Circuit Theory 2
Circuit Theory 2
Circuit Theory 2
REVITALISATION PROJECT-PHASE II
NATIONAL DIPLOMA IN
ELECTRICAL ENGINEERING TECHNOLOGY
THEORY
1
TABLE OF CONTENTS
Week 1:
1.1 Mathematical form of representing A.C signals
1.2 Conversion of a.c signal in polar form to the j-notation form
1.3 Subtraction, addition, multiplication and division of phasor using j
operator
1.4 Solved simple problems using j-notation
Week 2:
1.5 Phasor diagram for a.c circuits drawn to scale
1.6 Derivations with the aid of waveforms diagrams that the current in a capacitive
circuit leads voltage and the current in the inductive circuit lags the voltage
1.7 Inductive and capacitive reactances
1.8 Voltage and current waveforms on same axis showing lagging and
leading angles
Week 3:
1.9 Phasor diagrams for series and parallel a.c circuits
1.10 Voltage, current, power and power factor calculations in series and
parallel circuits
1.11 Series and parallel resonance
1.12 Conditions for series and parallel resonance
Week 4:
1.13 Derivations of Q-factor, dynamic impedance and bandwidth at
resonance frequency
1.14 Sketch of I and Z against F for series and parallel circuits
1.15 Calculation of Q-factor for a coil and loss factor for a capacitor
1.16 Bandwidth
1.17 Problems involving bandwidth and circuits Q-factor
Week 5:
2.1 Terms used in electric networks
WEEK 6:
2.2 Basic principles of mesh circuit analysis
2.3 Solved problems on mesh circuit analysis
Week 7:
2.4 Basic principles of nodal analysis
2.5 solved problems on nodal analysis
Week 8:
3.1 Reduction of a complex network to it series or parallel equivalent
3.2 Identification of star and delta networks
Week 9:
3.3 Derivation of formulae for the transformation of a delta to a star
network and vice versa
3.4 solved problems on delta/star transformation
Week 10:
3.5 Duality principles
3.6 Duality between resistance, conductance, inductance, capacitance,
voltage and current
Week 11:
3.7 Duality of a network
3.8 Solved network problems using duality principles
Week 12:
4.1 Thevenin’s theorem
4.2 Basic principles of Thevenin’s theorem
4.3 Solved problems on simple network using Thevenin’s theorem
4.4 Solved problems involving repeated used of Thevenin’s theorem
Week 13:
4.5 Norton’s theorem
4.6 Basic principles of Norton’s theorem
4.7 Comparison of Norton’s theorem with Thevenin’s theorem
4.8 Solved problems using Norton’s theorem
Week 14:
4.9 Millman’s theorem
4.10 Basic principles of Millman’s theorem
4.11 Solved network problems using Millman’s theorem
Week 15:
4.12 Reciprocity Theorem
4.13 Basic principles of Reciprocity theorem
4.14 Solved problems using Reciprocity theorems
A.C THEORY Week 1
At the end of this week, the students are expected to:
State different mathematical forms of representing a.c signals
Convert a.c signal in polar form to the j-notation
Subtract, add, multiply and divide phasor using j-operator
Solve simple problems using j-notation
1.1 MATHEMATICAL FORMS OF REPRESENTING A.C SIGNALS
Generally, A.C signal may be represented in the following mathematical form;
a) Trigonometric form, Z = r (Cos + sin ) (1.1)
b) Polar form, Z = r (1.2)
c) J – notation form, Z = x+jy (1.3)
1.2 CONVERSION OF A.C SIGNAL IN POLAR FORM TO THE
j – NOTATION FORM
Example 1.1: Express 18-56.3 in j – notation form
Solution:
0
Step I: Make a sketch as shown on fig 1.1 to measure = -56.3 with respect to the real axis and r =
OA = 18
Step II: Find x using trigonometric ratio,
Cos = x/18 j
0
i.e x = 18 cos = 18 Cos (-56.3 )
x 10 O x B
o
Example 1.2: Express 553.1 in j – notation form
Solution:
0
Step I: Make a sketch as shown on fig 1.2 to measure = 53.1 with respect to the real axis and r =
OA = 5
Step II: Find x using trigonometric ratio
1
A.C THEORY Week 1
Cos = x/5
0
i.e x = 5 cos = 5 Cos (53.1 ) x A
y
x3 S
N
Step III: Find y using trigonometric
0
= 53.1
ratio Sin = Y/5
B Re
O
i.e y = 5 sin = sin (53.1 ) Fig 1.2
y4
Step IV: finally obtain the required result as x + jy = 3 – j4
If two phasor are Z1 and a +jb and Z2 = c+jd, then addition of phasor gives: Z 1 + Z2 = (a+jb) + (c+jd)
= (a+c) + j(b+d)
Subtraction of phasor gives: Z1 – Z2 = (a-c) + j(b-d)
Example 1.3: If V1 = -10 + j20 and V2 = 20 +j30 , find the sum of V1 and V2, express the result in
polar form.
