz successfully generated and detected Electromagnetic waves. 9 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. 9 The unit of the frequency is given by his name Hertz

Introduction of Sinusoid Signal 9 DC sources were the main sources of providing electric power un-till the late 1800s.

9 AC ( Alternating Current ) is introduced in the beginning of 1900 century. 9 The biggest advantage of the AC is that is more efficient to transmit over long distances. 9 Sinusoid is a signal that has the form of the sine or cosine function. 9 Sinusoid Current is referred as AC, the AC current reverses at regular time interval and has alternative ( +ve ) and ( -ve ) values. 9 AC circuits are those which are driven by sinusoid Current or sinusoidal voltages

Why Sinusoid ? 9 9 9 9 ¾

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 9 Sinusoid signal is in the form of voltage generated throughout the world, then supplies to homes, factories, laboratories, universities and so on. 9 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,

v ( t ) = V m sin ω t

Vm = the amplitude of the sinusoid w = the angular frequency in radian/s

Time Period= Sinusoid Signal repeats itself every T seconds, the T is called period of the sinusoid.

=

T In sinusoid the replaced by, t+T

2 π

ω

repeat it selves every T seconds, so this can be

v (t + T ) = Vm sin ω (t + T ) v (t + T ) = Vm sin ω (t +

2π

ω v (t + T ) = Vm sin( ω t + 2π ) v (t + T ) = Vm sin ω t v (t ) = Vm sin ω t v (t ) = v (t + T ) Periodic Function ¾ A Periodic function is define as, the function that satisfies,

f (t ) = f (t + nT )

)

For all t and for all integers n.

Frequency, Time Period 9 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. 9 The reciprocal of the number of the cycles per second is known as frequency of sinusoid

f

=

1 T

9 f is in hertz (Hz)

General Expression for Sinusoid 9 The general Expression for the sinusoid signal is given as

v(t ) = Vm sin(ωt + φ ) • • •

The amplitude is Vm φ The phase is The angular frequency is

ω

Examine the two Sinusoids 9 The two sinusoid can be examine as,

v

1

=

V

m

sin

ω t

and

v2 = Vm sin(ωt + φ )

Leading and Lagging

v2 leads by v1 by φ v1 v2

9 The figure shows that

9 This can be said as 9

lags by

by

φ =0 then v1 andv 2 are said to be in phase. φ ≠ 0 then v1 and v 2 are said to be out of phase.

If

9 If

9 It is not necessary that apmlitude.

v1 and v2

have the same

Important Trigonometry Identities 9 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 ƒ ƒ

sin( A ± B ) = sin A cos B ± cos A sin B , cos( A ± B ) = cos A cos B m sin A sin B

By looking at the identities, it can be seen that

sin( ω t ± 180 ) = − sin ω t , cos( ω t ± 180 ) = − cos ω t sin( ω t ± 90 ) = ± cos ω t , cos( ω t ± 90 ) = m sin ω t

Sinusoids Addition

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

A cos ωt + B sin ωt = C cos(ωt − θ )

Here,

C = A +B 2

2

,

θ = tan

−1

B A

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

3 cos ω t

C = (3) + (4) = 5 2

2

and

− 4 sin ω t θ = tan

−1

,

(4) 3

3 cos ωt − 4 sin ωt 5 cos(ωt + 53.1)

Exercise Problems. Add the following sinusoids,

1 − , 5 cos ω t , 2 sin ω t 2 − , 4 cos ω t , − 3 sin ω t 3 − , 7 cos ω t , 5 sin ω t 4 − , 6 cos ω t , − 5 sin ω t 5 − , − 2 cos ω t , − 3 sin ω t 6 − , − 4 cos ω t , 3 sin ω t 7 − , 7 cos ω t , − 8 sin ω t 8 − , − 5 cos ω t , 2 sin ω t 9 − , − 6 cos ω t , sin ω t 10 − , 5 cos ω t , 4 sin ω t

Problem 2 Find the amplitude, phase , period and frequency of the sinusoid ?

v(t ) = 12 cos(50t + 10) • • • • •

The amplitude Vm is 12 The phase φ is 10 The angular frequency is = 50 rad/s The period T is 0.1257 s The frequency f is 7.958 Hz

ω

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

v ( t ) = 5 sin( 4π t − 60 0 ) • • • • •

The amplitude Vm is 5 The phase φ is -60 The angular frequency is The period T is 0.5 s The frequency f is 2 Hz

ω = 12.57 rad/s

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

v ( t ) = 10 sin( 33 t + 25 v ( t ) = 8 cos( 9 π t − 30 v ( t ) = 4 cos( 5 t − 10

0

0 0

)

)

)

v ( t ) = 16 sin( 3 π t + 40

0

)

v ( t ) = − 10 sin( 6 π t − 25 v ( t ) = 12 cos( 10 t − 15

0

0

)

)

Problem 2.2 Calculation of phase angle and leading sinusoid ?

