INTRODUCTION TO CIRCUIT THEORY (A PRACTICAL APPROACH)
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Transcript of INTRODUCTION TO CIRCUIT THEORY (A PRACTICAL APPROACH)
ii
INTRODUCTION
TO
CIRCUIT THEORY
(A PRACTICAL
APPROACH)
BY
ENGR. (DR) KAMORU OLUWATOYIN KADIRI
iii
DEDICATION
This book is dedicated to the Most High God and all Engineering Students.
iv
ACKNOWLEGDEMENTS
First and foremost I am grateful to the Almighty God, the beneficent and the Most Merciful who
makes all things possible.
I also appreciate my well-wishers for their support physically and morally in making this text a
success. I am also grateful to all the staff of Electrical/Electronics Department of the Federal
Polytechnic, Offa; the students, family members and friends.
Engr. Kadiri, K. O.
v
FORWARD
The book “Introduction to Circuit Theory” is designed to assist students of Electrical/Electronic
Engineering in acquiring necessary fundamentals of Circuit Theory. It was written to meet the
basic requirements of the National Board for Technical Education (NBTE) Syllabus for National
Diploma (ND) level in Polytechnic for Engineering Departments.
1
TABLE OF CONTENTS
Table of Contents DEDICATION iii
ACKNOWLEGDEMENTS iv
FORWARD v
TABLE OF CONTENTS 1
CHAPTER ONE 4
REVISION OF COMPLEX NUMBER 4
Introduction 4
Complex Equations 9
The Polar Form of a Complex Number 11
Multiplication and Derivation using Complex Numbers in Polar Form 11
De Moivre’s Theorem – Powers and Roots of Complex Numbers 12
Exercises 13
CHAPTER TWO 15
APPLICATION OF COMPLEX NUMBERS TO SERIES A.C CIRCUITS 15
Introduction 15
2.2 Series ac Circuits 15
Pure Resistance 15
Pure Inductance 16
Pure Capacitance 17
R – L Series Circuit 18
R-C Series Circuit 20
R – L – C Series Circuit 21
General Series Circuit 22
Exercise on Series ac Circuits 30
CHAPTER THREE 32
APPLICATION OF COMPLEX NUMBERS TO PARALLEL AC NETWORKS 32
Introduction 32
2
Admittance, Conductance and Susceptance 32
Series Connection 36
Parallel connection 37
Parallel ac Networks 37
Parallel R – L Network 38
Current Division in ac Circuits (Current Divider) 38
Exercises 46
CHAPTER FOUR 49
AC BRIDGES 49
Introduction 49
Balance Conditions for an AC Bridge 49
Types of AC Bridge Circuit 50
Maxwell’s Inductance Bridge 50
Hay Bridge 51
Owen Bridge 53
Maxwell – Wien Bridge or Mazwell 53
De Sauty Bridge 54
Schering Bridge 55
Wien Series Bridge 56
Wien Parallel Bridge 57
Exercise on ac Bridges 61
CHAPTER FIVE 63
INTRODUCTION TO NETWORK ANALYSIS 63
Introduction 63
Solution of Simultaneous Equations Using Determinants and Crammer’s Rule 63
Two unknowns 65
Three Unknowns 66
Exercises 75
CHAPTER SIX 78
MESH-CURRENT AND NODAL ANALYSIS 78
Mesh Current Analysis 78
3
Nodal Analysis 83
Exercises on Mesh-Current Analysis 92
Exercises on Nodal Analysis 93
CHAPTER SEVEN 96
SUPERPOSITION THEOREM 96
Introduction 96
Implementing the Superposition Theorem 96
Maximum Power Transfer Theorem 103
Exercises on Superposition Theorem 108
Exercises on Maximum Power Transfer Theorem 109
CHAPTER EIGHT 111
THEVENIN’S AND NORTON’S THEOREMS 111
Introduction 111
Thevenin’s Theorem 111
Norton’s Theorem 114
Thevenin and Norton Equivalent Networks 118
Exercises on Thevenin’s Theorem 122
Exercises on Norton’s Theorem 123
Exercise on Thevenin and Norton Equivalent Networks 124
CHAPTER NINE 126
DELTA – STAR AND STAR - DELTA TRANSFORMATIONS 126
Introduction 126
Star or Wye (Y) Connection 126
Delta/Star Transformation 126
Star/Delta Transformation 128
BIBLIOGRAPHY 139
4
CHAPTER ONE
REVISION OF COMPLEX NUMBER
Introduction
A Complex number is number that can be expressed in the form (a+jb), where a
and b are real numbers and “j” is the “imaginary unit” which satisfies the equation “j2 = -
1.” In this expression, “a” is the real part and “b” is the imaginary part of the complex
number. For example in (-3.5 + j2), the real number 2 is called the imaginary part of -3.5
+ j2. Hence, j2 is the imaginary part but 2 is the imaginary number, while 3.5 is the real
part
Complex numbers are mostly used in the analysis of series, parallel and series –
parallel electrical networks supplied with alternating sources, in deriving balance
equations with ac bridges, in analyzing ac circuits using Kirchhoff’s law, mesh and nodal
analysis, the superposition Theorem, Thevenin’s and Norton’s Theorems, Delta-Star and
Star-Delta Transforms and in many other aspects of Advanced Electrical Engineering.
The purpose of complex numbers is that the complex processes become simply easy
process.
Complex numbers extend the concept of the one – dimensional “number line” to
the two dimensional complex planes also called “Argand plane” by using horizontal axis
for the real part and vertical axis for the imaginary part. The complex number (a + jb) can
be identified with the point (a,jb) in the complex plane in Figure 1.1.
A complex number of the form (a + jb) is called a Cartesian or rectangular
complex number. A complex number whose real part is “zero” is said to be “purely
imaginary”, whereas, a complex number whose imaginary part is “zero” is a “real
number.” In this way, the complex number contains the ordinary real numbers while
extending them in order to solve problems that cannot be solved with real numbers alone.
Italian Mathematician “Gerolamo Cardano”, is the first known to have introduced
complex numbers. He called them “fictitious” during his attempts to find solutions to
“cubic equations” in the 16th
century. The number j8 (i.e., 0 + j8) is shown in Figure
1.2(a) drawn vertically upwards from the origin to the imaginary axis. Hence, multiplying
the number “8” by the operator “j” results in anticlockwise phase – shift of 900 without
altering its magnitude.
In the Figure 1.2(b), the number 7 (i.e., 7 + j0) is drawn as a phasor horizontally to
the right of the origin on the real axis. Multiplying j8 by j give j28, i.e. “- 8”, and is shown
in Figure 1.2(c) as a phasor eight units on the horizontal real axis to the left of the origin
– an anticlockwise phase shift of 900 compared with the position shown in Figure 1.2(b).
Thus, multiplying by “j” reverses the original direction of a phasor.
5
Multiplying j28 by j gives j
38, i.e. –j8, and is shown in Figure 1.2(d) as a phasor
eight units long on the vertical imaginary axis downward from the origin-an
anticlockwise phase shift of 900 compared with the position shown in Figure 1.2(c).
Multiplying “j38” by “j” gives j
48, i.e. 8, which give the original position of the
phasor shown in Figure 1.2 (a). Consequently, application of the operator j to any number
rotates it 900 anticlockwise on argand diagram, multiplying by j2 rotates to 180
0
anticlockwise, multiplying by a number by j3 rotates it 270
0 anticlockwise and
multiplication by j4 rotates it 3600 anticlockwise i.e. back to original position. In each
case, the phasor is unchanged in its magnitude.
If a phasor is operated on by –j then a phase shift of -900 (i.e., clockwise direction)
occurs, again without change of magnitude.
In electrical circuits, 900 phase shifts occurs between voltage and current with the
capacitors and inductors; this is the main reason why j notation is used so much in the
analysis of electrical networks.
Figure 1.1 The
Argand diagram
6
Figure 1.2: Operations Involving Cartesian Complex Numbers
(a) addition
(e + jf) + (g +jh) = (e + g) + j (f + h)
(e – jf) + (g – jh) = (e + g) –j (f –h)
Example If C1 = 3 +j6 and C2 = 4 +j 1, solve; C1 + C2
(ii) If C1 = 3 + j6 and C2 = -3 + j6, solve; C1 + C2
Solution
(i) C1 + C2
(3 + j4) + (4 + j1) = (3 + 4) +j (4 + 1)
C1 + C2 = 7 + j5
(ii) C1 = (3 +j6), C2 = - 3 +j6
C1 + C2 = (3+j6) + (-3 +j6)
C1 + C2 = (3 + (-3) +j (6+ 6)
C1 + C2 = 3 – 3 + j12
C1 + C2 = 0 + j12 = j12
(b) subtraction
In subtraction, the real and imaginary parts are again considered separately.
Example: C1 = (a + jb) and C2 = (C + jd), subtraction of the two, will give; C1-C2 =
(a +jb) – (c + jd)
Also, if C1 = (a – jb) and C2 = (c –jd), subtraction of the two, will give; C1 – C2 = (a
–jb) – (c – jd)
C1 – C2 = (a – c) – j(b – d)
Thus, if C1 = ±A2 ± jB2 and C2 = ± A3 ± jB3
7
C1 – C2 = [±A2 – (± A3)] +j (±B2 – (± B3)]
Example (1) subtract D2 = 1 + j4 from D1 = 4 +j 6
(iii) Subtract D2 = -3 + j5 from D1 = + 4 +j3
Solution:
(i) D1 – D2 = (4 + j6) – (1 + j4)
= (4 – 1) +j (6 – 4) = 3 +j2.
(ii) D1 – D2 = (+4 + j3) – (-3 + j5)
= (+4 – (-3) ±j (3 -5) = (4 +3) +j (-2)
=7 – j2.
Note: addition or subtraction cannot be performed in polar form, unless the complex
numbers have the same angle or differs only by multiples of 180 degree.
For example (1) 3450 + 445= = 745
0
(i) 3 00 – 4 180
0 = 7 0
0.
(C) Multiplication
In order to multiply two complex numbers in rectangular form, multiply the real and
imaginary parts of one by the real and imaginary parts of the other. For example, If
D1 = A1 + jB1 and D2 = A2 +jB2
D1 . D2 = A1A2 + jB1 A2 +jA1B2 + j2B1B2, but j
2= -1
= A1A2 + j (B1A2+A1B2) + B1B2(-1)
D1 . D2 = (A1A2 - B1B2) + j(B1A2 + A1B2)
For example, if D1 = (3 +j1) and D2 = (2+j4), multiplication of the two will give: D1D2
= (3 +J1) (2 + J4)
D1D2= 6+j12+j2+ j24
D1D2 = 6 + j14+4(-1) = 6 + j14 -4
D1D2 = 2 + j14
(D) Division
In order to divide two complex numbers in rectangular form, multiply the numerator and
denominator of the conjugate of the denominator and the resulting real, and imaginary
parts collected.
Example: if D1 = ( +j ) and D2 = ( +j ); Division of D2 by D1 will give;
8
Note: ( = - 1);
e.g. (1) If (2+j3) + (4+j6) and = (7+j7) – (3 –j3), evaluate
ii. If D1 = (60 300) (5 + j5) and D2 = 20 -30
0.
(E) Complex conjugate
The complex conjugate of (c-jd) is (c + jd). For example, the conjugate of (2 –j3) is (2 +
j3)
Note: The product of a complex number and its complex conjugate is always a real
number, and this is a major property used when dividing complex numbers. Thus;
(a + jb)(a – jb) = a2 – jab + jab – j
2b
2.
= a2 – (-b
2) = a
2+ b
2 (i.e. a real number)
For example; (2 + j1)(2-j1) = 22 – j
21
2 = 4– (-1) = 4 + 1 =5
Further examples
Example 1: In an electrical circuit, the total impedance ZT is given by ZT = Z3 +
Z1Z2/Z1+Z2.
Determine ZT in polar and rectangular form, correct to two decimal places, if Z1= 6, Z2.
= 4 + j3 and Z3 = 2.8 – j3.7
Solution: Z1Z2 = 6(4+j3) = 24+j18
Z1+Z2 = 6 + 4+j3 = 10 + j3
9
Hence:
ZT = 2.8 – j3.7 + 2.697 – j0.99
ZT = 5.50 –J4.69 (to2 d.p)
ZT = 7.23 - 40.460.
Example 2: Given: Z1 = 6 + j2 and Z2 = 4 + j3, determine in Cartesian form, correct to
three significant Figures(3 s.f); (a) 1/Z1 (b) 1/Z2 (c) 1/Z1 + 1/Z2 (d) 1/(Z1+
Z2).
Solution (a)
(b)
(c)
= 0.05 +j0.35
(d)
Complex Equations
Two complex equations is said to be equal if, the real part and imaginary parts of
both complex numbers are equal. That is, if two complex numbers are equal, their real
parts are equal likewise the imaginary part being equal.
Hence, if a +jb = c+jb, then a = c and b=d. complex equations are mostly used
when finding balanced equations of a bridges.
Example 3: solve the following complex equations: -
10
(a) 4(a+jb) = 6 – j2
(b) (3 +j)(-3 +j) = x + jy
(c) (a – j3b) + (b –j29) = 6 +j2
Solution:
(a) 4(a+jb) = 6 – j2. Thus 4a + j4b = 6 – j2
Equating the real parts gives 4a = 6,
a = 6/4 = 3/2
Equating the imaginary parts gives: 4b = -2,
b = -2/4 = -1/2.
(b) (3 + j) (-3 +j) = x+jy
Thus, -9+3j-j3+
-10 + j0 = x+jy
Equating the real and imaginary part gives x= -10 and y = 0
(c) (a –j3b) + (b – j2a) = 6 + j2
Thus; (a+b) +j (-3b – 2a) = 6 + j2
Hence; a + b = 6 ---- (i)
-3b – 2a = 2, -2a – 3b = 2 ---- (2)
We have two simultaneous equations to solve. Multiply equation to solve. Multiply
equation (1) by (ii) gives 2a + 2b = 12 (2)
-2a – 3b = 2 (3)
Addition of equation (2) and (3) gives
-b = 14, b = -14
Put b = -14 in equation (1)
a + b = 6
a +(-14) = 6, a = 6 –(- 14) = 6+14=20
Example 4: An equation derived from an A.C bridge network is given by:
R1R3 = (R2 + JwL2)
R1, R3 and R4 and C4 are known values. Determine expressions for R2 and L2 in terms
of known components.
Multiplying both sides of the equation by R4 + jwC4 gives:
(R1R3 (R4+jwC4) = R2 + jwL2
R1R3R4 + jR1R3wC4 = R2 + jwL2
Equating the real parts gives: R2 = R1R3R4
Equating the imaginary part gives: wL2 = R1R3wC4, from which L2 = R1R3C4
11
The Polar Form of a Complex Number
If Z = x + jy = r cos + jr sin form trigonometry, = r (cos + jsin )
The r(r cos + j sin ) is usually shorten to z = r0 and is called the polar form of a
complex number.
r is called the modulus (or magnitude of Z) and is written as mod Z or /z/ .r is determined
from pythagora’s Theorem
/Z/ = r =
For example, the Cartesian complex number (3 – j4) is equal to r < in polar form,
where; r = = and = arctan (-4/3) = -53.130.
Hence, (3 – j4) = 5 <-53.130
Simimarly, (-6+j4)
Where r = = 7.2, = arctan (4/-6) = -33.69
and = 1800 – 33.69 = 146.31
0.
Hence, (-6 + j4) = 7.2 <146.310.
Multiplication and Derivation using Complex Numbers in Polar Form
(a) Multiplication
(r1<1) (r2<2) = r1r2 < (1 + 2)
Thus; (1) 6 < 300 x 3 < 40
0 = (6 x 3) < (30
0 + 40
0)
= 18 < 700.
(a) 3 220 x 6 - 18
0 = (3 x 6) (22
0 – 18
0)
= 18 40.
(ii) 2 (π/4) x 6 (π /3) = (2 x 6) (π/4 + π/3)
= 12 (7π/12)
(b) Division
Thus;
Further examples
Example 5: convert 6 -1420 into a+jb form correct three significant Figures.
The polar complex number 6 -1420 lies in the third quadrant of the Argand diagram.
12
Using trigonometrical ratio
x = 6 cos 380 = 4.728 and y = 6 sin 38
0 = 3.694
hence, 6 - 1420 = - 4.728 –j3.694
Alternatively; 6 - 1420 = 6 (cos (-142
0) +j sin (-142
0)
= 6cos(-1420) + j sin(-142
0)
= -4.728 –j3.694 as above
N.B. The real and imaginary parts can be obtained directly, using an electronic calculator.
Example 6: Three impedances in an electrical network are given by Z1 = 3.7250, Z2 =
8.3 - 580 and Z3 = 2.4 60
0.
