ELCIAN1

Lecture 05-06 – Network Theorems

Elmer R. Magsino, MS EE

ELCIAN1 - p1
Outline
Delta to Wye, Wye to Delta Transformation
Linearity Property and Superposition
Theorem
Source Transformation
Maximum Power Transfer
Mesh Analysis
Nodal Analysis

ELCIAN1 - p2
 Equivalence
Two electric circuits are said to be equivalent with
respect to a pair of terminals if the voltages across
the terminals and currents through the terminals
are identical for both networks.

a x
+ +
I1 I2
V1 Circuit V2 Circuit
1 2
- -
b y

If V1=V2 and I1=I2, then with respect to terminals

ab and xy , circuit 1 and circuit 2 are equivalent.
ELCIAN1 - p3
 Resistors in Series and in Parallel

+ R1- + R2 -
Resistors in Series a …
+
I1 +
R eq  R1  R 2  ...  R n V1 Rn
- -
b

Resistors in Parallel a …
+
I1 + + +
1 1 1 1 V1 R1 R2 Rn
   ...  - -
R eq R 1 R 2 Rn - -
b
Special Case R1R 2
Two resistors in parallel:
R eq 
R1  R 2
ELCIAN1 - p4
 Delta-Wye Transformation
The transformation is used to establish equivalence
for networks with 3 terminals.
z z

Ra Rc R3

R1 R2
x Rb y
x y
Delta Wye
For equivalence, the resistance between any pair of
terminals must be the same for both networks.

ELCIAN1 - p5
Delta-to-Wye Transformation Equations
R aRb RbR c
R1  R2 
R a  Rb  R c R a  Rb  R c
R cR a
R3 
R a  Rb  R c
Wye-to-Delta Transformation Equations
R1R2  R2R3  R3R1 R1R2  R2R3  R3R1
Ra  Rb 
R2 R3
R1R2  R2R3  R3R1
Rc 
R1
ELCIAN1 - p6
 Delta-to-Wye Transformation Equations

Product of adjacent resistance s

RY 
Sum of all resistance s
 Wye-to-Delta Transformation Equations

Sum of the Products of resistance s taken two at a time

R 
Opposite Resistance

ELCIAN1 - p7
Proof
z z

R3
Ra Rc
R1 R2

x y x y
Rb

Find equivalent resistance with respect

to two terminals, x-z, z-y and x-y for both
networks.
ELCIAN1 - p8
Proof
The equivalent resistances as seen from
the respective terminals are:
Terminal x - y 1 Terminal z - y 2
R b // R a  R c   R 1  R 2 R c // R a  R b   R 2  R 3
R b R a  R c  R c R a  R b 
R R R R
Ra  Rb  Rc Ra  Rb  Rc
1 2 2 3

Terminal z - x 3
R a // R b  R c   R 1  R 3
R a R b  R c 
 R1  R 3
Ra  Rb  Rc
ELCIAN1 - p9
Proof

Subtract Eq. (2) from Eq. (1)

R b R a  R c  R c R a  R b 
  R 1  R 2  R 2  R 3 
Ra  Rb  Rc Ra  Rb  Rc
R b R a  R c   R c R a  R b 
 R1  R 3
Ra  Rb  Rc
RaRb  RaRc
 R1  R 3 4
Ra  Rb  Rc

ELCIAN1 - p10
Proof

Add Eq. (3) and Eq. (4)

RaRb  RaRc RaRb  RaRc
  R1  R 3  R1  R 3
Ra  Rb  Rc Ra  Rb  Rc
2R a R b
 2R 1
Ra  Rb  Rc Follow these steps
RaRb to verify the results
R1 
Ra  Rb  Rc shown in the slide 6.

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Example: Find the equivalent resistance across
terminals AB.

4 12 5
A
3 1
9 10
2 1.5 3
B
5
Starting from the right, we
get for resistors in series
10
3
R eq 1  5  10  3  18 

ELCIAN1 - p12
Req1 is in parallel with the 9-resistor.

