Electrical Systems 100
Lecture 3
(Network Theorems)
Dr Kelvin
1
Contents
•
Superposition Theorem
• Thevenin’s Theorem
• Norton’s Theorem
• Maximum Power Transfer Theorem
• Millman’s Theorem
• Reciprocity Theorem
2
Superposition Theorem
The Superposition theorem is very helpful in
determining the voltage across an element or current
through a branch when the circuit contains multiple
number of voltage or current sources
One big advantage is that we do not have to use
Cramer’s rule or complicated mathematical operations
but simply algebraically adding solutions obtained from
analysing the network with one source activated at a
time
3
Superposition Theorem
The superposition theorem states that:
“The current through, or voltage across, an
element in a linear bilateral network equal to the
algebraic sum of the currents or voltages
produced independently by each source”
In general number of networks to be analysed is equal to the
number of sources; however, it may be possible to treat the effect
of two sources at a time to reduce the number of network to be
analysed.
4
Superposition Theorem
In removing voltage sources from the network, the voltage source
is replaced by a short circuit (potential difference between the two
points set to zero)
In removing a current source from the network, the current source
is replaced by an open circuit between the two points (current set to
zero)
In doing so, the internal resistance of the voltage sources and shunt
resistance of current sources are preserved in the network as it was
in the original network.
All dependent sources must be left intact as they are controlled by
circuit variables
5
Removing the effect of ideal sources
Voltage source is
replaced by a S/C
Current source is
replaced by a O/C
Removing the effect of practical sources
6
Dependent Source
(a) Dependent Voltage Source
A voltage source whose parameters are controlled by voltage/current
else where in the system
v = ρix
v = µVx
CDVS
VDVS
(Current Dependent
(Voltage Dependent
Voltage source)
Voltage source)
(b) Dependent Current Source
A voltage source whose parameters are controlled by voltage/current
else where in the system
v = βix
v = αVx
CDCS
VDCS
(Current Dependent
(Voltage Dependent
Current source)
Current source)
For Superposition, All dependent sources must be left intact!!
You can’t apply O/C and S/C on dependent sources
7
An Example
Find i0 in the circuit shown below. The circuit involves a
dependent source. The current may be obtained as by using
superposition as :
i’0 is current due to 4A current source
i’’0 is current due to 20V voltage source
i0  i0'  i0''
8
i’0we
To obtain
20V sources
i2
short circuit the
i1
For loop 1
i1  4 A.
i3
For loop 2
3(i2  i1 )  2i2  5i '0 1(i2  i3 )  0
For loop 3
5(i3  i1 )  1(i3  i2 )  5i0'  4i3  0
i'0  i1  i3
For solving i1, i2, i3
52
i0 
A
17
'
9
To obtain i’’0 , we open circuit the 4A sources
For loop 4
6i4  i5  5i  0
''
0
i4
For loop 5
- i 4  10i5  20  5i0''  0
i5
i0''  i5
For solving i4 and i5
i '0'  
60
A
17
T herefore,i 0  i '0  i '0'
52 60


17 17
8
 A
17
10
Superposition is not applicable to Power
The superposition theorem does not apply to power calculations as the power is
proportional to current squared or voltage squared. Consider the following :
I T  I1  I 2
The power delivered to thecircuitsare
P1  I12 R, P2  I 22 R, and PT  I T2 R
PT  ( I1  I 2 ) 2 R  I12 R  I 22 R  2 I1 I 2 R
 P1  P2  I12 R  I 22 R
The total power must be determined using the total current
not by superposition
11
Thevenin’s Theorem
It often occurs in practice that a particular element in a circuit is variable while
the rest is fixed. Consider the household GPO which may be connected to
various appliances. Each time a different appliance is connected the entire circuit
may be required to analyse. To avoid this , Thevenin’s theorem gives a technique
where the fixed part of the circuit is represented by an equivalent circuit VTH and
RTH as shown:
VOC
VOC
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Thevenin’s Theorem
Thevenin’s theorem states that a linear two terminal circuit can be replaced by an
equivalent circuit consisting of a voltage source VTH in series with a resistor RTH where
VTH is the open circuit voltage at the terminals a-b and RTH is the input or equivalent
resistance looking at the terminals when all independent sources in the network are
turned off (Voltage sources set to zero and current sources are open circuited)
Rth
Rth=8Ω
Vth=20V
Vth
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Thevenin’s Theorem
If the circuit has dependent sources, then we need to turn off all
independent sources but not the dependent sources like
superposition theorem. In this case RTH can be determined as:
Case 1: Applying a known voltage source v0 and measuring i0 at the
terminals. The RTH is given by vo/i0.
Case 2: Applying a known current source i0 and measuring v0 and then RTh is
given by v0/i0
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Thevenin’s Theorem
The load current can then be obtained as:
VTH
IL 
RTH  RL
RL
VL 
VTH
RTH  RL
15
Thevenin’s Theorem
Example 1
Find the Thevenin equivalent circuit
of the shaded area in the bridge
network shown below.
1.Calculate VTh:
•
Calculate the open circuit voltage across terminal
ab
2.Calculate RTH:
•
Open circuit the current source and short circuit
the voltage source
•
Calculate the total resistor across terminal a b
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Thevenin’s Theorem-An Example
VTh is the open circuit voltage across a and b.
VTh is calculated as:
R1 E
6 * 72
V1 

