Ch3_Method of Analysis (Dc)

Chapter 3: Method of Analysis (dc)

BEE1133 : Circuit Analysis I

3.1 Nodal analysis

3.2 Application of Cramer’s rule

3.3 Nodal analysis with voltage sources: supernode

3.4 Mesh analysis

3.5 Mesh analysis with current sources: supermesh

3.6 Nodal versus mesh analysis

Method of Analysis : Syllabus

Nodal Analysis

• Without voltage source

• With voltage sources

• Understand the concept of nodal voltage

• Analyze dc circuit using nodal analysis

• Apply Cramer rule to solve simultaneous equations

• Understand the concept of supernode

Nodal Analysis:Lesson Outcomes

• Nodal analysis applies KCL to find unknown voltages value.

• To analyze a circuit using nodal analysis, node voltages are used as a circuit variables instead of element voltages to reduce number of equations.

• In the first section, circuits with no voltage sources are considered.

Nodal Analysis

Steps to determine node voltages:

1. Choose a reference node.

2. Assign node voltages to the other nodes.

3.Apply KCL to each node other than the reference node; express currents in terms of node voltages.

4. Solve the resulting system of linear equations.

Nodal Analysis : Steps

Nodal Analysis w/o voltage source

Example

Find the voltages at a and b for the circuit below

1 4

2

5 A 4 A

a b

Step 1 : Choose a reference node.

Step 2 : Assign node voltages to the other nodes.

Nodal Analysis w/o voltage source

2

1 4

V1

V0

V2

Reference node

Step 3 : Apply KCL to each node other than the reference node (in term of node voltage).

At node a :

i1 = i2 + i3 (apply KCL)

(in term of node voltage)

At node b :

i2 = i4 + i5 (apply KCL)

Nodal Analysis w/o voltage source

2

1 4

V1 V2

V0

i1 i2i3 i4

i5

125 0121 VVVV

442

0221

VVVV

5 A 4 A

i2

Step 4 : Solve the resulting system of linear equations.

Nodal Analysis w/o voltage source

)1(12

5 0121 VVVV

)2(442

0221

VVVV

0 0

125 121 VVV

121 210 VVV

442

221 VVV

16)(2 221 VVV

1632 21 VV21310 VV

103 21 VV

11

The above equations can be solved by using 2 methods:i) Simultaneous equations / elimination techniqueii) Cramer’s rule

For elimination technique: (x3) … (1)

… (2)

By subtracting eq.(2) from (1) and solving for V1 & V2

Nodal Analysis w/o voltage source

1632 21 VV

103 21 VV 3039 21 VV

14163029 11 VVVV 21

1632 21 VV4163 2 V VV 42

(1) – (2)

Application of Cramer’s rule

12

To use Cramer’s rule, we need to put the equations in matrix form as below:

Determinant of the matrix is:

Solve for V1 and V2

16

10

32

13

2

1

V

V

7)21()33(32

13

?7

1630316

110

11

V ?7

2048162

103

22

V

Nodal Analysis w/o voltage source

Exercise 1 (prob 4.3-4)

13

Consider the circuit below. Find the value of R1 and R2 if the voltages across those resistors are vR1 = 1 V and vR2 = 2 V .

500

3 mA 5 mA+VR1

_

+VR2

_R1 R2

Nodal Analysis w/o voltage source

Exercise 2 (prob 4.3-5)

14

Consider the circuit below. Find the voltage value at each non-reference node.

500

1 mA

500

250

250 125 v1 v3

v2

Ans: V1 = ? V2 = ? V3 = ?

V1 - V3 = 21.7 mV

Nodal Analysis with voltage source

There are 2 possibilities of a circuit with voltage source:i) A voltage source between reference node and a non-

reference nodeii) A voltage source between two non-reference nodes

(these two nodes form a supernode) For the case in (i), we simply set the voltage at the non-

reference node equal to the voltage source value. The rest of the non-reference nodes are treated like

previous steps for nodal analysis.

Nodal Analysis with voltage source

Example (case i)

Using nodal analysis, find vo for the circuit below

+_

8 3

6

10

90 V

45 V+_

+_

8 3

6

10

90 V

45 V+_

Step 1 : Choose a reference node.

Step 2 : Assign node voltages to the other nodes.

Nodal Analysis with voltage source

V1

V0

V2

+_

8

3

6

10

90 V45 V

Step 3 : Apply KCL to each node other than the reference node (in term of node voltage).

At node V1 :

i1 + i2 = i3 (apply KCL)

(in term of node voltage)

At node V2 and V3 :

V2 = 90 V

V3 = 45 V

Nodal Analysis with voltage source

V2

V1

V0

i2

i3

6183011213 VVVVVV

i1

+_

V3

Step 4 : Solve the resulting system of linear equations.

Nodal Analysis with voltage source

0

6183011213 VVVVVV

6

0

18

90

3

45 111

VVV

111 390)45(6 VVV

36010 1 V111 3906270 VVV

VV 361

Nodal Analysis with voltage source

Exercise 3 (prob 3.14)

20

Using nodal analysis, find v1 and v2 for the circuit below.

+_

+_

2 8

4

1

40 V20 V

5 A

+

_V1

Ans: V1 = 27.27 V2 = 42.73

V2V1

Nodal Analysis with voltage source

Exercise 4 (prob 4.4-8)

21

Using nodal analysis, find voltage value at each non-reference node for the circuit below.

