Chapter 4

Chapter 4

Methods of Analysis of Resistive Circuits

Seoul National University

Department of Electrical and Computer Engineering

Prof. SungJune Kim

Node Voltage Analysis of Circuits with Current Sources



The nodes of a circuit are the places where the elements are connected together.



n nodes will require n-1 KCL equations.



The voltage at any node of the circuit, relative to the reference node, is called a node voltage.

Figure 4.2-1 (p. 112)

Figure 4.2-1 (b) - 3 node (a, b, c)

- 2 node voltage (v_a, v_b) -

v

_c: reference node

Node Equations



To write a set of node equations, we do two things:

1. Express element currents as functions of the node voltage.

2. Apply Kirchhoff’s current law (KCL) at each of the nodes of the circuit, except for the reference node.

Figure 4.2-3 (p. 114)

Node voltages, v₁ and v₂, and element voltage, v_1-v₂, of a (a) generic circuit element, (b) voltage source, and (c) resistor.

Rules



Rule #1:Express currents in terms of node voltages: Count the number of

unknown node voltages and the required number of KCL equations to solve.

Example of node equations



Example of node equation

1. Express element currents as functions of the node voltage

(c) The resistor currents expressed as functions of the node voltages.

(b) The resistor voltages expressed as functions of the node voltages.

Figure 4.2-4 (p. 114)

(a) A circuit with three resistors.

Example of node equations



Example of node equation

2. Apply Kirchhoff’s current law (KCL) at each of the nodes of the circuit, except for the reference node.

1

2

R

v v R

i

_s

v

^{a a}



^b





<Node a>

3

1

R

v R

v

v

_a



_b



_b

<Node b>

Example 4.2-1 Node equations



Determine the value of the resistance R in the circuit shown in Figure 4.2-5a

Figure 4.2-5 (p. 116)

(a) The circuit for Example 4.2-1. (b) The circuit after the resistor currents are expressed as functions of the node voltages.

Solution

1. Apply KCL at node a to obtain

2. Using v

_b

=5V, gives

3. Solving for v

_a

, we get

4. Next, apply KCL at node b to obtain

5. Using v

_a

=-10V and v

_b

=5V gives 5 0

4  10 v

^a

 v

^a

 v

^b



) V (

 10 v

a

0 5    4 



 



  

R v v

v

_{a b b}

5 0 5 4  10 v

^a

 v

^a

 

0 5 4

5 5

10    



 



  

 R

Example 4.2-2 Node equations



Obtain the node equations for the circuit in Figure 4.2-6.

Figure 4.2-6 (p. 116)

The circuit for Example 4.2-2.

Solution

1. Apply KCL at node a to obtain

2. Apply KCL at node b to obtain

3. Node equation at node c

2 1 2

1 5

3 2

1

5 2

2 1

1

1 1 1

1 1

1

0

i i R v

v R v R

R R

R

R v i v

R v i v

R v v

c b

a

b a c

a c

a



 



 



 

 



 



 

 

 



  

 



 



  

 



 



  

 



 



  

2 3 3

5 4

3 5

3 4

3 5

2

1 1

1 1

1

0

i i R v

R v R

v R R

R i v R

v v R

v i v

c b

a

b c

b b

a



 



 



 

 

 



  

 



 



 



 



 



 

 

 



  

 

 



  



1 1

1 1

1 1

1    







 

 



 



Node Voltage Analysis Method with a Voltage Sources



Case 1



The voltage source connects node q and the reference node (ground).



Method:



Set v

_q

equal to the source voltage accounting for the polarities and proceed to write the KCL at the remaining nodes.

Figure 4.3-1 (p. 119)

Circuit with an independent voltage source and an independent current source.

Rules



Rule #1: Express currents in terms of node voltages: Count the number of unknown node voltages and the required number of KCL equations to solve.



Rule #2: If a node voltage is known, use it.: Then the number of unknown is

one less.

Node Voltage Analysis Method with a Voltage Sources



Case 1



The voltage source connects node q and the reference node (ground).