Determine in polar form, the total impedance ZT given that
Solution: Z1 = 3.7 3.7 250 = 3.7cos 25
0 + j3.7sin25
0 = 3.35 + j1.56
Z2 =8.3<- 580 = 8.3cos (-58
0) +j8.3sin (-58
0) = 4.398 –j7.039
Z3 = 2.4 < 600 = 2.4 cos 60
0 +j2.4sin 60
0 = 1.2 +j2.08
Z1 + Z2 = 3.35 +j1.56 + 4.398 –j7.039 = 7.75 –j5.48
Z1 Z2 = (3.35 + j1.56) (4.398-j7.039) = 25.76 –j 16.73
= 3.24 + j0.13 + 1.2 + j2.08 = 4.44 + j2.21 = 4.96 < 26.50.
De Moivre’s Theorem – Powers and Roots of Complex Numbers
[r <]n = r
n <n
De Moivre’s Theorem (aka De Moivre’s formula) named after Abraham de Moivre,
states that for any complex number (and, in particular, for any real number) x and integer
n it holds that (cos x +i sin x)n.
= cos(nx) +i sin (nx), where i is the imaginary unit (i2=-1). While the formula was
named after Demoivre, he never stated it in his works.
(3<160)
7 = 3
7< (17 x 16
0) = 21 187< 112
0 = 819.3 + j2.028
The formula is important because it connects complex numbers and trigonometry.
The expression cos x + i sin x is sometimes abbreviated to cis x.
A square root of a complex number is determined as follows;
However, it is useful to know that a real number has two square roots, equal in
size but opposite in sign.
13
Examples 7: determine (-3+j4)5 in polar and Cartesian (rectangular) form.
Solution: Z = -3+j4 can be found in the second quadrant of the argand diagram.
Thus, r= = 5 and
Hence, the argument = 180 – 53.120 = 126.87
0.
Thus -3 + j4 in polar form is 5 < 126.870.
(-3+j4)5 = [5 < 126.87
0]
5 = 5
5 < (5 x 126.87
0) = 3,125 <634.35
0.
= 3,125<634.350.
Exercises
1. Given the following two vectors: A=30<600
and B=10<300, perform the following
indicated operations and illustrate graphically (i) A x B (ii) (iii) (iv)A+ B (v)
A-B
300<900, 3<30
0, 0.3<-30
0, 38.44<53.25
0, 27<50.98
0.
2. Three vectors A,B,C are given as: 30<-1200
, 30+j30, 20+j0 respectively. Solve: (i)
(ii) (iii)
(63.65<-750, 28.29<165
0, 310.6<15
0, 14.14<-
1650).
3. Perform the following indicated operations: a.(80+ j60)+(40-j30) b.(6-j12)-
(20-j40)c.(8+j6)(4-j3) d.(8+j6)÷(4-j3)
120+j30, -18+j28, 50+j0,
0.56+j1.92
4. Express in rectangular and polar form a vector, the magnitude of which is 120units
and the phase of which with respect to reference axis is: a.+60 b. +120 c. -70 d.-120
e.+220.
5. The conjugate of (-a+jb) is?
(a jb)
6. Three impedance Z1= 6+j2 , Z2= 12-j6 and Z3= 4-j8 connected in a circuit so that they are
additive. Find the resultant impedance in polar form.
(ZT= 22-j12
7. Calculate: , leaving your answer in polar and rectangular form.
(1.58<-18.440
or 1.499-j0.5)
8. Evaluate in polar form; a. 3<500x6<20
0 b. 3.6<800
0x2.1<90
0
14
(18<700, 7.56<890
0)
9. Solve: 5(a+jb) = 8-j4
(a= , b= )
10. Determine the modulus of the following complex numbers;(a) 3-j4 (b) 5-j2 (c)-1+j4 (d)-3-
j5
(a)5,- 530.13
1 (b) 5.385,-21
0.80
1 (c)4.12,104
0.04
1 (d)5.83,-120
0.96
1
11. Solve the complex equations, leaving your answers in 3significant Figures.
(a) =x+jy
(b) 16<π/2 + 18<π/3 – 8<-π/2 = r<θ
(a) x= 30.03,y=17.34 (b) 40.6<77.190
12. Evaluate in Cartesian and polar form; a.(3+j2)2 b.(5-j4)
2 c.(5<33
0)2 d.(3<-π/2)
3 e.(1.6<-
1200)3 f.(-1-j2)
3
(a) 5+j12,13<67.380 (b) 9-j40,41<77.32
0 (c) 625<132
0 (d) j27,27<-270
0 (e) 4.096<-
3600,4.096+j0 (f) -1-j8,80.6<-97.13
0
13. Determine the two square root of the complex numbers in Cartesian form and show results
on the argand diagram. (a)-4+j3 (b)6<300 (c) 15<5π/2 (d) 2-j3 (e) 2+j4
14. Calculate the current flowing in the impedance in polar form: I= A
(4.67<39.7
15
CHAPTER TWO
APPLICATION OF COMPLEX NUMBERS TO SERIES A.C CIRCUITS
Introduction
Phasor diagrams can be used to analyze simple ac circuits. However, complex
numbers can be used when circuits become more complicated. It is important that the
basic operations used with complex numbers, as outlines in chapter one, are thoroughly
comprehended before proceeding with ac circuit analysis. The theories introduced in
electrical engineering science is relevant in this chapter; similar circuits will be analyzed
using j notation and argand diagrams.
2.2 Series ac Circuits
Pure Resistance
In an ac circuit containing resistance R only, the applied voltage V and current i
are in phase, and the magnitude. The phasor diagram super imposed on the Argand
diagram as shown in Figure 2.1 (b).
Im = Vm/R or Vm = ImR
diagram
In phasor form;
V = Vmsin wt V = V<00.
Where V = 0.7071Vm
Applying Ohm’s law and using phasor algebra, we have;
Since, i and v are in phase, the angle associated with “i" also must be 00. To satisfy the
condition, must be equal to 00. Substituting = 0
0, we have;
The fact that R = 00 will now be employed in the following polar format to ensure
the proper phase relationship between the voltage and current of a resistor.
The having both magnitude and an associated angle is referred to as the
‘’Impedance of resistive element”. It is measured in ohms and is a measure of how much
the elements will “impede” the flow of charge through the network.
16
Pure Inductance
In an ac containing pure inductance L only fig. 2.2 (a), voltage leads the current by 900,
and the reactance of the coil is determined by Figure 2.2 (b) shows phasor
diagram super imposed on the argand diagram and Figure 2.2 (c) shows the phasor
diagram.
V = Vm sin wt phasor form V = V<00.
By ohm’s law,
Since V leads i by 90
0, i must have an angle of -90
0 associated with it. To satisfy this
condition, must equal + 900. Substituting = 90
0, we obtain
= 90
0 will now be employed in the following polar format for inductive reactance to
ensure the proper phase relationship between the voltage and the current of an inductor.
Where is the inductive reactance given by = = 2
Where f is the frequency in hertz and L is the inductance in Henrys
ZL having both magnitude and an associated angle, is referred to as the impedance
of an inductive element. It is measured in ohms and is a measure of how much the
inductive element will “control or impede” the level of current through the network
(always keep in mind that inductive elements are storage devices and do not dissipate like
resistors). ZL is not a phasor quantity. The term phasor quantity are quantities that vary
with time and R and its associated angle of 00 are fixed, non varying quantities.
17
Figure 2.2: (a) circuit diagram (b) argand diagram (c) phasor diagram
Pure Capacitance
In an ac circuit having pure capacitance only fig 2.3 (a), the current leads the voltage by
900 and that the reactance of the capacitor Xc is determined by 1/Wc as shown in the
phasor diagram of Figure 2.3(b).Figure 2.3(c) shows phasor diagram superimposed on the
argand diagram.
V = Vmsin wt phasor form V = V<00.
Applying ohm’s law and using phasor algebra, we find
Since we know that i leads V by 900, i must have an angle of 90
0 associated with it and C
must equal – 900 substituting C = -90
0 yields
The fact that = - 90
0 will now be employed in the following polar form for capacitive
reactance to ensure the proper relationship between the voltage and current of a capacitor.
Where C is the capacitive reactance given by:
18
Where C is the capacitance in Farads.
[Note
Figure 2.3(a) circuit diagram (b) phasor diagram (c) argand diagram
R – L Series Circuit
An a. c. circuit comprising of resistance R and inductance L in series Figure
2.4(a), the applied voltage V is the sum of voltage across the resistance VR and voltage in
the inductor VL shown in the phasor diagram of Figure 2.4(b). The applied voltage leads
the current i by angle lying between 00 and 90
0 (the actual value depending on the values
of VR and V, which depend on the values of R and L. The angle between the current and
the applied voltage is the circuit phase angle shown as angle in the phasor diagram.
Current is the same through all components in series circuit and is taken as the reference
phasor in fig 2.4(b). Figure 2.4(c) shows phase diagram superimposed on the argand
diagram.
The supply voltage V given by; V = VR + jVL
19
Figure 2.4(a) circuit digram (b) phasor diagram (c) Argand diagram
Voltage triangle shown in Figure 2.5(a) is derived from the phasor diagram of
Figure 2.4(b) (i.e. triangle oab). Impedance triangle is derived by dividing current I
through each side of the voltage triangle shown in Figure 2.5(b). Figure 2.5(c) shows
impedance triangle superimposed on the argand diagram.
Impedance Z is given by;
Z = R + j x L
Thus, for example an impedance expressed as (4+j7) means that the resistance is
4 and inductive reactance is 7.
In polar form, Z = /Z/< , from the impedance triangle, the modulus of impedance
/Z/ = and the circuit phase angle = arc tan (XL/R) legging, that is tan-1
(XL/R) lagging
VR VLV
R
(a)
IVL V
VR
b
a
JXL V
I0 VR
( c)
Real axis
Imaginary axis
(b)
20
Figure 2.5 (a) voltage diagram (b) impedance diagram (c) argand diagram
R-C Series Circuit
In an a.c circuit comprising of resistance R and capacitance C in series (Figure 2.6(a), the
applied voltage is the phasor sum of voltage in the resistance VR and voltage in the
capacitance Vc as shown in the phasor diagram of Figure 2.6(b). the current I lead the
applied voltage V by an angle lying between 00 and -90
0, the actual value depending on
the values of VR and VC, which depend on the values of R and C. The circuit phase angle
is shown as angle in the phasor diagram. Figure 2.6(c) shows the phasor diagram being
superimposed on the argand diagram.
V=VR-jVc
Figure 2.6(a) shows the voltage triangle that is derived from the phasor diagram of
Figure 2.6(b), impedance triangle is derived by dividing current through each side of
voltage triangle shown in Figure 2.7(b) . Figure 2.7(c) shows impedance triangle
superimposed on the argand diagram where it can be seen that the impedance Z in
complex form given by;
Z = R – jXc
For example, an impedance expressed as (2 –j6) means that the resistance is 2 and
the capacitive resitance XC is 6.
VR=IR
V=IzVL = JXL
(b)
XL
z
JXL
real axis
Imaginary axis
R
0
Z
c.(a)
I R
VR Vc
(a)
VR I
VVc (b)
VRI
V-Jvc©
Realaxis
imaginaryaxis
21
In polar form, Z = /Z/< where, /Z/ = and tan-1
leading from
the impedance triangle.
VR=IR
V=IzVc = JVc
(a)
R
Xc(b)
z-Jxc©
real axis
Imaginary axis
R
0
Figure 2.7 (a) voltage diagram (b) impedance diagram (c) agrand diagram
R – L – C Series Circuit
In an ac circuit having resistance R, inductance L and capacitance C in series (Figure
2.8(a), the applied voltage V is the phase sum of voltage in resistance VR, voltage in
inductor VL and voltage in capacitor Vc as shown in the phase diagram of Figure 2.8(b)
(where the condition VL > Vc is shown). Figure 2.8(c) shows the phasor diagram being
superimposed on the argand diagram, complex form of the supply voltage given by;
V = VR + j(VL – Vc)
From the voltage triangle, the impedance Z triangle can be derived impedance
Z = R + j (XL – XC) or Z = /Z/<
Where, /Z/ = and = tan-1
(XL – XC)/R when; VL = Vc, XL = Xc
and the applied voltage V and the current I are in phase which can be termed as series
resonance.
Figure 2.8 (a) circuit diagram (b) phasor diagram (c) argand diagram
VR
VL
VC
I
L
C
V
(a)figure 2.8(a) circuit diagram
R
VL
V
VRI
Vc
(VL
- V
C)
(b)Phasor diagram
JVL
V
VR I
-Jvc
Real axis
J(V
L-
VC
)
Imaginary axis
©Argand diagram
22
General Series Circuit
An ac circuit having several impedance connected in series, Z1, Z2, Z3…Zn, the
total equivalent impedance ZT is given by ZT = Z1 + Z2 + Z2 +…+ Zn
Example 1: Determine the resistance R and series inductance or capacitor C for each of
the following impedance, assuming the frequency to be 50Hz;
(a) (4 +j7) (b) (3-j20) (c) j10 (d) –j3k
(e) 15<( /3) (f) 6<-450M
Solution:
(a) According to 2.2 (d) , for an R – L series circuit impedance Z = R + j x L
Thus, Z = (4+j7) represents a resistance of 4 and an inductive reactance of 7 in
series.
Since inductive reactance XL = 2 fL
Inductance L=
Thus, an impedance (4+j7) represents a resistance of 4 in series with an inductance
of 22.3mH.
(b) According to 2.2 (e), for an R – C series circuit, impedance Z = R –jXc. Thus Z = (3-
j20) represents a resistance of 3 and a capacitance reactance of 20 in series.
Since capacitive reactance Xc =
Capacitance C =
Thus, an impedance Z = (3 –j20) represents a resistance of 3 in series with a
capacitance of .
(c) j10
According to 2.2 (b) for a purely inductive circuit, impedance Z = j xL
Thus, Z = j10 represents zero resistance and an inductive reactance of 10
Since inductive reactance XL
Inductance
Thus an impedance (j10 ) represent a resistance of zero ohm in series with an inductance
of 31.8mH.
(d) –j3k
According to 2.2 (c), for a purely capacitive circuit, impedance Z = -jxc
Thus, Z = -j3000 represents zero resistance and a capacitive reactance of 3000.
23
Since, capacitance reactance Xc =
Capacitance
(e) 15<(
Recall;
15 (cos 600 + jsin 60
0) = 7.5 + j12.99
Thus; Z = 15<( represents a resistance of 7.5 and an
inductance reactance of 12.99 in series (According to 2.2(d)).
Since XL = 2
Inductance
(f) 6<-450M = 6 x 10
6 [cos(-45
0) + j sin (-45
0)]
= 4.243 x 106 –j4.24 x 10
6.
Thus, Z = 6<-45M = (4.243 x 106 – j4.24 x 10
6) represents a resistance of 4.243 x
106
) and a capacitive reactance of 4.24 x 106
in series [According to 2.2(e)]
Since capacitive reactance Xc =
Capacitance C =
Thus, an impedance 6<-450M represents a resistance of 4.23 X 10
6 in series
with 75nC capacitor.
Example 2: Determine in polar and rectangular forms, the current flowing in an inductor
of negligible resistance and inductance 50mH when it is connected to voltage of (30 +
j15)V, 60Hz supply
Solution:
Inductive reactance XL =2
= 18.85
Thus, circuit impedance Z = (0 + j18.85)
Supply voltage V = (30 +j15)V
Hence, current
I = 0.796 –j1.592A which is the same as 1.78<63.43
0A in polar form
24
Example 3: Determine (a) the resistance (b) the capacitance (c) the modulus of an
impedance and (d) the current flowing, its phase angle in an electrical circuit of
impedance (40 – j20) connected to a 230V, 50Hz supply.
Solution;
(a) Since impedance Z = (40-j20) , the resistance is 40 an the capacitive reactance is 20
(b) Since,
1 =1.59 F.
=2 x 3.142 x 50 x 20
(c) The modulus of impedance, /Z/ =
/Z/ = or 44.72<
(d)Impedance Z = (40 –j20) or 44.72<
Hence, current
That is; current I = 5.14<26.570A is leading the voltage by 26.57
0.
Example 4: Determine the value of the supply p.d. if 8 is connected to a supply of
frequency 2KHZ and current 2.69<900.
