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Topics to be DiscussedTopics to be Discussed
Superposition Theorem. Thevenins Theorem.
Nortons Theorem.
Maximum Power Transfer Theorem. Maximum Power Transfer Theorem for AC
Circuits.
Millmans Theorem.
Reciprocity Theorem.
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Network Theorems
Some special techniques, known as network
theorems and network reduction methods, have
been developed.
These drastically reduce the labour needed tosolve a network.
These also provide simple conclusions and good
insight into the problems.
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SuperpositionSuperposition
PrinciplePrinciple
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Superposition Theorem
The response (current orvoltage) in a linearnetwork
at any point due to multiple sources (current and/or
emf) (including linear dependent sources),
can be calculated by summing the effects of eachsource considered separately,
all other sources turned OFF or made
inoperative.
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Turning off the sources
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Linear Dependent Source
It is a source whose output current or voltage isproportional only to the first power of some
current or voltage variable in the network or to the
sum of such quantities.
Examples :
linear.notis6.06.0but,
linear,is166.0
21
2
1
21
viv
oriv
viv
s
s
s
!
!
!
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Application
Problem : Consider two 1-V batteries in
series with a 1- resistor. Let us apply the
principle of superposition, and find the
power delivered by both the batteries.
Solutions : Power delivered by only onesource working at a time is
P1
= 1 W
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Therefore, the power delivered by both thesources,
P= 2P1
= 2 W
The above answer is obviously wrong,
because it is a wrong application of
the superposition theorem.
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Example 1
Find the current Iin the network given,using the superposition theorem.
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Solution :
A0.375!
!
v!
4.0
15.0
3.01.0
3.05.01I
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A0.1
A0.2
!!@!
v
!
21
3
2
I
3.01.0
1080
II
I
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Example 2
Using superposition theorem, find current ix in thenetwork given.
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Solution :
A05.015050
101 !
!i
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A3015050
150402 !v!i
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A3015050
501203 !
v!i
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A0.05!
!
!
303005.0
321 iiiix
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BenchmarkExample 3
Find voltage v across 3- resistor by applyingthe principle of superposition.
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Solution :
Using current divider,
A
3
2
)32(1
14 !
v!i
V2.0)(3A)(2/34
!v!v!@ Riv
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Using current-divider, the voltage v5 across 3-
5
15 A (3 ) 2.5V1 (2 3)
v ! v v ; ! -
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By voltage divider,
V3.0321
366 !v!v
4 5 6 2.0 2.5 3.0v v v v@ ! ! ! 2.5 V
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Find current i2 across R2 resistor by applying the
principle of superposition. Where R1=R2=R3=1-
and S=10 , b= 5 , = .
Example 4
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Thevenins Theorem
It was first proposed by a French telegraph
engineer, M.L. Thevenin in 1883.
There also exists an earlier statement of the
theorem credited to Helmholtz.
Hence it is also known as Helmholtz-Thevenin
Theorem.
It is useful when we wish to find the responseonly in a single resistance in a big network.
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Thevenins Theorem
Any two terminals AB of a network
composed of linear passive and active
elements may by replaced by a simple
equivalent circuit consisting of
1. an equivalent voltage source Voc,and
2. an equivalent resistance th in series.
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The voltage Voc is equal to the potential
difference between the two terminals AB causedby the active network with no externalresistance connected to these terminals.
The series resistance Rth is the equivalentresistance looking back into the network at theterminals AB with all the sources within the
network made inactive, ordead.
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Illustrative Example 3
Using Thevenins theorem, find the current inresistorR2 of 2 .
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Solution :
1. Designate the resistor R2
as load.
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2. Pull out the load resistor and enclose the remaining
network within a dotted box.
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3. Temporarily remove the load resistor R2, leaving the
terminals A and B open .
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4. Find the open-circuit voltage across the terminals A-
B,
11.2!v!
!!
!
12.47
A;4.25
21
14
728
ABV
I
5. This is called Thevenin voltage, VTh = VAB = 11.2 .
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6. Turn OFF all the sources in the circuit
Find the resistance between terminals A and B. This is
the Thevenin resistance, RTh. Thus,
1 41 || 4
1 4ThR
v! ; ; ! !
0.8
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7. The circuit within the dotted box is replaced by the
Thevenins equivalent, consisting of a voltage source of
VTh in series with a resistor RTh,
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8. The load resistor R2 is again connected to Theveninsequivalent forming a single-loop circuit.
The current I2 through this resistor is easily calculated,
Th2
Th 2
11.2
0.8 2
VI
R R! ! !
4 A
Important Comment
Theequivalent circuitreplacesthe circuit withinthe
box only
forthe
effects
externaltothe
box.