1. at node a

2. at node b

3. Solving for the unknown node voltage v

_b

s

a

v

v 

2

3

R

v v

R

i

_s

 v

^b



^b



^a

2

3

R

v v R

i

_s

v

^{b b}



^s





3 2

3 3

2

R R

v R i

R

v

_b

R

^{s s}



 

s

a v

v 

Circuit with an independent voltage source and an independent current source.

Node Voltage Analysis Method with a Voltage Sources



Case 2



The voltage source lies between two nodes, a and b



Method:



Create a(one) supernode that incorporates a and b and equate the sum of

all the currents into the supernode to zero.

Node Voltage Analysis Method with a Voltage Sources



Case 2



of a Supernode 1. voltage source

2. At supernode

3. Solving for the unknown node voltage v

_b

b s

a s

b

a

v v v v v

v     

s b

a

i

R v R

v  

2 1

s b

b

s

i

R v R

v

v   

2 1

2 1

2 2

1

R R

v R i

R

v

_b

R

^{s s}



 

b s

a v v

v  

Circuit with a supernode that incorporates v_aand v_b.

A supernode consists of two nodes

connected by an independent or a

dependent voltage source

Rules



Rule #1: Express currents in terms of node voltages: Count the number of unknown node voltages and the required number of KCL equations to solve.



Rule #2: If a node voltage is known, use it.: Then the number of unknown is one less.



Rule #3: Supernode makes the KCL equation simpler by removing the need

to introduce another variable, a current through the voltage source whether

it is an independent or a dependent source.



Determine the node voltages, v1, v2, in the circuit shown in Figure 4.3-3a.

Example 4.3-1 Node Equations

Figure 4.3-3 (p. 112)

The circuit considered in Example 4.3-1.

Solution



80 0 60 50

65

1 1 2

1

 v  v  v  

v

80 ) 60

65 ( 1 80 )

1 65

1 50

( 1   v

₁

 v

₂



1 . 75 0

60 65

2 1

2

 

 v  v

v ) 0 . 1

75 1 65 ( 1 65 )

( 1

₁

 

₂



 v v



Determine the values of the node voltages, v

_a

and v

_b

, for the circuit shown in Figure 4.3-4

Example 4.3-2 Supernode

Solution

Method 1:

the KCL equation at node a

the KCL equation at node b

Combining two equations

12

12   





_{a b a}

b

v v v

v

3 0 5 .

3  

 v

^b

i

5 6 .

1 v

^a

i 



3 0 6

. 6 2

5 3 . 3 5 .

1 v

^b

  v

^a

   v

^a

 v

^b



 



 



Solution

Method 2:

Apply KCL to the supernode

Applying KCL to the supernode is a shortcut for doing three things:

1. Labeling the voltage source current as i

2. Applying KCL at both nodes of the voltage source 3. Eliminating i from the KCL equations

In summary,

V 0 V

12 





_b

a

and v

v

3 0 6

. 3 2

5 . 6 3 5 .

1 v

^a

v

^b

v

^a

v

^b















0 . 3 2

6

12











b a

a b

v v

v

v Solving the equation



Determine the node voltages for the circuit shown in Figure 4.3-7.

Example 4.3-3 Node Equations for a Circuit Containing Voltage Sources

Figure 4.3-7 (p. 114)

Solution

1. First notice that

2. KCL equation for the supernode

3. Using v

_a

=v

_c

+10 and v

_b

=-12 10

) V ( 12









c a

b

v v

v

) V ( 4

120 )

12 ( 5 )

10 (

4













c c c

v v v

120 5

4

40 5 10 2







 



 

b c

a

b c

b a

v v

v

v v

v

v supernode



When a circuit contains a dependent source, the controlling

current or voltage of that dependent source must be expressed as a function of the node voltage

Node Voltage Analysis with Dependent Source

Rules



Rule #1: Express currents in terms of node voltages: Count the number of unknown node voltages and the required number of KCL equations to solve.