Solution
Capacitance
= 9.95
Hence, circuit impedance Z = (0 – j9.95) = 9.95< - 900
Current I = 2.69<900A (or(0 + j2.69)A)
Supply p.d., V = IZ = (2.69<900)(9.95<- 90
0)
p.d. V = 26.8<00V
Alternatively; V = IZ = (0 + j2.69) (0 – j9.95)
= - j2 (2.69)(9.95) = 26.8v≈27V
Example 5: Determine the value of impedance and components forming a series circuit of
current I (9 + j17)A, voltage (100 + j220)V. If the frequency of supply is 20MHZ
Solution:
Impedance
25
Or 13.12<3.47
Thus, the series circuit consists of a 13.12 resistor and a capacitor of capacitive
reactance 3.47
Since
C = 2.29 X 10-9
F.
Example 6: A 400V, 50HZ supply is connected across a coil of negligible resistance and
inductance 0.20H connected in series with a 40 resistor. Find (a) the impedance of the
circuit (b) the current and circuit phase angle (c) the p.d. across the 40 resistor and (d)
the p.d. across the coil.
Solution
(a) Inductive reactive XL = 2fL = 2 (50) x 0.20 = 62.84
Impedance Z = R x JXL = (40 + j62.84) or 74.49 < 57.520
The circuit diagram is shown in Figure 2.9
VR VLV
R=0I
v+ -400v, 50Hz
L - 0.20H
(b) Current
i.e. the current is 5.37A lagging the voltage by 57.520V
(c) The p.d. across the 40 resistor VR = IR
VR = (5.37 <- 57.52) (40 < 0) = 214.8 < - 57.520V.
(d) The p.d across the coil, VL = IXL = (5.37 < - 57.52) (62.84 < 900)
VL = 337.45 < 32.480V.
The phasor sum of VR and VL is the supply voltage V as shown in the phasor diagram of
26
V =
214
.8v.
R
32.480
V = 400v57.52
0
I = 0.10A
X =
337.4
5
L
Figure 2.10
VR = 214.8 < - 57.520 = (115.35 –j 181.20)V
VL = 337.45 < 32.480 = (284.67 + J 181.21)V
Hence, V = VR + VL =( 115.35 –j 181.20 + 284.67 + j181.21) V
V = (400 + j0) V or 200 < 00 V, correct in three significant Figures
Example 7: Determine the value of impedance Z2 in the circuit shown in Figure 2.11
Solution
Total circuit impedance Or (4.7 + j8.14)
Total impedance Z = Z1 + Z2 (see section 2.12(g)
Hence, (4.7 + j8.14) = (9.36 – j3.10) + Z2 from which,
9.36
-j3.10
Z2
80
<3
0v
0
I=8.5<-30 A0
Figure 2.11
27
Impedance Z2 = (4.7 + j8.14) – (9.36 – j3.10)
Z2 = (-4.66 + j11.24) or
Multiply through by – 1. Which equal to Z2 = (4.66 – j11.24) Or 12.17 < 67.460
Example 8: A coil of resistance 40 and inductance 30mH has an alternating voltage
given by V = 262.4sin (620t + (/3)) volts applied across it. Determine (a) the rms value
of voltage (in polar form) (b) the circuit impedance (c) the rms current flowing and (d)
the circuit phase angle.
Solution
(a) Voltage V = 262.4sin (620t + /3) volts means Vm = 262.4V, hence, rms voltage
Vr.m.s =
Vr.m.s=0.707×262.4 ; Vr.m.s=186v.
In complex form the rms voltage may be expressed as 186 < /3V or 186 < 600V.
(b) w = 2f = 620 rad/s, hence, frequency f;
Inductive reactive XL = 2fL = 2x3.142 x 98.7 x 30 x 10-3
.
XL = 18.6
Hence, circuit impedance Z = R x jXL; Z= (40 + j18.6) or 44.11 < 24.940
(c) Rms current ;
(d) Circuit phase angle is the angle between current I and voltage V, i.e. 600 – 35.06
0
= 24.940 lagging.
Example 9: Determine the values of R and L in a circuit coil of resistance R ohms and
inductance L henrys connected in series with a 60µf capacitor, if the supply voltage is
222V at 50Hz and the current flowing in the circuit is 2.5 < -200A. Also, determine the
voltage across the coil and the voltage across the capacitor.
Solution
Circuit impedance (83.4+30.4)
Capacitance reactance
XC = 53.04
Circuit impedance; Z = R + j(XL – XC)
83.4 + J30.4 = R + j (XL – 53.04)
Equating the real parts gives:- resistance = 83.4
28
Equating the imaginary parts gives: 30.4 = XL = 53.04
XL = 30.4 + 53.04; XL = 83.44
Since XL = 2fL, inductance
L = 8.27H. The circuit diagram is shown in Figure 2.12
Voltage across coil, Vcoil = IZcoil
Zcoil = R + jXL = (83.4 + j83.44) 118 < 450
Hence, Vcoil = (2.5 < -200)(118 < 45
0)
= 295 < 250V or (267.36 + j124.67) V
Voltage across capacitor , VC = IXC = (2.5 < -200) (53.04 < -90
0)
= 132.6 < -1100V OR (- 45.35 - j124.60)V
(check; supply voltage V = Vcoil + VC
V = (267.36 + j124.67) + (- 45.53 – j124.60)
V = (222 + j0)V or 222 < 00V
Example 10: Below is a circuit shown in Figure 2.12, determine the values V1 and V2 if
the supply frequency is 3KHZ. Determine also the value of the supply voltage V and the
circuit phase angle. Draw the phasor diagram.
R
-
222v,
50H
z
2.5<-20 A0
Figure 2.12
Vcoil
coil
L
60UfVc
29
For impedance Z1, XC =
XC = 19.77
Hence, Z1 = (10 – j19.77) or 22.16 < - 63.170
And voltage V1 = IZ1 = (9 < 0) (22.16 C <- 63.170)
V1 = 199 < - 63.170V or (89.82-j 177.58)V
For impedance, Z2, XL = 2fl = 2x3.142 x 3x103 x 0.487 x 10
-3.
XL = 9
Hence, Z2 = (6 + j9) or (10.8 < 56.30)
And voltage V2 = IZ2 = (9 < 0)(10.8 < 56.30)
= 97.2<56.30V OR (53.9 + J80.9)V
Supply voltage, V = V1 + V2 = (89.82 – j177.58) + (53.9 + j80.9)V
V = (143.72 – j96.68) V or (173.21 < - 33.93)V
Circuit phase angle = 33.930 leading. The phasor diagram is shown in Figure 2.14
V2 = 97.2
uuuuu
v2Z2
I = 9
<0A0
v1
6 0.487mH2.683ufi0Z1
Figure 2.13
63.170
I = 9A
56.30
143.72
=33.930
Figure 2.14
V =1
30
Exercise on Series ac Circuits
1. In a series circuit containing pure resistance and a pure inductance, the current and
the voltage are expressed as:
i(t) = 6sin (314t + 4/3) and v(t) = 20sin (314t + 6/7)
(a) Calculate the impedance of the circuit
(b) Find the value of the resistance
(c) What is the value of the inductance in Henrys?
(d) Calculate the total power drawn by the circuit and the power factor
(3Ω, 0.22Ω, 9.52mH, 3.96W and 0.075 leading).
2. Potential difference measured across a coil is 5.4V, when it carries a direct current
of 10A. The same coil when carrying an alternating current of 10A at 30Hz, the potential
difference is 30V. Find the current, the power and the power factor when it is supplied by
60v, 60Hz supply.
(10.14A, 55.52W).
3. In an R – L series circuit, a voltage of 20v at 50Hz produces a current of 600mA,
while the same voltage at 65Hz produces 400mA. What are the values of R and L in the
circuit? (0.14H)
4. A 49Hz sinusoidal voltage = 121sin wt is applied to a series R – C circuit. The values
of the resistance and capacitance are 6 and 0.2f respectively. Calculate:
i. The r.m.s. value of the current in the circuit and its phase angle with respect
to the voltage.
ii. Write the expression for the instantaneous current in the circuit.
iii. R.m.s. value and the phase of the voltages appearing across the resistance
and the capacitance.
(5.27x10-3
<-89.980
lagging, i =7.45x10-3
Sin (wt-89.980), 31.62x10
-3<-89.98
0v,
85.58<179.980v).
5. Three impedances are connected in series across a 130V, 20KHz supply. The
impedances are:
i. Z1, a coil of inductance 300f and resistance 9
ii. Z2, a resistance of the 14
iii. Z3, a 0.60f capacitor in series with a 12 resistor. Determine:
a. The circuit impedance
b. The circuit current
c. The circuit phase angle and
d. The p.d. across each impedance
( 4.46<-10.760Ω, 29.15<10.76
0A, 10.76
0leading).
31
32
CHAPTER THREE
APPLICATION OF COMPLEX NUMBERS TO PARALLEL AC
NETWORKS
Introduction
Parallel networks can be analysed using Phasor diagrams. Moreover, when parallel
networks having more than two branches, use of phasor diagram may become very
complicated. Parallel ac networks may be analysed with the use of complex numbers.
Impedance may be easily used to represent series ac circuit while admittance,
conductance and susceptance may be easily used to represent parallel ac circuit having
more than two parallel impedances.
Admittance, Conductance and Susceptance
Admittance may be defined as the measure of how well an ac circuit will “admit
or allow” current to flow in the circuit. The larger its value, the heavier the current flows
for the same applied potential.
It can also be defined as the current I flow in an ac circuit divided by the supplied
voltage V. The unit of measurement for admittance as define by SI system is Siemens
represented symbolically as “S”. Admittance can also be said to be the reciprocal of
impedance (i.e. Y = 1/Z). Admittance is represented by “Y” in a circuit
In series ac circuit, impedance may be resolved into a real part R and imaginary part X,
which gives; Z = R ± jX. Similarly, in an ac parallel circuit, admittance may be resolved
into two parts (i.e. the real part and the imaginary part). The real part called conductance
“G” and the imaginary part called susceptance “B” which can be represented in complex
form Y = G ± JB
Figure 3.1: Parallel ac network
33
Reciprocal of resistance (1/R) is called conductance represented symbolically by
“G” measured in Siemens “S”.
Reciprocal of reactance (1/X) is caked susceptance represented symbolically by
“B” and defined as the measure of how susceptible an element is, to the passage of
current through it. It can be measured in Siemens “S”
If an ac circuit contains
(a) Pure resistance, then
(b) Pure inductance, then
Note: Negative sign is associated with inductive susceptance, BL. For inductance,
an increase in frequency or inductance will result in a decrease in susceptance or
correspondingly in admittance.
(c) Pure capacitance, then
Note: Positive sign is associated with capacitive susceptance, BC. For capacitor,
an increase in frequency or capacitance will result in a increase in its
susceptibility.
(d) Resistance and inductance in series, then
34
Thus, conductance, G = R//Z/
2 and inductive susceptance, BL = -XL//Z
2/.
(Note, in an inductive, circuit, the imaginary term of the impedance, XL is positive while
the imaginary term of the admittance, BL is negative).
(e) Resistance and capacitance in series, then;
Thus, conductance G = R//Z/2 and capacitive susceptance, BC = XC//Z/
2.
(f) Resistance and inductance in parallel
(g) Resistance and capacitance in parallel
35
Example 1: Determine the admittance (in polar form), conductance and
susceptance of the following impedance; (a) J10 (b) –j40 (c) 40<-600
(d) (6+j8) (e) (18-j20)
Solution
(a) J10; Z=j10
Y = G ± JB; Y = 0.1<-90
0S.
G=0s, BL=0.1s
(b) –j40 ; Z = -J40
(c) 40<-600, Z = 40<-60
0
=0.025<600
G = 0.0125s, Bc = 0.0217s
36
(d) (6 + j8)
Y = 0.06-j0.8s. or Y = 0.1< - 53.13
0s
G = 0.06s, BL = 0.08s
(e) (18-J20)
Y = (0.025 + J0.028)s or Y = 0.04 < 48.24
0S
G = 0.025s, BL= 0.028s
Example 2: Derived expressions in polar form, for the impedances of the following
admittances: (a) 0.03<500S (b) (0.015 – j0.02)S
Solution:
(a) Y = 0.03 < 500S
Z=
(b) Y = (0.015 – J0.02)S
Y = 0.025 < - 53.1305
Z=
Example 3: The admittance of a circuit is (0.02+j0.06)S. Determine the values of the
resistance and the capacitive reactance of the circuit if they are connected (a) In series (b)
In parallel. Draw the phasor diagram for each of the circuit connection.
Solution;
Series Connection
Admittance Y = (0.02 + j0.06)S, therefore
Impedance
Thus, the resistance R = 5 and capacitive reactive, Xc = 15
37
The circuit and phasor diagrams are shown Figure 3.2
Parallel connection
Admittance Y = (0.02 + j0.06)S therefore; conductance G = 0.02s and capacitance
susceptance, Bc = 0.06s from eqn. (3.1) when a circuit consists of resistance and the
capacitive reactance in parallel, then Y = (1/R) + (J/XC)
Hence, resistance
=
The circuit and phasor diagrams are shown in Figure 3.2.1
The circuits shown in Figure 3.2(a) and 3.21(a) are equivalent in that they take the same
supply current I for a given supply voltage V; the phase angle between the voltage and
the current is the same in each of the phasor diagrams shown in Figure 3.2(b) and 3.2(b)
Parallel ac Networks
Circuit diagram of an ac network having four impedances, Z1, Z2, Z3 and Z4
connected in parallel is shown in Figure 3.3. it can be deduced that the same potential
difference (voltage) passes through four impedances but different current passes through
them.
VR
VCV
I
(b) phasor diagram
+-
V
VRVC
XC = 15R5.0 = 15I
Figure 3.2(a) circuit diagram
IR
IC
I
V
V
IRIC
I
X = 16.67C
R = 50
3.21(a) Circuit diagram (b) Phasor
38
The current through each part can be found by application of ohm’s law V =IR; V=IZ;
I=V/Z
If ZT is the total equivalent impedance of the circuit, then or or
YT. where E is the supply voltage and YT is the equivalent circuit admittance.
The total or source current, I = I1 + I2 + I3 + I4 (phasorially)
Thus,
Or total admittance, YT = Y1 + Y2 + Y3 + Y4.
In general, for n impedances connected in parallel,
YT = Y1 + Y2 + Y3 + Y4 + - - - - - + Yn
The use of admittance has its great advantage in parallel circuit.
Power = E Icos T, where T is the phase angle between E and I
Note: T for inductive networks is negative while T for capacitive network is positive.
Parallel R – L Network
Current Division in ac Circuits (Current Divider)
In the case of two impedances connected in parallel (i.e. Z1 and Z2) in Figure 3.4, current
divider rule can be used to calculate the current through each impedance in the circuit.
IR
IIL
XL
+
-R
I1
X1V
I
Figure 3.3
I2
X2
I3
X3
I4
X4
39
From Figure 3.4
Supply voltage V = IZT = I1Z1 = I2Z2.
Z1
Z2
Example 4: using current divider rule, find the current through each parallel branch of
Figure 3.5
Fig
ure
3.5
V
I2I1
Z2Z1
I
Figure 3.4
I =61<20T
0
R XL
ZR-L
XC
ZC3
92
40
Using current divider rule
(a)
Note: IT = I1+ I2 = 2.85 < -141.57
0 + 8.73<25.9
0
IT = - 2.23 – j1.77 + 7.85 + J3.81
IT = (5.62 + J2.04) A = 6 < 200 A.
Example 5: using current divider rule, find the values of current IT, I1 and I2 shown in the
network of Figure 3.6
Total circuit impedance,
+ = (0.51+J1.91) or (1.98
Supply current
7
60<00
Figure 3.6
I2
J79
I1
I
41
I2 = 29.6 < -87.58
0A
To check; IT =I1 +I2= (6.59 < 2.420) + (29.6 < -87.58
0)A
IT = 6.58 + j0.28 + 1.25 –j29.57 = (7.83 –j 29.29)A
IT = (30.3 < -75.030)A
Example 6: Determine the value of supply current I and phase angle of the network of
four impedances connected in parallel shown in Figure 3.7 when connected across a
250V ac supply.
I1I3I2
I4
R =101R =302 R =23
R =254
X =20C X =0LX =15L
X =12L
250V
I
Figure 3.7
42
Impedance Z1 = (10 –j20), Z2 = (30 + j0), Z3 = (2 –j15) and Z4=(25+j12)
Supply current I =
YT =total circuit admittance. YT = Y1 + Y2 + Y3 + Y4
YT = (0.092 + j0.09)S or 0.13 < 44.37
0S
Current, I = VYT = (250 < 00)(0.13 < 44.37
0) = 32.5<44.37
0A
Hence, the current I is 32.5A and is leading the 250V supply by 44.370.
Example 7: An ac network consists of a coil, of inductance 89.28 mH and resistance 12,
in parallel with a capacitor of capacitance 54.86µf. If the supply voltage is 240<00V at
60Hz, determine (a) the total equivalent circuit impedance (b) the supply current (c) the
circuit phase angle (d) the current in the coil and (e) the current in the capacitor. The
circuit is shown in Figure 3.8.