Solution
V1 = -10+j20
V2 = 20 + j30
V3 = V1 + V2 = -10+ 20+j20 + j30 = 10 +j50
2 2 -1 50 0
Finally, V3 = 10 +50 Tan ( /10) = 5178.7
2
A.C THEORY Week 1
0
Example 1.4 Subtract I1 = 3-56.3 from I2 =
0
5.830 Solution
Step I: Convert first to j – notation form to get (using the trigonometric form)
0 0
I1 = 3 {cos(-56.3 ) + j sin (-56.3 )}
= 3{0.5548 + j(-0.8320)} = 1.66 – j2.50
0 0
similarly, I2 = 5.8 (cos 30 + j sin 30 ) 5.02 + j2.90
3
A.C THEORY Week 1
Example 1.6: perform the following operation and the final result may be given in polar for; (8
+j6)x(-10-j7.5)
Solution
2
(8+j6)x(-10-j7.5)= -80-j60-j60-j 45=-80+45-j120 = -35-
2 2 -1 120 0
j120 = (-35) + (-120) tan ( /35) = 12573.7
4
A.C THEORY Week 2
At the end of this week, the students are expected to:
Draw to scale phasor diagram for a.c circuits
Show with the aid of waveforms diagrams that the current in a capacitor circuit leads voltage
and the current in the inductive circuit lags the voltage.
Distinguish between inductive and capacitive reactances.
Draw voltage and current waveforms on same axis to show lagging and leading angles.
1.5 PHASOR DIAGRAMS FOR A.C CIRCUITS DRAWN TO SCALE
For a pure inductive circuit, the voltage across the inductor (V L) leads the currents (I)
0
flowing through it by 90 . Taking a scale of 1cm:2V, and 1cm:3A, the Phasor diagram is
drawn to scale as shown in figure1.3 (b)
I V
V,I
VL 2 6
(a) I
4
2 6
3
0 90 0
90
I VC
3 (b) 6 2 4
(c)
where t is the time, in seconds, after the current has passed through zero from negative to positive
values, as shown in fig 1.4.
Suppose the current to increase by di ampere in dt seconds, then instantaneous value of
induced e.m.f is
e = L di
dt
= LIm d (sin2πft)
dt
= 2πfL Imcos2πft = Vmcos2πft, where Vm = 2πfLIm
e = Vmsin(2πfL +π/2) (1.5)
v,i
Current Applied voltage
di v Im
i
π 2
t / t
dt
0
(90 )
Fig 1.4: voltage and current waveforms for a purely inductive circuit
The induced e.m.f is represented by the curve in fig 1.4, leading the current by a quarter of a cycle.
6
A.C THEORY Week 2
Since the resistance of the circuit is assumed negligible, the whole of the applied voltage is
equal to the induced e.m.f, therefore instantaneous value of applied voltage is
V=e
V = Vmsin(2πft + π/2) (1.6)
Comparism of expressions (1.4) and (1.6) shows that the applied voltage leads the current by
a quarter of a cycle.
1.6.2 Current and voltage in a capacitive circuit
v,i
dv
i Vm
v
t
t dt
Fig 1.5: voltage and current waveforms for a purely capacitive circuit
Suppose that the instantaneous value of the voltage applied to a capacitor having capacitance C
farad is represented by
V = Vmsint = Vmsin2πft ` (1.7)
If the applied voltage increases by dv volts in dt seconds (fig,1.5) then, instantaneous value of
current is
i = C dv/dt
d
= C /dt(Vmsin2πft)
Vm
= 2πfCVmcos2πft = /Xc cos2πft
i = Imsin(2πft + π/2) (1.8)
where Im = 2πfCVm
7
A.C THEORY Week 2
Comparism of expression (1.7) and (1.8) shows that the current leads the applied voltage by
a quarter of a cycle.
1.7 DISTINGUISH BETWEEN INDUCTIVE AND CAPACITIVE
REACTANCES
1.7.1 Inductive Reactance
From the expression Vm = 2πfLIm, Vm/Im = 2πfL
If I and V are the r.m.s values, then
inductive reactance
The inductive reactance is expressed in ohms and is represented by the symbol XL.
