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 3: Nodal and Mesh Analysis

Topic 3 Content
This lecture covers:
• Nodal Analysis
• Supernode Concept
• Mesh Analysis
• Supermesh Concept
Corresponds to Chapter 3 of your textbook
Page 1

Topic 2 recap
• Electric resistance 𝑅 is the property of materials
resisting against the flow of current.
• Ohm’s Law establishes the relationship between the voltage across a resistive element (called resistor) and the current through it.
𝑣 = 𝑅𝑖
• Ohm’s law uses passive sign convention, i.e., current must enter the positive terminal of the resistor for positive sign in the formula, otherwise compensate it with negative sign.
• Shortcircuit:𝑅 = 0 → 𝑣=𝑅𝑖=0,
• Open circuit: 𝑅 = ∞ → 𝑖 = 𝑣 = 0
𝑅
• Conductance: 𝐺 = 1 𝑅
• Power dissipation on resistor:
• Branch, node, mesh
𝑝=𝑅𝑖2 =𝑣2 𝑅
Page 2

Topic 2 recap I
• Kirchhoff’s Current Law (KCL)
• The sum of all current entering and leaving a node is zero • Kirchhoff’s Voltage Law (KVL)
• Thesumofallvoltagesinaloopiszero 𝑀 𝑣 =0 𝑚=1 𝑚
𝑁 𝑖 𝑛=1 𝑛
= 0
• Ohm’s Law, KCL, and KVL can be applied to analyse any linear circuit
• Seriesresistors,𝑅𝑒𝑞 =𝑅1+𝑅2+⋯+𝑅𝑁
•Parallelresistors,1 =1+1⋯+1 𝑅eq 𝑅1 𝑅2 𝑅𝑁
• Voltage division in series resistors (2-resistor voltage divisor) •𝑣1=𝑅1𝑣, 𝑣2=𝑅2𝑣
• Current division in parallel resistors (2-resistor current divisor) •𝑖1=𝑅2 𝑖, 𝑖2=𝑅1 𝑖
• Simplify circuits by finding the equivalent resistance from a given set of terminals
• Series, parallel, or wye – delta transformation
𝑅1 +𝑅2 𝑅1 +𝑅2
𝑅1+𝑅2 𝑅1+𝑅2
Page 3

Nodal Analysis
• It is based on KCL and Ohm’s
law
• We use node voltages (potential of each node) as the main circuit variables
• It reduces the number of equations that must be solved simultaneously
• We first consider circuits with no voltage source
The reference node is a node where it is assumed to have zero potential 𝑣0 =0V,mostlyitisthe ground node
Commonly used symbols for the ground
• (a) & (b): Earth ground • (c): Chassis ground
Page 4

Nodal Analysis II
• Given a circuit with 𝒏 nodes, the nodal analysis is accomplished via the following steps:
• Select a node as the reference node.
• Assign voltages 𝑣1, 𝑣2, … , 𝑣𝑛 to the remaining 𝑛 − 1 nodes, voltages are relative to the reference node.
• Apply KCL to each of the 𝑛 − 1 non- reference nodes. Use Ohm’s law to express branch currents in terms of node voltages
𝑣−𝑣 𝑎𝑏
𝑅
• Solve the resulting 𝑛 − 1 simultaneous
equations to obtain node voltages.
𝑖=
Common choices for the reference node:
• The ground node • Top or bottom
node
• The node connected to the highest number of branches
Page 5

Nodal Analysis III
• Consider the following circuit
• Choose ground as reference node with 𝑣0 = 0 𝑉
• Assign voltage 𝑣1 and 𝑣2 to nodes 1 and 2 • Voltages across each resistor in terms of
node voltages:
𝑣𝑅1 = 𝑣1 − 0
𝑣 =𝑣−𝑣 𝑅2 1 2
𝑣𝑅3 = 𝑣2 − 0
• Also, since resistor is a passive element, by passive sign convention for Ohm’s Law, current must always flow from a higher potential to a lower potential
node 1
node 2
𝑣1 𝑣2
𝑣 𝑅2
𝑣𝑅1 𝑣𝑅3
𝑣
0
Ref. node Ground
𝑖 = 𝑣higher − 𝑣lower 𝑅
Page 6