+_

2 2

3 4

12 V

1 A

Ans: V1 = ? V2 = 12.0 V3 = ?

V1 - V3 = 3.33

V1 V3V2

Nodal Analysis with voltage source

A supernode is formed by enclosing:i) A voltage source connected between 2 non-

reference nodesii) and any elements connected in parallel with it.

Below are the properties of supernode:i) voltage source inside supernode provides constraint

equation to solve for node voltagesii) Supernode has no voltage of its owniii)Supernode requires the application of both KCL &

KVL.

Nodal Analysis with voltage source

Example (case ii)

Using nodal analysis, find vo for the circuit below

500 3 mA

8 V

500 5 mA

_+

+

_Vo

500 3 mA

8 V

500 5 mA

_+

+

_Vo

Step 1 : Choose a reference node.

Step 2 : Assign node voltages to the other nodes.

Nodal Analysis w/o voltage source

Reference node

V0

V1 V2

Step 3 : Apply KCL to each node other than the reference node (in term of node voltage).

At supernode :

i1 + i2 = i3 + i4 (apply KCL)

3 mA + 5 mA (in term of node voltage)

Apply KVL at supernode :

V1 = V2 + 8 (apply KVL)

Nodal Analysis w/o voltage source

i1i3 i4

i2

5005000201 VVVV

500 3 mA

8 V

500 5 mA

_+V1 V2

Step 4 : Solve the resulting system of linear equations.

8 mA

…(1) …(2)

Solve above equations by using elimination technique:

(1) – (2) :

Nodal Analysis w/o voltage source

0 0

5005000201 VVVV

421 VV

821 VV

821 VV

84)( 2121 VVVV

02 2 VV 61 V

Nodal Analysis with voltage source

Exercise 5 (exp 8.22)

27

Using nodal analysis, find voltage value at V1 and V2 for the circuit below.

10

4

12 V

6 A 2 4 A

_+V1 V2

Ans: V1 = -10.67 V2 = -1.33

Nodal Analysis with voltage source

Exercise 6 (prob 4.4-4)

28

Using nodal analysis, find voltage value at each non-reference node for the circuit below.

+_

125 250

500 500

12 V

8 V

_+

Ans: V1 = ? V2 = 12.0 V3 = 4.0

V1 V2 V3

Mesh Analysis

• Without current source

• With current sources

• Understand the concept of mesh current

• Analyze dc circuit using mesh analysis

• Understand the concept of supermesh

• Evaluate the better method between nodal and mesh analysis for a particular circuit problem

Mesh Analysis:Lesson Outcomes

• Mesh analysis applies KVL to find unknown currents value.

• Using mesh currents instead of element currents as circuit variables is easier because it reduces the number equations to be solved simultaneously.

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

Mesh Analysis

Steps to determine mesh currents:

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.

Mesh Analysis : Steps

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.

Example

For the circuit below, find the branch currents i1, i2 and i3 using mesh analysis

Mesh Analysis w/o current source

5

10 15

12 V

i1 i2i3

+_

Mesh Analysis w/o current source

5

10 15

12 V +_

Step 1 : Identify mesh (loops).

Step 2 : Assign a current to each mesh.

Mesh 2Mesh 1

Ia Ib

Mesh Analysis w/o current source

5

10 15

12 V

+_

Step 3 : Apply KVL around each loop to get an equation in terms of the loop currents.

For mesh 1: -Vs + VR15 + VR5 = 0

-12 + 15Ia + 5(Ia - Ib) = 0

20Ia - 5Ib = 12

For mesh 2: VR10 + VR5 = 0

10Ib + 5(Ib - Ia) = 0

-5Ia + 15Ib = 0

Ia Ib

Step 4 : Solve the resulting system of linear equations.

• The above equations can be solved by using 2 methods:

i) Simultaneous equations / elimination technique

ii) Cramer’s rule

• For elimination technique:

20Ia - 5Ib = 12 (x3) 60 Ia - 15 Ib = 36 …(1)

-5 Ia + 15 Ib = 0 …(2)

By adding eq.(2) from (1) and solving for Ia & Ib

60 Ia - 5 Ia = 36 Ia = 36/55 = 0.65 A

From eq. (2) Ib = (5 x 0.65)/15 = 0.22 A

Mesh Analysis w/o current source

Application of Cramer’s rule

37

To use Cramer’s rule, we need to put the equations in matrix form as below:

Determinant of the matrix is:

Solve for Ia and Ib

0

12

155

520

b

a

I

I

275)55()1520(155

520

AI aa 65.0

275

180150

512

AI bb 22.0

275

60005

1220

Mesh Analysis w/o current source

Exercise 7 (exa 8.18)

38

Find the current through the 10 resistor of the circuit below by using mesh analysis.

+_

10

5

2 3

8

15 V

IR10

Ans: IR10 = 1.22 A

Mesh Analysis w/o current source

Exercise 8 (exer 4.6-1)

39

Use mesh analysis to determine the current, IR2 for the circuit shown below.

6

2 4 12 28 V

8 V

_

+

+_

I

Ans: IR2 = 1 A

Mesh Analysis with current source

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.

There are 2 possibilities of a circuit with current source:i) A current source exist only in one meshii) A current source exist between two meshes. (these

requires a supermesh analysis) For the case in (i),

Mesh Analysis with current source

Exercise 2 (prob 2.35)

41

Given voltage supply VS = 50 V, calculate the voltages across 70 resistor, V1 and 5 resistor, V0.

Conclusion

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