The inductive reactance is proportional to the frequency and the current produced by a given
voltage is inversely proportional to the frequency, as shown in fig 1.6
x,i
Current
Inductive
reactance
f
Fig 1.6: variation of reactance and current with frequency for a purely inductive circuit
8
A.C THEORY Week 2
x,i
Capacitive
reactance
Current
f
Fig 1.7: variation of reactance and current with frequency for a purely capacitive circuit
t
Fig 1.8
9
A.C THEORY Week 3
At the end of this week, the students are expected to:
Draw phasor diagram for series and parallel a.c circuits
Calculate voltage, current, power and power factor in series and parallel circuits
Explain series and parallel resonance
State conditions for series and parallel resonance
1.9 PHASOR DIAGRAMS FOR SERIES AND PARALLEL A.C CIRCIUT
1.9.1 Phasor diagrams for series a.c circuits:
VL
VL Vs VR I Vs
VL - Vc
I Vs
VR Vc I
(a) (b) VR
fig 1.9: Phasor diagram for series a.c circuits: Vc (c)
(a) R-L series circuit (b) R-C series circuit (c) R-L-C series circuit
1.9.2 Phasor diagram for parallel a.c circuits: Ic
IR Vs Ic I Ic - IL I
V
I Vs S
IR
IL (d) (e) IR
(f)
IL
(d) R-L parallel circuit (e) RC parallel circuit (f) RLC parallel circuit
10
A.C THEORY Week 3
11
A.C THEORY Week 3
(d) power factor = cos
-1 -1 0
Where = cos (VR/VS) = cos (61.8/100) = 51.8
P.f = cos51.8 = 0.6184
Example 1.9: A circuit consists of a 115Ω resistor in parallel with a 41.5µF capacitor and is
connected to a 230V, 50Hz supply (fig 1.11). Calculate: (a) the branch currents and the supply
current (b) the power factor (c) the power consumed
I Ic I
IR IC
VS 230V 115Ω 41.5µF
50Hz
Vs
IR
Fig 1.11: circuit and phasor diagrams for example 1.9
Solution
(a) IR = VS/R = 230/115 = 2A
XC = 1 = 1 = 76.7Ω
2πfC -6
2πx50x41.5x10
IC = VS/XC = 230/76.7 = 3A
2 2 2 2
I = (IR + IL ) = (2 + 3 ) = 3.6A
(b) P.f = cos
-1 -1
and = cos (IR/I) = cos (2/3.6) =
0 0
56.3 P.f = cos56.3 = 0.5548
2 2
(c) P = I R = 3.6 x 115 = 1490.4W
Example 1.10 Three branches, possessing a resistance of 50Ω, an inductance of 0.15H and a
capacitance of 100µF respectively, are connected in parallel across a 100V, 50Hz supply. Calculate:
(a) the current in each branch (b) the supply current (c) the phase angle between the supply current
and the supply voltage (d) the power factor of the circuit.
12
A.C THEORY Week 3
Solution
The circuit diagram for example 1.10 is shown in fig 1.12
Ic
I I
100V IR IL IC IC – IL
50Hz 50Ω 0.15H 100µF
V
(a) IR = 100/50 = 2A
IL = 100 = 2.12A
2 x 3.142 x 50 x 0.15
-6
and IC = 2 x 3.14 x 50 x 100 x 10 x 100 = 3.14A
(b) The resultant of IC and IL is
IC – IL = 3.14 – 2.12 = 1.02A
2 2 2 2
I = [IR + (IC – IL) ] = (2 + 1.02 ) = 2.24A
(c) From fig 1.13:
cos = IR/I = 2/2.24 = 0.893
-1 0
= cos (0.893) = 26.7
(d) P.f = cos = cos26.7 = 0.8934
13
A.C THEORY Week 3
This condition where by XL = XC in a series R-L-C circuit is called series resonance, and the
frequency at which it occurs is called resonant frequency, f0. The phasor diagram of the R-L-C
series circuit at resonance is shown in fig 1.14
VL = IXL
I
VS = IR
VC = IXC
ILcos=I
R ILL
I
ILsin IL
VS
(b) Phasor diagram
(a) Parallel circuit
Fig 1.15: parallel resonance
15
A.C THEORY Week 4
At the end of this week, the students are expected to:
Prove the relevant formulae for Q-factor, dynamic impedance and bandwidth at resonance
frequency
Sketch current and impedance against frequency for series and parallel circuits
Calculate the Q-factor for a coil; loss factor for a capacitor
Explain with the aid of a diagram, bandwidth
Solve problems involving bandwidth and Q-factor
1.13 DERIVATIONS OF Q-FACTOR, DYNAMIC IMPEDENCE AND
BANDWIDTH AT RESONANCE FREQUENCY
1.13.1 Q-Factor of a Series Resonance
1
A.C THEORY Week 4
1.13.4 Dynamic Impedance of a Parallel Resonance
Consider the phasor diagram of fig 1.15(b)
S
I = I1Cos1 , where I1 = V and Cos1 =R
Z1 Z1
I = VS .