Inductive reactance, XL = 2fl = 2 (60) (89.28 x10-3
) = 34
Hence, the impedance of the coil
Zcoil= (R + JXL) = (12 + j34) or (36.06 < 70.560)
Capacitance reactance,
In complex form, the impedance presented the capacitor, Zc is –JXC, i.e. –j48.35 or
48.35 < 900
Icoil
R =12
L = 89.28mH
IC240<0 v0
60Hz
Figure 3.8
43
(a) Total equivalent circuit impedance
(b) Supply current,
(c) Circuit phase angle = 30.660 lagging i.e. the current I lags the voltage V by 30.66
0.
(d) Current in the coil, I coil =
(e) Current in the capacitor,
Example 8: Determine the series equivalent circuit for the network of Figure 3.9
RP=8k
Xp (resultant) = (XL – Xc) = (9k - 4k) = 5k
And
With (inductance)
Equivalent series circuit for parallel network of Figure above
R
RP
8K
Xc 4K
Xp
9KXL
Figure 3.9
44
Example 9: (a) Determine YT (b) Find E and IL (c) compute the power factor of the
network and the power delivered to the network (d) equivalent series circuit as far as the
terminal characteristics of the network are concerned (e) using equivalent circuit
developed in part (d), calculate E and compare it with the result of part (b). (f) Determine
the power delivered to the network and compare it with the solution of part (c). (g)
Determine the equivalent parallel network from the equivalent series circuit and calculate
the total admittance YT. Compare the result with part (a) in the Figure 3.10 shown
below.
Solution
(a) Combining common elements and finding the reactance of the inductor and
capacitor. I =
We obtain,
F
XL = wL = (1000rad/s)(4.3mH) = 4.3
Total Y; YT = YR + YL + YC.
= G<00 +BL< - 90
0 + Bc < +90
0
YT =
YT = (0.21)s - j(0.23)s + (j0.05)s = (0.21 – j0.18)s or (0.28<- )s
I = 12A<00
5
+
-
ZT
RI R2L1 L2
C1C2
IL ICIR
YT
90 7mH
11mH
e
Figure 3.10
45
DIA
(b)
(c)
Power = EI cos ; P = (42.86 x 12) cos 40.60= 390.51W
(d)
The equivalent circuit diagra
=5.49
(e) E = IZT = (12A<00)(3.57<40.6
0) = (42.84 < 40.6
0)V
(f) P = I2R = (12A)
2 (2.71
(g)
Parallel equivalent circuit
46
Exercises
1. Determine the admittance in polar form, conductance and susceptance of the
following impedances.
(a) J20 (b) -30 (c) (16-j20)
(a. Y=0.05<-90S, G=0S, BL=-0.05S, b. Y=0.03<90S, G=0S, Bc=0.03S,c.
Y=0.014+j0.05S, G=0.014S, Bc=0.05S).
2. Derive expression, in polar form, for the impedances of the following admittance:
(a) 0.002<-25O6 (b) (0.0030 –j0.045)s (c) 0.02<40
Os (d) (0.02 + j0.5)s
(50<25OΩ, 22.22<86.19
OΩ, 50<-40
OΩ, 1.85<-68.2
OΩ).
3. The admittance of a two branch network is (0.04 + j0.09)s. Determine the circuit
component if the frequency is 30Hz
(R=4.142Ω and C=0.049F).
4. Determine the total admittance, in rectangular and polar forms pf each of the
networks show in below:
30
j60
j16
30
12
j6
-j12
20
47
(0.015<-63.43OS, 0.083<56.6
OS, 0.11<-7.22
OS or0.109-j0.014S).
5. Determine the equivalent circuit impedances of the parallel networks shown
below:
(11.14<-68.2OΩ, 21.98<-38.48
OΩ)
6. Determine for the network shown below,
(a) The total network admittance
(b) The total network, impedance
(c) The supply current I
(d) The network phase angle
(e) Currents I1, I2, I3 and I4.
(25.67<-0.16OS, 0.039<0.16
OΩ, 770<-0.16
OA).
-j15
j14
13
-j12 30
30<0O
V
3KH
z
6f
I4
I3
36 30
4m
H
I2
I1
I
48
49
CHAPTER FOUR
AC BRIDGES
Introduction
AC Bridges are circuit configuration of Wheatstone bridge principle used for
measuring unknown values of resistance, accurate measurement of unknown impedances
and frequency. AC bridges consist of an ac power supply, a balance detector which is
sensitive to alternating currents and four impedance arms. AC bridges are formed by
replacing the DC battery with an ac source and galvanometer by detector of Wheatstone
bridge. Bridge circuit can be constructed to measure any device value desired, example
inductance, capacitance, storage factor, dissipation factor, etc.
Balance Conditions for an AC Bridge
AC bridges are categorized as four-arm bridges, consisting of four resistors or
impedances in complex from (i.e. Z = R+ jX), R being the resistance, jx, the imaginary
side which can be inductive reactance XL or capacitive reactance XC. Bridge circuit
operates as a pair of two component voltage dividers connected across the same source
voltage, with a null detector meter movement connected across them to indicate a
condition of “balance” at zero volts. Anyone of the four resistors or impedances in a
bridge circuit can be the resistor/impedance of unknown value, and its value may be
calculated by a ratio of the other three, which are calibrated or whose
impedances/resistances are known to be precise.
In bridge circuits, the unknown quantity is always “balanced” against a known
standard, obtained from a high quality, calibrated component that can be adjusted in
value until the null detector device indicates a condition of balance. Depending on how
the bridge is set up, the unknown components value may be determined directly from the
setting of the calibrated standard, or derived from value may that are standard through a
mathematical formula.
In AC bridges, instead of using a source of direct current, alternating current is
used and galvanometer is replaced by a vibration galvanometer (for commercial
frequencies or for frequency higher than 500 to 200Hz).
50
Condition for balance;
There will also be a phase balance for the impedances. In polar form,
Z1<1.Z3<Z3=Z22.Z44
Z1Z3<1 + 3 = Z2Z4<2 + 4
There are two balance conditions which must be satisfied simultaneously in a four – arm
ac impedance bridge, which are
Z1Z3= Z2Z4 ………………………. for magnitude balance
1 + 3 = 2+ 4 ……………………… for phase angle balance
Types of AC Bridge Circuit
There are numerous types of a .c. bridges developed, each of which has its own
advantage over some ac bridges under some conditions. Most of important AC bridges
are: Maxwell’s inductance bridge, Maxwell-Wien bridge, Hay bridge, Schering bridge,
De Sauty bridge, Wien series bridge, Wien Parallel Bridge, Anderson Bridge and many
others.
Maxwell’s Inductance Bridge
Maxwell’s Inductance Bridge was named after James Clerk Maxwell. Maxwell’s
Inductance Bridge is used for measuring unknown inductances in an AC bridge. In
Maxwell inductance bridge, products of opposite arm resistances are equal. There are two
pure resistance in two arms of bridge for balance relations, phase balance depend on the
remaining two arms. If a coil of an unknown impedance Z1 is placed in one arm, then its
positive phase angle can be compensated for either of two ways, namely:
R1 R2
R3R4
A A
D
B
Figure 4.1(a)
Z1 Z2
Z3Z4
A A
D
B
Figure 4.1(b)
51
i. A known impedance with an equal positive phase angle may be used in either of
the adjacent arms (so that 1 = 3 or 2 = 4) the remaining two arms having
zero phase angles (pure resistances). This type of network is known as
Maxwell’s AC bridges or L1/L4 bridge.
ii. Or an impedance with an equal negative phase angle (i.e., capacitance) may be
used in opposite arm (so that 1 + 3 = 0). Such a network is referred to as
Maxwell - Wien bridge/Maxwell L- C bridge.
Z1 = R1 + JXL = R1 + Jwl1 ………..unknown
Z4 = R4 + jX4 = R4 + jwL4 ……….. known
R2, R3 = known pure resistances, D = detector
Balance condition is that; Z1Z3 = Z2Z4
(R1+jWL1)R3 = (R4 + JwL4)R2
Equating the real and imaginary parts on both sides, we have R1R3 = R2R4 or R1/R3 =
R2/R4 (products of resistances of opposite arms are equal)
and WL1R3 = WL4R2 or L1 = L4R2/R3.
Hay Bridge
Hay Bridge is very useful in measuring the resistance and inductance of a coil having a
very high Q – factor.
R2
R3Z4
A
D
B
C
R1
I1
L1 I2
I3
R4
L4
I4
R 1
R2
R3R4
E
L3
C 1
52
Z3 = R3 + jwLX Z2=R2, Z1 = R1 – JXc and Z4 = R4.
At balance; Z1Z3 = Z2Z4.
(R1 – JXc)(R3 + jXL) = R2R4.
R3 + JXL3 =R2R4/(R1 –JXc)
Rationalizing the right hand side give,
Since XC1 = 1/wC;
Since XL3 = WL3;
53
Thus, at balance, the unknown components in the Hay bridge are given by:
Owen Bridge
Owen Bridge is used to measure inductance in term of resistance and capacitor.
At balance; (ZX)(Z1)=(Z2)(Z4);
i.e. (Rx + jXLx)(- jXC1) =(R2) (R4-jXC4)
separating real and imaginary we get;
RX = R2 C1/C4 and L3 = C1R2R4.
Since ‘w’ does not appear in final balance equation, the bridge is unaffected by frequency
variations and wave forms.
Maxwell – Wien Bridge or Mazwell
Positive phase angle of inductance impedance may be compensated by the
negative phase angle of a capacitive impedance then becomes known in the terms of this
capacitance.
R1
R2
R3R4
A C
D
BC
D
L3
R2
RxR4
E
Lx
C 1
C4
Figure 4.5
54
Combine impedance of arm 1;
Z3 = R3 + jwL3 and Z4=R4
Balance condition is Z1Z3 = Z2Z4
Or
Separating the real and imaginaries, we get;
C
De Sauty Bridge
De sauty bridge is a very simple method of measuring a capacitance by comparing
with another known capacitance.
CX = capacitor whose capacitance is to be measured
C3 = a standard capacitor
R1
I1
I3R2
A C
D
B
D
C3
I1
CX
55
R1,R2 = non- inductive resistors
Balance can obtain by varying either R1 or R2.
At balance; point B and D are at the same potential
IR1 = I2R2 and (J/CX )I1 = (j/C3 )I2.
Dividing one equation by the other we get;
De Sauty bridge has maximum sensitivity when Cx = C3.
Simplicity of this method is offset by the impossibility of obtaining a perfect balance if
both capacitors are not free from the dielectric loss. Perfect balance can be obtained if air
capacitors are used.
Schering Bridge
Schering bridge is used for measuring capacitor and dielectric loss of a capacitor (i.e.
capacitance and equivalent series resistance of a capacitor). It is also used for comparing
imperfect capacitor Cx in terms of a loss free standard capacitor C1 (figure below). The
imperfect capacitor is represented by its equivalents loss free capacitor Cx in series with a
resistance r (figure below).
For balance Z1Z3 = ZXZ4.
C1
Cx
R4
A C
D
B
R3
D
C4
C1
Cx
R4
E
C
D
R3
Rx
56
Separating,
Quality of a capacitor is usually expressed in terms of its phase effect angle and
dielectric loss angle which is defined as the “angle by which current leave exact
quadrature from the applied voltage, i.e. complement of the phase angle”. If is the
actual phase angle and defect angle, then For smaller values of
(approximate). Tan usually called the dissipation factor
of the R-C circuit.
For low power factor, dissipation factor is approximately equal to the power
factor.
Dissipation factor = power factor
Putting the value of r Cx from above
Dissipation factor = r Cx, C4R4 = Power factor
Wien Series Bridge
Wien Series Bridge is a simple ratio bridge and is used for measuring audio
frequency of capacitors over a wide-range.
r
Xc
R1
R2
R3C4
A A
D
B
Rx
Cx
R4
57
At balance;
Wien Parallel Bridge
Wien parallel bridge is a parallel bridge used mainly for measuring audio
frequency although it is not as accurate as the mode in digital frequency meters.
Under balance condition;
Separating real from imaginary, we get;
And
RR4
R3C1
B C
D
A
R1
58
Examples 1: Arms of a bridge are arranged as follows: XY and YZ are non-reactive
resistors of 50 each, DX is a standard variable reactor L1 of resistance 33.7 and CD
comprises a standard variable R in series with a coil of unknown impedance. Balance was
obtained with L1 = 37.8mH, R = 1.26, find resistance, inductance of coil.
Products of opposite arm resistance are equal.
33.7 x 50 = (1.26 + R4) 50
1685 = 63 + 50 R4
1685 – 63 = 50 R4
1622 / 50 = R4
R4 = 32.44
L1 x 50 = L4 x 50; L4 = L1 = 37.8mH
Or because time constants are the same, hence;
L4 = 37.8MH
Example 2: Using Maxwell-Wien formula, determine
the value of R3 and L3 in the bridge circuit shown below.
R2
R1L4
L1
R1
37.8m
H
33.7
1.26
Coil
50
R1
L3
R4
200
400
500
R3
59
0.6 x 10
-6 x 200 x 500
Example 3: A Schering bridge network is as shown in Figure 4.7, given C2 = 0.3µf, R4 =
400, R3 = 200, C3 = 2000pf and the supply frequency is 1.2kHz, determine, when the
bridge is balanced. Determine (a) the value of resistance Rx, (b) the value of capacitance
CX, (c) the phase angle of the unknown arm (d) the power factor of the unknown arm and
(e) its loss angle.
(a) Resistance
Rx
C 3(2000pf)E
Cx
R2(200 )
R 4=400
Ix
60
(b) Capacitance
= 0.15µF
(c) The phasor diagram for Rx and Cx in series is shown = (1xRx)
Phase angle, Ø = =
i.e,
Power factor of capacitor = Cos
Loss angle, shown in Figure 4.7, is given by
Alternatively, loss angle wCXRX.
Example 4: For Wien bridge shown in Figure below, R2 = R3 = 40K and C2 = C3 =
2nf. Determine, when the bridge is balanced, determine (a) the value of resistance R1 and
(b) frequency of the bridge.
IX
I (X I Z )X XV (CX= I z )X CX
R1
E
R 4
C1
C3
R (40K ) 2
(40K )
R3
61
i. e. 1+1 =1000/R1, since R2 = R3 and C2 = C3, from which;
resistance R1 = 1000/2 = 500
(b) frequency f =
Frequency f = 1.99 x 103 = 1.99KHz
Exercise on ac Bridges
1. In a four arm De Sauty ac bridge, arm 1 contains a 3K non-inductive resistor,
arm 3 contains a loss free 3.2µf capacitor, arm 4 contains a 8k non –inductive
resistor. When the bridge is balanced, determine the value of the capacitor
contained in arm 2.
(Cx=1.2µF)
2. At balance, an ac bridge ABCD used to measure the inductance and resistance of
an inductor has the following values AB – a non-inductive 300 resistor, BC- the
inductor with unknown inductance Lx in series with resistance Rx; CD – a 2µf
capacitor in series with non inductive 300, DX-a 12 capacitor. A detector is
connected between B and D, and the ac supply is connected between A and C.
derive the balance equations for Rx and Lx and determine their values.
(Rx=1800Ω, Lx=1.08H).
3. The conditions at balance of a Schering bridge PQRS used to measure the
capacitance and loss angle of a paper capacitor are as follows: PQ – a pure
capacitance of 1 QR – a pure capacitance of 200 in parallel with a 200
resistance, RS – a pure resistance of 100, DP – the capacitance under test which
may be considered as a capacitance Cx in series with a resistance Rx. If the supply
frequency is 3KHz, determine: (a) the value of Rx (b) the value of Cx (c) the
power factor of the capacitor and (d) its loss angle.
(Rx=20KΩ, Cx=2µF, p.f=1, 89.9Oloss angle).
4. An ac bridge has an arm Rx, a pure capacitor of 2 ; in arm XY, a pure resistor of
600; in arm YZ, a coil of 20 resistance and 0.2H inductance, in arm ZB an
unknown impedance consisting of Rx and capacitance Cx in series. If the
frequency of the bridge at balance is 50Hz, determine the value of Rx and Cx.
(Rx=166.67Ω, Cx=60µF).
62
5. In an ac bridge circuit shown in Figure below, is balanced when the values of the
components as shown in fig below. determine at balance, the values of Rx and Lx
(Rx=1,333.33Ω, Lx=1.6H).
63
CHAPTER FIVE
INTRODUCTION TO NETWORK ANALYSIS
Introduction
Network analysis refers to analysis of general networks (i.e. ac network/ d.c.
network). It also refers to any structured technique used to analyze a circuit (a network of
interconnected circuit).