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Example 4
Using Thevenins Theorem, find the current in the
ammeterA of resistance 1.5 connected in an
unbalanced Wheatstone bridge shown.
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Solution :
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V6!vv!
!!@
!
!
!
!
65.1475.0
A5.162
12
andA7
5.0412
12
2
1
BDADABocVVVV
I
I
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Ans. -1 A
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BenchmarkExample 5Again consider ourbenchmark example to determine
voltage across 3- resistor by applying Thevenins
theorem.
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Solution :
We treat the 3- resistor as load.
Thevenin voltage VTh is the open-circuit voltage(withRL removed).
We use source transformation.
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V5!v!@ 15ThV
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To compute RTh, we turn off all the sources in the
circuit within box and get the circuit
Thus, RTh = 3 .
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V2.5!
v!33
35
LV
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Nortons Theorem
It is dual of Thevenins Theorem.
A two terminal network containing linear
passive and active elements can be replaced
by an equivalent circuit of a constant-current source in parallel with a resistance.
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The value of the constant-current source is
the short-circuit current developed whenthe terminals of the original network are
short circuited.
The parallel resistance is the resistance
looking back into the original network with
all the sources within the network madeinactive (as in Thevenins Theorem).
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Example 6
Obtain the Nortons equivalent circuit with respect tothe terminals AB for the network shown, and hence
determine the value of the current that would flow
through a load resistor of 5 if it were connected
across terminals AB.
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Solution : When terminals A-B are shorted
1 2
10 5
5 10 I I I @ ! ! ! 2.5
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Turning OFF the sources,
3
10!
v!@
105
105NR
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A1!
v!
!
5)3/10(
)3/10(
5.2LN
N
NL RR
R
II
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Power Transferred to the Load
Consider the circuit :
Source Load
r
ERLp
(Variable)
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p
RL0
pmax
RL= r
Maximumpoweristransferredwhen
RL = r.
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Proof
? A 0
02)()(
zero.toequalnumeratorpute,maximizingor
)(
1)(21)(4
2
2
2
!!
vvv!@
!
L
LLL
L
LLL
L
L
L
RrRrRrR
rR
rRRrRE
dR
dp
RrR
Ep
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Maximum Power Transfer Theorem
Maximum power is drawn form a sourcewhen the Load Resistance is equal to the
Source Internal Resistance.
When maximum power transfer condition issatisfied, we say that the load is matched
with the source.
Under maximum power transfer condition,
the efficiency of the source is only 50 %.
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What is the maximum power that a sourceof emfE and internal resistance r can
ever deliver ?
Ans.
r
E
4
2
Available Power
Next
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Prove that under maximum power transfer
condition, the efficiency of the source is only
50 %.
%50
%100)(2
2
!
v!| rRI
RI
P
P
L
L
in
o
L
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Example 7The open-circuit voltage of a standard car-battery is 12.6
V, and the short-circuit current is approximately 300 A.
What is the available power from the battery ?
Solution : The output impedance of the battery,
oco
sc
12.60.042
300
VR
I! ! ! ;
Therefore, the available power22 2
ocThavl
Th o
(12.6)
4 4 4 0.042
VVP
R R! ! ! !
v
945 W
Next
Click
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Millmans Theorem
A number of parallel voltage sources V1, V2, V3 ,Vnwith internal resistances R1, R2, R3, Rn,
respectively can be replaced by a single voltage
source Vin series with equivalent resistance R.
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n
nn
GGGG
GVGVGVGVV
!
...
...
321
332211
1 2 3
1 1
...n
RG G G G G
! !
and
Equivalent Circuit
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Reciprocity Theorem
In a linear bilateral network, if a voltage source Vin a branch A produces a current Iin any other
branch B, then the same voltage source Vacting
in the branch B would produce the same current I
in branch.
The ratio V/Iis known as the transfer
resistance.
Letusverifythereciprocitytheorem byconsideringanexample.
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Example 8
In the network shown, find the current in branch B dueto the voltage source of 36 V in branch A.
Now transfer the voltage source to branch B and find
the current in branch A.
Is the reciprocity theorem established ?
0Also, determine the transfer resistance from branch A
to branch B.
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Solution : The equivalent resistance for the voltage
source,;!!! 94324)]13(||12[2eqR
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The current supplied by the voltage source = 36/9 =4 A.
Using current divider, the current Iin branch B,
A3!
v!
412
124I
Now, transferring the voltage source to branch B,
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ReviewReview
Superposition Theorem.
Thevenins Theorem.
Nortons Theorem.
Maximum Power Transfer Theorem.
Maximum Power Transfer Theorem for AC Circuits.
Millmans Theorem.
Reciprocity Theorem.
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