S
R = V R (1.15)
2
Z1 Z1 Z1
Also, IC = I1 Sin 1, where Sin 1 = X L , and IC = V S
Z1 XC
2
VS =V S . XL or Z1 = XLXC where XL = L, XC = 1/C
Xc Z1 Z1
2 (1.16)
Z1 = L/C
Putting eqtn (1.16) in (1.15) gives
I = V S RC VS =Z= L = dynamic impedance
L I RC
Z = L/RC (1.17)
Z=R
f
f0
Fig 1.16
2
A.C THEORY Week 4
1.14.2 Sketch of current (I) and impedance (Z) against frequency (F) for
parallel circuit
The sketch of I and Z against f for parallel circuit is shown in fig 1.17
f0 f
Fig 1.17:
In practice, a capacitor has an equivalent resistance (Rse) either in series or in parallel with it [fig
1.18(a)]
From the phasor diagram [fig 1.18(b)],
3
A.C THEORY Week 4
= actual angle
= loss angle
and loss factor = tan
0 0 0 0 0
+ = 90 , = 90 - = 90 - 80 = 10
0
and tan = tan10 = 0.1763
1.16 BANDWIDTH
When the current in a series RLC circuit is plotted as a function of (or f) we obtain the curve as
shown in fig 1.19. We notice that the points where the current is 0.707 of the maximum (as indicated
on the graph), the corresponding frequencies are 1 and 2.
Io
BW
2
1 0 (f)
(f1) (f0) (f2)
Fig 1.19
The distance between these points 1 and 2 is known as bandwidth BW. We define the
bandwidth, BW, of the resonant circuit to be the difference between the frequencies at which the
circuit delivers half of the maximum power. The frequencies 1 and 2 are called half power
frequencies.
4
A.C THEORY Week 4
Solution
0 =1300 = 10
Qf = f
BW 130
Example 1.14 obtain the bandwidth in example 1.13 if the Q-factor is reduced by 50%
Solution
BW = f0 , where Q.f = 50 x 10 = 5
Q.f 100
BW = 1300 = 260Hz
5
5
Method of Analysis Week 5
At the end of this week, the students are expected to:
Explain
the following terms used in electric networks:
Ideal and practical independent current and voltage sources
Branch
Node
Loop
Network
A A
+
Is _ Vs
B B
(a) (b)
Fig 2.1: Ideal independent sources
A practical independent current source exhibits an internal resistance in parallel with the
ideal independent current source. A practical independent voltage source exhibits an internal
resistance in series with the ideal independent voltage source.
The schematic representation of a practical independent voltage and current source is
shown in fig. 2.2 (a and b) respectively.
1
Method of Analysis Week 5
R A
A
+ Is R
_
Vs
B B
(a) (b)
Fig 2.2: practical independent sources
2.1.2 Branch
A branch is part of a network which lies between two junctions. The circuit of fig. 2.3 has
three branches.
2.1.3 Node
A node is a junction in a circuit where two or more circuit elements are connected together.
The circuit of fig. 2.3 has two nodes.
2.1.4 Loop
A loop is any closed path in a circuit. For example, the circuit of fig. 2.3 has three loops:
abefa, bedcb and acdfa
a b node 1 c
R
V C
branch 3
d
e branch 2
f node 2
branch 1
Fig 2.3
2
Method of Analysis Week 6
At the end of this week, the students are expected to:
Explain the basic principle of mesh circuit analysis
Solve problems on mesh circuit analysis
2.1.5 Network
A combination of various electric elements, connected in any manner whatsoever, is called a
network. This is shown in fig. 2.3
has two meshes. Let the two mesh currents be designated as I 1 and I2 and all the two may be
assumed to flow in the clockwise direction for obtaining symmetry in mesh equations.
R1 R2
I1 R3 I 2 E2
E1
It would be seen that the first item in the first two row i.e (R1 + R3) represents the self
resistance of mesh (i) which equals the sum of all resistance in mesh (i). Similarly, the second item 1
Method of Analysis Week 6
in the first row represents the mutual resistance between meshes (i) and (ii) i.e the sum of the
resistances common to mesh (i) and (ii).