Rule #2: If a node voltage is known, use it.: Then the number of unknown is one less.



Rule #3: Supernode makes the KCL equation simpler by removing the need to introduce another variable, a current through the voltage source whether it is an independent or a dependent source.



Rule #4: If there is a controlled source : Express its controlling source with

node voltages. You probably have to do this first.

Example 4.4-1 Node Equations for a Circuit Containing a Dependent Source



Determine the node voltages for the circuit shown in Figure 4.4-1

Figure 4.4-1 (p. 123) A circuit with a CCVS.

Rules



Rule #1: Express currents in terms of node voltages: Count the number of unknown node voltages and the required number of KCL equations to solve.



Rule #2: If a node voltage is known, use it.: Then the number of unknown is one less.



Rule #3: Supernode makes the KCL equation simpler by removing the need to introduce another variable, a current through the voltage source whether it is an independent or a dependent source.



Rule #4: If there is a controlled source : Express its controlling source with

node voltages. You probably have to do this first.

Solution

1. Controlling current of the dependent source (i

_x

)

2. Node voltage at node a

3. Node voltage at node c

4. Apply KCL at node b 6

b a x

v i v 



4 2 6

3 8

3

_x ^{b b}

c

v i v

v   



 



  



6 V 8

8

_x ^b

a

i v

v 







2 4 4 8 2

2 3 6

8





 

 

  





 

 

 

b b

b b

c b b

v v v

v

v v v

5. Solving for v

_b

and v

_c

2 V 1 4 2

V 7









b c

b

v v

v

Example 4.4-2



Determine the node voltages for the circuit shown in Figure 4.4-2

Figure 4.4-2 (p. 124) A circuit with a VCVS.

Rules



Rule #1: Express currents in terms of node voltages: Count the number of unknown node voltages and the required number of KCL equations to solve.



Rule #2: If a node voltage is known, use it.: Then the number of unknown is one less.



Rule #3: Supernode makes the KCL equation simpler by removing the need to introduce another variable, a current through the voltage source whether it is an independent or a dependent source.



Rule #4: If there is a controlled source : Express its controlling source with

node voltages. You probably have to do this first.

Solution

1. Controlling voltage of the dependent source (v

_x

)

2. Node voltage at node a and b

3. Apply KCL to the supernode

a

x

v

v  

a b

a a

x b

a

v v

v v

v v

v

5

4 )

( 4 4















a a

a

b a

v v v

v v

4 3 10

5 3 4

10 3 4









 4. Solving for v

_a

and v

_b

V 20 5

V 4







a b

a

v v

v

Example 4.4-3



Determine the node voltages corresponding to nodes a and b for the circuit shown in Figure 4.4-3

Figure 4.4-3 (p. 124) A circuit with a CCCS.

Rules



Rule #1: Express currents in terms of node voltages: Count the number of unknown node voltages and the required number of KCL equations to solve.



Rule #2: If a node voltage is known, use it.: Then the number of unknown is one less.



Rule #3: Supernode makes the KCL equation simpler by removing the need to introduce another variable, a current through the voltage source whether it is an independent or a dependent source.



Rule #4: If there is a controlled source : Express its controlling source with

node voltages. You probably have to do this first.

Solution

1. Controlling current of the dependent source (i

_a

) Apply KCL at node a

v

_a

=0

2. Apply KCL at node b

3. Solving for v

_b

20 10

6

_{a b}

a

a

v v

v  i  



20 12

_b

a

i   v

 

 



 

 

 

20 5 12 20

0

20 5 0

b b

a b

v v

v i

V

 10

b



v



A mesh is a loop that does not contain any other loops within it.

Mesh Current Analysis with Independent Voltage Source

Figure 4.5-1 (p. 126)

Nonplanar circuit with a crossover.

Figure 4.5-2 (p. 126)

Circuit with four meshes. Each mesh is identified by dashed lines.



Loop



A closed path drawn by starting at a node and tracing a path such that we return to the original node

without passing an intermediate node more than once



Mesh



A loop that does not contain any other loops within it.