In an ac network comprising of resistors, voltage sources in series/parallel, current
flows in each branch of the series/parallel circuit can be determined using any of the laws
stated below.
The laws are (laws that determine the currents and voltage drops in a .c. network):
(1) Ohm’s law; V=IZ, I=V/Z, where Z is the complex impedance, V is the voltage
drop across the impedance and I is the current flowing in the impedance.
(2) Laws for impedances connected in series and parallel, total impedance ZT =
Z1Z2/Z1+Z2--- for two impedances connected in parallel.
If more than two impedances are connected in parallel, the total impedance is
written as follows -----+
where Zn indicate numbers of impedances in the current.
and ZT = Z1 + Z2 + Z3 + Z4 + -----+Zn for series connection of impedance
(3) Kirchhoff’s laws which are two : Kirchhoff voltage low and Kirchhoff current
law, stated as follows
Kirchhoff voltage law (KVL) states that “the sum of voltage entering a node,
junction or a point is equal to the amount of voltage leaving a node, junction”
Kirchhoff current law (KCL) states “the algebraic sum meeting at a junction is
practically equal to zero”.
Other forms of analyzing ac network are : Mesh – current analysis , Nodal
analysis, superposition Theorem, Thevenin’s Theorem, Norton’s Theorem, Maximum
power transfer Theorem and Star-Delta Transformation
Solution of Simultaneous Equations Using Determinants and Crammer’s Rule
Simultaneous equations erupt when electrical circuit are solved using Kirchhoff’s
laws. Simultaneous equations may contain more than two unknowns depending on the
number of loops involved in the electrical circuit. Methods of elimination/substitution can
be used for solving the simultaneous equations. However, when complex numbers are
involved, more appropriate method to use is determinant and crammer’s rule.
64
Determinant and crammer’s rule make equations of complex numbers simpler to
solve.
Two unknowns
When solving linear simultaneous equations involving two unknowns using
determinants.
(a) The equations can be written as:
a1+x1 + b1x2 = E1 ……… (i)
a2+x1 + b2x2 = E2 …….. (ii)
(b) The solution is given by
Each of the expressions, ( , ( , and ( is called a
determinant.
They are denoted respectively as;
The above example is “2 x 2” determinant (i.e. it has 2 rows and 2 columns). The
quantities a1, a2, b1, b2, E1, E2 are called the elements of the determinants.
Three unknowns
When solving linear simultaneous equations in three unknowns using determinant;
(1) The equations are written as:
a1x1 + b1x2 + c1x3 = E1
a2x1 + b2x2 + c2x3= E2
a3x1 + b3x2 + c3x3= E3
(2) The denominator of each solution, x1, x2, x3 is obtained as;
a1b2c3 – a1b3c2 +a2b3c1 – a2b1c3 +a3b1c2 – a3b2c1
which can be denoted by,
The above example is “3x3” determinant (i.e. it has 3 rows and 3 columns).
“3x3” determinant can be expressed in terms of “2x2” determinant.
65
=a1(b2c3 – b3c2) – a2(b1c3 – b3c1) + a3(b1c2 – b2c1) = a1 - a2 + a3
Where 1, 2 and 2 are called the minor of a1, a2 and a3 respectively, and the “2x2”
determinant obtained when the row and column containing a1, a2, a3 respectively are
deleted.
The “3x3” determinant as expressed above is said to have been expanded through
the first column. Since, the elements a1, a2 and q3 (which are respectively the coefficient
of the 1, 2, and 3) are elements in the first column of the third order determinant.
When expanding determinants, the following procedures needed to be taken:
(a) A determinant has as many minor as the elements inside it. (i.e. every element aij
has a minor αij numbers of column) where ‘’I’’ indicate no of row and “j’’
indicate no of column.
(b) Determinant can be expanded along any of its rows or down through the columns.
(c) Signs (positive and negative) should be assigned properly. Signs to be assigned to
“3x3” determinant, “4x4” and “5x5” determinant is given as follows:
+ - + + - + - + - + - +
- + - - + - + - + - + -
+ - + + - + - + - + - +
- + - + - + - + -
+ - + - +
“3 x 3” “4 x 4” “5 x 5’’
Two unknowns
When solving simultaneous equation involving two unknowns using crammer’s
rule;
i. The equation is written as
a1x1 + b1x2 = E1.
a2x2 + b2x2 = E2.
ii. The solution is given by:
66
The coefficient of x1 and x2 in the determinant can be denoted by . That is,
x1 is the determinant obtained by replacing a1 and a2 (i.e. the coefficients of x1) by E1
and E2 respectively in , then x2 is also the determinant obtained by replacing the
coefficients of x2 in the equation (that is b1 and b2) by E1 and E2 respectively. The
solution of the system of equation becomes,
Three Unknowns
(i) The equations are written as:
a1x1 + b1x2 + c1x3 = E1.
a2x1 + b2x2 + c2x3 = E2.
a3x1 + b3x2 + c1x3 = E3.
(ii) The solution is given by
= defined as: =
And x1, x2 , x3 are obtained from the determinant “” by substituting the E’s in
place of the coefficient of x1, x2, and x3 respectively.
Network analysis using Kirchhoff’s laws
Kirchhoff’s laws are two I.e. Kirchhoff’s current law and Kirchhoff’s voltage
law) which are stated as follows:
Kirchhoff’s current law states that “the amount of current entering anode, junction
or a point is equal to the amount of current legging anode, junction or point”.
Kirchhoff’s current law states that “the algebraic sum of current meeting at the
same junction is practically equal to zero. I1 = I2 =I3 = 0.
Kirchhoff voltage law states that “the algebraic sum of voltage entering anode is
equal to the algebraic sum of voltage leaving the node”
To analyze a circuit, example shown in Figure 5.1, branch currents with the
direction will be indicted in each loop. The direction of flow of current may be assume to
67
flow from the positive terminals of the source voltage. To each loop, Kirchhoff voltage
law can then be apply which will result to a simultaneous equation which can be solved.
Using determinant or crammer’s rule.
Example 1: Find the current flowing in each branch of the network shown in Figure 5.2
using Kirchhoff’s law.
Solution:
The branch currents and their direction in each loop can be labeled as shown below in
Figure 5.3
From loop ABEF,
10I1 + 5(I1+I2) = 40<00.
10I1 + 5I1+5I2 = 40<00.
15I1 + 5I2 = 40<00.
V1
r1
V2
r2R
Figure 5.1
40<0 V0
105
20<90 V0
20
Figure 5.2
40<0 V0
105
20<90 V0
20
A B CI1 I2
I1
(I1+I )1
I2
F E DFigure 5.3
68
From loop BCDE;
20I2 + 5(I1+I2) = 20<900.
20I2 + 5I1+5I2 = 20<900.
5I1 + 25I2 = 20<900.
Equating the two equations together gives:
15I1 + 5I2 = 40<00 ---- (i)
5I1 + 25I2 = 20<900 --- (ii)
Using crammer’s rule.
=
= 375 – 25 = 350
= 1,000 < 0
0 =100 < 90
0.
= 1,000 + j100
= 300 < 90
0 - 200
= -200 + j300
Or 1.03 < 123.5A
Example 2: Determine the current flowing in the 3 resistor of the circuit shown in
Figure 5.4 using Kirchhoff voltage law. Calculate also the power in the 5 resistance.
69
Solution:
Direction of current flown is indicated below:
There are three loops in the circuit above, which give rise the three unknown currents.
From loop ABCDE, moving in anticlockwise direction;
I1 + 2I2 + 4(I2 – I3) = 7
I1+ 2I2 + 4I2 – 4I3 = 7
I1 + 6I2 - 4I2 = 7 ---- (i)
From loop EDGF, moving in clockwise direction,
2I2 + 3I3 - 6(I1 – I2) = 0
2I2 + 3I3 - 6I1 + 6I2 = 0
- 6I1 + 8I2 + 3I3 = 0 ----- (ii)
Figure 5.4
5
4
1
6 3
7V
2
Figure 5.5
5
4
1
63
7V
2I1
E
(I )1+I1
F
I2
(I )2+I3
H
(I =I )1 2+I3
B
C
A
G
D
I3
70
From loop DCHG, moving in anticlockwise direction;
3I3 + 5 (I1 – I2 + I3) – 4 (I2 – I3) = 0
3I3 + 5I1 – 5I2 + 5I3 – 4I2 + 4I3 = 0
5I1 - 9I2 + 12I3 = 0 ----- (iii)
Putting the three equations together gives;
I1 + 6I2 - 4I3 = 7 ----- (ii)
- 6I1 + 8I2 + 3I3 = 0 ----- (ii)
5I1 - 9I2 + 12I3 = 0 ----- (iii)
Hence, using crammer’s rule;
=+1
71
Thus, the current flowing in the 3 resistance is 0.17A
Current in the 5 resistance = I1 – I2 + I3.
= (1.46 – 1.03 + 0.17) = 0.6A
Hence, power in the 5 resistance, I2(R) = (0.6)
2 x 5 = 1.8W
Example 3: Using Kirchhoff law voltage and crammer’s rule; determine the current
flowing in the ac network shown in Figure 5.6 below:
Solution:
Indicate the direction of flow of current in the two loops
E = (3+j6)V2
Figure 5.6
Z = (3 - j7)2
Z = (7 + j2)3
Z = (7 + j2)1
E = (6+j0)V1
72
From loop ABEF, moving in clockwise direction,
(6 + j0) = I1 (4 +j3) + (I1 – I2) (7 + j2)
6 = 4I1 +j3I1 + 7I1 + j2I1 – 7I2 – j2I2.
6 = I1 (11+j5) – I2(7+j2) ------------------------- (i)
From loop BCDE, moving in clockwise direction,
3+j6 = I2 (3-j7) – (I1- I2)(7+j2)
3+j6 = 3I2 - j7I2 + 7I1 + j2I1 – 7I2- j2I2.
3+j6 = I1(7+j2) – I2(4+j7)
Join the two equations together
6 = I1(11+j5) – I2(7+j2) ------------- (i)
3+j6 = I1(7+j2) – I2(4+j7) ------------- (ii
Using crammer’s rule;
= 36 - j71
E = (3+j6)V2
Z = (3 - j7)2
Z = (7 + j2)3
(4 + j3)
A CB
(I +I )1 2
(6+j0)VDEF
I1 I2
73
=
The current flowing in the (7+j2) impedance is given by:
= 0.35 < 230.580 - 0.996 < 230.58
0
= -0.22 - j0.27 + 0.63 + j0.77
= (0.41 + j0.5) A or 0.65 < 50.650A
Example 4: Use Kirchhoff law and crammer’s rule to determine the magnitude of the
current in the (2+j3) impedance shown in the current below:
Solution:
Figure 5.9
A CB
EGH
I1I2
F
5
20<0 V0
14V 13V 2
-j6 6 j3
I2
(I - I - I )31 2
I1(I + I )1 2
I3 D
I3
Figure 5.8
A CB
(6+j0)VDEF
I1 I2
E
5
20<0 V0
14<0 V0
13<0 V0
2
-j6 6 +j6
74
Direction of flow current in each loop is shown in Figure 5.9 current in the (2+j3)
impedance is (I3). There are 3 loops in the above circuit, from loop ABGH, moving in
clockwise direction
5I1 – j6I2 = 20 + 14 ;5I1 – j6I2 = 34 ---------------- (i)
From loop BCFG, moving in anticlockwise direction;
- j6I2 – 6(I1-I2-I3) = I3+14
- j6I2 – 6I1+ 6I2 +6I3 = 27
- 6I1 + I2(6 – j6)+ 6I3 = 27 ----------------- (ii)
From loop CDEF, moving in clockwise direction;
- 6(I1 – I2 - I3) + (2+j3) I3 = 13
- 6I1 + 6I2 + 6I3 + 2I3 + j3I3 = 13
- 6I1 + 6I2 + I3(8 + j3) = 13 ----------------- (iii)
Hence, the three equations are:
5 I1 – j6I2 = 34 ---------------- (i)
- 6I1 + I2(6 – j6) + 6I3 = 27 ----------------- (ii)
- 6I1 + 6I2 + I3(8 + j3) = 13 ----------------- (iii)
+ 0
= 84- j222
+0
)
75
Hence, the magnitude of current flowing in the (2+j3) impedance is 5A.
Exercises
1. Determine the current flowing in each branch using Kirchhoff law and
determinant in the circuit shown in Figure 5.10 below.
(60v source discharges at 1.89A, 30v source discharges at 0.32A and current through
10Ω resistor is 2.21A).
2. Find the value of currents I1, I2 and I3 in the circuit shown in Figure 5.11 below
(-0.98A, -0.21A and -0.77A).
3. Determine (a) the current flowing in the 12V source (b) the potential difference in
the 2 resistance and (c) real power dissipated in the 6 resistance in the circuit
shown in Figure 5.12 below.
25
30V
Figure 5.10
60V
20 10
6
Figure 5.11
12V
3
I2
4V
I3
I1=
2
I1
76
(
4. Evaluate the value of current flowing in the bridge circuit shown in Figure 5.13 for
(a) the 3 resistance (b) 30 resistance and (c) 4 resistance.
5. Using Kirchhoff voltage law and crammer’s rule, find the current flowing in the
inductive branch for the network shown in Figure 5.14.
8
Figure 5.12
12V
20
65
2
a
Figure 5.13
3
30
4615A
7
60V
Figure 5.14
30<0 V0
j4
6
12<90 V0
16<0 V0
37
13
77
(
6. In Figure 5.15 circuit shown below, determine; (a) current flowing in 15 and 3
resistance respectively (b) the p.d. across the 7 resistance and (c) power in 12
resistance. Using Kirchhoff law/crammer’s rule
(-0.31A, 8.25A, 2.17v,4.6W).
7. Determine the magnitude of p.d. across the (7 – j8) impedance shown in Figure
5.16 and the power dissipated across it. (47.99v, 216.7W).
60<80 V0
R
Figure 5.16
20<80 V0
3
-j8
7
1
Figure 5.15
7
3
515
40V
8
D
12
2
78
CHAPTER SIX
MESH-CURRENT AND NODAL ANALYSIS
Mesh Current Analysis
Mesh-current method of analyzing system of networks is also known as “loop
current method”. Mesh current method is an extension of Kirchhoff voltage law which is
similar to branch current method in that it uses simultaneous equation, Kirchhoff voltage
law and ohm’s law are to determine unknown currents in a network but does not use
Kirchhoff current law as used by branch current.
Mesh-current of analyzing system of networks is more useful than branch current
method because it allows for the solution of a large network with fewer unknown values
and fewer equations. Example can be seen in Figure 6.1
Figure 6.1
In mesh-current analysis, loop currents are arranged to rotate in the same direction,
i.e. clockwise direction, in which there are three loops and three equations will be
derived.
I1(Z1+Z2) – I2Z2 = E1.
I2(Z2+Z3+Z4) – I2Z2 – I3Z4 = E1.
I3(Z4+Z5) – I2Z4 = - E1.
Example 1: Determine the current flowing (a) the 6 resistance and (b) 3 resistance
using mesh-current analysis.
I1
I1
I2 I3
E3E1
E2Z1Z2
Z4
Z3
Z5
79
Figure 6.2
Solution:
The mesh currents I1I2 and I3 are shown in Figure 6.2 using Kirchhoff voltage law:
For loop1, 2 = I1 (6+4) – 6I2.
For loop2, 0 = (6+2+3+1) I2 – 6I1– 3I2.
For loop3, 6 = (3+7)I3 – 3I2.
Thus,
10I1 – 6I2 = 2 …………………….. (i)
- 6I1 + 12I2 – 3I3 = 0 ……………. (ii)
0I1 – 3I2 + 10I3 = 6 …………….. (iii)
Using crammer’s rule
I2
I3
2V
I1 6V
4
1
2 7
3
80
Hence, current in 6 resistance = I1 – I2 = 0.44 – 0.40 = 0.04A
Current the 3 resistance = I2 – I3 = 0.40 – 0.672 = -0.272A
Example 2: Determine the ac network using mesh current analysis (a) current I1 and I2 (b)
current flowing in the inductor and (c) power dissipated in the 200<00 voltage source.
Figure 6.3
Solution
For loop1, 200<00 = (2+j6)I1 – J6(I2)
For loop2, 0 = (3 – j2 + j6)I2 – j6I1.
Putting equation 1 and 2 together gives;
(2 + j6)I1 – j6(I2) = 200 ………………….. (i)
-j6I1 + (3 +j4)I2 = 0 ………………………….(ii)
Using crammer’s rule
I2200<0 V
0
I1 6V
2
1
3
j6
-j2
81
(b) current flowing in inductor = I1 – I2.