 48V
R1  R3 6  3
R2 E
12 * 72
V2 

 54V
R2  R4 12  4
Applying KVL we get,
VTh  V1  V2  0
or, VTh  V2  V1  54  48  6 V
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Thevenin’s Theorem
Finding RTh:
Short circuiting the voltage source we get the RTh as:
RTh  Ra b
 R1 // R3  R 2 // R4
 6 // 3  4 // 12
 5
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Thevenin’s Theorem-An Example
VTh  6 V
R TH  5
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Thevenin’s Theorem
Example 2
Find the Thevenin equivalent circuit with respect to
the terminal a and b.
Finding RTh:
Applying test voltage
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Thevenin’s Theorem
Finding RTh:
IT
All independent sources set to
zero
Apply the test voltage VT
Using Node
Substitute Eq2 into Eq1
VT
 i x  IT  0
2
V
I T  4ix  T
Eq(1)
2
3ix 
ix 
VT
8
 VT  VT
I T  4
 VT

 8  2
VT
 R TH 
 1Ω
IT
Eq(2)
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Finding VTh, Open circuit
VTh
VTh
VTh  24
 4  3ix 
0
8
2
VTh
ix 
8
Substituting ix into the first equation
VTh = 8V
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Norton’s Theorem
We have seen earlier that every voltage source with an internal
resistance has a current source equivalent. The current source
equivalent of the Thevenin’s equivalent network is the Norton’s
equivalent network and is determined by Norton’s Theorem.
Norton’s Theorem states that:
Any two terminal linear bilateral dc network can be replaced by
an equivalent circuit consisting of a current source IN and a
parallel resistance RN
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Norton’s Theorem
Figure below show a Linear two terminal network and its Norton’s
equivalent. In the Figure it is the terminals a-b across which the
Norton equivalent is to be found.
Is/c
IN
Is/c
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Norton’s Theorem
Steps to determine Norton’s equivalent:
• Remove the portion of the network across which Norton
equivalent is to be found
• Calculate RN by setting all voltage sources to zero and current
sources to open circuit but keeping all internal series and shunt
resistances intact. Keep all dependent sources in the circuit like
superposition theorem as well. You will, note that RN = RTh
• Calculate IN by returning all sources to their original positions
and then finding current through the short circuited terminals ab as mentioned before.
• Draw the Norton equivalent circuit with IN as current source
and RN as parallel resistor and the portion of the circuit returned
between the terminals a-b.
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Norton’s Theorem-An Example
Find the Norton equivalent circuit for the portion of the
network to the left of a-b in Figure given below?
Identifying the terminals of interest for Norton’s equivalent
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Norton’s Theorem-An Example
Finding RN :
RN  R1 R2  4 6  2.4
Finding IN:
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Norton’s Theorem-An Example
Using Superposition Theorem,
E1 7 V
I 

 1.75 A
R1 4 
'
N
Now find the contribution to IN from the current source :
Looking at circuit below,
I  I 8A
''
N
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Norton’s Theorem-An Example
The Norton equivalent circuit is then:
IN  I  I
''
N
'
N
 8 A  1.75A  6.25A
and R N  2.4 
29
Maximum Power Transfer Theorem
The maximum power transfer theorem states that:
A Load will receive maximum power from a linear bilateral dc
network when its total resistance value is exactly equal to the
RTH of the network
Maximum power transfer is extremely important for maximum efficiency of a
transmission and distribution network of an electric utility such as Western Power.
The theorem also find application in electronic circuits such as matching input
impedance of a speaker system to the output impedance of an amplifier.
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Maximum Power Transfer
Power is max when RL = RS
RL = RS
31
Maximum Power Transfer Theorem
Proof:
VTh
PL  i RL  (
) 2 RL
RTh  RL
2
Taking derivativew.r.toR L we get,
2
(
R

R
)
 2 RL ( RTh  RL )
dPL
L
 VTh2 { Th
dR L
( RTh  RL ) 4
( RTh  RL  2 RL )
V {
0
3
( RTh  RL )
2
Th
This imply that(R Th  RL )  0
Therefore, R Th  RL
V 2Th
Pmax 
4 RTh
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Maximum Power Transfer Theorem
Designing a Speaker System for your Amplifier:
Consider an amplifier and a speaker and their equivalent
circuit as below:
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Maximum Power Transfer Theorem
In the first circuit the power delivered to the speaker is 4.5
Watts
In the second circuit, the power delivered to each speaker is 2
Watts
In the third circuit power delivered to each speaker is also 2
Watts! Which one is better arrangement?
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Maximum Power Transfer Theorem
In the third circuit power delivered to each speaker is also 2
Watts! Which one is better arrangement?
35
Maximum Power Transfer Theorem
Speakers are also available with 4 Ohm and 16 Ohm input
impedance. You can use them to obtain an input impedance
equal to 8 ohm to match the amplifier’s output impedance as
below:
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