9

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 ,

v1 = −10 cos(ωt + 50), v2 = 12 sin(ωt − 10) v1 = − 10 cos( ω t + 50 ) = 10 cos( ω t + 50 − 180 ) v1 = 10 cos( ω t − 130 ), or v1 = 10 cos( ω t + 230 )

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

v 2 = 12 sin( ω t − 10 ) = 12 cos( ω t − 10 − 90 ) v 2 = 12 cos( ω t − 100 )

We can write V2 as,

v 2 = 12 cos( ω t − 130 + 30 )

v 1 = 10 cos( ω t − 130 ) v 2 = 12 cos( ω t − 130 + 30 ) 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,

v1 = −10 cos(ωt + 50) = 10 sin(ωt + 50 − 90) v1 = 10 sin(ωt − 40) = 10 sin(ωt − 10 − 30) but , v2 = 12 sin(ωt − 10) Comparing V2 and V1 , it shows that V1 lags V2 by 30.

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

Method 3:

v1 = −10 cos(ωt + 50), v2 = 12 sin(ωt − 10) From the above it can be express V1 as phase shift of +50. with the phase shift of -10

− 10 cos ω t , with

“PHASORS” ‰ A Phasor is complex number that represent the amplitude and phase of a sinusoid. 9 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. 9 The idea of solving AC circuits using phasors was first introduced by Charles Steinmetz in 1893 9 Phasors provides a simple means of analyzing linear circuits excited by sinusoidal sources, solution of such circuits would be difficult otherwise.

Mathematics of Phasors ƒ A complex number z can be written in rectangular form as z = x + jy ƒ

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

ƒ The complex form z can be written in polar or exponential form as,

z = r ∠ φ = re

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

Mathematics of Phasors (cont’) Representation of Phasors.

jφ

z = x + jy

o Rectangular Form o Polar form

z = r∠φ

o Exponential Form

z = re jφ

Where, r = So,

x 2 + y 2 , φ = tan −1

y x

,

x = r cos φ , y = r sin φ

z = x + jy = r∠φ = r (cos φ + j sin φ ) Mathematics operations on Phasors 9 when dealing with phasors, addition and subtraction of complex numbers are better perform in rectangular form. 9 Multiplication and division are better done in polar form. ™ Addition ™ Subtraction

z1 + z 2 = ( x1 + x2 ) + j ( y1 + y2 ) z1 − z 2 = ( x1 − x2 ) + j ( y1 − y2 )

™ Multiplication ™ Division ™ Reciprocal ™ Square root

z1 z 2 = r1r2 (∠φ1 + ∠φ2 )

z1 r = 1 (∠ φ1 − ∠ φ z2 r2 1 1 = (∠ − φ ) z r

z = r (∠ φ ) 2

2

)

Phasors Representation 9 The idea of Phasor representation is based on the Euler’s Identity. cos φ = Re(e jφ ) ± jφ given as,

e

Given a sinusoid,

= cos φ ± j sin φ

sin φ = Im(e jφ )

v (t ) = V m cos( ω t + φ )

v ( t ) = V m cos( ω t + φ ) = Re( V m e j ( ω t + φ ) ) v ( t ) = Re( V m e j φ e j ω t ) thus , v ( t ) = Re( Ve where , V = Vm e

jφ

jω t

)

= Vm ∠ φ

Phasors Representation 9 V is the phasor representation of sinusoid of v(t). 9 A phasor is a complex representation of magnitude and phase of sinusoid. 9 Phasor has magnitude and phase “ direction”, it behaves as vector. 9 Phasors , V = Vm ∠φ and I = I m ∠ − θ , Graphical representation of phasor is known as a Phasor.

