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
• Additivity
• 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
y1 and x2 y2 Additivity
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
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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
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Example 1
• Using the superposition theorem, find 𝑣𝑜 in the circuit below. • (Worked solutions of examples are given in class)
Answer:
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)
Answer:
𝑣 = 31.25 V 𝑥
Example2 (Web view) Page 10

Practice problem 1
• Using the superposition theorem, find 𝑖𝑜 in the circuit below.
Answer:
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)
Answer:
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)
Answer:
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)
Answer:
𝑣𝑥 = 7.5 V
Example5 (Web view) Page 17

Practice problem 2
• Use source transformation to find 𝑖𝑥 in the circuit below.
Answer:
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
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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
• The current through the load 𝑹𝑳 (load
current 𝐼𝐿) and the voltage across the load
(load voltage 𝑉 ) is obtained using simple 𝐿
voltage division or KVL & KCL
Important !!!!
• It is possible for Thevenin resistance to become negative, 𝑅Th < 0. • This implies that the circuit is supplying power • Negative 𝑅Th is possible in circuits with dependent sources 𝑉 Th 𝑅Th + 𝑅𝐿 𝑉= 𝑅𝐿 𝑉 L 𝑅Th + 𝑅𝐿 Th 𝐼= L Page 22 Example 6 • Using Thevenin’s theorem, find the equivalent circuit to the left of the terminals in the circuit below, and then find the load current 𝐼𝐿. • (Worked solutions of examples are given in class) Example6 (Web view) Page 23 Answer: 𝑉 =90V Th 𝑅Th = 45 Ω 𝐼 = 1.5 A Example 7 • Find the Thevenin equivalent of the circuit below at the terminals a-b. • (Worked solutions of examples are given in class) Example7 (Web view) Page 24 Answer: 𝑉 =20V Th 𝑅Th = 6 Ω Practice problem 3 • Find the Thevenin equivalent of the circuit below at the terminals a-b. Answer: 𝑉 =16=5.333V Th 𝑅Th = 444.4 mΩ 3 Page 25 Example 8 • Find the Thevenin equivalent of the circuit below at the terminals a-b. • (Worked solutions of examples are given in class) Example8 (Web view) Page 26 Answer: 𝑉 =0V Th 𝑅Th = −4 Ω Practice problem 4 • Find the Thevenin equivalent of the circuit below at the terminals a-b. Answer: 𝑉 =0V Th 𝑅Th = −7.5 Ω Page 27 Norton’s theorem • Norton’s theorem is in fact the dual form of Thevenin’s theorem • It provides a similar technique to simplify the analysis by replacing the fixed part of the circuit having a variable load, with an equivalent one known as Norton’s equivalent circuit Norton’s theorem states that a linear two-terminal circuit (Fig. (a)) can be replaced by an equivalent circuit consisting of a current source 𝑰𝑵 in parallel with a resistor 𝑹𝑵 (Fig. (b)) Page 28 Norton’s theorem II The current value known as Norton current 𝑰𝑵 is equal to the short-circuit current at the terminals The Norton resistance 𝑹𝑵 is equal to the input resistance measured at the terminals when all independent sources are turned off 𝑅𝑁 = 𝑅in = 𝑅eq IN R a isc = IN N RN b a Rin = RN b 𝐼=𝑖 𝑁 𝑠𝑐 Page 29 Thevenin-Norton transformation • Thevenin’s and Norton’s theorems are related to each other through source transformation Thevenin and Norton resistances are exactly the same and equal to input resistance measured from the terminals of the circuit RTh = RN aa 𝑅Th = 𝑅𝑁 IN RTh = RN bb VTh 𝑉=𝑅𝐼 Th 𝑁𝑁 𝐼= 𝑁 𝑉 Th 𝑅Th Page 30 Thevenin-Norton transformation II • Dependent sources in Norton equivalent circuit are handled in the same way as Thevenin equivalent circuit (Case 2 in Thevenin’s theorem) aa voc − isc Linear two- terminal circuit Linear two- terminal circuit bb Th𝑁 Th𝑁 a •Since𝑉 ,𝐼 and𝑅 =𝑅 arerelated based on source transformation, finding the Thevenin or Norton equivalent circuit requires at least two of the main three variables as follows: Req = Rin Linear two- terminal circuit with all independent sources set equal to zero b 1. The open-circuit voltage 𝑣𝑜𝑐 across terminals 𝑎 and 𝑏 2. The short-circuit current 𝑖𝑠𝑐 through terminals 𝑎 and 𝑏 3. The equivalent or input resistance 𝑅eq = 𝑅in at terminals 𝑎 and 𝑏 when all independent sources are turned off 𝑉=𝑣 Th 𝑜𝑐 𝐼=𝑖 𝑁 𝑠𝑐 𝑣 𝑅Th= 𝑜𝑐=𝑅𝑁 𝑖𝑠𝑐 Alternative method for finding 𝑅Th = 𝑅𝑁 Page 31 + Example 9 • Find the Norton equivalent of the circuit below at the terminals a-b and confirm Thevenin-Norton transformation by calculating 𝑣𝑜𝑐 at the same terminals. • (Worked solutions of examples are given in class) Answer: 𝐼𝑁 = 1 A 𝑅𝑁 = 𝑅Th = 4 Ω 𝑉 =4V Th Example9 (Web view) Page 32 Practice problem 5 • Find the Norton equivalent of the circuit below at the terminals a-b and confirm Thevenin-Norton transformation by calculating 𝑣𝑜𝑐 at the same terminals. Answer: 𝐼𝑁 = 4.5 A 𝑅𝑁 = 𝑅Th = 90 Ω 𝑉 =405V Th Page 33 Example 10 • Find the Norton equivalent of the circuit below at the terminals a-b and confirm Thevenin-Norton transformation by calculating 𝑣𝑜𝑐 at the same terminals. • (Worked solutions of examples are given in class) Answer: 𝐼𝑁 = 7 A 𝑅𝑁 = 𝑅Th = 5 Ω 𝑉 =35V Th Example10 (Web view) Page 34 Practice problem 6 • Find the Norton equivalent of the circuit below at the terminals a-b and confirm Thevenin-Norton transformation by calculating 𝑣𝑜𝑐 at the same terminals. Answer: 𝐼𝑁 = 10 A 𝑅𝑁 = 𝑅Th = 1 Ω 𝑉 =10V Th Page 35 Maximum Power Transfer • In many practical applications, a circuit is designed to provide power to a load • It is desirable to maximize the power transferred to the load in those applications • Unlike an ideal source, internal resistance of circuits restricts the amount of power that can be transferred to a load • The power consumption of those internal resistance is known as power loss • Using the Thevenin equivalent circuit we can find the maximum power that can be RTh a VTh RL i b transferred based on fixed 𝑉 variable 𝑅𝐿 𝑝(𝑅𝐿) = 𝑅𝐿𝑖2 = 𝑅𝐿 and 𝑅 and Th 𝑉 𝑅𝑇h + 𝑅𝐿 Th 2 𝑇h Page 36 Maximum Power Transfer II • To find for which value of 𝑅𝐿 the power would be maximum, differentiate 𝑝 with respect to 𝑅𝐿 and set the result to zero. 