The sign of the e.m.f’s, while going along the current, if we pass from negative to the
positive terminal of a battery, then, its e.m.f is taken positive. If it is the other way around, then
battery e.m.f is taken negative.
In the end, it may be pointed out that the directions of mesh currents can be selected
arbitrarily.
+ 40Ω
I1 9Ω I2 V
20V _
13 -9 I1 20
-9 81 I2 0
0= 13 -9
= = 1053-81 = 972
-9 81
-9
20
I1 = 0 81 = 1620A
2
Method of Analysis Week 6
I2 = 20 = 18OA
13
-9 0
I1 = ∆I1/∆0 = 1620/972 = 1.67A, I2 = ∆I2/∆0 = 180/972 =
0.185A V = 40I2 = 40 x 0.19 7.4V
EXAMPLE 2.2 Determine the current in the 4Ω resistor in the circuit shown in figure 2.6 using
mesh analysis
2Ω 2Ω
10V
12V I2 12Ω
(2) I3
1Ω 3 (3)
Ω
(1) I1
24V 4Ω
1 -15 12 I2 = -12
3 12 -17 I3 10
3
Method of Analysis Week 6
0 = 8 -1 -3
= 2664 + 84 – 18 = 2730
4
Method of Analysis Week 7
At the end of this week, the students are expected to:
Explain the basic principle of nodal analysis
Solve problems on nodal analysis
2.4 BASIC PRINCIPLE OF NODAL ANALYSIS
Consider the circuit of fig 2.7 which has three nodes. One of these i.e node 3 has been taken in as
the reference node. VA represents the potential of node 1 with reference to the reference node (or
zero potential node). Similarly, VB is the potential difference between node 2 and node 3. Let the
current directions which have been chosen arbitrary be as shown
R1 Node R2 Node R3
1 2
I1 VA I2 VB I3
R4 R5
E1 I4 I5 E2
Node
3
Reference node
Fig 2.7
For node 1, the following current equation can be written with the help of KCL.
I1 = I4 + I2 (2.8) Now E1 = I1R1 + VA I1 = (E1 – VA)/R1
E 1– V A =V A + V A - V B
R1 R4 R2
Simplifying the above, we have
1
VA 1 + 1 + 1 _V B = E (2.9)
R1 R2 R3 R2 R1
The current equation for node 2 is I5 = I2 + I3 (2.10)
1
Method of Analysis Week 7
Or V B = V A – V B + E 2 - VB
R5 R2 R3
Or -V A + VB 1 + 1 + 1 =E 2 (2.11)
R2 R2 R3 R5 R3
The node voltages in equation (2.9) and (2.11) are the unknowns and when determined by a suitable
method, result in the network solution. After finding different node voltages, various current can be
calculated by using ohm’s law.
2.5 SOLVED PROBLEMS ON NODAL ANALYSIS
Example 2.3 obtain the node voltage and the branch current in the circuit shown in fig 2.8
5ΩV1 2Ω
I1 I3 I2
20V 10Ω
8V
Solution Vref
Fig 2.8
For node 1, applying KCL gives
I3=I1+I2
Or V 1= 20 - V 1 + 8 - V 1
10 5 2
V1 + V1 + V1 = 20 + 8 = 8
10 2 5 5 2
Or V1 + 5V1 + 2V1 = 8 x 10 = 80
8V1 = 80
80
V1 = /8 =10V
I1 = 20 – 10 = 2A, I2 = 8 – 10 = -1A
5 2
10
I3 = /10 = 1A
Example 2.4 Determine the voltages at nodes b and c in the network shown in figure 2.9
using nodal analysis
0.2Ω b 0.3Ω c 0.1Ω
30A 20A
116V
120V
2
Fig 2.9
Method of Analysis Week 7
Solution:
Let the voltages at nodes b and c be Vb and Vc respectively.
Using nodal analysis for node b, we get
b
3
Network Transformation Week 8
At the end of this week, the students are expected to:
Reduce a complex network to its equivalent
Identify star and delta networks
3.1 REDUCTION OF A COMPLEX NETWORK TO ITS SERIES OR
PARALLEL EQUIVALENT
The network of fig. 3.1 (a) and (b) can be reduced to its equivalent as shown below.
R2
e
R4
R1 R R3
e R1 R3 q 2
q
Fi.g. 3.2
R3
The equivalent resistance of fig. 3.1(b), is shown in fig.