(18)(9)
R eq 2  6 9 Req1=18
18  9

The resulting network becomes

4 12
A
3 1
Req2=6

2 1.5
B
ELCIAN1 - p13
 Ra
Convert wye
3 1
into delta Rc Rb
1.5
9
Rb   3 
(3)(1)  (1)(1.5)  (1.5)(3) 9 3
Ra   6
1.5 1.5 9
Rc   9 
1
Replace the wye with its
delta equivalent and simplify. 12
4
We get A
6
R eq 3  12 // 6  4  9 3 6
R eq 4  3 // 6  2  B 2

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 Req3=4
4
Re-draw the network A
and simplify further.
9 Req4=2
R eq 5  4  2  6 B 2

Req5 is in parallel with 4

A
the 9-resistor.
9 Req5=6
R eq 6  9 // 6  3.6  B 2

4
Finally, we get A
Req6=3.6
RAB = 4+3.6+2 = 9.6
B
2
ELCIAN1 - p15
 Linear Element
A linear element is a passive element whose
voltage-current relationship is described by a linear
equation; i.e. if the current through the element is
multiplied by a constant k, then the voltage across
the element is likewise multiplied by k.

Consider a resistor R with current R

i=i1. From Ohm’s Law, we get i + v -
v  v1  Ri1
Suppose the current is increased by a factor k; i.e.
i2=ki1. The new voltage is
v2  Ri2  Rki1  kv1 (R is a linear element)
ELCIAN1 - p16
Consider an inductor L with current L
i=i1. The inductor voltage is: i + v -
di1
v  v1  L
dt
Suppose the current is increased by a factor k; i.e.
i2=ki1. The new voltage is:
di2 d(ki1 )
v2  L L  kv1
dt dt
Thus, L is a linear element.

ELCIAN1 - p17
Consider a capacitor C with current C
v=v1. The capacitor current is:
dv1 i + v -
i  i1  C
dt
Suppose the voltage is increased by a factor k; i.e.
v2=kv1. The new current is:
dv 2 d(kv1 )
i2  C C  ki1
dt dt
Thus, C is a linear element.

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 Linear Dependent Source
A linear dependent source is a current or voltage
source whose output current or voltage is
proportional only to the first power of some
current or voltage variable in the circuit, or to the
sum of such quantities.

+ +
kvx kiy kvw kiz
- -

Linear Dependent Linear Dependent

Voltage Sources Current Sources
ELCIAN1 - p19
 Linear Electric Circuit
An electric circuit is linear if it consists of
 independent sources
 linear dependent sources
 linear elements

The response of a linear circuit is proportional to the

sources; that is, if all independent sources are
multiplied by a constant k, all currents and voltages
will likewise increase by the same factor k.

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 Principle of Superposition
In an electric circuit containing N independent
sources, the current (or voltage) in any branch is
equal to the algebraic sum of N components, each
of which is due to one independent source acting
alone.

Note: Reducing an independent source to zero:

1. For a voltage source, remove the source and
replace with a short circuit;
2. For a current source, remove the source and
replace with an open circuit.

ELCIAN1 - p21
Example: Find I1 and R1 +Vx
I2 using nodal analysis.
+ I1
From KCL, we get Vs R2 I2 Is
-
Vx  Vs Vx
Is  
R1 R2 REF

which can be simplified as

1 1 1
Vx     Is  Vs
 R1 R 2  R1
or
R1R 2 R2
Vx  Is  Vs
R1  R 2 R1  R 2
ELCIAN1 - p22
Thus, we get
Vs  Vx
I1 
R1
1 R2
I1  Vs  Is
R1  R 2 R1  R 2
and
Vx
I2 
R2
1 R1
I2  Vs  Is
R1  R 2 R1  R 2
ELCIAN1 - p23
Solution using Superposition:
R1
Assume Vs is acting alone.
Replace Is by an open circuit. + I1a
Vs R2 I2a
1
I1a  I2a  Vs -
R1  R 2
Next, assume Is is acting alone. Replace Vs with a
short circuit. We get
R1
R2
I1b  Is
R1  R 2 I1b
R2 I2b Is
R1
I2b  Is
R1  R 2
ELCIAN1 - p24
Finally, we apply superposition to get I1 and I2.