Mesh current



The current that flows through the

Loop, Mesh, Mesh current

Figure 4.5-3 (p. 126)



To write a set of mesh equations, we do two things:

1. Express element voltages as functions of the mesh currents.

2. Apply Kirchhoff’s voltage law (KVL) to each of the meshes of the circuit.

Mesh equations

Figure 4.5-4 (p. 127)

Mesh currents, i₁and i₂, and element current, i₁ – i₂, of a (a) generic circuit element, (b) current source, and (c) resistor.



Example of mesh current analysis

1. Express element voltages as functions of the mesh currents.

Example of Mesh Current Analysis

Figure 4.5-5 (p. 128) (a) A circuit.

(b) The resistor currents expressed as functions of the mesh currents.

(c) The resistor voltages expressed as functions of the mesh currents.



Example of mesh current analysis

2. Apply Kirchhoff’s voltage law (KVL) to each of the meshes of the circuit.

Example of Mesh Current Analysis

0 ) (

_{1 2}

3 1

1

  



 v

_s

R i R i i

<Mesh 1>

0 )

(

_{1 2 2 2}

3

  

 R i i R i

<Mesh 2>



A circuit that contains only independent voltage sources and resistors results in a specific format of equations that can readily be obtained.

Mesh 1:

Mesh 2:

Mesh 3:

These equations can be written as Mesh 1:

Mesh 2:

Mesh 3:



The coefficient of the first mesh current for the first mesh:

 The sum of resistances in the loop



The coefficient of the second mesh current for the first mesh:

Mesh Current Analysis with Independent Voltage Source

0 )

(

0 ) (

) (

0 ) (

3 3 2

3 5

1 2 4 3

2 5 2 2

2 1 4 1 1





























g s

v i R i

i R

i i R i

i R i R

i i R i R v

g s

v i

R R i

R

i R i R R R i

R

v i R i R R





























3 5 3 2

5

3 5 2 5 2 4 1

4

2 4 1 4 1

) (

0 )

( ) (

Figure 4.5-6 (p. 128) Circuit with three mesh and two voltage sources



In general, the equation for the nth mesh with independent voltage sources only is obtained as follows:

Mesh Current Analysis with Independent Voltage Source



All mesh currents flow clockwise.



The general matrix equation for the mesh current analysis for independent voltage sources present in a circuit is

Mesh Current Analysis with Independent Voltage Source

R i = v

_s

i =

i

₁

i

₂

. . . i

_N

v

_s

=

v

_s1

v

_s2

. . . v

_sN



Where R is a symmetric matrix with a diagonal consisting of the sum of resistances in each mesh and the off-diagonal elements are the negative of

 







 

























) (

0

) (

0 )

(

5 3

5

5 5

4 2

4

4 4

1

R R

R

R R

R R

R

R R

R

R



For linear systems of three equations in three unknowns



Cramer’s rule is

with the “determinant of the system” D given by

Cramer’s rule

 ⁰ 

,

,

₂ ² ₃ ³

1

1

   D 

D x D

D x D

D x D

3 3

33 2

32 1

31

2 3

23 2

22 1

21

1 3

13 2

12 1

11

b x

a x

a x a

b x

a x

a x a

b x

a x

a x a



















 

 

 

 



 

 

 

 



 

 

 

 



_{21 22 2}

1 12

11 3

23 2

21

13 1

11 2

23 22

2

13 12

1

1

, , a a b

b a

a D

a b

a

a b

a D

a a

b

a a

b D

33 32

31

23 22

21

13 12

11

a a

a

a a

a

a a

a

D 

Mesh Current Analysis with Current and Voltage Source



Case 1



A current source appears on the periphery of only one mesh, n



Method:



Equate the mesh current i

_n

to the current source current, accounting for the direction of the current source.