𝑑𝑝 =𝑉 2 𝑅Th+𝑅𝐿 2−2𝑅𝐿(𝑅Th+𝑅𝐿) 𝑑𝑅𝐿 𝑇h 𝑅Th+𝑅𝐿 4 =𝑉 2 (𝑅Th+𝑅𝐿−2𝑅𝐿) =0 Maximum power is transferred to the load when the load resistance 𝑹𝑳 is equal to the Thevenin resistance 𝑹𝐓𝐡 as seen from the load terminals 𝑇h ⇒ 𝑅Th+𝑅𝐿−2𝑅𝐿 =0 𝑅𝐿 = 𝑅Th 𝑅Th+𝑅𝐿 3 It can be shown that this point is the global maximum extremum since 𝑑2𝑝 <0 𝑑𝑅2 𝐿 𝑉 𝑝max = Th 2 4𝑅Th 𝑣𝐿 = 𝑉 Th 2 Page 37 Example 11 • Find the value of load resistance 𝑅𝐿 for maximum power transfer in the circuit below, and then find the maximum power 𝑝 • (Worked solutions of examples are given in class) Answer: 𝑅𝐿 = 9 Ω 𝑝max = 13.44 W Example11 (Web view) Page 38 Practice problem 7 • Find the Norton equivalent of the circuit below at the terminals a-b and confirm Thevenin-Norton transformation by calculating 𝑣𝑜𝑐 at the same terminals. Answer: 𝑅𝐿 = 38 = 126.67Ω 9 𝑝max = 96.71 mW Page 39 Source Modelling • Thevenin and Norton equivalent circuits can be useful in modelling realistic voltage and current sources to deliver maximum power to a load • The internal resistance 𝑅𝑠 of the voltage source 𝑣𝑠 in series with the load 𝑅𝐿 acts as voltage divider • The load voltage will be constant if 𝑅𝑠 is zero or very small compared to the load 𝑅𝑠 ≪ 𝑅𝐿 • Without load, 𝑅𝐿 → ∞, 𝒗𝒐𝒄 = 𝒗𝒔 which is known as unloaded voltage source • With load connected to the source, the terminal voltage drops in magnitude, which is known as loading effect 𝑣 = 𝑅𝐿 𝑣 𝐿 𝑅𝑠+𝑅𝐿𝑠 Note that 𝑣 = 𝑣 = 𝑉 , 𝑜𝑐 𝑠 Th Thevenin voltage, they are all equivalent Page 40 Source Modeling II • Similarly, with a realistic current source, the internal resistor 𝑅𝑝 in parallel with the load 𝑅𝐿 acts as current divider • The load voltage will be constant if 𝑅𝑝 is infinitive or very large compared to the load 𝑅𝑝 ≫ 𝑅𝐿 • Without load, 𝑅𝐿 = 0, 𝑖𝑠𝑐 = 𝑖𝑠 which is known as unloaded current source • With load connected to the source, the terminal current drops in magnitude, which is known as loading effect 𝑖𝐿= 𝑅𝑝 𝑖𝑠 𝑅𝑝 + 𝑅𝐿 Note that 𝑖𝑠𝑐 = 𝑖𝑠 = 𝐼𝑁, Norton current, they are all equivalent Page 41 Source Modeling III • Now follow these steps to determine the unloaded source voltage 𝑣𝑠 and internal resistance 𝑅𝑠 for voltage source model in practice: 1. Measure the ope-circuit voltage (Fig. (a)) 2. Connect a variable load 𝑅𝐿 (Fig. (b)) and adjust its resistance until 𝑣 = 1 𝑣 𝑣=𝑣 𝑠 𝑜𝑐 𝐿2𝑠 3. Disconnect 𝑅𝐿 and measure its resistance as it should be equal to internal resistance 𝑅𝑠 based on maximum power transfer principle Use source transformation to identify 𝑖𝑠 and 𝑅𝑝 for current source model 𝑣 𝑣𝐿= 𝑠 ⇒ 𝑅𝑠=𝑅𝐿=𝑅Th 2 In the laboratory, you will be doing a similar experiment but with constant load 𝑅 to identify 𝑣 and 𝑅 and not 𝐿𝑠𝑠 necessarily following maximum power transfer principle Page 42 Source Modeling – Example The open-circuit voltage across a certain amplifier is 16 V. The voltage drops to 8 V when an 8-W speaker is connected to the amplifier. Determine the internal resistance of the amplifier and calculate the load voltage when a 24-Ω speaker is used instead. 𝑣=𝑣 =16V 𝑠 𝑜𝑐 =8Ω (a) (b) AmplifierModel voc = 16 V − vL = 8 V − Amplifier Amplifier 𝑣2 64 𝑅𝐿 = 𝐿 = 𝑝𝐿 8 Rs Vs Since 𝑣𝐿 = 𝑣𝑠 , we have the case of maximum 2 vL = 8 V − power transfer, because the voltage can only be divided equally across series resistors if they have the same resistance. Thus: (c) Amplifier Model 𝑅𝑠 =𝑅𝐿 =8Ω Now, the load voltage for the new 𝑅𝐿 = 24 is: 𝑣 = 24 𝑣 =3×16=12V 𝐿 8+24𝑠 4 vL RL = 24 Ω − (d) 8W RL 8W RL 8Ω 16 V Page 43 ++++ Questions!? Page 44