I1  I1a  I1b
I2  I2a  I2b
Substitution gives

1 R2
I1  Vs  Is
R1  R 2 R1  R 2

1 R1
I2  Vs  Is
R1  R 2 R1  R 2
ELCIAN1 - p25
 6A

Example :
Find all currents using
superposition. 6 12

+
81V 9A 3
-

6

We have three independent sources.

Solve for the currents with each source
acting alone.

ELCIAN1 - p26
Consider the 81-V source 6 12
acting alone.
+
81 3
Ia  3A 81V Ia
27 -

6
Consider the 9-A source
acting alone. 6 12
15
Ib  (9A)  5 A
15  12 Ib 9A Ic 3

Ic  9  Ib  4 A
6
ELCIAN1 - p27
 6A
Finally, consider the 6-A
source acting alone.

9 6 12
Id  (6A)  2 A
9  18 Id
Ie 3
Ie  6  Id  4 A

Apply superposition to get 6

the current in any resistor.
For example, the current in the 3 resistor is

I3  Ia  Ic  Ie  3  4  4  11 A 
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 4V
Example: Use + - 3
superposition to + 1 I
find the current I. Vx 2
+
5Vx
-
- 2A

Consider the 4-V source acting alone. From KVL,

we get
4V
 4  3Ia  5Vx  2Ia + - 3

Vx  2Ia + Ia +
Vx 2 - 5Vx
which gives -

Ia  0.8 A
ELCIAN1 - p29
Consider next the 2-A 3
source acting alone. + 1 Ib +
We get Vx 2 - 5Vx
- 2A
Vx  2(2 - Ib )
and

Vx  3Ib  5Vx

Solving simultaneously, we find Ib=3.2 Amps.

Applying superposition, we get

I  Ia  Ib  4 A
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 1.5A
Example: Find the voltage
Vx using the principle of
superposition. 20 10V
+ -
+ - Vx +
16V 3A 80
-
Consider the 16-V source
acting alone. Using voltage
division, we get 20
20 + - Vx1 +
Vx1   (16V) 16V 80
20  80 -
 3.2 V
ELCIAN1 - p31
 20
Consider next the 3-A
source acting alone. - Vx2 +
80 Ix 3A 80
Ix  (3 A)
20  80
 2.4 A

From Ohm’s Law, we get Vx 2  20 I x  48 V

Next, consider the 10-V
source acting alone. 20 10V
+ -
20
Vx3  (10 V) - Vx3 +
20  80 80
2V
ELCIAN1 - p32
Finally, consider the 1.5-A 1.5A
source acting alone.
Because of the short
20
circuit, we get
- Vx4 +
Vx4  0 80

Applying superposition, we get

Vx  Vx1  Vx2  Vx3  Vx4

 3.2  48  2  0  46.8 V
ELCIAN1 - p33
 2
Example: Calculate I
10A
using superposition.
1
Consider the 20 volt
source alone. 3
5A I 4

2 + -
20V
1 3
Req  4  2 //(1  3)  5.33 

20V
I1 4 20
+ - I1   3.75 A
R eq

ELCIAN1 - p34
Consider the 5 ampere 2
source alone.
1 +Va 3
Node a: +Vb
Va Va  Vb
5   5A I2 4
1 3
Node b: REF
Vb Vb Vb  Va
0  
2 4 3
Solving simultaneously, we get Vb=-1.25 volts.
Thus
Vb
I2   0.3125 A
4
ELCIAN1 - p35
Consider the 10 ampere 2 10A
source alone.
Node a: 1 3
+Vb
Va Va  Vb +Va
 10   I3 4
1 3
Node b: REF
Vb Vb Vb  Va
10   
2 4 3
Solving simultaneously, we get Vb=7.5 volts and
I3=1.875 Amps. Finally, we get