3.3 1/Req= 1/R1 + 1/R2 +1/R3
Req = R1R2R3 / (R1R2 + R1R3 + R2R3)
R q
R1 R2 R3 eq
q
Fig 3.3
1
Network Transformation Week 8
Also the circuit of fig 3.1(c) can be reduced to its equivalent as under
R2
R1 R
R
3
6
R5 Fig 3.1(c)
R4 R 7
R5 R7
Fig 3.1(d)
From fig 3.1(d),
RS1 = R1 + RP1 + R6
The circuit becomes as shown in fig 3.1(e)
R
s1
R5 R7
Fig 3.1(e)
RS2 = R5 + R7
The circuit becomes as shown in fig 3.1(f)
R
s1
R
S2
Fig 3.1(f)
2
Network Transformation Week 8
Req = RS2 + RS1
The circuit is finally reduced as shown in fig 3.1(g)
R
eq
Fig 3.1(g)
1
1 R1
R31 R12 R3
R2
3 R23 2
3
2
Fig. 3.4 Delta-network Fig. 3.5 Star-network
3
Network Transformation Week 9
At the end of this week, the students are expected to:
Derive the formula for the transformation of a delta to a star network and vice versa
Solve problems on Delta to star transformation
3.3 DERIVATION OF FORMULAE FOR THE TRANSFORMATION OF A
DELTA TO A STAR NETWORK AND VICE VERSA
3.3.1 Derivation of formulae for the transformation of a delta to a star network.
The circuit of fig. 3.4 and 3.5 can be redrawn as shown in figure 3.6 and 3.7 respectively.
RC 1 3
1 3
R1
Rb Ra R2
a
2 4 R3
Fig 3.6 4
2
Fig 3.7
Consider figure 3.6 and 3.7, the resistance between terminal 1 and 2 is
R12 (Y) = R1 + R3 (3.1)
Ra + Rb +Rc
and R34 = R2 + R3 = Ra(Rb+Rc) (3.5)
Ra + Rb +Rc
Subtracting eqtn. (3.5) from (3.3), we get
C b a
R1 – R2 = R (R – R ) (3.6)
Ra + R b + R c
1
Network Transformation Week 9
Adding eqtn. (3.4) and (3.6) gives
R1 = RbRc (3.7)
Ra + R b + R c
Subtracting eqtn. (3.6) from (3.4) gives
R2 = RcRa (3.8)
Ra+Rb+Rc
Subtracting eqtn. (3.7) from eqtn. (3.3) gives
a b
R3 = R R (3.9)
Ra+Rb+Rc
Fig 3.8
Network Transformation Week 9
Solution
R1 = R aR b = 1 X 9 = 0.75Ω
Ra + Rb + Rc 2+9+1
R2 = Ra Rc = 2X1 = 2 = 1/6Ω
Ra + Rb + Rc 2+9+1 12
b c
R3 = R R = 9X2 = 1.5Ω
Ra + R b + R c 9+2+1
The equivalent star network is shown in fig 3.9
R1 R2
0.75Ω 1/6Ω
R3 1.5Ω
Fig. 3.9
Example 3.2 Transform the network of fig 3.9 to an equivalent delta network
Solution
1 2 2 3 1 3
Ra = R R + R R + R R
R1
1 6 16
0.75 x / + / x 1.5 + 1.5 x 0.75 = 2Ω
0.75
1 2 2 3 3 1
Rb = R R + R R + R R
R2
1 1
0.75 x / 6 + / 6x 1.5 + 1.5 x 0.75 = 9Ω
1
/6
Rc = R 1R 2+ R 2R 3+ R 1 R3
R3
1 1
0.75 x / 6 + / 6x 1.5 + 1.5 x 0.75 = 1Ω
1.5
3
Network Transformation Week 10
At the end of this week, the students are expected to:
Explain the meaning of duality principle
Establish duality between resistance, conductance, inductance, capacitance, voltage and current.
3.5 DUALITY PRINCIPLE
Consider for example, the relationship between series and parallel circuits. In a series circuit,
individual voltages are added and in a parallel circuit, individual currents are added. It is
seen that while comparing series and parallel circuits,voltage takes the place of current and
current takes the place of voltage. Such a pattern is known as duality principle.
