Loop (Mesh) Analysis (3.2)

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Loop (Mesh) Analysis (3.2)

• Prof. Phillips
• February 17, 2003

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Loop Analysis

• Nodal analysis was developed by applying KCL at
  each non-reference node.
• Loop analysis is developed by applying KVL around
  loops in the circuit.
• Loop (mesh) analysis results in a system of
  linear equations which must be solved for unknown
  currents.

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Example A Summing Circuit

• The output voltage V of this circuit is
  proportional to the sum of the two input voltages
  V1 and V2.
• This circuit could be useful in audio
  applications or in instrumentation.
• The output of this circuit would probably be
  connected to an amplifier.

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Summing Circuit
1kW
1kW



V1
Vout
1kW
V2

• Solution Vout (V1 V2)/3

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Steps of Mesh Analysis

• 1. Identify mesh (loops).
• 2. Assign a current to each mesh.
• 3. Apply KVL around each loop to get an equation
  in terms of the loop currents.
• 4. Solve the resulting system of linear equations.

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Identifying the Meshes
1kW
1kW
1kW
Mesh 2
Mesh 1


V1
V2
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Steps of Mesh Analysis

• 1. Identify mesh (loops).
• 2. Assign a current to each mesh.
• 3. Apply KVL around each loop to get an equation
  in terms of the loop currents.
• 4. Solve the resulting system of linear equations.

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Assigning Mesh Currents
1kW
1kW
1kW


V1
V2
I1
I2
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Steps of Mesh Analysis

• 1. Identify mesh (loops).
• 2. Assign a current to each mesh.
• 3. Apply KVL around each loop to get an equation
  in terms of the loop currents.
• 4. Solve the resulting system of linear equations.

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Voltages from Mesh Currents
VR


I2
R
R
I1
I1
VR I1 R
VR (I1 - I2 ) R
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KVL Around Mesh 1

• -V1 I1 1kW (I1 - I2) 1kW 0
• I1 1kW (I1 - I2) 1kW V1

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KVL Around Mesh 2

• (I2 - I1) 1kW I2 1kW V2 0
• (I2 - I1) 1kW I2 1kW -V2

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Steps of Mesh Analysis

• 1. Identify mesh (loops).
• 2. Assign a current to each mesh.
• 3. Apply KVL around each loop to get an equation
  in terms of the loop currents.
• 4. Solve the resulting system of linear equations.

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Matrix Notation

• The two equations can be combined into a single
  matrix/vector equation.

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Solving the Equations

• Let V1 7V and V2 4V
• Results
• I1 3.33 mA
• I2 -0.33 mA
• Finally
• Vout (I1 - I2) 1kW 3.66V

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Another Example
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1. Identify Meshes
2kW
Mesh 3
2mA
1kW
Mesh 2
Mesh 1
2kW

12V
4mA
I0
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2. Assign Mesh Currents
2kW
2mA
1kW
2kW

12V
4mA
I0
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Current Sources

• The current sources in this circuit will have
  whatever voltage is necessary to make the current
  correct.
• We cant use KVL around the loop because we dont
  know the voltage.
• What to do?

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Current Sources

• The 4mA current source sets I2
• I2 -4 mA
• The 2mA current source sets a constraint on I1
  and I3
• I1 - I3 2 mA
• We have two equations and three unknowns. Where
  is the third equation?

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The Supermesh!
The Supermesh does not include this source!
2kW
The Supermesh surrounds this source!
2mA
I3
1kW
2kW

12V
4mA
I1
I2
I0
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KVL Around the Supermesh

• -12V I3 2kW (I3 - I2)1kW (I1 - I2)2kW 0
• I3 2kW (I3 - I2)1kW (I1 - I2)2kW 12V

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Matrix Notation

• The three equations can be combined into a single
  matrix/vector equation.

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Solve Using MATLAB

• gtgt A 0 1 0 1 0 -1
• 2e3 -1e3-2e3 2e31e3
• gtgt v -4e-3 2e-3 12
• gtgt i inv(A)v
• i 0.0012
• -0.0040
• -0.0008

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Solution

• I1 1.2 mA
• I2 -4 mA
• I3 -0.8 mA
• I0 I1 - I2 5.2 mA

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Class Examples