= 31.61 – j1.198 – 31.2 –j21.6
=(0.41 –j22.8)A or (22.8< - 88.970)A
(c) source power (real power) “p” =
= (200)(31.63) cos (-2.170)
= 5,321.5W = 6,3200W, correct to three
significant Figure
(check; power in 2 resistor = I12(2)
= (31.63)2 (2) = 2,000.91W
Power in 3 resistor = I22(3) =(37.95)
2(3)= 4,320.6W
Thus, total power dissipated = 2,000.9 + 4,320.6 = 6,321.5W
= 6,3200W(to 3s.f)
Example 3: Using mesh current analysis, calculate the value of current IA, IB and IC in
Star connected 3 – phase load shown in Figure 6.4 below
82
Figure 6.4
I1 and I2 are chosen as the mesh current in Figure 6.4
From loop1, I1 (100<00) + I1 (62.25<18.4
0) – I2(62.25<18.4
0) = 215<0
0.
215 = (159.07 +j19.65) I1 – I2 (-j50)
From loop2, I2(62.25<18.40) - I1 (62.25<18.4
0) + I2(50<-90
0) = 215<15
0.
I1(- 59.07 – j19.65) + I2(59.07 –j30.35) = 207.67 + j55.65
215 = (159.07 +j19.65) I1 – I2 (-j50) ……………………….. (i)
207.67 + j55.65 = I1(-59.07 – j19.65) + I2 (59.07 – j30.35) ………. (ii)
= 9,992.5 – j3,666.6 – 2,953.5 + 982.5
= 8,021.5 – j3,666.6
(100<0 )0
(50<-9
0)
0
(215<15 V)0
(215<0 )0
(62.
25<18
.4)0
IC
IB
IA
I2
I1
83
Thus, current IA = I1 = 2.599 < -22.96
0A
IB = - I2 = - (5.42<45.580)A or (-379 –j3.87)A
IC = I2 – I1 = 3.79 + j3.87 – 2.98 + 1.07 = (0.86 + 4.88)A or 4.96 < 800A
Nodal Analysis
Node is a junction in a circuit where two or more circuit elements are connected
together. Node analysis uses Kirchhoff current law unlike loop current which uses
Kirchhoff’s voltage law. Node analysis is most suited for networks having many parallel
circuits with common circuit connected and it gives solutions to a problem using less
number of equations to solve for unknown parameters. A point where three or more
branches meet is regarded as a principle node or junction and in which one of the
branches is referred to a reference node/a datum node or zero – potential node.
Node analysis involves substituting voltage sources with equivalent current
sources. Resistors values in (ohm) are also substituted by equivalent conductances in
Siemen(s); G = 1/R
Guidelines to be taken when solving network analysis using node voltage analysis
1. Voltage source in series with a resistance must be substituted for by an equivalent
current source in parallel with the resistance.
2. Select a reference node [i.e. 1, 2 --------N]. the positive coefficient of the first
voltage in the first voltage in the first equation is the sum of conductance
connected to the node. Coefficient of second equation is the sum of conductance
connected to that node.
84
3. Resistors values in (ohm) must be substituted with equivalent conductance in
siemen(s); G = 1/R
4. All other coefficients for all equations are negative, representing conductance
between nodes.
5. The RHS side of the equation is the current source connected to the reference
nodes.
6. Solve system of equations for unknown node voltages at circuit node using KCL
equation (Kirchhoff’s current law)
Example 4: Using the node method, find the current I in the j20 branch of circuit given
in Figure.
Figure 6.5
Figure 6.5 contains two principal nodes of which, 2 is taken as the reference node
Example 5: In Figure (a) determine voltage VBC in the circuit. Also find the value of V1
if the polarity of the second voltage source is reversed as shown in Figure (b)
Z1 Z2
50<00
50<-200
Z3 j20
85
Figure 6.6: (a) (b)
In the circuit given, there are no principal nodes. Hence, let’s take point (as the reference
node and points as node 1, then using nodal analysis;
When source polarity is reserved; V1=
Example 6: Determine, using nodal equation V1 and V2 for the network showing Figure
6.7
20<0 V0 20<30 V
0
2 6
4 Bj3
+-
C
+-
20<0 V0 20<30 V
0
2 6
4 Bj3
+-
C
+-
86
Figure 6.7
It can be seen that the current of the second source is flowing away from nodal, therefore
it will be taken as negative, but the term containing this source will become positive
because it has been reversed twice. Node 3 can be taken as reference node.
Considering node 1, we have;
Considering node 2, we have;
From equation 1;
In equation 2;
Joining the two equation together gives;
…………………….(i)
Using crammer’s rule
10<200
j20
j4
10<300
10<0 V0
j225
6
2
7
1
+-
+-+
-
87
Example 7: Using nodal analysis, determine current flowing through 3 resistance in
circuit shown in Figure 6.8
Figure 6.8: (a) (b)
The current passing through 3 resistance can be found by finding the voltage VB of
node B with the help of nodal analysis. Points C in fig. (a) has been taken as the reference
node.
-j550<00
2 C
6 A
+-
B
j2-j5
50<00
I2
6
+- j2I1
I3
88
Similarly, for node B, we have
---------------------(i)
-0.25VA + VB (0.70 – j0.03) = 0
In equation (i);
Put in eqn.---------------------------(ii)
-0.25(
(-0.31 + j0.13) VB = 2.08
Put value of VB = -5.64-j2.37 in eqn.
I = 2.04<-157.25
0A
Example 8: Determine the voltage VY2 using nodal analysis in circuit shown in Figure 6.9
89
Figure 6.9
Node 3 is taken as the reference node
At nodal,
(0.33<-28.89)V1 - 0.17V2 = 40……………………..(i)
At node 2;
0.17V1 + (-J0.05 –J0.03 – 0.17)V2 = 0
0.17V1 + (-0.17 – J0.08)V2 = 0……………………..(i)
Putting both equations gives;
(0.33< -28.89) V1 – 0.17V2 = 40……………….(ii)
0.17 + (-0.17 – j0.08) V = 0 …………………….(iii)
Using crammer’s rule;
j4
312
j20
6
3
j30
I=40<0 A0
Y
90
Current flowing in the (3+j4) branch is V1/(3+j4). Hence, the voltage between point *
and node 3 is
V2 = 200<-290.53
0V
Thus, the voltage;
VYZ = VY – VZ = VY – VZ = 200<-290.530 – [-(226.7<-172.6
0]
VYZ = 200<-290.530 + 226.7<- 172.6
0
VYZ = 70.14 + j187.3 – 224.8 - j29.2 = -154.66 + j158.1V
Example 9: Determine the voltage at nodes 2 and 3 in Figure and hence, determine the
current flowing in the 3 resistor and power dissipated in the 4 resistor using nodal
analysis.
Figure 6.10
Solution” the reference node is point A
At node 1,
0.84V1 – 0.14V2 – 0.5V3 = 3
At node 2,
-0.14V1 + 0.72.V2 – 0.025V3 = 0
At node 3,
-0.5V1 - 0.25V2 – 0.92V3 = 3
3
7
2 4
5
2 6V
6A
3
91
Putting crammer’s rule;
Power in the 4 resistor =
Current in the 3 resistor = flowing from node 2 to node A.
92
Ω Ω 2Ω
8 Ω 20 < 0 o V
16 < 90 o V
Exercises on Mesh-Current Analysis
1. Determine current, IA, IB and IC for the network shown in Figure 6.11, Using
mesh-current analysis. 3Ω 2Ω IB
Ic
60<00v 5Ω j5
-j6Ω 20<90v
Figure 6.11
(5.92<7.470A, 3.75<55.49
0A and 4.4<148.48
0A).
2. Determine the current flowing in the (3+j4) in impedance of circuit using mesh
current analysis
Figure 6.12
3. In network of Figure below, determine (a) current in the capacitor (b) current in
the inductor (c) current in the 2 resistor (d) active power output of the 20<00V
source and (d) magnitude of the p.d. across the j5 inductance using mesh-current
analysis.
-j20Ω J5Ω
J2Ω
Figure 6.13
( 8394.07<-7.080A, 8331.35-j1,033.98A, 162KW, 7.15<-63.7
0v).
j 4 Ω
16 Ω
12 Ω
Ω j 2 Ω
j 3 Ω
30 < 0 o V
93
4. In the circuit shown in Figure below, using mesh current analysis, determine (a)
current in the capacitor IC (b) current in the resistance 3 IR (c) p.d. across the 6
resistance and (d) total active (real) circuit power
Figure 6.14
5. Using mesh current analysis, determine current IA, IB and IC shown in balanced 3-
phase Delta connected load in Figure below and hence, find the line current IX, IY
and IZ.
Figure 6.15
(32.09<123.430A, 32.09<63.43
0A, 16.02<63.43
0A).
Exercises on Nodal Analysis
6. Determine for the network shown in Figure below, the voltage to node A and the
voltage VXY.
Figure 6.16
(5.25<81.020v, 0.63<-2.14
0v).
7. Determine voltage VXY in the network circuit shown in fig below. Using nodal
analysis.
I =
20<90OA
20
3
2
A Y
-j16
X
I C
215 < 60 O V
Iz 215 < 0 O V
Iy 32
32
3 - j 6 )
J6
(3-j6)Ω
I B
I X
I A
- j 6 Ω
Ω Ω
50 < 0 o V j 6 Ω
Ω
50 < 90 o V
94
Figure 6.17
(94<131.50v, 20<17.2
0v).
8. Determine voltage VRS in the network circuit shown below, using nodal analysis.
Figure 6.18
(22.57<88.240v).
9. In the network diagram shown below, determine (a) voltages at nodes 1 and 2 (b)
current in the 30 resistance (c) current in the 15 resistance (d) magnitude of the
active power dissipated in the 12 resistance.
(a)8.6<84.020v,12.14<124.06
0v (b)1.63<17.55
0A (c)1.33A (d)33.33<180
0W
10. Using nodal analysis, solve for value of current I1, I2 and I3 in the circuit shown in
Figure.
(-6.25A, 10.84A and 4.8A)
20<90OA
20
10
2
A
Y
-j6
B
j12
j13
20<
0O
V
20<
90
O
V
15 1
2
12 j12
-J30 30
-j8
j6
60<0OA
2
12
60<90OV
95
I1 I2
I3
Figure 6.20
6
90v
3
12
20v
96
CHAPTER SEVEN
SUPERPOSITION THEOREM
Introduction
Superposition Theorem is only applicable to linear networks where current is
linearly related to voltage as per ohm’s law.
Superposition Theorem states that “In a network of linear resistance containing
more than one source of e.m.f, the current which flows at any point is the sum of all the
currents which would flow at that point if each source or e.m.f. where considered
separately and all the other source e.m.f. replaced for the time being by resistances equal
to their internal resistance.”
Implementing the Superposition Theorem
Superposition Theorem may be applied to both d.c. and ac networks. If there are a
number of e.m.f’s acting simultaneously in any linear bilateral network, then each e.m.f
acts independently of the others as if other e.m.f’s did not exist.
Value of current/voltage in any conductor is the algebraic sum of the
currents/voltages which each e.m.f would have produced while acting singly i.e.
current/voltage across conductor of the network can be obtained by superimposing the
currents and voltages due to each e.m.f. in the network.
Example 1: To find current flowing in each branch of the circuit in fig shown below, the
following steps can be taken;
Figure 7.1 (a)
7
V
2.6
3
0.2
2
4
I
I1 I2
I1 A
B
97
Label and redraw the original circuit with one of the sources removed and replaced by
corresponding resistance in series with it and current in each branch and its direction.
Figure 7.1(b)
Figure 7.2
Figure 7.1(b) shows the current values which would have obtained if left hand side
battery had acted alone. Figure 7.1(2) represents the current values which would have
obtained if right side acts alone.
Combining current values of fig. 7.1(b) and fig. 7.2, actual values of Figure 7.1(a)
and be obtained as:
I1 = I1’ – I1
’’
I2 = I2’ – I2
’’
I = I’ – I
’’
It can be seen that there are two parallel paths between point A and B, having resistance
of 4 and (2+3) = 5
equivalent resistance,R1R2/R1+R2 = (5×4)/(5+4) = 2.2
Total resistance, Rt = Rp + Rs = 0.2 + 2.6 + 2.2 = 5
I11 =V/R= 7/5 = 1.4A
This current divides at point A inversely in the ratio of the resistances of the two parallel
paths.
7
V
2.6
3
0.2
2 4
I1
I11
I21
I1
A
B
I11 I2
1
I211
I111
3 A
B
0.2
2.6
10
V2
4
I11
98
I’ = 1.4 x (2.8/9) = 0.44
12’ = 1.4 x (4/9) = 0.62A
In Figure 7.2, 7volt battery has been removed but not its internal resistance. The
various current and their directions are also shown.
The equivalent resistance to the left to points A and B is Rp = (R1R2)/ (R1+ R2)
(2.8×4)/(2.8 + 4)
= 1.65
total resistance = 2+ 3+1.65 = 6.65
I2’’ = 10/6.65 = 1.50A
At point A, this current is divided into two parts,
I’ = 1.50 x 2.8/9 = 0.47A
I1’’ = 1.50 x 4/9 = 0.67A
Actual current values of fig. 7.1 (a) can be obtained by superposition of these two
sets of current values.
I1 = I1’ – I1
’’ = 1.4 – 0.67 = 0.73A
I2 = I2’ – I2
’’ = 1.50 – 0.62 = 0.88A
I = I’ – I
’’ = 0.44 + 0.47 = 0.91A
Voltage drop across 4 resistor = 4 x 0.9 = 3.64V
Example 2: Using Superposition Theorem, find the current in resistance R shown in fig.
7.3 (a)
R1 = 0.02, R2 = 0.01, R = 2, E1 = 2V, E2 = 5V.
(Internal resistances of cells are negligible)
Figure 7.3a
Solution:
Redraw the current, removing voltage source E2;
0.02
0.01
R=2
2V
5V
B
E1 A B
D
R2 E2
99
Figure 7.3(b)
In Figure 7.3(b) E2 has been removed. Resistance of 2 and 0.01 are in parallel across
points A and C. Rp= (R1R2)/ (R1+ R2) = (2×0.01)/(2+0.01) = 9.995 x 10-3. This
resistance is in series with 0.02. Hence, total resistance offers to the battery; E1 = 0.02 +
0.00995 = 0.03
I =V/R= 2/0.03 = 66.7A
Current through 2 resistance, I1 = 66.7 x 0.00995 = 0.66A from C to A
If E2 is removed, it can be shown in Figure 7.3 (c)
Figure 7.3c
Combined resistance of CBA and CDA is; Rp= (R1R2)/ (R1+ R2 ) = 2//0.02 = 0.02.
Total resistance offered to E2 = 0.01 + 0.02 = 0.03
Current I = 5/0.03 = 166.7A
Current I2 = 166.7 x (0.02) = 166.7 x 0.02 = 3.33A
Current through 2 resistance when both batteries are present
= I1 + I2 = 0.66A + 3.33 = 3.99A
Example 3: Using superposition Theorem, determine the current in the 3 resistor of the
network shown in Figure 7.4a
0.01
0.02
B
A C
D
I
2
5V
I2
E2
0.01
0.02 B
A C
D
I
2
2V
I1
E2
100
Figure 7.4a
(1) Removing the 25V source gives the network shown in Figure 7.4b
Figure 7.4b
4 and 1 are in parallel; Rp= (R1R2)/ (R1+R2) =(4×1)/ (4+1) =0.8
2 is in series with 0.8 = 2+ 0.8=2.8
2.8 is in parallel with 3; Rp= (R1R2)/ (R1+R2) = (2.8×3)/ (2.8+3) = 1.5
1.5 in series with 6 gives;1.5+6 = 7.5
Figure 7.4c
Thus, current I1 = 13/7.5 = 1.73A
Current I2 = (1.73) = 0.09A
6 2 1
3 4 13
V
25
V
A
6 2 1
3 4 13
V
A
12
6 2
3 0.8
13V
101
(2) Removing the 13V source from the original network gives the network shown in
Figure 7.4d.
Figure 7.4d
(3) Current I3, I4 and I5 are shown and labeled in Figure 7.4d.
From Figure 7.4d, 6 is in parallel with 3 = (6×3)/(6+3) = 2, 2 is in series
with 2 ; 2+2 = 4,
4 is in parallel with 4 = (4× 4) / (4+4) = 2
2 is in series with 1; 2+1 = 3
(4) Superimposing Figure 7.4d on Figure 7.2 shows that the current flowing in the 3
resistor is given by I5 – I2.
(5) I5 – I2 = 2.78 – 0.89 = 1.89A.
Example 4: Using Superposition Theorem, determine for the network shown in Figure
7.5 (a) the magnitude of the current flowing in the capacitor (b) the p.d. across the 5
resistance (c) active power dissipated in the 10 resistance and (d) the total active power
taken from the supply.
