Download Ch. 3 Techniques of Circuit Analysis

Electric Circuits
(EELE 2312)
Chapter 3 Techniques of Circuit Analysis
Dr. Basil Hamed
3.1 Terminology
Planar and Nonplaner Circuits
2
Describing a Circuit
The Vocabulary
3
Example 3-1
Identifying Node, Branch, Mesh and Loop
For the figure shown identify:
a) All nodes and all essential nodes
b) All branches and all essential branches
c)
All meshes
d) Two paths that are not loops or essential branches
e) Two loops that are not meshes
4
The Systematic Approach
An Illustration
i 1  i 2  i 6  I  0
i 1  i 3  i 5  0
i 3  i 4  i 2  0
R1i 1  R 5i 2  i 3 (R 2  R 3 )  1  0
i 3 (R 2  R 3 )  R 6i 4  R 4i 5  2  0
 R 5i 2  R 7 i 6  i 4 R 6  0
5
3.2 Introduction to the Node-Voltage Method
1  10 1 1  2
 
0
1
5
2
2  1 2
 20
2
10
100
1 
 9.09
11
120
2 
 10.91
11
V
V
6
Example 3.2
Using the Node-Voltage Method
the node-voltage method to find the branch currents i , i ,
a)1 Use
50
1 1
and i
  3  0
b) Find the power associated with each source and state whether
5the source
10is delivering
40
or absorbing power.
1  40 V
a
b
c
P50V  50i a  100 W
50  40
ia 
 2 A P3A  31  3(40)
5
P


120
W
3
A
40
Pdel  100  120  220 W
ib 
4 A
10
40
P

4(5)

16(10)

1(40)
abs
ic 
1 A
Pabs  220 W
40
7
Example 3.3
Node-Voltage Method with Dependent Sources
1  16
2  10
Use the node-voltage method to find the power dissipated in the
5Ω resistor in the circuit.
1  20 1 1  2
 
0
2
20
5
2  1 2 2  8i 
 
0
5
10
2
1  2
i 
5
V
V
16  10
i 
5
i   1.2 A
2
P5  1.2  5
P5  7.2 W
8
3.4 The Node-Voltage Method
Some Special Cases
1  50 V
2  1
2

V
i 0
1 5 100
50
2  

1
i  4  02

5  0
100
2 10
 1 2 503

 V 40
25 125
50 100
3
9
3.4 The Node-Voltage Method
The Concept of Supernode
2  60 V
60  50
i 
5
i  2 A
3  60  20
2  1 2 3


 4  0   80 V
3
5
50 100
3  2  10i 
2  50
i 
5
10
3.5 Introduction to the Mesh-Current Method
i1  i a
i1  i 2  i 3
i 2  ib
1  i 1 (R1  R 3 )  i 2 R 3
i 3  i a  i b 2  i 1R 3  i 2 (R 2  R 3 )
1  i a R1  (i a  i b )R 3
Unknowns
i 2 ia i)R
2 i 1(
icurrents=3
b 
2 i3
3 i31i bi 2R
be=3 ne=2
11 current

ia1(equation
R1of
R13 )
Ri
R

i 1(Mesh
)

i
R
Number
Equations=b
b
3
i
R
R3
e-(n
3
2
3i e3-1)
1
1 1 
2voltage
i aof
RMesh
R
i bEquations=7-(4-1)=4
(
R 2i (iR
RR
Number
2 
3 i
3 ) 

i
R
equations



R
)
2
2
2
3
3
2
1 3
2
2
3
11
Example 3.4
Using the Mesh-Current Method
i  5.6
ib  2
i c  0.8
a) Use the mesh-current method to determine the power
associated with each voltage source. a
b) Calculate the voltage across the 8Ω resistor
A
A
A
P40V  40i a
40  2i a  8(i a  i b )  0
W
8(i b  i a )  6i b  6(i b  i c )  0 P40V  225
P20V  20i c
6(i c  i b )  4i c  20  0
P20V  16
W
10i a  8i b  0i c  40
o  8(i a  i b )
o  8(3.6)
8i a  20i b  6i c  0
o  28.8
V
0i a  6i b  10i c  20
12
3.6 Mesh-Current Method with Dependent Sources
Example 3.5
i  29.6
i 2  26
i 3  28
i   1.6
Use the mesh-current method to determine the power dissipated
in the 4Ω resistor.
1
50  5(i 1  i 2 )  20(i 1  i 3 )
0  5(i 2  i 1 )  1i 2  4(i 2  i 3 )
0  20(i 3  i 1 )  4(i 3  i 2 )  15i 
i   i1  i 3
25i 1  5i 2  20i 3  50
5i 1  10i 2  4i 3  0
5i 1  4i 2  9i 3  0
A
A
A
A
i 4   28  26
i 4  2
A
2
P4   2  4
P4   16 W
13
3.7 The Mesh-Current Method
Some Special Cases
i a  1.75
i b  1.25
i c  6.75
A
A
A
100  3(i a  i b )    6i a
50  4i c    2(i c  i b )
50  9i a  5i b  6i c
0  3(i b  i a )  10i b  2(i b  i c )
ic  ia  5
14
3.7 The Mesh-Current Method
The Concept of Supermesh
100  3(i a  i b )    6i a
50  4i c 50
  92(
i a ic 5i bib )6i c
i ai a5i bi )b 2(6iic c i b )  50  4i c  6i a  0
50
10093(
15
3.8 Node-Voltage Versus Mesh-Current
16
Example 3.6
Understanding Node-Voltage/Mesh-Current Method
Find the power dissipated in the 300Ω resistor in the circuit shown.
1 1  2a 3a  256
3  2 a 3 b(2 a128)
 c 3  256