1
Network Transformation Week 11
Consider the circuit shown in fig. 3.12. The voltage across the two resistors is given by
V=I(R1+R2 ) (3.14)
R1 R2
V
Fig 3.12
V=I
I=V
R1 + R2 = G1 + G2
Also, the dual of series circuit is parallel. Hence, the dual circuit of fig. 3.12 is redrawn as
shown in fig. 3.13
V G1 G2
Fig. 3.13
Network Transformation Week 11
+
20V
_
Solution
Step 1: The dual of 20V voltage source is 20A current source, as shown in fig. 3.15(a)
20A
Fig. 3.15(a)
Step 3: Since the 20V voltage source is in series with 5Ω resistor, it dual counterpart is 20A current
source in parallel with 1/5Ω. This is shown in fig. 3.15(b)
20A
1/5Ω
30A
2F 3H
2
Fig. 3.16
Network Transformation Week 11
Solution
Since the current source, capacitor and inductor are all connected in parallel, their dual counterparts
will be connected in series.
The dual of 30A current source is 30V voltage source
The dual of 2F capacitor is 2H inductor
The dual of 3H inductor is 3F capacitor
2H
30mV + 3F
3
Network Theorems Week 12
At the end of this week, the students are expected to:
State Thevenin’s theorem
Explain the basic principle of Thevenin’s theorem
Solve problems on simple network using Thevenin’s theorem
4.1 THEVENIN’S THEOREM
Thevenin’s theorems state that a linear two terminal network may be reduced to an equivalent circuit
consisting of a single voltage source in series with a single resistor as shown in figure 4.1
Rth a
Eth
b
Fig 4.1: Thevenin’s equivalent circuit
1
Network Theorems Week 12
4.3 SOLVED PROBLEMS ON SIMPLE NETWORK USING THEVENIN’S
THEOREM
Example 4.1 Obtain the Thevenin’s equivalent circuit for the active network in fig 4.2(a).
a a
3 3 6
3
3 I
6
18V b (b) b
(a)
Fig 4.2
Solution
With terminals ab open, the current through the 3 and 6 resistor is
I = 18 = 2A
3+6
The Thevenin’s voltage VTh, is the voltage across terminal a-b.
Hence, Vab = Vth = I X 6 = 2 X 6 = 12V
The Thevenin’s resistance can be obtained by shorting out the 18V sources [fig 4.2(b)] and finding
the equivalent resistance of this network at terminals ab:
RTh = 3 + 3 x 6 = 5
3+6
The Thevenin’s equivalent circuit is shown in fig 4.3
Rth a
Eth
b
Fig 4.3: Thevenin’s equivalent circuit
Example 4.2 Applying Thevenin’s theorem to find the current through the resistance R as shown in
figure 4.4(a)
2
Network Theorems Week 12
5 3 5 3
10
b
Fig 4.4(c)
To find the current through R, we replace the left side of ab by its Thevenin’s equivalent and connect
the load resistance as shown in fig 4.4(d)
RTh
R=10
V
Th
Fig 4.4(d)
I= V Th = 20 = 1.224A
19
RTh + R /3 + 10
3
Network Theorems Week 13
At the end of this week, the students are expected to:
State Norton’s theorem
Explain the basic principle of Norton’s theorem
Compare Norton’s theorem with Thevenin’s theorem
Solve simple problems on Norton’s theorem
4.5 NORTON’S THEOREM
Norton’s theorem state that a linear two terminal network may be reduced to an equivalent circuit
consisting of a single current source and a single shunt resistor as shown in fig 4.8
a a
Isc
IN RN IN RN
b b
Fig 4.8: Norton’s equivalent circuit Fig 4.8: (a)
As seen from fig 4.8(a), to find the Norton’s current IN, we determine the short circuit current
flowing from terminal a and b. It is evident that the short circuit current i sc in fig 4.8(a) is IN
(Norton’s current).
We find Norton’s resistance (RN) in the sane way we find Thevenin’s resistance (RTh). Thus
RN = RTh (4.2)
Finally, after obtaining IN and RN we draw the Norton’s equivalent circuit as shown in fig 4.8
2 a 2 a
10A 8 5 8 5
Fig 4.9 b b
Solution
To find RN, we open circuit the 10A current source as shown in fig 4.10(a)
10
RN = (8 + 2) x 5 = 50 = /3Ω
8 + 2 + 5 15
To find IN, we short circuit terminal ab as shown in fig 4.10(b).
2Ω a
Isc=IN
10A 8Ω 8A 10
/3 Ω
Example 4.6 Determine the current IL in the circuit shown in figure 4.11(a) by using the Norton’s
theorem
5 2 a I1 5 2 a
2 isc
12V RL 5 12V 2 isc RN 5
2
Network Theorems Week 13
Solution
With RL disconnected and replaced by short circuit [fig 4.11(b)],
I1 = 12 .
5 + 2 x 2 = 2A
2+2
isc = 2 x 2 = 1A =
IN 2 + 2
To find RN, the voltage is replaced by a short circuit,
RN = 2 + 5 x 2 = 24
5+2 7
The Norton’s equivalent circuit with RL connected across terminal ab is shown in fig 4.11(c).