The network redrawn with the 30V source removed as shown in Figure 7.5a
6 2 1
3 4 13
V
A
25
V 15
14 13
15 -
j2 2
8 6 20
V
A
30
V
102
Figure 7.5a
(1) Current I1 to I5 are labeled in Figure 7.5a,
From Figure 7.5a, 6 resistor is in parallel with 6 resistor;
Rp = (R1R2)/(R1+ R2) =(6×6)/(6+6) = 36/12=3.
Hence, =0.89<2.660A
(2) The original network is redrawn with the 20V source removed, as shown in Figure
7.6a
Figure 7.6a
(3) Current I6 to I10 are shown in Figure 7.32 from Figure 7.32, 15 is in parallel with
8 =. Rp= (R1R2)/ (R1+ R2 ) =(15×8) / (15+8) = 5.22
Hence; = 3.66<0.590A
1.43<-20.610A
2.38<13.090A
1.55<13.090A
15 -
j2 6
8 6 20
V
I2
I3 I4
I5
I1
15 -
j2 6
8 6
I10
I8
I9 I7
I5
I6
103
0.83<13.090A
(a) Current flowing in the capacitor is given by;(I3 – I8)= 0.62+j0.14-2.32+j0.54= (-
1.7+j0.68)A
i.e. the magnitude of current in the capacitor is 1.7A
(b) The p.d. across the 8 resistance is given by (I2 + I9) (8)
(I2 + 19=) =0.26-j0.1+1.5+j0.35 = (1.76+j0.25)A
Hence, the magnitude of the p.d. across the 5 resistance is
(c) Active power dissipated in the 15 resistance is given by (I1 – I10)2(15)
=(0.89<2.66 - 0.83<13.090A)
2 x15
Hence, the active power dissipated in the 15 resistance is 44.4<-15.40W
(d) Active power developed by the 30V source,
P1 = V (I1 – I10) Cos 1 =34.4 Cos(-7.7) = 34.09W
Active power developing by 30V source,
P2 = 30 (I6 – I5) Cos2
(I6 – I5) = 3.66<0.590 – 0.32<12.7
0 = 3.35-j0.03 0r 3.35<-0.5
0A
Hence, P2 = (30)(3.35<-0.50) = 100.5W
Total power developed; P = P1 + P2 = (100.5 + 34.09)W = 135W.
Maximum Power Transfer Theorem
Maximum power transfer Theorem can be used when the source has fixed
complex impedance and delivers power to a load consisting of a variable resistance or
variable complex impedance. Maximum power transfer Theorem states that “to obtain
maximum external power from a source with a finite internal resistance, the resistance of
the load must equal to the resistance of the source as viewed from its output terminal.
Maximum power transfer Theorem was introduced and published by Moritz Von
Jacobi around 1840, it is also referred to as “Jacobi’s law”
Maximum power transfer Theorem results in maximum power transfer not
maximum efficiency. Maximum power transfer from the source to the load depends on
the following conditions.
(a) If load consists of only a variable resistance R1 (shown below)
The circuit current is
Power delivered to RL is PL =
To derive the value of RL form maximum transfer of power, dPL/dRL is zero
104
If Xy is zero, maximum transfer RL = Ry
(b)When load impedance consist of both variable resistance and variable reactance
(fig)
The circuit current is
Power delivered to the load is
Now, if RL is held fixed, PL is maximum when Xy = XL.
In that case, PL max =
(c) ZL with variable resistance and fixed reactance (fig below). The equations for
current I and power RL are the same as in case (b) except that we will consider
consider XL to remain constant when first derivative of PL with respect to RL is
set equal to zero, it is found that;
RL2 = Ry
2 + (Xy+XL)
2 and RL = (/Zy + jZL/
Since Zy and XL are both fixed quantities, these can be combined into
single impedance. Then with RL variable, case C is reduced to case (a) and the
maximum power transfer takes place when RL equals the absolute value of the
network impedance.
I
V
y
I
Z
y
R
L
Ry +
Jxy
A
B
I
V
y
Z
y
X
L
Ry +
Jxy
A
B
R
L
105
Figure 7.7a Figure 7.7b
Figure 7.7c
Example 5: In the circuit shown below, find the values of load to be connected
across terminals Xy consisting of variable resistance RL and capacitance reactance
XL which would result in maximum power transfer.
Figure 7.8(a) (b) (c)
We will firstly use Thevenin’s equivalent circuit between terminal X and Y. when
the load is removed, the circuit becomes as shown in fig…. (b).
Vth = drop across = 53.43<16.22 or 51.30+j14.92
Rth = = 3.56<16.22 0r 3.42+j0.99
Thevenin’s equivalent source circuit is shown ; since for maximum power transfer,
conjugate match is required hence, XC=0.99Ω and RL= 3.42Ω.
Examples 6: For the circuit shown in Figure 7.9, load impedance Z is a pure
resistance. Determine (a) the value of R formation maximum power to be
transferred from the source to the load and value of the maximum power transfer
delivered to R.
I
V
y
Z
y
R
L
Ry +
Jxy
A
B
X
L
+
- 60V
-
jxc R
L
4
3
j12
X
Y
+
- 60V
j12
3
4 X
Y
E =
140V Z =
(16+J10)
I
Z
106
Figure 7.9
Solution
From case (s) maximum power transfer occurs when R = /Z/, i.e. when R =
/16+j10/ =
Current I flowing in the load is given by I = E/ZT, where the total circuit
impedance ZT = Z + R
16 + j10 + 18.87 + (34.87 +j10) or 36.28 < 160
Hence,
Thus, maximum power delivered, P = I2R = (3.86)
2(18.87)
= 281W
Example 7: If the load impedance Z in Figure 7.9 above of example 6, consists of
variable resistance R and variable reactance X, determine (a) the value of Z that
results in maximum power transfer and (b) value of the maximum power.
(a) Maximum power transfer occurs when X = x and R = r, thus, if Z = r +jx ; Z =
(16 +j10), then;
Z = (16 – j10) or 18.87 < -320
(b) Total circuit impedance at maximum power transfer condition, ZT = Z + Z1
i.e.
ZT = (16 +j10) + (16 –j10) = 32
Hence, current in load,
And maximum power transfer in the load; P = I2R = (4.38)
2(16)
= 306.95W
Example 8: Determine (a) the value of the load resistance R required for maximum power
transfer and (b) the value of the maximum power transferred in the network shown in
Figure 7.10.
E=100<0v
200Ω 2µF I
Load R
107
Figure 7.10
Solution
i) Maximum power is achieved when R = /Z/, source impedance Z is composed of a
200 resistance in parallel with a capacitor of 2µf.
Capacitive reactance,
Hence, source impedance,
Z = 198.4 < - 7.20 or (196.8 – j24.87)
Thus, the value of load resistance for maximum power transfer is 198.4 i.e.
/Z/
ii) With Z = (196.8 –j24.87) and R = 198.4 for maximum power transfer, the total
circuit impedance,
ZT = 196.8 – j24.87 + 198.4 = 395.2 – j24.87 or 395.98 < - 3.60
Current flowing in the load,
Thus, maximum power transfer red, P = I2R = (0.25)
2(198.4) = 12.4W
Example 9: In the network shown in Figure below, the load consists of a fixed
capacitive reactance 8 and a variable resistance R. Determine (a) the value of
R for which the power transferred to the load is a maximum and (b) the value
of the maximum power.
J6Ω
Figure 7.11
(a) Maximum power transfer can be achieved when
(b) Current
I = 13.05 < 16.860.
90<0V
3
R
J8Ω
108
Hence, the maximum power transferred,
Exercises on Superposition Theorem
1. Using superposition Theorem, determine the circuit I in the network shown in
Figure 7.12:
3Ω 7Ω
2Ω
+ +
8<20v 6<20v
- -
Figure 7.12
(1.79<19.830V)
2. Using superposition Theorem, find the magnitude of current flowing in the branch
YZ of the circuit shown in Figure.
Y
7Ω j8Ω
2+j8Ω
- +
10<50v 10<-50v
+ -
Z
Figure 7.13
(0.92<16.490A)
3. Use the superposition Theorem to determine the magnitude of current flowing in
the inductive branch of the network shown in Figure 7.14.
60v -j5Ω
7Ω
2Ω
J8Ω 90v
Figure 7.14
(3<18.660A)
109
4. A – C sources of 10 < 1800V and internal resistance of 12 and 20 < 70
0V and
internal resistance 9 are connected in parallel across a 2 load. Use
superposition Theorem, determine (a) the current in the 2 load and (b) current in
each voltage source.
(1.52<91.90A, I1=1.01<-171.31
0A,I2= 2.07<68.25
0A)
5. Three batteries each e.m.f. 12V are connected in parallel to supply a load of
resistance 3. The internal resistance of the batteries are 0.2 and 0.4.
determine, using the superposition Theorem, the current in the load and the current
supplied by each battery.
(-8v, 48A and 18A)
Exercises on Maximum Power Transfer Theorem
6. Circuit shown below consists of fixed inductance having a reactance of 20 and a
variable load resistor RL. Find the value of RL for maximum power transfer and
the value of the power. –j30
Figure 7.15
(60.83Ω)
7. Circuit shown below, consists of source resistance Ry is variable between 6 and
60 but RL has a fixed of 30. Find the value of Ry for which maximum power
is dissipated in the load and the value of this power.
(27.5Ω)
Ry -j12
Figure 7.16
I
50
60
X
L
J20
R
L
I
50
30
ry
110
8. For the circuit shown below, determine the value of load impedance Z1 for
maximum load power if (a) ZL comprises a variable resistance R and variable
reactance X and (b) ZL is a pure resistance R. Determine the values of load power
in each case.
-j3 3 j2 load
20
Figure 7.16
(3.16-j3)Ω , 3.16Ω, 567.82<43.50W and 790.86W
9. The output stage of an amplifier has an output resistance of 3. Determine the
optimum turn’s ratio of a Transformer that would match a load resistance of 80
in the output resistance of the amplifier for maximum power transfer.
(138.45Ω)
10. A single phase, 240V/3000V, deal Transformer is supplied from a 240V source
through a capable of resistance 2. If the load across the secondary winding is
2.8k. Determine (a) primary current flowing and (b) power dissipated in the load
resistance.
(0.09A, 3205.72W)
111
CHAPTER EIGHT
THEVENIN’S AND NORTON’S THEOREMS
Introduction
Network analysis can be analyzed more quickly using Thevenin’s or Norton’s
Theorem than some other Theorems mentioned in the previous chapters. These Theorems
in values replace what may be a complicated network of sources and linear impedances
with simple equivalent circuit.
Thevenin’s Theorem
Thevenin’s Theorem was independently derived in 1853 by the German scientist
“Herman Von Helmholtz and in 1883 by “Leon Charles Thevenin (1857 – 1926), an
Electrical Engineer with France’s National Postes et Telegraphes Telecommunication
organization.
Thevenin’s Theorem can be stated as “the current through a load impedance ZL
connected across two terminals A and B of a linear network is given by Vth/(Zth + ZL),
where Vth is the open-circuit voltage across A and B and Zth is the internal impedance of
the network as viewed from the open-circuited terminals A and B with all voltage sources
replacing their internal impedances (if any) and current sources infinite impedance.
Figure 8.1 (a) 8.1(b)
Figure 8.1 shows Thevenin’s equivalent circuit
Example 1: In the network shown in Figure 8.2
Z1 = (6+j6); Z2 = (6 – j6); and Z3 = (1 + j10); V = 20 < 00 and ZL = j20. Find the
current through the load ZL using Thevenin’s Theorem.
Network
containing
voltage
sources and
impedances
A
B
E
Z
A
B
Z1 Z3
Z4 Z2 20<0O
112
Figure 8.2(a)
Solution:
When the load impedance ZL is removed, the circuit becomes, as shown in Figure
8.2(b). The open-circuit voltage which appears across the terminals A and B represents
the Thevenin’s voltage, Vth. This voltage equals the drop across Z2 because there is no
current flowing through Z3.
Current flowing through Z1 and Z2 is
I = V/(Z1 + Z2) = 20 < 00/ [(6 + j6) + (6 - j6) = 20 < 0
0/12 = 1.67 < 0
0A
Vth = IZ2 = 1.67 (6 – j6) = (10.02 – j10.02) = 14.17 < - 450V
The Thevenin impedance Zth is equal to the impedance as viewed from open
terminals A and B with voltage source shorted.
Zth = Z3 + (Z1×Z2)/ (Z1 + Z2) = (1 + j10) + (6 + j6)// (6 – j6) = (6 + j60)
The equivalent Thevenin circuit is shown in Figure 8.2 (c) a cross which the
impedance has been reconnected. The load current is given by:
IL = 0.35< - 126.47
OA or (- 0.21 - j0.28)A
Figure 8.2 (b) 8.2(c)
Z1 Z3
Vth Z2
6
+j60
14.17<-
45O
-j20
113
Example 2: Find the Thevenin’s equivalent circuit at terminals AB of the circuit in Figure
8.3.
Figure 8.3(a)
Solution:
Vth = VAB. We have to find the phasor sum of the voltages available on the way as
we go from point B to point A because VAB means voltage of point A with respect to
point B. the value of current I = 50 < 00/(10 –j6) = 4.29 < 30.96
0A or (3.7 + j2.2)A
Figure 8.3(b)
Drop across the 7 resistor (3.7 + j2.2) = 25.9 + j15.4
Vth = VAB = - (25.9 +j15.4) + (50 +j0) – 40 (0.5 + j0.866)
= 4.1 – j19.26 or 19.69 < - 77.980V
ZAB = Zth = (20 + [711(3- j6)] = 20 +
Thevenin equivalent circuit is shown Figure 8.3(b) above.
Example 3: Find the Thevenin’s equivalent of the circuit shown in Figure 8.4 and hence,
calculate the value of the current which will flow in an impedance of (4 + j20)
connected across terminals A and B. Also calculate the power dissipated in the
impedance
+
-
3
A
B
7
50<0O
-j6 I
+
-
23.4 –
j2.16
19.69< -
77.98OV
50<0O j2
-j6 7
A
B
114
Figure 8.4
Solution:
First find the value of Thevenin voltage across open terminals A and B. (Vth). With
terminals A and B open, there is no potential drop across the capacitor. Hence, Vth is the
drop across the pure inductor j2.
Drop across the inductor =
= (3.77 + j13.2)V
Finding impedance of the circuit as viewed from terminals A and B after replacing the
voltage source by a short circuit as shown in fig. 8.4(a)
Zth = - j6 + (j14/7+j2) = - j6 +
= -j6 + j1.85 + 0.53 = 0.53 – j4.15
The equivalent Thevenin circuit along with the load impedance of (4+j20) is
shown Figure 8.4(c).
The current in the load is 0.83A and leads the supply voltage by 0.030
power in the load
impedance is (0.83)2(4) = 2.76W.
Norton’s Theorem
Norton’s Theorem was independently derived in 1926 by Siemens and Halske
researcher Han Ferdinard Mayer (1895 – 1980) and Bell laboratory Engineer Edward
Lavry Norton (1898 – 1983).
Norton’s Theorem allows a source of electrical energy to be represented by a
constant current source, which may be alternating or direct, in parallel with an
impedance.
50<0O j2
-j6 7
A
B
3.77 +
J13.21
0.53 –
J4.15
4 +
J20
115
Norton’s Theorem can be stated as “under any two terminal linear network
containing voltage sources and impedances when viewed from its output terminals is
equivalent to a constant current source and a parallel impedance. The constant current is
equal to the current which would flow in short circuit placed across the terminals and the
parallel impedance is the impedance of the network when viewed from open-circuited
terminal after voltage sources have been replaced by their internal impedances (if any)
and current sources by infinite impedance.
Figure 8.5 (a) Figure8.5(b)
Figure 8.5: Norton’s equivalent circuit
Example 4: Find the Norton’s equivalent of the circuit shown in Figure 8.6. Also find the
current which will flow through an impedance of (5 –j10) across the terminals A and B
(5 + j10).
Figure 8.6(a) (b)
As shown in Figure 8.6(b), the terminals A and B have been short circuited
When voltage source is replaced by a short, then the internal resistance of the circuit, as
viewed from open terminals A and B is RN = (5 + j10). Hence, Norton’s equivalent
circuit becomes as shown in Figure 8.6(c).
Network
containing
voltage
sources and
impedances
A
B
A
B
Z
Isc
+
-
-
20<00
(5+j10
) A
B
B
+
-
-
20<00
(5+j10
) A
Isc
B
+
-
-
(5 -
j10)
A
Isc
(5 +
j10)
1.7
9<
-
63.4
3O
116
Figure 8.6c
When the load impedance of (5 – j10) is applied across the terminals A and B, current
through it can be found with the help if current divider rule.