I  I1  I2  I3  5.3125 A
ELCIAN1 - p36
Source Transformation
R
a a
+ + +
I I
Vs V Is R V
- - -
b b
From KVL, From KCL,
V
Vs  RI  V Is   I or RIs  RI  V
R
If the two networks are equivalent with respect to
terminals ab, then V and I must be identical for
both networks. Thus
Vs
Vs  RIs or Is 
R
ELCIAN1 - p37
Nodal Analysis
General Procedure
1. Label all nodes in the circuit. Arbitrarily select
any node as reference (GND).
2. Define a voltage variable from every remaining
node to the reference. These voltage variables
must be defined as voltage rises with respect
to the reference node.
3. Write a KCL equation for every node except
the reference.
4. Solve the resulting system of equations.
ELCIAN1 - p38
Example: Find the voltage VX using nodal analysis.

40

30  +Vb 10 + Vc
+Va
+ Vx -
4.8V + 15 20 0.2A

REF

For node a, the voltage of the node is dictated

by the voltage source. Thus, Va=4.8 Volts.

ELCIAN1 - p39
The KCL equations for nodes b and c are
Vb  4.8 Vb Vb  Vc
node b: 0  
30 15 10
Vc  Vb Vc  4.8 Vc
node c: 0.2   
10 40 20
Solving simultaneously, we get
Vb = 2.4V Vc = 3.2V
Finally, we get the voltage Vx
Vx = 4.8 - Vb = 2.4V
ELCIAN1 - p40
Example: Find the voltages Va, Vb and Vc using
nodal analysis (a voltage source between 2 nodes).
8

6 +Vb + 6V - + Vc
+Va

3A 3 4 5A

REF

The KCL equations for node a and the supernode

Va  Vb Va  Vc
node a: 3  
6 8
ELCIAN1 - p41
 Vb Vc Vb  Va Vc  Va
supernode: 5    
3 4 6 8
For the voltage source, we get Vb-Vc=6 volts.

The equations can be simplified into

72  7Va  4Vb  3Vc

6  Vb  Vc
120  7Va  12Vb  9Vc
Solving simultaneously, we get

Va  24 V Vb  16.3 V Vc  10.3 V
ELCIAN1 - p42
Example: Find the voltages Va, Vb and Vc using
nodal analysis (dependent voltage source between
two nodes).
v
+ x -
8
6 + Vb + Vc
+Va - +

3A 3 2vx 4 5A

REF
The KCL equations for node a and the supernode
Va  Vb Va  Vc
node a: 3  
6 8
ELCIAN1 - p43
 Vb Vc Vb  Va Vc  Va
supernode: 5    
3 4 6 8
For the dependent voltage source, we get
Vc  Vb  2vx  2(Va  Vc )
The equations can be simplified into
72  7Va  4Vb  3Vc
0  2Va  Vb  3Vc
120  7Va  12Vb  9Vc
Solving simultaneously, we get

Va  24 V Vb  9.6 V Vc  19.2 V
ELCIAN1 - p44
 Mesh Analysis
General Procedure
1. Count the number of “window panes” in the
circuit. Assign a mesh current to each window
pane.

2. Write a KVL equation for every mesh whose

current is unknown.

3. Solve the resulting equations.

Mesh - a loop that does not contain an inner loop.