Phasors Representation Sinusoid-Phasor Transformation Time-Domain Representation

Phasor-domain Representation

V m cos( ω t + φ )

V

m

∠ φ

V m sin( ω t + φ )

V

∠ φ − 90

I m cos( ω t + θ )

m

I

m

∠ θ

I

m

∠ θ − 90

I m sin( ω t + θ ) dv dt

∫ vdt

jω V V jω

Problems. Evaluate these complex number. 1

(1); ( 40 ∠ 50 + 20 ∠ − 30 ) 2 solution , 40 ∠ 50 = {40 (cos 50 + j sin 50 )} = 25 . 708 + 30 . 64 20 ∠ − 30 = {20 (cos (− 30 ) + j sin (− 30 ))} = 17 . 32 − j10 ( 40 ∠ 50 + 20 ∠ − 30 ) = {(25 . 708 + 30 . 64 ) + (17 . 32 − j10 )} ( 40 ∠ 50 + 20 ∠ − 30 ) = 43 . 03 + j 20 . 64 where , r =

θ = tan

−1

x2 + y2 =

⎛ y⎞ ⎜ ⎟ = tan ⎝x⎠

−1

(43 . 028 )2 + (20 . 64 )2

= 47 . 72

⎛ 20 . 64 ⎞ ⎜ ⎟ = 25 . 63 ⎝ 43 . 03 ⎠

Taking square root.

( 40 ∠ 50 + 20 ∠ − 30 )

1

2

= 6 . 91 ∠12 . 81

Problems 2 Evaluate these complex number 10 ∠ − 30 + ( 3 − j 4 ) ( 2 + j 4 )( 3 − j 5 ) * 10 ∠ − 30 + ( 3 − j 4 ) 10 (cos (− 30 ) + j sin( − 30 ) ) + ( 3 − j 4 ) = * ( 2 + j 4 )( 3 − j 5 ) ( 2 + j 4 )( 3 + j 4 ) 10 ∠ − 30 + ( 3 − j 4 ) ( 8 . 66 − 5 j ) + ( 3 − j 4 ) 11 . 66 − 9 j = = * ( 2 + j 4 )( 3 − j 5 ) ( 2 + j 4 )( 3 + j 5 ) − 14 + 22 j ( 2 ),

⎛ y ⎞ x 2 + y 2 ⎜ ⎟, r = ⎝ x ⎠ 10 ∠ − 30 + ( 3 − j 4 ) 14 . 73 ∠ − 37 . 66 = ( 2 + j 4 )( 3 − j 5 ) * 26 . 08 ∠ 122 . 48

but , θ = tan

−1

= 0 . 565 ∠ − 160

. 31

Exercise Problems Evaluate these complex number

(1 ), [( 5 + j 2 )( − 1 + j 4 ) − 5 ∠ 60

]

*

10 + j 5 + 3 ∠ 40 ( 2 ), + 10 ∠ 30 − 3 + j4 Problems 2 Transformation ansformation of sinusoids to Phasors.

(1), v = −4 sin( 30t + 50 ) where , sin(ωt ± 180) = − sin ωt , cos(ωt ± 180) = − cos ωt sin(ωt ± 90) = ± cos ωt , cos(ωt ± 90) = m sin ωt .

v = −4(30t + 50 ) = 4 cos( 30t + 50 + 90 ) v = 4 cos( 30t + 140 )

The phasor form of v is,

v = 4 cos( 30t + 140 ) = V = 4∠140

Problems 2.1 Transformation of sinusoids to Phasors.

i = 6 cos( 50 t − 40 ) The phasor form of I is,

i = 6 cos( 50 t − 40 ), I = 6 ∠ − 40 Problems 2.2 Transformation of sinusoids to Phasors

v = − 7 cos( 2 t + 40 ) solution : sin(ωt ± 180 ) = − sin ωt , cos(ωt ± 180 ) = − cos ωt sin(ωt ± 90) = ± cos ωt , cos(ωt ± 90) = m sin ωt

v = − 7 cos( 2 t + 40 ) = 7 cos( 2 t + 40 + 180 ) v = − 7 cos( 2 t + 40 ) = 7 cos( 2 t + 220 ) The phasor form of V is,

V = 7 ∠ 220 Problems 2.3 Transformation of sinusoids to Phasors.

i = 4 sin( 10 t + 10 ) solution , sin( ω t ± 180 ) = − sin ω t , cos( ω t ± 180 ) = − cos ω t sin( ω t ± 90 ) = ± cos ω t , cos( ω t ± 90 ) = m sin ω t

i = 4 sin( 10 t + 10 ) = 4 cos( 10 t + 10 − 90 ) i = 4 sin( 10 t + 10 ) = 4 cos( 10 t − 80 ) i = 4 cos( 10 t − 80 ), I = 4 ∠ − 80