Figure 4.6-1 (p. 130)

Mesh Current Analysis with Current and Voltage Source



Case 1



A current source appears on the periphery of only one mesh, n 1. Mesh current

2. KVL for the 1

^st

mesh

3. Since i

₂

=-i

_s

,

i

s

i

₂

 

v

s

i R i

R

R

₁



₂

)

₁



_{2 2}

 (

2 1

2

1

R R

v R i v

^{s s}



 

Circuit with an independent voltage

source and an independent current source.

Mesh Current Analysis with Current and Voltage Source



Case 2-A



A current source is common to two meshes



Method:

A. Assume a voltage v

_ab

across the terminals of the current source, write the KVL equations for the two meshes, and add them to eliminate v

_ab

B. Create a supermesh as the periphery of the two meshes and write one KVL equation around the periphery of the supermesh. In addition, write the

constraining equation for the two mesh currents in terms of the current

source.

Mesh Current Analysis with Current and Voltage Source



Case 2-A



A current source is common to two meshes 1. Mesh current

2. KVL at node a mesh 1:

mesh 2:

3. Since i

₂

=i

_s

+i

₁

i

s

i i

₂



₁



s

ab

v

v i

R

₁₁

 

v

s

i R R

i

R

₁₁

 (

₂



₃

)

₂



s

s

i v

i R R

i

R

₁₁

 (

₂



₃

)( 

₁

) 

0 )

( R

₂

 R

₃

i

₂

 v

_ab



Circuit with an independent current source common to both meshes.

Example 4.6-1 Mesh Equations



Consider the circuit of Figure 4.6-3 where R

₁

=R

₂

=1Ω and R

₃

=2Ω. Find the

three mesh current.

Solution

1. 4-A source 2. 5-A source

3. KVL for mesh 2 and mesh 3

4. Solve equations

5. Using the values for the resistors, we obtain

1

 4 i

3

5

2

 i  i

0 )

( :

mesh3

10 )

( :

mesh2

3 3 1

3 2

1 2 1















ab ab

v i R i

i R

v i

i R

4 A 5 33

4 A 13

3 2

3

 and i   i 

i

10 )

4 (

) 4 5

(

10 )

4 (

) 4 (

3 3 3

2 3

1

3 3 3

2 2

1























i R i

R i

R

i R i

R i

R

Mesh Current Analysis with Current and Voltage Source



Case 2-B



A current source is common to two meshes



Method:

B. Create a supermesh as the periphery of the two meshes and write one KVL equation around the periphery of the supermesh. In addition, write the

constraining equation for the two mesh currents in terms of the current source.

Figure 4.6-4 (p. 132)

Mesh Current Analysis with Current and Voltage Source



Case 2-B



A current source is common to two meshes 1. supermesh

2. Mesh 3

3. Mesh current

0 2

) (

3 ) (

10 

₁



₃



₂



₃



₂



 i i i i i

0 ) (

3 2

)

( i

₃

 i

₁

 i

₃

 i

₃

 i

₂

 0 6

3

_{2 3}

1

  

 i i i

2

5

1

 i  i

10 4

5

_{2 3}

1

 i  i 

i

Circuit with a supermesh that incorporates mesh 1 and mesh 2. The supermesh is indicated by the dashed line.

A supermesh is one larger mesh created from two meshes that have an independent or dependent

current source in common.

Example 4.6-2 Supermesh



Determine the values of the mesh currents, i

₁

and i

₂

, for the circuit shown in Figure 4.6-5

Figure 4.6-5 (p. 132)

The Circuit for example 4.6-2

12V 1.5A

9Ω 3Ω

i

₁

i

₂

6Ω

Solution

Method 1:

the KVL equation for mesh 1

the KVL equation for mesh 2

Combining two equations

5 . 1 5

.

1

_{1 2}

2

1

 i   i  i 

i

0 12 9 i

₁

 v  

0 6

3 i

₂

 i

₂

 v 

12 9

9 0

12 ) 6 3 (

9 i

₁

 i

₂

 i

₂

   i

₁

 i

₂



Solution

Method 2:

Apply KVL to the supermesh to get

Applying KVL to the supermesh is a shortcut for doing three things:

1. Labeling the current source voltage as v

2. Applying KVL to both meshes that contain the current source 3. Eliminating v from the KVL equations

In summary, 5 .

