Download - 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 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