Basic Concepts - Oakland University

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Transcript Basic Concepts - Oakland University

Circuit Theorems
Discussion D2.5
Sections 2-9, 2-11
1
Circuit Theorems
•
•
•
•
Linear Circuits and Superposition
Thevenin's Theorem
Norton's Theorem
Maximum Power Transfer
2
Linear Circuits
• A linear circuit is one whose output is
directly proportional to its input.
• Linear circuits obey both the properties of
homogeneity (scaling) and additivity.
3
4
Superposition Principle
Because the circuit is linear we can find the response of the
circuit to each source acting alone, and then add them up to find
the response of the circuit to all sources acting together. This is
known as the superposition principle.
The superposition principle states that the voltage across (or
the current through) an element in a linear circuit is the
algebraic sum of the voltages across (or currents through) that
element due to each independent source acting alone.
5
Turning sources off
Current source:
a
i  is
is
We replace it by a current
source where is  0
b
An open-circuit
Voltage source:
+
DC
vs
v  vs
-
We replace it by a voltage
source where v s  0
i
An short-circuit
6
Steps in Applying the Superposition Principle
1. Turn off all independent sources except
one. Find the output (voltage or current)
due to the active source.
2. Repeat step 1 for each of the other
independent sources.
3. Find the total output by adding
algebraically all of the results found in
steps 1 & 2 above.
In some cases, but certainly not all, superposition can simplify
the analysis.
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Example: In the circuit below, find the current i by superposition

24V
DC

DC
12V
Turn off the two voltage sources
(replace by short circuits).


3A

i

12V

v1

v2
3A
i1
8
1 4  1 3  1 4

1 4

1 4
  v1   0 
    
1 4  1 8   v2   3 

5
6

12V

v1

i1
i1  1
v2

v1 
1
4
3A
v2 
10
3
v1
1
4
v1 
v2  0
3
8
v2  3
 10 2 
v1 
 3
8
 8
v1  3
9
Example: In the circuit below, find the current i by superposition

24V
DC

DC
12V
Turn off the 24V & 3A sources:


3A

i
i1


DC
12V
i2

O.C.
i2
10



DC


O.C.
12V
DC

12V

O.C.
i2
i2
12  4
3
16

i2 
12
6
DC
2
12V

O.C.
i2
11
Example: In the circuit below, find the current i by superposition

24V
DC

DC
12V
Turn off the 3A & 12V sources:


3A

24V
i
DC
i2

i3


O.C.
i3
12
48 4

4


 4   i2    24 
   

4  3   i3   0 
24V
16 i2  4 i3   24
DC
i2

i3


i3
 4 i2  7 i3  0
O.C.
i2 
7
4
i3
i3  2 8  4    2 4
i3   1
13

24V
DC

DC

12V
i  i1  i2  i3  1A  2A  1A  2A

3A
i



24V
DC
i1

12V

v1

i1
i1  1
i2

v2
DC
3A

12V
i2

i2
i2  2

O.C.
i3


O.C.
i3
i3   1
14
Circuit Theorems
•
•
•
•
Linear Circuits and Superposition
Thevenin's Theorem
Norton's Theorem
Maximum Power Transfer
15
Thevenin's Theorem
In many applications we want to find the response to a particular
element which may, at least at the design stage, be variable.
a
+
Linear
Circuit
V
b
Variable
R
Each time the variable element
changes we have to re-analyze the
entire circuit. To avoid this we
would like to have a technique that
replaces the linear circuit by
something simple that facilitates the
analysis.
A good approach would be to have a simple equivalent circuit to
replace everything in the circuit except for the variable part (the load).
16
Thevenin's Theorem
Thevenin’s theorem states that a linear two-terminal resistive
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, and RTh is the input or
equivalent resistance at the terminals when the independent
sources are all turned off.
i
i
a
a
RTh
Linear
Circuit
RL
DC
RL
VTh
b
b
Rin
Rin
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Thevenin's Theorem
Thevenin’s theorem states that the two circuits given below are
equivalent as seen from the load RL that is the same in both cases.
i
i
a
a
RTh
Linear
Circuit
RL
DC
RL
VTh
b
b
Rin
Rin
VTh = Thevenin’s voltage = Vab with RL disconnected (= ) = the
open-circuit voltage = VOC
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Thevenin's Theorem
i
i
a
a
RTh
Linear
Circuit
RL
DC
RL
VTh
b
b
Rin
Rin
RTh = Thevenin’s resistance = the input resistance with all
independent sources turned off (voltage sources replaced by short
circuits and current sources replaced by open circuits). This is the
resistance seen at the terminals ab when all independent sources are
turned off.
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Example

DC
10V

RT h 
a
vOC 

VT h

i SC
5
 2
2.5
RTh  2
2
10V  5V  VTh
22
DC
a
VTh  5V
b

DC
10V

a
iSC 

b
10 2 10

 2.5A
23 4
2
3

b


a
RTh  1 
2 2
 2
22
20
b
Circuit Theorems
•
•
•
•
Linear Circuits and Superposition
Thevenin's Theorem
Norton's Theorem
Maximum Power Transfer
21
Norton's Theorem
Norton’s equivalent circuit can be found by transforming the
Thevenin equivalent into a current source in parallel with the Thevenin
resistance. Thus, the Norton equivalent circuit is given below.
i
IN 
VTh
RTh
a
RN  RTh
RL
b
Formally, Norton’s Theorem states that a linear two terminal
resistive circuit can be replaced by an equivalent circuit consisting
of a current source IN in parallel with a resistor RN, where IN is the
short-circuit current through the terminals, and RN is the input or
equivalent resistance at the terminals when all independent sources
22
are all turned off.
Circuit Theorems
•
•
•
•
Linear Circuits and Superposition
Thevenin's Theorem
Norton's Theorem
Maximum Power Transfer
23
Maximum Power Transfer
In all practical cases, energy sources have non-zero internal
resistance. Thus, there are losses inherent in any real source. Also,
in most cases the aim of an energy source is to provide power to a
load. Given a circuit with a known internal resistance, what is the
resistance of the load that will result in the maximum power being
delivered to the load?
Consider the source to be modeled by its Thevenin equivalent.
i
a
RTh
DC
RL
VTh
b
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i
a
RTh
DC
RL
VTh
b
The power delivered to the load (absorbed by RL) is
p  i R L  VTh
2
 RTh  R L  
2
RL
This power is maximum when  p  R L  0
p
RL
 VT h
2
  R  R  2  2 R  R  R  3   0
L
L
Th
L
 Th

25
dp
dR L
 VT h
  R  R  2  2 R  R  R  3   0
L
L
Th
L
 Th

2
RTh  R L  2 R L
R L  RTh
Thus, maximum power transfer takes place when the resistance of
the load equals the Thevenin resistance RTh. Note also that
p m ax  VT h
 RT h  R L  
p m ax  VTh
 2 RTh  
2
2
RL
R L  RT h
RTh  VTh
2
4 RTh
Thus, at best, one-half of the power is dissipated in the internal
resistance and one-half in the load.
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