Nodal Analysis IV
• Write KCL equations at nodes 1 and 2
node 1
node 2
𝑣2
𝑣𝑅3
Ref.node Ground
Node1: Node2:
𝐼 =𝐼 +𝑖 +𝑖 1212
𝐼 +𝑖 =𝑖 223
𝑣1
𝑣𝑅1
𝑣0
𝑣 𝐼=𝐼+1+1 2
• Apply Ohm’s Law for 𝑅1, 𝑅2, and 𝑅3
𝑖1 = 𝑣𝑅1 = 𝑣1 − 0 , 𝑅1 𝑅1
𝑖2=𝑣𝑅2 =𝑣1−𝑣2, 𝑅2 𝑅2
𝑖 =𝑣𝑅3 =𝑣2−0, 3 𝑅 𝑅
𝑣𝑅2
𝑣 −𝑣 33 12𝑅𝑅
Substitute 𝑖1, 𝑖2, and 𝑖3 back into the node equations
12
𝑣−𝑣𝑣
𝐼+1 2=2
Solve these two equations for 𝑣1 and 𝑣2
2
𝑅2 𝑅3
Page 7

Example 1
• Obtain the node voltages in the circuit below and find the power
dissipated on 4-Ω resistor
• (Worked solutions of examples are given in class)
Answer:
𝑣1 = 30 V 𝑣2 = −2.5 V
𝑝4Ω =900=255W 4
Example1 (Web view) Page 8

Nodal analysis with voltage source
• There are 2 cases for nodal analysis with voltage source (independent or dependent)
• Case 1: Voltage source is between v1 reference node and a non-reference node
Supernode
i4 v3
i3 6 Ω
4Ω
2Ωi1v2 5V i2
Set the voltage at the non-reference node equal to the voltage of the voltage source (beware the polarity)
10 V
8 Ω
Case 2
𝑣1 = 10 V
• Case 2: Voltage source is between two non-reference nodes
Case 1
A supernode is formed by enclosing a voltage source connected between two non-reference nodes and any element connected in parallel with it
• The two nodes form a supernode
• The voltage source can be expressed in terms of node voltages using definition of potential difference (or KVL in bottom-right mesh)
𝑣2 − 𝑣3 = 5 V
Page 9

Nodal analysis with voltage source II • Why do we need supernode?
• Nodal analysis requires applying KCL
• The current through the voltage source cannot be known in advance (Ohm’s law does not apply)
Supernode
i4
2Ωi1v2 5V
i2 i3
4Ω
v1 10 V
v3
Supernode properties
• Voltage source inside the supernode provides a constraint equation (extra equation needed for finding node voltages)
• A supernode has no voltage of its own
• We need to apply KCL to supernode using the already assigned node voltages inside it.
8 Ω
6 Ω
Page 10

Example 2
• For the circuit below, find the node voltages.
• (Worked solutions of examples are given in class)
Answer:
𝑣1 = − 22 = −7.333 V 3
𝑣2 = − 16 = −5.333 V 3
Example2 (Web view) Page 11

Practice problem 1
• In the circuit below, find the voltage 𝑣 and the current 𝑖 using nodal analysis
Answer:
𝑣 = −400 mV 𝑖 = 2.8 A
Page 12

Mesh Analysis
• It is based on KVL
• We use mesh currents instead of element currents as the main circuit variables
• It reduces the number of equations that must be solved simultaneously
• We first consider circuits with no current source
• Mesh is a loop that does not contain any other loop within it
(a)
A nonplaner circuit:
The branch with the 13-Ω resistor prevents the circuit from being drawn without crossing branches
Mesh Analysis is only applicable to planar circuits
A planer circuit can be drawn in a plane with no branches crossing one another
(b)
A planer circuit
(a) Original, (b) Redrawn circuit
Page 13

Mesh Analysis II
• Given a circuit with 𝒎 meshes, the mesh analysis is accomplished via the following steps:
• Identify all meshes in the circuit and assign mesh currents 𝑖1, 𝑖2, … , 𝑖𝑚 to the 𝑚 meshes with a direction (clockwise (CW) or counter clockwise (CCW).
• Apply KVL to each of the 𝑚 meshes following the same direction as mesh currents. Use Ohm’s law to express branch voltages in terms of mesh currents
𝑣 = 𝑅𝑖
(𝑖 can be a mesh current or the difference between
two mesh currents)
• Solve the resulting 𝑚 − 1 simultaneous equations to obtain mesh currents.
Mesh currents are not always the same as branch currents depending whether the branch is an isolated one or a shared one between two adjacent meshes.
Page 14