ELCIAN1 - p45
 I1+
16 Vx 40 40
Example: Find the 2V -
voltage VX using I2

+
_
+
mesh analysis.
1A I3 20 - 5V

30
The KVL equations for meshes 1 and 2 are
Mesh 1: -2 = 40(I1- I2) + 16I1
Mesh 2: 5 = 40I2 + 40(I2 -I1) + 20(I2- I3)
In mesh 3, the current source dictates the value of
the mesh current. Thus, I3=1 A.
ELCIAN1 - p46
The two equations can be simplified into

-2 = 56I1 - 40I2

25 = -40I1 + 100I2

Solving simultaneously, we get

I1 = 0.2A I2 = 0.33A

Finally, we get the voltage Vx

Vx = 40(I2 - I1) = 5.2V

ELCIAN1 - p47
Example: Find the 5V
5 + -
currents I1, I2 and
I3 using mesh I2
analysis (current 1 3
source between
+
two meshes).
36V I1 3A I3 2
-

4
We cannot write a KVL equation for mesh 1 or for
mesh 3 because of the current source. Form a
supermesh and write a KVL equation for it.

supermesh: 36  1(I1  I2 )  3(I3  I2 )  2I3  4I1

ELCIAN1 - p48
The KVL equation for mesh 2 is unchanged.

 5  5I 2  3( I 2  I 3 )  1( I 2  I1 )

The third equation is dictated by the current source.

I1  I3  3 A

Solving simultaneously, we get

I1  5.45 A I2  0.86 A I3  2.45 A

ELCIAN1 - p49
Example: Find the currents I1, I2 and I3 using
mesh analysis
(dependent source I2
1 2
included). 3
15A I1 1 + vx -
v
9 x
I3 1
2

The current in mesh 1 is dictated by the current

source. Thus, I1=15 Amps.
The KVL equation for mesh 2 is

0  2I2  3(I2  I3 )  1(I2  I1 )

ELCIAN1 - p50
We cannot write a KVL equation for mesh 3. Can’t
form a supermesh either. However, we can write an
equation for the dependent source.

1 1
I3  I1  vx  [ 3 ( I3  I2 )]
9 9

Solving simultaneously, we get

I1  15 A I2  11 A I3  17 A

ELCIAN1 - p51
Example: Using + I0
mesh analysis, find i1 i2
6V 2k 4k
I0. -
2k
Mesh 2:
2mA
1mA
0  4kI2  2k(I2  I4 ) 6k i3 i4
2k

 2k(I2  I1 )
Supermesh: 6  2k(I1  I2 )  2k(I4  I2 )
 2kI4  6kI3
Current Source: 2 mA  I3  I1
Current Source: 1 mA  I3  I4

ELCIAN1 - p52
Solving simultaneously, we get

I1  0.9 mA
I2  0.2 mA
I3  1.1 mA
I4  0.1 mA
Thus
I0  I1  I2  0.9  0.2  0.7 mA

ELCIAN1 - p53
 Choice of Method
Given the choice, which method should be used?
Nodal analysis or mesh analysis?

Nodal analysis: The number of voltage variables

equals the number of nodes minus one. Every
voltage source connected to the reference node
reduces the number of unknowns by one.
Mesh Analysis: The number of current variables
equals the number of meshes. Every current
source in a mesh reduces the number of
unknowns by one.

Note: Choose the method with less unknowns.

ELCIAN1 - p54
 10
Example: Write the nodal
and mesh equations that
describe the circuit shown. 4A
+Va +Vb + Vc
+ Vd
We need 4 4 6 8
voltage 3A 2 5 5A
variables.
REF
The nodal equations are
Va Va  Vb
node a: 3 
2 4
Vb  Va Vb Vb  Vc Vb  Vd
node b:  4    
4 5 6 10
ELCIAN1 - p55
 Vc  Vb Vc  Vd
node c: 4 
6 8
Vd  Vb Vd  Vc
node d:  5  
10 8
10
There are 5 meshes but the
3A and 5A current sources 4A
flow in distinct meshes. We I3
need to define 3 current I2
variables.
4 6 8
3A 3A 2 I1 5 5A 5A

ELCIAN1 - p56
The mesh equations are

mesh 1: 0  2(I1  3)  4I1  5(I1  5)

supermesh: 0  6(I2  5)  10I3  8(I3  5)
4A source: 4  I2  I3

Note: We need either three current variables or

four voltage variables to describe the circuit. It is
preferable to use mesh analysis.

