# CS计算机代考程序代写 Prepared by Dr. Majid MALEKPOUR

Prepared by Dr. Majid MALEKPOUR
In conjunction with School of Electrical Engineering and Telecommunication, UNSW, and McGraw-Hill Education
Electrical Engineering and Telecommunications EE1111
Topic 4: Circuit Theorems

Topic 4 Content
This lecture covers:
• Linearity and Superposition
• Source Transformation
• Thevenin’s and Norton’s theorems • Maximum Power Transfer
Corresponds to Chapter 4 of your textbook
UNSW Diplomas | Page 1

Topic 3 recap
• The ground is mostly chosen as the reference node • Voltage nodes are measured with respect to the ground
• Steps in Nodal analysis
• Set the reference node and node voltages
• Apply KCL and use Ohm’s law to have currents in terms of node voltages
• Solve the set of linear equations for node voltages • Nodal analysis with voltage source
• Case 1: Voltage source is connected to the reference node
• Case 2: Voltage source is between two non-reference nodes
• Form a supernode by enclosing the voltage source and any parallel element with it
• Apply KCL to supernode using the already assigned node voltages inside it
• Write the extra equation relating the node voltages inside supernode with voltage source (KVL or voltage definition)
𝑣=𝑣
𝑘 source
𝑣−𝑣=𝑣
𝑎 𝑏 source
UNSW Diplomas | Page 2

Topic 3 recap II
• Planner and non-planner circuits
• Steps in Mesh analysis
• Set the mesh currents with arbitrary direction (mostly clockwise)
• Apply KVL and use Ohm’s law to have voltages across elements in terms of mesh currents
• Solve the set of linear equations for mesh currents
• The sum of all current entering and leaving a node is zero 𝑁 𝑖 = 0
• Mesh analysis with current source
• Case 1: current source is only in one mesh (isolated current source)
𝑖𝑘 = 𝑖source
• Case 2: Current source is shared between two meshes
• Form a supermesh by excluding the current source and any
series element with it and merging the two meshed as one
• Apply KVL to the supermesh with the already assigned mesh
currents inside it
• Write the extra equation relating the mesh currents inside supermesh
with current source (KCL at the node containing shared current source)
𝑛=1 𝑛
𝑖𝑎 − 𝑖𝑏 = 𝑖source
Page 3

Linear property
• A system is called linear if it follows
two main properties:
• Homogeneity
Homogeneity
x y kx ky
System
• If the input 𝑥 is multiplied by a constant 𝑘, the output/response 𝑦 is multiplied by the same constant
• The response to the sum of the inputs is the sum of the individual responses to each input
• Together, the response to the linear combination of inputs is equal to the
x1
x3 = x1 + x2 y3 = y1 + y2
Two properties combined
System
System
System
System
linear combination of individual responses to each input
x4 = k1x1 + k2x2 y4 = k1y1 + k2y2
Linear System
𝑘1 and 𝑘2 are constant
Page 4

Linear property II
• Resistor is a linear electrical element since the voltage-current relationship satisfies both the homogeneity and additivity properties 𝑣=𝑅𝑖
A linear circuit is the one whose output is linearly related (or directly proportional) to its input
i = k1 i1 + k1i2
A linear circuit consists of only linear elements, linear dependent sources, and independent sources
R
𝑣 = 𝑅 𝑘1𝑖1 + 𝑘2𝑖2
v = R(k1 i1 + k1i2) −
Power dissipated on a resistor has nonlinear relationship with both voltage and current
2 𝑝 = 𝑣2
𝑝=𝑅𝑖 1 𝑅𝑣2𝑣2
𝑅(𝑖 +𝑖)2≠𝑅𝑖2+𝑅𝑖2 2 1 2 1 2 1 2 𝑅 (𝑣1 + 𝑣2) ≠ 𝑅 + 𝑅
𝑣 = 𝑘1 𝑅𝑖1 + 𝑘2 𝑅𝑖2 𝑣1 𝑣2
𝑣 = 𝑘1𝑣1 + 𝑘2𝑣2
Page 5
+

Linear property III
• In following circuit, for instance, the voltage𝑣𝑜 when𝑖𝑠 =30Aand𝑖𝑠 =60Ais obtained as below:
Using current division
𝑖2 = 4 𝑖𝑠 = 1 𝑖𝑠 4+(12+8) 6
𝑣 =8𝑖 =8×1𝑖 =4𝑖 𝑜26𝑠3𝑠
𝑖= 𝑣=4𝑖= 𝑠 𝑜3𝑠
𝑖= 𝑣=4𝑖= 𝑠 𝑜3𝑠
i
2
i1
if if
if
𝑖𝑠 =
𝑣 =4𝑖 =4 𝑘30+𝑘60 =𝑘40+𝑘80V 𝑜3𝑠31212
30A
40V
60A
80V
𝑘130 + 𝑘260 A
Page 6

Superposition
• If there are two or more independent sources in a circuit, there are two ways to solve for the circuit variables
• Nodal or mesh analyses • Use superposition
• It is based on linear property of circuits
• Only one independent source is considered at a time
• The rest of the independent sources are set to zero (turned off, disabled, shut down)
• Dependent sources are left intact since they are controlled by circuit variables
The superposition principle states that the voltage across (or 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.
Page 7

Superposition II
• Steps to apply superposition principle:
Turning off current source means replacing it with an open circuit
1. Turn off all independent sources except one by setting them to zero
2. Find the output (voltage or current in question) using methods covered in Topics 2 and 3
3. Repeat step 1 for each of the other independent sources
4. Find the total contribution by adding algebraically all the contributions due to the independent sources
is
is = 0 A
is
is ≡ 0
Turning off voltage source means replacing it with a short circuit
Question: Is it possible to use superposition to find the power absorbed by a resistor? NO!
Power has a nonlinear relationship with voltage and current of a resistor
v
s
vs ≡ 0
vs = 0 V

