Chapter 3 Methods of Analysis

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Chapter 3 (Methods of Analysis)
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Contents

• Nodal analysis
• Mesh analysis

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Nodal analysis
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• Node is the point of connection between two or
  more branches.
• The above circuit have 4 node.

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Basic Step

• Mark the nodes
• Reference node
• Decide on number equations required
• Perform KCL at the selected nodes
• Solve the equation

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Example 3.1- Calculate node voltages and currents
in circuit shown

• Applying KCL Ohms law
• Node1
• .
• .
• Node 2
• .
• .

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Method 1- Elimination technique

• (1) (2)
• 4V2 80
• V2 20V ----(3)

• Insert equ (3) into (1)
• 3V1 20 20
• V1 13.333V

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Method 2- Crammers rule

• 3V1 V2 20
• -3V1 5V2 60
• Determinant

• V1?1/ ?
• V2?2/ ?

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• For current
• i1 5A
• i4 10A
• i2 (V1-V2)/4 -1.6667A
• i3 V1/ 2 6.666A
• i5 V2/6 3.333A

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P.P 3.1- Find node voltages
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Example

• Find the node voltages

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Example 3.2 Find node voltages

• Node 1
• 12 3V1 2V2 V3
• Node 2
• 0 - 4V1 7V2 V3
• Node 3
• where
• 0 2V1 3V2 V3

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• 12 3V1 2V2 V3
• 0 - 4V1 7V2 V3
• 0 2V1 3V2 V3

V3?3/ ?
V1?1/ ?
V2?2/ ?
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• 3(7-3) 4 (- 2 3) 2 (2 7)
• 10
• 48

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• 24
• - 24
• V1 48/10 4.8V V2 24/10 2.4V
• V3 -24/10 -2.4V

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P.P 3.2 Find node voltages

• Node 1
• Node 2
• Node 3

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Nodal analysis with voltage sources
A supernode is formed by enclosing a (dependent
or independent) voltage source connected between
two nonreference nodes and any elements connected
in parallel with it
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Nodal analysis with voltage sources

• Apply both KCL KVL
• to determine the node
• voltage

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• At supernode
• KCL

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• KVL
• -V2 5 V3 0
• V2 V3 5

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Example 3.3 Find node voltages

• Supernode 2V 10?
• KCL
• .
• .
• .
• 2V1 V2 - 20 ---- (1)

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• KVL
• -V1 2 V2 0
• V2 V1 2 ----(2)
• From equ (1) (2)
• V1 -7.333 V
• V2 - 5.333 V

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Example 3.4 Find node voltage in the circuit

• Node 1 2 supernode
• Node 3 4 supernode
• Supernode 1 2
• 10 i3 i1 i2
• 60 5V1 V2 V3 - 2V4
• ----- (1)

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• Supernode 3 4
• i1 i3 i4 i5
• 4V1 2V2 5V3 -16V4 0
• ---- (2)

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• KVL
• Loop 1
• -V1 20 V2 0
• V1-V2 20 ---- (3)
• Loop 2
• -V3 3Vx V4 0
• but Vx V1-V4
• 3V1 V3 2V4 0 ---- (4)
• Loop 3
• Vx 3Vx 6i3 -20 0
• but 6i3 V3 V2
• -2V1 V2 V3 2V4 20 ---- (5)

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• Four voltages V1, V2, V3 V4 require four
• out of five equation. Use equ (1) to (4), equ
• (5) used to check results
• From equ (3)
• V2 V1 20
• Substitute in equ (1) (2)
• 60 5V1 (V1-20) V3 2V4
• 80 6V1 V3 2V4 ---- (6)

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• 4V1 2 (V1 - 20) 5V3 16V4 0
• 6V1 5V3 -16V4 40 ---- (7)
• From equ. (4), (6) (7), we get
• 3V1 V3 2V4 0
• 80 6V1 V3 2V4
• 6V1 5V3 -16V4 40

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• By using Crammers rule or calculator, the
• answers are -
• V1 26.667V
• V3 173.333V
• V4 -46.667V
• V2 6.667V

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Mesh analysis
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• Nodal analysis applies KLC to find unknown
  voltages
• Mesh analysis- applies KVL to find unknown
  currents.

A mesh is a loop which does not contain any other
loops within it
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Basic Steps

• Label all meshes in the circuit with a clockwise
  mesh current
• Decide on number equations required
• Perform KVL at the each meshes
• Solve the equation

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• KVL
• Mesh 1
• -V1 R1i1 R3 (i1-i2) 0
• (R1R3)i1 R3i2 V1
• Mesh 2
• R2i2 V2 R3(i2-i1) 0
• -R3i1 (R2R3)i2 -V2

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Example 3.5 Find current branch i1, i2 i3

• KVL
• Mesh 1
• -155i110(i1-i2)10 0
• 3i1 2i2 1 ---- (1)
• Mesh 2
• 6i2 4i2 -10 10 (i2-i1) 0
• i1 2i2 -1 ---- (2)

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Method 1 Substitution method

• 3i1 2i2 1 ---- (1)
• i1 2i2 -1 ---- (2)
• Substitute equ (2) into (1)
• 6i2 3 2 i2 1
• i2 1A

• From equ (2)
• i1 2(1) 1 1A
• For i3
• i1 i2 i3
• i3 i1 i2 0

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Method 2 Cramers rule
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Example 3.6 use mesh analysis to find i0

• KVL
• Mesh 1
• -24 10(i1-i2) 12(i1-i3) 0
• Mesh 2
• 24i2 4(i2-i3) 10(i2-i3) 0
• Mesh 3
• 4i0 12(i3-i1) 4(i3 i2) 0
• For i0
• io i1 i2

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P.P 3.6
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Problem 3.51- Applying mesh to find V0
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Mesh analysis with current sources
A supermesh results when two meshes have a
(dependent or independent) current source in
common
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Example
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• Applying KVL to supermesh
• -20 6i1 10i2 4i2 0
• 6i1 14i2 20

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• Applying KCL to node 0
• i2 i1 6

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• Solving equation, we get
• i1 -3.2A
• i2 2.8A

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P.P 3.7

• Use mesh analysis to determine i1,i2 i3

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answer

• i1 3.474A
• i2 473.7 mA
• i3 1.1052A

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Using mesh analysis to find currents i1 i2
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answer

• i1 1.072A and i2 2.041A

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Using mesh analysis, find V0 and i0 for fig.
shown below
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Find the mesh currents i1 to i3 in the network of
fig. below
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For fig. shown below, find i1 to i4 using mesh
analysis