ELCIAN1 - p57
 Example 1
Determine the nodal voltages of the circuit
below.

ELCIAN1 - p58
Using KCL :
@ Supernode 1 - 2
v3  v2 v v v
 10  1 4  1  5v1  v2  v3  2v4  60 (1)
6 3 2
@ Supernode 3 - 4
v3  v2 v v v
 10  1 4  1  4v1  2v2  5v3  16v4  0 (2)
6 3 2
ELCIAN1 - p59
Using KVL :
@ Loop 1
 v1  20  v2  0  v1  v2  20 (3)
@ Loop 2
 v3  3vx  v4  0
But,
vx  v1  v4
Equation (4) becomes :
3v1  v3  2v4  0 (4)ELCIAN1 - p60
The four important equations are :
5v1  v2  v3  2v4  60 Solving the simultaneo us equations, we will get :
1
4v1  2v2  5v3  16v4  0  v1   5 1  1  2   60
v   4 2  5  16  0
v1  v2  20  2     
v3   1  1 0 0  20
3v1  v3  2 v4  0        
 v4    3 0  1  2   0
In matrix form :
 v1   26.67 
5 1  1  2   v1  60 v   6.67 
4 2  5  16 v   0   2   
  2     v3   173.33 
   
1  1 0 0  v3  20  4 
v  46. 67 
    
 3 0  1  2  v4   0 

ELCIAN1 - p61
 Example 2

Find the currents I1, I2, and I3, then find the
currents through the resistors.

ELCIAN1 - p62
Circuit Analysis Theory and Practice by A. H. Robbins and W. Miller
ELCIAN1 - p63
 Example 3

Determine the nodal voltages of the circuit

below.

ELCIAN1 - p64
Circuit Analysis Theory and Practice by A. H. Robbins and W. Miller
ELCIAN1 - p65
ELCIAN1 - p66
Problem Solving
If R5=1k, find the voltage across R5.
What value of R5 will make the voltage
across it equal to zero? Use KVL.

ELCIAN1 - p67
Problem Solving
 For the given figure below, answer the following.
Resistor values in ohms.
 Use nodal analysis to determine V1, V2, and V3.
 Use mesh analysis to determine IS and Ix.

- V1 +
8A
IS
14  4
Ix
+ +
24  V2 8
100 V
V3
- 10 A
-

ELCIAN1 - p68
Problem Solving
 Set up the node
equations and solve for
6
the node voltages VA, 8A
1
VB, and VC. Use node
D as reference node.
4 7

B
What is Vx and the A C
+ Vx -
power delivered by the
voltage source? 8 0.75Vx 60 V

 Confirm your answers in

item b using mesh
D
analysis.

ELCIAN1 - p69
Problem Solving
 Find the maximum power that can be delivered to the
resistor R below. Vs1
R4 5V
2

+
R3 R
3

+
Vs2 R1 Is1
10V 5 3A
-

ELCIAN1 - p70
Problem Solving
 Given the resistor network below, find the equivalent
resistance Rab, if all the resistors equal 10 ohms.

ELCIAN1 - p71
Problem Solving
 Use source transformation to find V1 and V2.
V1 V2

R3 R R2
10 120 20

+ +
Vs2 R1 Is1 Vs1
25V 5 4A 5V
- -

ELCIAN1 - p72
Problem Solving
Use superposition to solve for the current
through the 4-ohm resistor.
Is1
8A

R3 R R2
4 3 2

+
Vs2 R1 R4 Is2
24V 10 5 4A
-

ELCIAN1 - p73
Problem Solving
 Use superposition to solve for the voltage across
the 10-ohm resistor.
R4
5

Is1
8A

R3 R
4 3

+
Vs2 Is2 R1
24V 4A 10
-

ELCIAN1 - p74