Page 8
+

Example 1
• Using the superposition theorem, find 𝑣𝑜 in the circuit below. • (Worked solutions of examples are given in class)
𝑣 =16V 𝑜
Example1 (Web view) Page 9

Example 2
• Using the superposition theorem, find 𝑣𝑥 in the circuit below. • (Worked solutions of examples are given in class)
𝑣 = 31.25 V 𝑥
Example2 (Web view) Page 10

Practice problem 1
• Using the superposition theorem, find 𝑖𝑜 in the circuit below.
𝑖𝑜 = −6.44 A
Page 11

Source transformation
• Similar to the delta-wye transformation, it is possible to transform a source from one form to another
• This can be useful for simplifying circuits
• The principle behind all of these transformations is the concept of equivalence, where two circuits are called equivalent if their 𝒊-𝒗 characteristics are identical at the terminal
• Note that direction of current and polarity of voltage should follow the passive sign convention for sources as active elements
A source transformation is the process of replacing a voltage source 𝒗𝒔 in series with a resistor 𝑹 by a current source 𝒊𝒔 in parallel with a resistor 𝑹 (preferably the same) or vice versa
Page 12

Source transformation II
R
aa
Req =R vs =0V Req =R

bb aRa
• Terminalequivalency:
• Turn off the sources, the equivalent resistance (𝑅eq) at terminal a-b in both
circuits must be the same
• Short-circuit current (𝑖𝑠𝑐) flowing from a to b and open-circuit voltage (𝑣𝑜𝑐) in both circuits must be the same
• The equivalent source is obtained using Ohm’s Law
is =0A
R
i =vs
i
isc=is
s vsscR
bb aRa
R
is
is R voc=Ris vs voc=Ris
−−
R
bb aa
𝑣
𝑖=𝑠 or𝑣=𝑅𝑖 𝑠𝑅𝑠𝑠
vs isR
bb
Page 13
+
+
+

Source transformation III
• The source transformation also applied to dependent sources, provided that the dependent variable is handled carefully
R
a
a
𝑣
𝑖=𝑠 or𝑣=𝑅𝑖 𝑠𝑅𝑠𝑠
vs
is R
b
b
Some limitations:
• Source transformation is not possible when 𝑹 = 𝟎 for an
ideal voltage source. (In a practical voltage source 𝑹 ≠ 𝟎)
• Source transformation is not possible when 𝑹 = ∞ for an ideal current source. (In a practical current source 𝑹 ≠ ∞)
Page 14

Example 3
• Use source transformation to find 𝑣0 in the circuit below. • (Worked solutions of examples are given in class)
𝑣𝑜 = 3.2 V
Example3 (Web view) Page 15

Example 4
• Use source transformation to find 𝑖0 in the circuit below. • (Worked solutions of examples are given in class)
𝑖𝑜 = 1.78 A
Example4 (Web view) Page 16

Example 5
• Use source transformation to find 𝑣𝑥 in the circuit below. • (Worked solutions of examples are given in class)
𝑣𝑥 = 7.5 V
Example5 (Web view) Page 17

Practice problem 2
• Use source transformation to find 𝑖𝑥 in the circuit below.
𝑖𝑥 = 7.05 𝑚A
Page 18

Thevenin’s theorem
• In many circuits, one element can be
variable
• An example of this case is the household appliances consuming different powers (hairdryer, fridge, etc.)
• This variable element is called the load
• The circuit would have to be analyzed
again for any change in the load
• Thevenin’s theorem provides a technique to simplify the analysis by replacing the fixed part of the circuit with an equivalent one known as Thevenin equivalent circuit
Thevenin’s theorem states that a linear two- terminal circuit (Fig. (a)) can be replaced by an equivalent circuit consisting of a voltage source 𝑽𝐓𝐡 in series with a resistor 𝑹𝐓𝐡 (Fig. (b))
Page 19

Thevenin’s theorem II
The voltage source known as Thevenin voltage 𝑽𝐓𝐡 is equal to the open-circuit voltage at the terminals
The resistance known as Thevenin resistance 𝑹𝐓𝐡 is equal to the input resistance measured at the terminals when all independent sources are turned off
i=0A RTh
RTh
a
Rin = RTh b
a
VTh
voc −
𝑉=𝑣 Th 𝑜𝑐
𝑅Th = 𝑅in = 𝑅eq
b
Page 20
+

Thevenin’s theorem III • There are two cases in finding 𝑅Th:
• Case 1: The network has no dependent sources
• Set all independent sources to zero
• 𝑹𝑻𝒉 is the input resistance seeing from
terminal a and b (equivalent resistance) • Case 2: The network has at least one
dependent sources
• Set all independent sources to zero
• Leave dependent sources intact
• Attach/connect a voltage source (𝑣 ) to the 𝑜
terminals a-b and find the generated
current (𝒊𝒐 in Fig. (a)), or attach a current
source 𝒊𝒐 and find the generated voltage
(𝑣 inFig.(b)) 𝑜
• They can be set 𝑣𝑜 = 1 V in Fig. (a), or 𝑖𝑜 = 1 A in Fig. (b) for simplicity
• Use Ohm’s law to find 𝑅Th
𝑣 𝑅Th = 𝑜
𝑖𝑜
Page 21

Thevenin’s theorem IV
• Thevenin’s theorem is a powerful technique
in circuit analysis with variable loads
• It allows to simplify a large linear circuit
• The equivalent circuit behaves externally exactly the same as the original circuit