Problems 2.4 Find the sum of Sinusoids The The

i1 (t ) i2 (t )

i 1 ( 1 ) = 4 cos( ω t + 30 ) i 2 ( 2 ) = 5 sin( ω t − 20 )

can be written as, i1 (t ) = 4 cos( ω t is change from sine to cosine form

+ 30 ), I 1 = 4 ∠ 30

i 2 ( t ) = 5 sin( ω t − 20 ) = 5 cos( ω t − 20 − 90 ) i 2 ( t ) = 5 cos( ω t − 110 ) Phasor can be written as, I 2 = 5 ∠ − 110

The addition of i = i1 + i2 I = I 1 + I 2 = 4 ∠ 30 + 5 ∠ − 110

I = {4 cos( 30 ) + j sin( 30 ) } + {5 cos( − 110 ) + j sin( − 110 ) } I = 3 . 464 + j 2 − 1 . 71 − j 4 . 698 I = 1 . 754 − j 2 . 698 I = 3 . 218 ∠ − 56 . 97 Transforming into time domain, i ( t ) = 3 . 218 cos( ω t − 56 . 97 )

Problems 2.5 Find the sum of Sinusoids.

v 1 = − 10 sin( ω t + 30 ) v 2 = 20 cos( ω t − 45 )

The sinusoid is change from sine to cosine.

v1 = − 10 sin( ω t + 30 ) = 10 cos( ω t + 30 + 90 ) v1 = 10 cos( ω t + 120 ) = − 5 + 8 .66 j v 2 = 20 cos( ω t − 45 )

v 2 = 20{cos( − 45 ) + j sin( − 45 )} v1 = 14 .14 − 14 .14 j v = v1 + v 2 v = ( − 5 + 14 . 14 ) + j (8 . 66 − 14 . 14 ) v = 9 . 14 − 5 . 48 j ⎛ y⎞ 2 2 ⎟ = 30 . 94 , r = x + y = 10 . 65 ⎝x⎠ v ( t ) = 10 . 66 cos( ω t − 30 . 94 )

φ = tan −1 ⎜

Problems 2.6 Using the phasor approach, determine the current describe by integrodifferential equation.

i (t )

in a circuit

4 i + 8 ∫ idt − 3

di = 50 cos( 2 t + 75 ) dt

Transfer domain from time to phasor.

8I − 3 j ω I = 50 ∠ 75 jω Where , ω = 2 I ( 4 − j 4 − j 6 ) = 50 ∠ 75 50 ∠ 75 50 ∠ 75 I = = 4 . 642 ∠ 143 . 2 = 4 − j10 10 . 77 ∠ − 62 . 2 4I +

Now convert to time domain.

i ( t ) = 4 . 642 cos( 2 t + 143 . 2 )

Practice Problems

Using the phasor approach, determine the current v(t ) in a circuit describe by integrodifferential equation.

dv + 5 v + 10 ( a ), 2 dt

∫ vdt

= 20 cos( 5 t − 30 )

dv ( b ), + 5 v + 4 ∫ vdt = 20 sin( 4 t + 10 ) dt

Phasor Relationship for Circuit Element 9 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. 9 The current through resistor R is,

9 by applying Ohm’s l aw, the voltage across it is given by, 9 the phasor form of the voltage is, 9 the current I is represented in phasor form as,

Phasor Relationship Current-Voltage for Resistor VI relationship for R in time domain.

VI relationship for R in Phasor domain.

9 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

9 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

9 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

v = iR

V = IR

L

di v=L dt

V = jωLI

V Problems 3.

di i=C dt

V =

I jω C

The voltage v = 12 cos( 60 t + 45 ) is applied to a 0.1H inductor, Find the steady state current through the inductor. Solution: For the inductor, V = jωLI , whereω = 60

rad/sec.

v = 12 cos(60t + 45) = 12∠45

12 ∠ 45 12 ∠ 45 = jω L j × 60 × 0 . 1 j×6 V 12 ∠ 45 I = = = 2 ∠ − 45 A 6 ∠ 90 jω L I = 2 ∠ − 45 , V

I =

=

Converting to time domain,

i ( t ) = 2 cos( 60 t − 45 ) A

Problems 3.1

The voltage v = 6 cos( 100 t − 30 ) applied to a 50uF capacitor, Find the current through the capacitor. Solution: I For the capacitor, V = , where ω = 100 rad/sec.

