Circuit Analysis Methods

Circuit Analysis Methods

1. Branch Current Analysis 2. Loop (Mesh) Analysis

3. Node Analysis

“Where more than one current or voltage source exist which are difficult to solve by using simple circuit analysis,”

2

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.

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

4

Summing Circuit

Solution: V_out = (V₁ + V₂)/3

+ – V_out

1kW 1kW 1kW

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.

6

Mesh 2 1kW 1kW 1kW

Identifying the Meshes

V₁ Mesh 1 + 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.

8

1kW 1kW 1kW

Assigning Mesh Currents

V₁ I₁ I₂ + 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.

10

Voltages from Mesh Currents

R

I₁

+ V_R –

V_R = I₁ R

R

I₁

+ V_R – I₂

V_R = (I₁ - I₂ ) R

KVL Around Mesh 1

-V₁ + I₁ 1kW + (I₁ - I₂) 1kW = 0 I 1kW + (I - I ) 1kW = V

1kW 1kW 1kW

V₁ I₁ I₂ + V₂

– +

–

12

KVL Around Mesh 2

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

1kW 1kW 1kW

V₁ I₁ I₂ + 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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Matrix Notation

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

úû ê ù

ë é

= - úû ê ù

ë úé û ê ù

ë é

W +

W W

-

W -

W +

W

2 1 2

1

k 1 k

1 k

1

k 1 k

1 k

1

V V I

I

Solving the Equations

Let: V₁ = 7V and V₂ = 4V Results:

I₁ = 3.33 mA I₂ = -0.33 mA Finally

) 1kW

16

Example 2

Step 3: Apply KVL in each loop

Loop 1:

Loop 2:

18

Example 3

1kW 2kW 2kW

12V 4mA

2mA

I₀ +

–

Mesh 2 Mesh 3

Mesh 1

1. Identify Meshes

1kW 2kW

2kW

12V 4mA

2mA

I₀ +

–

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2. Assign Mesh Currents

I₁ I₂

I₃

1kW 2kW

2kW

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?

22

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?

1kW 2kW

2kW

12V 4mA

2mA

I₀

I₁ I₂

I₃ The

Supermesh surrounds this source!

The Supermesh does not include this source!

+ –

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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

Matrix Notation

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

úú ú û ù êê

ê ë é-

= úú ú û ù êê

ê ë é úú ú û ù êê

ê ë é

W +

W W

- W -

W

-

V 12

mA 2

mA 4

1k 2k

2k 1k

2k

1 0

1

0 1

0

3 2 1

I I I

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Solve Using MATLAB

>> A = [0 1 0; 1 0 -1;

2e3 -1e3-2e3 2e3+1e3];

>> v = [-4e-3; 2e-3; 12];

>> i = inv(A)*v i = 0.0012

-0.0040 -0.0008

Solution

I₁ = 1.2 mA I₂ = -4 mA I₃ = -0.8 mA

I₀ = I₁ - I₂ = 5.2 mA

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Exercise 1

Exercise 2

Exercise 3

Q. Determine each branch currents/loop currents.