IL = i R
sc N = 1 x 24/ 7 = 0.407A
24
RN + 5 /7 + 5
3
Network Theorems Week 14
At the end of this week, the students are expected to:
State Millman’s theorem
Explain the basic principle of Millman’s theorem
Solve network problems using Millman’s theorem
4.9 MILLIMAN’S THEOREM
This states that any number of parallel voltage sources E1, E2,……,En having internal resistance R1,
R2,…….,Rn respectively can be replaced by a single equivalent source E in series with an equivalent
series resistance R.
4.10 BASIC PRINCIPLE OF MILLIMAN’S THEOREM
In circuits of the type shown in figure 4.12, the voltage sources may be replaced with a single equivalent source as
shown in figure 4.13
a a
R1
R2 In R n RL R RL
eq
I1 I2
E
E2 En
eq b
E1
b Fig 4.13
Fig 4.12
The values of I1, I2 ……In would be determined using ohm’s law:
I1 = E1/R1, I2 = E2/R2, …… In = En/Rn
Hence the equivalent current in the circuit is given by
Ieq = I1 + I2 + ….. + In (4.5)
The equivalent resistance Req is obtain by short circuiting the voltage sources. The resistance seen
via terminal ab when RL is removed is given by
Req = R1//R2 // ….//Rn (4.6)
Which may be determined as
Req = 1
(4.7)
1 1 1
/R1 + /R2 + …. + /Rn
The general expression for the equivalent voltage is
Eeq = IeqReq = E1 + E2 + E3 + ….. + En
R1 R2 R3 Rn . (4.8)
1 1 1 1
/R1 + /R2 + /R3 + …..+ /Rn
1
Network Theorems Week 14
Example 4.7 Using Milliman’s theorem to find the common voltage across terminals a and b in the
circuit of fig 4.14
a
4 3 4
12V 6V 4V
b
Fig 4.14
Solution
Vab = Veq = 12 + 6 + 4
4 3 4 = 3.79v
1 1
¼ + /3 + /1
Example 4.8 Using the Milliman’s theorem, determine the current through RL in the circuit
shown in fig 4.15(a)
I
2 3 4
RL= 5
R 0.923 RL5
eq
8V 7V V 2.23V
eq
3V
3
Network Theorems Week 15
At the end of this week, the students are expected to:
State Reciprocity theorem
Explain the basic principle of Reciprocity theorem
Solve network problems using Reciprocity theorem
4.12 RECIPROCITY THEOREM
This state that in any linear bilateral network , if a source of e.m.f E in any branch produces a
current I in any other branch, then the same e.m.f E acting in the second branch would be the same
current I in the first branch.
4.13 BASIC PRINCIPLE OF RECIPROCITY THEOREM
When applying the reciprocity theorem, the following principle must be followed:
1. The voltage sources is replaced by a short circuit in the original location
2. The polarity of the source in the new location is such that the current direction in that branch
remains unchanged.
R1
I
E22V R2 8 R3 12
Fig 4.16
Solution
(a)V(12) =8//12 (22)
= 4.8 x 12 = 12V
4 + (8//12) 8.8
1
Network Theorems Week 15
I = V(12) = 12 = 1A
12 12
(b) Removing the voltage sources E and placing it into the branch with R3 gives the
circuit shown in figure 4.17
R1 I
4 22V
R2 E
8
12
Fig 4.17
V(4) = 4//8 (22)
= 2.6 x 22 =4V
4 4
Hence, current I is the same in both cases
Example 4.10 In the network of fig 4.18(a), find (i) ammeter current when battery is at A and
ammeter at B and (ii) when battery is at B and ammeter at point A
C
2 3 2 3
1 A 1
E=36V 36V
A 12 12
4 B 4 B
D
Fig 4.18(b)
Fig 4.18(a)
2
Network Theorems Week 15
Solution
(i) Effective resistance between points C and B [fig4.18 (a)]
4
is RCB = 12 x /16 = 4
Total circuit resistance = 2 + 3 + 4 = 9
36
Battery current = /9 = 4A
12
Ammeter current = 4 x /16 = 3A
(ii) Effective resistance between points C and D [fig 4.18(b)] is
6
RCD = 12 x /18 = 4Ω
Total circuit resistance = 4 + 3 + 1 = 8Ω
36
Battery current = /8 = 4.5A
12
Ammeter current = 4.5 x /18 = 3A
Hence, ammeter current in both cases is the same.