LOOP ANALYSIS

VSR

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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

Solution: Vout = (V1 + V2)/3

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+

–

Vout

1k

1k

1k

V1 V2

+–

+–

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

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Mesh 2

1k

1k

1k

V1 V2Mesh 1+–

+–

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

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1k

1k

1k

V1 V2I1 I2+–

+–

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

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R

I1

+ –VR

VR = I1 R

R

I1

+ –VRI2

VR = (I1 - I2 ) R

KVL Around Mesh 1

-V1 + I1 1k + (I1 - I2) 1k = 0

I1 1k + (I1 - I2) 1k = V1

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1k

1k

1k

V1 V2I1 I2+–

+–

KVL Around Mesh 2

(I2 - I1) 1k + I2 1k + V2 = 0

(I2 - I1) 1k + I2 1k = -V2

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1k

1k

1k

V1 V2I1 I2+–

+–

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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2

1

2

1

k1k1k1

k1k1k1

V

V

I

I

Solving the Equations

Let: V1 = 7V and V2 = 4V

Results:I1 = 3.33 mA

I2 = -0.33 mA

FinallyVout = (I1 - I2) 1k = 3.66V

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Another Example

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1k

2k

2k

12V 4mA

2mA

I0

+–

1. Identify Meshes

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Mesh 2

Mesh 3

Mesh 1

1k

2k

2k

12V 4mA

2mA

I0

+–

2. Assign Mesh Currents

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I1 I2

I31k

2k

2k

12V 4mA

2mA

I0

+–

Current Sources

The current sources in this circuit will

have whatever voltage is necessary

to make the current correct.

We can’t use KVL around the loop

because we don’t 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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1k

2k

2k

12V 4mA

2mA

I0

I1 I2

I3

The Supermesh surrounds this source!

The Supermesh

does not include this

source!

+–

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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V12

mA2

mA4

1k2k2k1k2k

101

010

3

2

1

I

I

I

Solution

I1 = 1.2 mA

I2 = -4 mA

I3 = -0.8 mA

I0 = I1 - I2 = 5.2 mA

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THANK YOU

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