Chapter 3

Methods of Analysis

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So far, we have analyzed relatively simple circuits by applying Kirchhoff’s laws in combination with Ohm’s law. We can use this approach for all circuits, but as they become structurally more complicated and involve more and more elements, this direct method soon becomes cumbersome. In this chapter we introduce two powerful techniques of circuit analysis:

Nodal Analysis

and

Mesh Analysis .

These techniques give us two systematic methods of describing circuits with the minimum number of simultaneous equations. With them we can analyze almost any circuit by to obtain the required values of current or voltage. SJTU 2

Nodal Analysis

• Steps to Determine Node Voltages: 1. Select a node as the reference node(ground), define the node voltages  1 ,  2 ,…  n-1 to the remaining n-1nodes . The voltages are referenced with respect to the reference node.

2. Apply KCL to each of the n-1 independent nodes. Use Ohm’s law to express the branch currents in terms of node voltages.

3. Solve the resulting simultaneous equations to obtain the unknown node voltages.

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Fig. 3.2 Typical circuit for nodal analysis

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i



v higher



v lower R

So at node 1 and node 2, we can get the following equations.

I I

1 2 

I

2 

i

2  

i i

1 3 

i

2

I I

2 1  

I v

1 2 

v

1

R

1 

v

2  

R

2

v v

1

R

3 2 

R

2

v

2 SJTU 5

In terms of the conductance, equations become

I

1

I

2  

I

2

G

2  (

v G

1

v

1  1 

v

2 )

G

2 (

v

1 

G

3

v

2 

v

2 ) Can also be cast in matrix form as  

G

1  

G G

2 2

G

2 

G

2 

G

3    

v v

2 1    

I

1

I

2

I

2   Some examples SJTU 6

Fig. 3.5 For Example 3.2: (a) original circuit, (b) circuit for analysis SJTU 7

Nodal Analysis with Voltage Sources(1)

• Case 1 If a voltage source is connected between the reference node and a nonreference node, we simply set the voltage at the nonreference node equal to the voltage of the voltage source. As in the figure right:

v

1  10

V

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Nodal Analysis with Voltage Sources(2)

• Case 2 If the voltage source (dependent or independent) is connected between two nonreference nodes, the two nonreference nodes form a supernode; we apply both KVL and KCL to determine the node voltages. As in the figure right:

i

1

or



and i

4

v

1 

v

2

i

2

v

 2

i

2 

v

3 3 

v

 1 5 

v

3 4 

v

2 8  0 

v

3  6 0 SJTU 9

Nodal Analysis with Voltage Sources(3)

• Case 3 If a voltage source (dependent or independent) is connected with a resistor in series, we treat them as one branch. As in the figure right:

i



V

11 

V

22 

V

1

R

1 R1 V 1 1 i V 11 V 22 SJTU 10

Nodal Analysis with Voltage Sources(3)

Example P113 SJTU 11

Mesh Analysis

• Steps to Determine Mesh Currents: 1. Assign mesh currents i 1 , i 2 ,…i n to the n meshes.

2. Apply KVL to each of the n meshes. Use Ohm’s law to express the voltages in terms of the mesh currents.

3. Solve the resulting n simultaneous equations to get the mesh currents.

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Fig. 3.17 A circuit with two meshes SJTU 13



or V R

2

i

2 1  

R

1

i

1

V

2  

R

3 (

R

3 (

i

2

i

1  

i i

1 ) 2 )   0 0 ( 

R

1

R

3 

i

1

R

3  )

i

1 (

R

2  

R

3

i

2

R

3 )

i

2 

V

1  

V

2 In matrix form:  

R

1  

R

3

R

3

R

2 

R

3 

R

3    

i i

2 1     

V

1 

V

2   SJTU 14

Fig. 3.18 For Example 3.5

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Mesh Analysis with Current Sources(1)

• Case 1 When a current source exists only in one mesh: Consider the figure right.

i

2   10   2

A



i

1 4

i

1   6  2

A

(

i

1 

i

2 )  0 SJTU 16

Mesh Analysis with Current Sources(2)

• Case 2 When a current source exists between two meshes:Consider the figure right.

2 solutions: 1. Set v as the voltage across the current source, then add a constraint equation.

2. Use supermesh to solve the problem.

R1 R2 V 1

i a

R4 + v -

i b

R3 Is

i c

R5 V 2 SJTU 17

R1 Solution 1: V 1 R2

i a

R4 + v -

i b

Is R3

i c

R5 V 2

R

2 (

i a R

1

i b

 

i b

)

R

3 (

i b



v

 

i c

)

R

4

i a



R

2  (

i b V

1 

i a

)

R

5

i and c



v i a

 

R

3 (

i c i c

 

Is i b

)  

V

2  0 supermesh Solution 2:

R

2 (

ia



ib

) 

R

3 (

ic



ib

) 

V

2 

R

5

ic



R

4

ia



V

1  0 Note: 1, The current source in the supermesh is not completely ignored; it provides the constraint equation necessary to solve for the mesh current.

2, A supermesh has no current of its own.

3, A supermesh requires the application of both KVL and KCL.

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Fig. 3.24 For Example 3.7

2

i

1  4

i

3  8 (

i

3 

i

4 )  6

i and

2

i

4

i

2  8 (

i

4 

i

1  5 

i

3 )  10

i

2   0

i

3 2  0  3

i

0

i

0  

i

4 SJTU 19

Fig. 3.31 For Example 3.10

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Fig. 3.32 For Example 3.10; the schematic of the circuit in Fig. 3.31.

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Nodal Versus Mesh Analysis

• Both provide a systematic way of analyzing a complex network.

• When is the nodal method preferred to the mesh method?

1. A circuit with fewer nodes than meshes is better analyzed using nodal analysis, while a circuit with fewer meshes than nodes is better analyzed using mesh analysis.

2. Based on the information required.

Node voltages required--------nodal analysis Branch or mesh currents required-------mesh analysis SJTU 22