 1

 i i

Solving the equation

12 9

9 0

12 ) 6 3 (

9 i

₁

 i

₂

 i

₂

   i

₁

 i

₂





When a circuit contains a dependent source, the controlling

current or voltage of that dependent source must be expressed as a function of the mesh currents.

Mesh Current Analysis with Dependent Sources

Example 4.7-1 Mesh Equations and Dependent Sources



Consider the circuit shown in Figure 4.7-1a. Find the value of the voltage measured by the voltmeter.

Figure 4.7-1 (p. 135)

Solution



Fig. 4.7-1b (circuit after replacing the voltmeter) Fig. 4.7-1c (circuit after numbering the meshes)

1.

Controlling current of the dependent source, i

_a

Solving for i

₂

gives

2.

Apply KVL to mesh 1 to get

3.

Apply KVL to mesh 2 to get

16 A 15 4

3 4 5

4 A 0 3

24 32

2

1 1

 



 



 







 i

i i

2 2 1

5 i i i i i

a a









15

32 0

32

_{2 2}











 v

_m

v

_m

i i

1

2

4

5 i

i 

Example 4.7-2 Mesh Equations and Dependent Sources



Consider the circuit shown in Figure 4.7-2a. Find the value of the gain, A, of the CCVS

Figure 4.7-2 (p. 136)

Solution

1. Ai

_a

2. Short-circuit current of the dependent source

3. KVL to mesh 1

4. KVL to mesh 2

5. Finally,

A 8 . 1 0

) 2 . 7 (

4 i

₂

    i

₂

 V 2 . 7 ) 2 . 7

(  



a

 Ai

A 6 . 3 0

36

10 i

₁

   i

₁



) V/A ( 8 4

. 1 6 . 3

2 . 7

2 1

 

 



 i i

Ai i

A Ai

^a

a a

2 1

i i i

_a

 

Figure 4.7-2



The advantage of using node voltage method and mesh current method



Systematic procedure



Simultaneous equation



In some cases one method is preferred over another.



The circuit contains only voltage sources: mesh current method



The circuit contains only current sources: node voltage method



The circuit has fewer nodes than meshes: node voltage method



The circuit has fewer meshs than nodes: mesh current method



You need to know several currents: mesh current method



You need to know several voltages: node voltage method

The Node Voltage Method and Mesh Current

Method Compared

Example 4.8-1 Mesh Equations



Consider the circuit shown in Figure 4.8-1. Find the value of the resistance, R.

Figure 4.8-1 (p. 136)

The circuit considered in Example 4.8-1.

Solution

1. The mesh current i

₁

is equal to the current in the 1- A current source.

2. The mesh current i

₂

is equal to the current in the 3- A current source.

3. The mesh current i

₃

is equal to the current in the short circuit that replaced the ammeter

4. Apply KVL to mesh 3

5. Finally,

0 ) (

) ( 12 ) (

2 i

₃

 i

₁

 i

₃

 R i

₃

 i

₂

 A

1

 1 i

A 5 .

3

 0 i















 1 ) 12 ( 0 . 5 ) ( 0 . 5 3 ) 0 2 5

. 0 (

2 R R

A

2

 3

i

Example 4.8-2 Node Equations



Consider the circuit shown in Figure 4.8-3. Find the value of the resistance, R.

Figure 4.8-3 (p. 137)

The circuit considered in Example 4.8-2.

Solution

1. Node voltage at node 1

2. Node voltage at node 2

3. Node voltage at node 3

4. Apply KCL at node 3

5. Finally,

R v v

v

_{1 3 3}

2  2 



V 16 0

16  v

₁

  v

₁



V

3

 16 v









 

16 8 16 2

16 R

V 2 16

18

18  v

₁

 v

₂

   v

₂

 v

₂

 