Example 5: Using Norton’s Theorem, find current in the load connected across terminals
A and B of the circuit shown in Figure 8.7
Figure 8.7(a)
Solution
The first step is to short circuit terminals A and B as shown in Figure 8.7. The short
across A and B not only short circuits the load but the (5 + j5) impedance as well.
Since the impedance of the Norton and Thevenin equivalent circuits is the same ZN = 5 –
j5)
Figure 8.7(b) Figure 8.7(c)
+
-
-
100<00
j5 A
B
j5
I
N
j5
20<90O 5 –
j5
A
B
A
B
20<90O 5 –
j5
I
L
117
The Norton’s equivalent circuit is shown in Figure 8.7(b). In Figure 8.7(c), the load has
been reconnected across the terminals A and B since the two impedances are equal,
current through each is half of the total current i.e.
Example 6: Use Norton’s Theorem to find the magnitude of the p.d across the 2
resistance of the network shown in Figure 8.8
Figure 8.8
Solution
(1) Branch containing the 2 resistance is initially short circuited, as shown in Figure
8.8(b)
Figure 8.8(b)
3 is parallel with – j13 in parallel with 0 (i.e. the short circuit) is equivalent to 0,
giving the equivalent circuit of Figure 8.8(c)
Hence; Isc = 20/3 = 6.67A
Figure 8.8(c)
10V source is removed from the network of Figure 8.8, and the impedance Z, ‘pointing
in’ at a break made in AB is given by;
3 3 2
-
j3
20V
A
B
3 3 Isc
-
j3
20V
3 3 -
j3
A
B
3
20
V
Isc
118
Figure 8.8(d)
Norton equivalent circuit is shown in Figure 8.8 (e), from which current is given as:
Figure 8.8(e)
Hence, the magnitude of the p.d. across the 2 resistor is given by (5.53)2 = 11.06V
Thevenin and Norton Equivalent Networks
It was deduced that when Thevenin and Norton’s Theorems are applied to the
same circuit, identical results are obtained. Thus, Thevenin and Norton equivalent
network shown in Figure 8.9 are equivalent to each other. The impedance pointing in at
terminal Y2 is the same of each networks i.e. Z
Isc
=
6.6
7A
2
2
=
(1.2
-
j0.6
1)
E
Z
Y
Z
Y
Z
Z
Is
c
119
Figure 8.9(a) Figure 8.9(b)
If terminal YZ in Figure 8.9(a) are short circuited, the short – circuit current is given by
E/Z.
If terminals YZ in Figure 8.9(b) are short – circuited, the short circuit current is Isc. Thus,
Isc = E/Z
Figure 8.10 shows source of e.m.f E in series with an impedance Z feeding a load
impedance ZL
Figure 8.10
From Figure 8.10,
Figure 8.11
Example 7: (a) Convert circuit shown in Figure 8.12(a) to an equivalent Thevenin circuit
(b) convert the network shown in Figure 8.63 (b) to an equivalent Norton circuit.
Z
E
IL
Z
L
Y
Z
IL
Z
L
Z
Isc
Y
Z
3
4
A 10
V
5
Y
Z
120
Figure 8.12(a) Figure 8.12(b)
Solution:
Open circuit voltage E across terminals YZ in Figure 8.12(b) is given E = (Isc)(2) = (4)
(3) = 12V. The impedance pointing in a terminal YZ is 3,
Hence, the equivalent Thevenin circuit is as shown in Figure 8.13(a)
Figure 8.13(a)
(b) If the terminals YZ of Figure 8.63(b) are short circuited, the short circuit current,
Isc = 10/5 = 2A. The impedance pointing in at terminal YZ is 5. Hence, equivalent
Norton network is shown in Figure 8.64(b)
Figure 8.13(b)
Example 8: Determine by successive conversion between Thevenin’s and Norton’s
equivalent networks, a Thevenin equivalent circuit for terminals YZ of Figure 8.14.
Hence, determine the magnitude of the current flowing in the inductive part connected to
YZ
E=12
V
Z=3
Y
Z
Isc =
2A Z = 5
Y
Z
15V
2K
5V
3K
100
j4K
Y
Z
2m
A
1K
121
Figure 8.14
Solution:
For the branch containing the 15v source, converting to Norton equivalent network gives
Isc = 15/2000 = 7.5mA and Z = 2k. for the branch containing the 5V source, converting
to a Norton equivalent network gives 5/3000 = 1.7mA and Z = 3k. Thus, the circuit of
Figure 8.14 converts to that of 8.15.
Figure 8.15
The above two Norton equivalent networks shown in Figure 8.15 may be combined,
since the total short circuit current is (7.5 + 1.7) = 9.2mA and the total impedance Z is
given by (2 x 3) = 1.2k. This results in the network of Figure 8.16
Figure 8.16
Both of the Norton equivalent networks shown in Figure 8.16 can be converted to
Thevenin equivalent circuit. Open circuit voltage across AB gives (9.2 x 10-3
) = 11.04V
and impedance pointing in is 1.2k. Open circuit voltage across (i) is (1 x 103)(2x10
-3) =
2V and the impedance pointing in CD is 1k. Thus, 8.16 convert to 8.17
7.5m
A 2K
Y
Z
2m
A
1.7m
A 3K
1K
9.2m
A 1.2K
A 2m
A
B
C D Y
Z
11
– 0
4V
1.2
k
1k
2
V Y
Z
122
Figure 8.17
Combing the two Thevenin circuit gives e.m.f. E = 11.04 – 2 = 9.04V and impedance Z =
(1.2 + 1) = 2.2k. Thus, Thevenin’s equivalent circuit for terminals YZ of Figure 8.18 is
shown in Figure 8.19
Figure 8.19
If an impedance (100 + j4000) is connected across terminals YZ, then the current I
flowing is given by:
i.e. the current in the inductive branch is 1.96mA.
Exercises on Thevenin’s Theorem
1. Use Thevenin’s Theorem to determine the current flowing in the 20 resistor of
the network shown in Figure 8.20
Figure 8.20
(2.38A)
2. Determine the Thevenin equivalent circuit with respect to terminals YZ of the
network shown in Figure 8.21. Hence, determine the magnitude of the current
flowing in a (4 + j6) impedance connected across terminals AB and the power
delivered to this impedance.
9.40
V
2.2k
Y
Z
60
V
3
40
V
4
20
- j5
j4
4
20
V
Y
Z
123
Figure 8.21
(0.94A, 3.53W)
3. Derive for the network shown in Figure 8.22 the Thevenin equivalent circuit at
terminals AB and hence, determine the current flowing in a 3 resistance
connected between A and B.
4.
Figure 8.22
(1.52<42.550A)
Exercises on Norton’s Theorem
5. Determine current flowing in the 5 resistance of the network shown in Figure
8.23 by using Norton’s Theorem.
Figure 8.23
(60A)
6. Determine for the network shown in Figure 8.24 the Norton equivalent network at
terminals YZ. Hence, determine the current flowing in a (2-j4) impedance
connected between Y and Z
40
V
-j5 6.5 15
15V 3 A
B
Isc =
12A
3 15
4 5 6
J3
-j3
6
Y
Z
20<90OV
124
Figure 8.24
(0.15<-126.860A)
Exercise on Thevenin and Norton Equivalent Networks
7. Convert the circuits shown in Figure 8.25 to Norton equivalent networks
Figure 8.25(a) Figure 8.25(b)
(0.87<16.76A)
8. Determine by successive conversions between Thevenin and Norton equivalent
networks, a Thevenin equivalent circuit for terminals YZ of Figure 8.26. Hence,
determine the current flowing in a (2 +j6) impedance connected across Y and Z.
Figure 8.26
(0.89A)
9. Convert the networks shown in Figure 8.27 to Thevenin equivalent circuits.
10
5<30O
V
5
2V
Y
Z
2
15 10 4
2V 2V 3V
125
Figure 8.27
(4A,3Ω and 12V, 3Ω)
10. Derive an equivalent Thevenin circuit for terminals AB of the network in Figure
8.28. Hence, determine the p.d. across AB when (4 + j3) impedance is connected
between these terminals.
Figure 8.28
(14.95mV)
Isc = 3 <
50OV
5 4 Isc =
4A
2K
V
j10
1K
3m
A
A
B
126
CHAPTER NINE
DELTA – STAR AND STAR - DELTA TRANSFORMATIONS
Introduction
Delta - Star/Star - Delta Transformation also written as wye-Delta which was
introduced and published by Arthur Edwin Kennedy in 1899. Delta-Star/Star-Delta
Transformation is widely used in analysis of three phase electric power circuits.
Star or Wye (Y) Connection
In this method of interconnection, the similar ends say “Star” ends of three coils (it
could be “stop” end also) are joined together at point N as showing fig.9.1 the point N is
known as “Star pointer neutral point”. The three conductors meeting at point N are
replaced by a single conductor known as “neutral conductor” as shown in fig. 9.1(b) such
an inter connection is known as four-wire, 3 – phase system.
Figure 9.1 (a) Figure9.1 (b)
Delta/Star Transformation
In an ac network, complicated network having large numbers of simultaneous
equation to solve can be simplified by successively replacing Delta meshes by equivalent
Star system and vice-versa.
Assuming we are given tree impedances, Z1, Z2 and Z3 connected in Delta-fashion
between terminals 1, 2 and 3 in Figure 9.2(a). So far, as the respective terminals are
F
N
I
R
E
R
S S
S
I
B
F
E
Y E
B
I
Y
Load
s
E
B
E
Y
E
R
N
I
B I
Y
Neutra
l Wir
e
127
concerned, these three given impedances can be replaced by the three impedances Z1, Z2
and Z3, connected in Star as shown in Figure 9.2(b).
Figure 9.2(a)
Figure 9.2(b)
First, take Delta connected: between terminals 1 and 2, there are two parallel
paths, one having impedance of Z1 and other having in impedance of (Z2 + Z3)
Impedance between terminals 1 and 2 is
Now, take Star connection, the impedance between the same terminals 1 and 2 is
(Z11 + Z21). As terminals impedances have to be the same.
……………………… (i)
Similarly, for terminals 2 and 3 and terminals 3 and 1, we get
……………………… (ii)
Z3 Z1
Z2
1
3 2
Z31
Z11
Z21
1
3 2
128
And ……………………… (iii)
It can be deducted that for Delta-Star Transformation, impedance of each arm of
the Star is given by the product of the impedances of two Delta sides that meet at its end
divided by the sum of the three Delta impedances.
Star/Delta Transformation
Star-Delta Transformation can easily be done by using equations derived above in
Delta/Star Transformation. Multiplying (i) and (ii), (ii) and (iii) and (iii) and (i) and
adding them together and then simplifying them, we have.
It can be deduced that, equivalent Delta impedance between any two terminals is
given by the sum of Star impedances between those terminals plus the product of these
Star impedances divided by the third Star impedance.
Example 1: Replace the Delta-Star connected network shown in Figure 9.3 by an
equivalent Star connection.
Figure 9.3(a)
ZA=10
ZC =-
J50
J12
Z
B 5
129
Solution:
Le the equivalent Star network be as shown in Figure 9.3(b)
Figure 9.3(b)
Then, from:
Z1 = 3.18 < 135.84
Example 2: In network shown in Figure 9.4, determine (a) equivalent circuit impedance
across terminals AB, (b) supply current I and (c) power dissipated in the 20 resistor.
Figure 9.4(a)
Z1
Z2 Z3
-
j20
j20 j5
20
J15
20<
0O
V
A
B
1
2 3
130
Figure 9.4(b)
The network of Figure 9.4 is redrawn in Figure 9.4(b), showing more clearly part of the
network 1, 2, 3 forming a Delta connection. This may be Transformed into a Star
connection as showing fig. 9.4 (c)
Solution:
Figure 9.4(c)
-
j20
ZB = j5
20 20<
0O
V
A
B
1
2 3
ZA =
j20 ZC = j15 20<
0O
V
A
B
20
-
j20
2 3
1
Z1
Z2 Z3
131
The equivalent network is shown in Figure 9.4(d) and is further simplified in 9.4 (e)
Figure 9.4(d) Figure 9.4(e)
(20 + j7.5) in parallel j5 gives an equivalent circuit impedance of:
Hence, total circuit equivalent impedance across terminal AB is given by;
ZAB = (0.9 + j4.4) +j1.88 = 0.9 + j6.28 or 6.34 < 81.80
(b) Supply current
(c) Power P dissipated in the 20 resistance is given by: (I1)2(20)
Hence, power p= (I2)R = (0.67)
2(20) = 8.98W.
Example 3: Determine the Delta connected equivalent circuit network for the Star Delta
connected impedances shown in Figure 9.5
The network of fig. 9.4 is redrawn in Figure 9.4(b), showing more clearly part of the
network 1, 2, 3 forming a Delta connection. This may be Transformed into a Star
connection as showing fig. 9.4 (c)
j5
J1.88
20<
0O
V (20+j7.5
20<
0O
V
-j20
j25
20
J1.88 J7.5
A
B
132
Figure 9.5
Solution:
Figure 9.5(a) shows the networks of Figure 9.5 redrawn and Figure 9.5 (b) shows the
equivalent Delta containing impedances ZA, ZB and ZC.
Example 4: Three impedances, Z1 = 50, Z2 = 60, 23 = 70 are connected in Star
convert the Star to an equivalent Delta connection.
5 15
j12
ZA ZB
ZC
Z1=5
Z2=15
Z3=j12
133
The Star connected network and the equivalent Delta network comprising impedances
ZA, ZB and ZC are shown in fig. 9.6
Figure 9.6
Example 5: Three impedance, Z1 = 50 <00, Z2 = 60 < 80
0 and Z3 = 50 < - 90
0 are
connected in Star. Convert the Star to an equivalent Delta connection.
The Star – connected network and the equivalent Delta network comprising impedances
ZA, ZB and ZC shown in Figure 9.7.
Z1=5
Z2=15
Z3=j12
ZA ZB
ZC
134
Figure 9.7
ZA ZB
ZC
Z1=50<O
0
Z2=60<80O
Z3=50<-
90O
135
Exercises
1. Transform the Delta connected networks shown in Figure 9.8 to their equivalent
networks.
Figure 9.8(a)
(0.3Ω, 1.8Ω, 0.6Ω)
2. Determine the Delta connected equivalent networks for the Star connected
impedance shown in Figure 9.8
Figure 9.8(a) (b)
3 1
6
1
2 3
-j12
j12
1
2
3
(6-
j2)
(2-j4)
1
2
3
(4+j2)
7
4 3
3 2
1
136
(2Ω, 1.5Ω and 0.5Ω)
3. For the network shows in Figure 9.9 determine (a) current I and (b) power
dissipated in the 20 resistance.
Figure 9.9
(-j7.5Ω, -j1.5Ω and j3.75Ω)
4. Derive the Star-connected network of three impedances equivalent to the network
shown in Figure 9.10
Figure 9.10
(1.5Ω, 0.27Ω and 1.61Ω)
1. Transform the Delta connected networks ABC shown in Figure 9.11 and hence,
determine the magnitude of the current flowing in the 10 resistance
60<
0O
V
-j30
-j12 -j6
20 j12
2.3 12 6.25
5 3
60<
0O
V
-j30
-j12
j81
20 j3
j12
-
j3
137
Figure 9.11
(0.25<78.930Ω, 2.7<-64.81
0mΩ, 0.32<-76.12
0Ω,5A)
2. For the network shown in Figure 9.12, determine (a) the current supplied by the 20
< 00 V source, and (b) power dissipated in the (3.00 – j0.619) impedance.
Figure 9.12
(133.33<-87.64A)
3. For the A. C. bridge network shown in Figure 9.13, Transform the Delta
connected network XYZ into an equivalent Star, and hence, determine the current
flowing in the capacitor.
Figure 9.12
(3.72<30.330Ω, 9.36<-8.33
0Ω, 9.75<30.33
0Ω)
(0.40 –
j2.10)
(2+j5)
(5 –
j10) (5
.5 +
j2
(30 –
j16)
(300
–
j0.6
19
20<
0OV
16 42
j20 20
j20 10
D
X Y
Z
50<-
90OV
-
j32
138
4. A Delta connected network contains three 42 < 800 impedances. Determine the
impedances of the equivalent Delta connected network.
(14<800Ω)
5. Three impedances, each of (3+j2), are connected in Star. Determine the
impedances of the equivalent Delta-connected network.
(1.2<33.70Ω)
6. Transform the network shown in Figure 9.13 to its equivalent Star-connected
network.
(13.62<1.770Ω, 20.43<-24
0Ω, 10.21<6.77
0Ω
30<10O
60< -
20O
45 < -
15O
139
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