jω C

v = 6 cos(100t − 30) I v= ,6 ∠ − 30 = jω C j × 100 × 50 × 10 − 6 I

I = 6 ∠ − 30 × j 5 × 10 − 3 = 6 ∠ − 30 × 5 × 10 − 3 ∠ 90 I = 30 ∠ 60 mA Converting to time domain,

i ( t ) = 30 cos( 100 t + 60 )

Impedance ‰ The impedance Z of a circuit is the ratio of the phasor voltage V to the phasor current I measured in ohms Ώ. 9 We obtained the voltage-current relationship for passive elements as, 1 V= V = j ω LI , V = IR , jω C 9 These equation can be written in terms of the ratio of the phasor voltage to the phasor corrent.

V = R I

V = jω L I

V 1 = I j ωC

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

Z =

V I

OR

V = ZI

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

Impedance 9 The impedance represent the opposition, which the circuit exhibit to the flow of sinusoidal current. 9 The impedance is the ratio of two phasor it is not a phasor. 9 The impedance for inductor Z L = jωL and for capacitor is 9 Two cases for ω = 0 ( for DC sources), − j Z

ZC = 0

ZL = ∞

C

=

ω C

9 The inductor act like short circuit , where as capacitor acts like open circuit. 9 The two cases for ( for high frequencies)

ω=∞

ZL = ∞

ZC = 0

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

Impedance

Impedance 9 Impedance is a complex quantity, the impedance may be expressed in rectangular form as,

Z = R + jX 9 R is Real part of Z is the resistance, X is the Imaginary part of of Z which is reactance. 9 X may be negative or positive, when impedance is inductive X is positive, when impedance is capacitive X is negative. 9 The impedance is said to be inductive or lagging since current lags voltage.

Z = R + jX 9 The impedance is said to be capacitive or leading since current leads voltage.

Z = R − jX Impedance 9 The impedance may be expressed in the rectangular form as,

Z = R + jX = Z ∠θ 9 The impedance may be expressed in polar form as,

Z = Z ∠θ Z = R + jX = Z ∠θ Where,

Z = R +X 2

2

θ = tan

−1

X R

Admittance ‰ The admittance Y is the reciprocal of impedance, admittance measured in siemens (S). ‰ The admittance Y of an element (or a circuit) is the ratio of the phasor current through it to phasor voltage

Y =

I 1 = Z V

9 As a complex quantity admittance can be write as,

Y = G + jB

9 G is the real part of admittance called conductance, B is the imaginary part of the admittance called susceptance. 1 R G + jB = G = 2 R + X 2 R + jX R By rationalization, B = − 2 R + X R − jX R − jX 1

G + jB =

R + jX

•

R − jX

=

R2 + X 2

Impedances and admittances of Passive Elements

Element

Impedance

R

Z=R

Z = j ωL

L

Z =

C

Problem 4 Find i ( t ) and

v(t )

Admittance

1 Y= R 1 Y= j ωL

1 jω C

Y = jω C

in the circuit shown in Figure?

2

From the voltage source 10 cos 4t , ω = 4

vs = 10 cos 4t , VS = 10∠0

The impedance is, Z = R + jX = R +

Z = 5 +

1 jω C

1 1 = 5 + = 5 − j 2 .5 Ω j × 4 × 0 .1 j × 0 .4

Hence the current is,

10 ( 5 + j 2 . 5 ) VS 10 ∠ 0 = = Z 5 − j 2 .5 5 2 + 2 .5 2 I = 1 . 6 + j 0 . 8 = 1 . 789 ∠ 26 . 57 A I =

The voltage across capacitor is,

V = IZ

=

I

=

1 . 789 ∠ 26 . 57 j × 4 × 0 .1

jω C 1 . 789 ∠ 26 . 57 V = = 4 . 47 ∠ − 63 . 43 V 0 . 4 ∠ 90 V = 4 . 47 ∠ − 63 . 43 V I = 1 . 789 ∠ 26 . 57 A C

i ( t ) = 1 . 789 cos( 4 t + 26 . 57 ) A v ( t ) = 4 . 47 cos( 4 t − 63 . 43 )V It should be noted that the Current leads the voltage by 90.

Problem 4 Find

i (t )

and

v(t ) in the circuit shown in Figure?