 
 

 0
0
100
250
200 200150 400 100 500 300
150
2 2 c1 2 c31282 128
b3 c  a
c 


 
 0
0
300
250
400 400
500
500
250
300
2 50(c  a ) (c  a )
3  1  50
bi 50
1i  
for the
 supernode
6
300
6
P300  16.57 Watt
17
Example 3.7
Comparing Node-Voltage & Mesh-Current Methods
Find the voltage vo in the circuit shown.
  b
a  (b  0.8 ) 
  2 

10


o  173 V
o 193
io 0.8
a 
193
10i a0.4
10
i b 10
supermesh
c


0

i b 10
i a  0.4  0.8i c 2.5


(


0.8

)
a  
7.5
i
o
a
b

b
 0.5

0
i c 2.5
 i b  0.5
10
b  80i a b i a0.8
 a
160
 2 A
 0.5 
0
7.5
o  173 V
10
18
3.9 Thevenin and Norton Equivalents
a Thevenin equivalent circuit is an independent
voltage source V in series with a resistor R ,
which replaces an interconnection of sources and
resistors. This series combination of V and R , is
equivalent to the original circuit in the sense that,
if we connect the same load across the terminals
a,b of each circuit, we get the same voltage and
current at the terminals of the load. This
equivalence holds for all possible values of load
resistance
Th
Th
Th
Th
T
19
3.9 Thevenin and Norton Equivalents
Finding a Thevenin Equivalent
V Th
i sc 
RTh
V Th
RTh 
i sc
20
3.9 Thevenin Equivalent Circuit
Finding a Thevenin Equivalent
2  
25  25
2
2

3 1  0
2  16 V
1


3

0
5
20
4
V Th 32
5
20
16
i sc   4 
A V RTh32 V  8 
ab
Th
4 1
i sc
4
21
3.9 Norton Equivalent Circuit
A Norton equivalent circuit consists of an
independent current source in parallel with
the Norton equivalent resistance, We can
derive it from a Thevenin equivalent circuit
simply by making a source transformation.
Using Source Transformation
22
Example 3.8
Thevenin Equivalent with Dependent Source
Find the Thevenin equivalent for the circuit containing dependent
sources.
5  500i
V Th isc ab20i (20i i)(25)
 2.5 mA
5  3 5  3V Th 2000
VVThTh 5 5 V3
i 

i sc  20(2.5) 2000
 50 mA 2000
RTh 

10  100 
i
50
sc
23
3.10 More on Deriving a Thevenin Equivalent
5  20
R ab  RTh  4 
8 
25
24
Example 3.9
Thevenin Equivalent Using a Test Source
T 60T
iT  
25 2000
iT
1
6
 
T 25 200
iT
50
1


T 5000 100
T
RTh 
 100 
iT
Find the Thevenin resistance RTh for the circuit shown.
T
RTh 
iT
T
iT   20i
25
3T
i
mA
2
25
3.11 Maximum Power Transfer
• Circuit analysis plays an important role in the
analysis of systems designed to transfer power
from a source to a load.
• We discuss power transfer in terms of two
basic types of systems.
• The first emphasizes the efficiency of the
power transfer (Power utility systems)
• The second basic type of system emphasizes
the amount of power transferred.
26
3.11 Maximum Power Transfer
Maximum power transfer can best be described
with the aid of the circuit shown in Fig.
We assume a resistive network containing independent and
dependent sources and a designated pair of terminals, a,b,
to which a load, RL, is to be connected.
27
3.11 Maximum Power Transfer
The problem is to determine the value of RL that
permits maximum power delivery to RL
The first step in this process is to recognize that a
resistive network can always be replaced by its
Thevenin equivalent.
28
3.11 Maximum Power Transfer
Replacing the original
network by its Thevenin
equivalent
greatly
simplifies the task of
finding RL
29
3.11 Maximum Power Transfer
Next, we recognize that for a given circuit, VTh and RTh
will be fixed. Therefore the power dissipated is a
function of the single variable RL, To find the value of
RL that maximizes the power, we use elementary
calculus. We begin by writing an equation for the
derivative of p with respect to RL:
30
3.11 Maximum Power Transfer
The derivative is zero and p is maximized when
Solving above eq. yields
To find the maximum power delivered to R
L
31
3.11 Maximum Power Transfer
2
 V Th 
p  i RL  
 RL
 RTh  R L 
2


(
R

R
)
dp
2
Th
L  R L 2(RTh  R L )
V Th 

2
dR L
(RTh  R L )


(RTh  R L )2  2R L (RTh  R L )
2
R L  RTh
2
V Th R L
Pmax 
2
(2R L )
2
V Th
Pmax 
4R L
32
Example 3.10
Condition for Maximum Power Transfer
a) For the circuit shown, find the value of RL that results in maximum
R L  25power
 being
transferred to RL.
 300 
b) Calculate the maximum power delivered to RL.
ab  
25  150 V

c) When RL is adjusted for maximum power transfer, what
percentage
of the
 50

power delivered by the 360 V source reaches RL?
360  150
 
is 
7 A
30
Ps  360  i s  2520 W
900
PL 
100  35.71%
2520
V Th
150

 360

300
V
2
180
 300 
Pmax  
25   900 W


150
 30
50

  25 
RTh 
180
33
End of
Chapter Three
Basil Hamed
34