Transcript ES100_Lecture 3
Electrical Systems 100
Lecture 3 (Network Theorems) Dr Kelvin
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Contents
•
Superposition Theorem • Thevenin’s Theorem • Norton’s Theorem • Maximum Power Transfer Theorem • Millman’s Theorem • Reciprocity Theorem
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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
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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.
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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
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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
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Dependent Source
(a) Dependent Voltage Source A voltage source whose parameters are controlled by voltage/current else where in the system
v = µVx v = ρix
VDVS (Voltage Dependent Voltage source) CDVS (Current Dependent Voltage source) (b) Dependent Current Source A voltage source whose parameters are controlled by voltage/current else where in the system
v = αVx
VDCS (Voltage Dependent Current source)
v = βix
CDCS (Current Dependent Current source)
For Superposition, All dependent sources must be left intact!!
You can’t apply O/C and S/C on dependent sources
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An Example
Find i 0 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
i
0
i
0 '
i
0 ''
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To obtain i ’ 0 we short circuit the 20V sources i For loop 1 1 4 A.
For loop 2 3 (
i
2
i
1 ) 2
i
2 5
i
' 0 1 (
i
2
i
3 ) 0 For loop 3 5 (
i
3
i
1 ) 1 (
i
3
i
2 ) 5
i
0 ' 4
i
3 0 i' 0 i 1 i 3 For solving i 1 , i 2 , i 3 i ' 0 52 A 17 i 1 i 2 i 3
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To obtain i ’’ 0 , we open circuit the 4A sources For loop 4 6i 4 i 5 5i 0 '' 0 For loop 5 i 4 10
i
5 20 5
i
0 '' 0
i
0 ''
i
5 For solving i 4 i 0 '' 60 17 A and i 5 i 4 i 5 Therefore, i 0 i ' 0 i 0 '' 52 17 60 17 8 17 A
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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
I
1
I
2
The power
delivered to the circuits are P
P T
1
I
1 2
R
, P 2 (
I
1
I
2 ) 2
I
2 2
R
,
and
P T
R
I
1 2
R
I
2 2
R I T
2
R
2
I
1
I
2
R
P
1
P
2
I
1 2
R
I
2 2
R
The total power must be determined using the total current not by superposition
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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 V TH and R TH as shown:
V OC V OC 12
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 V TH in series with a resistor R TH where V TH is the open circuit voltage at the terminals a-b and R TH 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)
R th V th R th =8Ω V th =20V
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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 R TH can be determined as: Case 1: Applying a known voltage source v 0 terminals. The R TH is given by v o /i 0 .
and measuring i 0 at the Case 2: Applying a known current source i 0 and measuring v 0 and then R Th is given by v 0 /i 0
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Thevenin’s Theorem
The load current can then be obtained as:
I L
R TH V TH
R L V L
R TH R L
R L V TH
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Thevenin’s Theorem Example 1 Find the Thevenin equivalent circuit of the shaded area in the bridge network shown below. 1.Calculate V Th: • Calculate the open circuit voltage across terminal
a b
2.Calculate R TH : • 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 V Th is the open circuit voltage across a and b.
V Th is calculated as:
V
1
R
1
R
1
E
R
3 6 * 72 6 3
V
2
R
2
R
2
E R
4 12 12 * 72 4 48
V
54
V
Applying KVL we get,
V Th or
,
V
1 V Th
V
2 V 2 0
V
1 54 48 6 V
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Thevenin’s Theorem Finding R Th: Short circuiting the voltage source we get the R Th as:
R Th
R a
b R
1 // 6 //
R
3 3
R
2 4 //
R
4 // 12 5
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Thevenin’s Theorem-An Example V Th R TH 6 V 5
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Thevenin’s Theorem
Example 2 Find the Thevenin equivalent circuit with respect to the terminal a and b.
Finding R Th : Applying test voltage
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Thevenin’s Theorem
Finding R Th :
I T
All independent sources set to zero Apply the test voltage V T Using Node 3i x V T 2 I T i x I T 0 4i x V T Eq(1) 2 i x V T Eq(2) 8 Substitute Eq2 into Eq1 I T 4 V T 8 V T 2 R TH V T I T 1Ω V T
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Finding V Th , Open circuit i V Th 8 x 4 3i x V Th 2 24 0 V Th 8 Substituting i x into the first equation V Th = 8V V Th
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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 I N and a parallel resistance R N
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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 I N 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 R N 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 R N = R Th
• Calculate I N by returning all sources to their original positions and then finding current through the short circuited terminals a b as mentioned before.
• Draw the Norton equivalent circuit with I N as current source and R N 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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Finding R N
:
Norton’s Theorem-An Example
R N
R
1
R
2 4 6 2 .
4 Finding I N :
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Norton’s Theorem-An Example
Using Superposition Theorem, I N
'
E
1
R
1 7 V 4 1 .
75 A Now find the contribution to I N Looking at circuit below,
I
''
N
from the current source :
I
8 A
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Norton’s Theorem-An Example
The Norton equivalent circuit is then: I N
I
''
N
I
'
N
8 A 1.75
A and R N 2 .
4 6.25
A
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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 R TH 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
Proof:
Maximum Power Transfer Theorem
P L
i
2
R L
(
R Th V Th
R L
) 2
R L Taking derivative
w.r.to
R L we get, dP L dR L
V
2
Th
{ (
R Th
R L
( ) 2
R Th
2
R L R L
( ) 4
R Th V Th
2 { (
R Th
(
R Th R L
R L
2
R L
) 3 ) 0
This
imply that
Therefore
, R Th (R Th
R L
R L
) 0
R L
)
P
max
V
2
Th
4
R Th
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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?
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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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