Mesh Analysis III
• Consider the following circuit
• Assign currents 𝑖1 and 𝑖2 to meshes 1 and 2
• The current for each branch is obtained as follow:
𝐼=𝑖 11
𝐼=𝑖 22
𝐼=𝑖−𝑖 312
For shared branches, mesh current is the difference between the two adjacent mesh currents (depending on which mesh you are applying KVL or the direction of the
sharedbranchcurrent,𝐼 =𝑖 −𝑖) 312
For isolated branches, mesh current is the same as branch current
(𝐼 =𝑖 and𝐼 =𝑖) 11 22
Page 15

Mesh Analysis IV
• Write KVL equations in meshes 1 and 2
Mesh1: −𝑉+𝑅𝑖+𝑣 =0 1 11 𝑏𝑒
Mesh2: 𝑉+𝑣 +𝑅𝑖=0 2 𝑒𝑏 22
• Apply Ohm’s Law 𝑅3
𝑣𝑏𝑒 = 𝑅3(𝑖1 − 𝑖2) Purple arrow
𝑣𝑒𝑏 = 𝑅3(𝑖2 − 𝑖1) Blue arrow Substitute 𝑣𝑏𝑒 and 𝑣𝑒𝑏 back
into the mesh equations:
−𝑉 + 𝑅 𝑖 + 𝑅 (𝑖 − 𝑖 ) = 0 111312
−𝑅 (𝑖 − 𝑖 ) + 𝑅 𝑖 + 𝑉 = 0 312222
+−
𝑣𝑏𝑒 𝑣𝑒𝑏
−+
Choose the direction of current through shared branch based on the direction of mesh current for which you are writing KVL so that passive sign convention enables you to use positive sign for voltage drop across shared elements
Solve these two equations for 𝑖1 and 𝑖2
Page 16

Example 3
• Calculate the mesh currents 𝑖1 and 𝑖2 in the circuit below, and find the
power supplied by 40-𝑉 voltage.
• (Worked solutions of examples are given in class)
Answer:
𝑖1 = 23 = 4.6 A 5
𝑖2 = 1 = 0.2 A 5
Example3 (Web view) Page 17

Mesh analysis with current source • There are 2 cases for mesh analysis with
current source (independent or dependent)
Case 1
10 A
• Case 1: Current source exists only in one mesh (or in an isolated branch)
i2 Supermesh 2 Ω
Exclude them 5V
Set the mesh current equal to the current of the current source (beware the direction)
8Ω
2Ωi1 i36Ω
𝑖2 = 10 A
• Case 2: Current source is between two meshes (shared current source)
i1
6 A Case 2 i3
A supermesh is formed by merging two meshes and excluding the shared current source and any element connected in series with it
• The two meshes form a supermesh
• The current source can be expressed in terms of mesh currents using KCL (bottom-right node)
𝑖3 = 𝑖1 + 6 A
Page 18

Mesh analysis with current source II • Why do we need supermesh?
10 A
• Mesh analysis requires applying KVL • The voltages across the current
source cannot be known in advance (Ohm’s law does not apply)
i2 2 Ω
8Ω
2Ωi1 i36Ω
6A
i1 i3
Supermesh
5V
Supermesh properties
• Shared current source inside the supermesh provides a constraint equation (extra equation needed for finding mesh current)
• A supermesh has no current of its own
• We need to apply KVL to supermesh using the already assigned mesh currents inside it.
Intersecting supermeshes in a circuit must be combined to form a larger supermesh
Page 19

Example 4
• For the circuit below, find the mesh currents 𝑖1 to 𝑖4.
• (Worked solutions of examples are given in class)
4Io
Answer:
𝑖1 = −6.5 A 𝑖2 = −1.5 A 𝑖3 = 2.5 A 𝑖4 = 1 A
Example4 (Web view) Page 20

Practice problem 2
• In the circuit below, use mesh analysis to determine 𝑖1, 𝑖2 , and 𝑖3.
Answer:
𝑖1 = 4.8 A 𝑖2 = 0.8 A 𝑖3 = 1.6 A
Page 21

Questions!?
Page 22