ELT221

1

Nodal & Loop Analysis - Nodal Analysis - Loop Analysis - Mesh Analysis

Enzo Paterno

EP Revision 12/9/14

ELT221

2

NODAL ANALYSIS

Enzo Paterno

In a nodal analysis, the variables in the circuit are selected to be the node voltages. The node voltages are defined with respect to a common point in the circuit (i.e. ground) and assumed to be positive voltages. For example the circuit below has four node voltages, (VS , Va , Vb , Vc ) with respect to the ground node.

Once the node voltages in a circuit are defined , we can calculate the current through any resistive element using Ohm’s law:

RVV

I NN −+ −=

VS Va Vb Vc + - I R

+

ELT221

3

NODAL ANALYSIS

Enzo Paterno

In a nodal analysis, KCL equations are used so that the variables are the unknown node voltages. For a network consisting of N nodes, N 1 Linearly Independent, (LI), KCL equations are required to determine the N 1 node voltages (i.e. Nth node is ground). Example: 3 nodes 2 LI equations required

B

B

BB

iRR

VRV

iRV

RVV

iiiiiiv

=

+−

=−−

=−⇒+=

322

2

1

3

2

2

21

32322

11

:@

A

A

A

iR

VRR

V

iR

VVRV

iiiv

=

−

+

=−

+

+=

22

211

2

21

1

1

211

111

:@

ELT221

4

NODAL ANALYSIS

Enzo Paterno

Let R1 = 12 kΩ, R2 = 6 kΩ, R3 = 6 kΩ, IA = 1 MA, and IB = 4 mA.

321

321

22

211

1

1016

141

10161

61

121

111

:@

−

−

=−

+

=

−

+

=

−

+

xVk

Vk

xk

Vkk

V

iR

VRR

V

v

A

321

321

322

21

2

1043

161

10461

61

61

111

:@

−

−

=−

+

=

+−

=

+−

xVk

Vk

xkk

Vk

V

iRR

VR

V

v

B

ELT221

5

NODAL ANALYSIS

Enzo Paterno

321 101

61

41 −=

−+ xV

kV

k3

21 1043

161 −=

−+ xV

kV

k

We have a 2x2 linear system. Solve for V1 and V2. How do we solve an n x n linear system?

TI-83 Plus

Method of substitution Method of elimination Gaussian elimination Cramer’s rule Matrix SW calculator MATLAB software tool JAVA Applet

ELT221

Enzo Paterno 6

GAUSSIAN ELIMINATION

Two simultaneous linear equations form a 2x2 linear system:

This 2x2 linear system can be represented in matrix form:

[ ][ ] [ ]KxMKK

DCBA

x

x=→

=

2

1

2

1

Using matrix theory, the solution to this linear system can be found by:

[M]-1 is the inverse of matrix [M]

M is a square matrix, x and K are column matrices (vectors)

A x1 + B x2 = K1 C x1 + D x2 = K2

[ ] [ ] [ ]KMx1−

=

ELT221

Enzo Paterno 7

Below is a general system of m equations with n unknowns:

This system can be expressed as a matrix equation: Ax = b

The system solution is given by: x = A-1 b

GAUSSIAN ELIMINATION

ELT221

8

MATRIX ALGEBRA REVIEW

Enzo Paterno

( ) ( ) ( )

[ ]TCAAdj

AAAdjA

MijjiCij

caAA

abBTAaA

ba

m

kkjk

nxmjiijmxnij

m

i

r

j

n

kkjik

=

=

+−=

==

== →=

−

=

= = =

∑

∑∑∑

)(

)()1(

)det(

1

11

1 1 1

(Amn Bnr = Cmr)

)()()det( 12212211

2221

1211

AAAAAA

AAAA

A

−==

=

Matrix Multiplication

Transpose

Determinant

Cofactor

Inverse

Adjoint

−

−=

=

−

1121

12221

2221

1211

1AAAA

AA

AAAA

A

ELT221

9

NODAL ANALYSIS – GAUSSIAN ELIMINATION

Enzo Paterno

321 101

61

41 −=

−+ xV

kV

k3

21 1043

161 −=

−+ xV

kV

k

=

−

−

−

−

3

3

104101

21

31

61

61

41

xx

VV

kk

kk

−

−

=

−

−

−

3

3

1

104101

31

61

61

41

21

xx

kk

kkVV

−−

=

−−

=

−

−

156

104101

5.4336

21

3

3

xx

VV V1 = - 6 V

V2 = - 15 V

ELT221

10

NODAL ANALYSIS – CRAMER’S RULE

Enzo Paterno

V

kk

kk

kx

kx

V 6

31

61

61

41

31104

61101

3

3

1 −=

−

−

−

−

=

−

−

321 101

61

41 −=

−+ xV

kV

k3

21 1043

161 −=

−+ xV

kV

k

V

kk

kk

xk

xk

V 15

31

61

61

41

10461

10141

3

3

2 −=

−

−=

−

−

ELT221

11

NODAL ANALYSIS

Enzo Paterno

SV aV bV cV

0:@

321

321

=++−+=

IIIIIIVa

0:@

543

543

=++−+=

IIIIIIVb

kV

kkkV

kV

kVV

kV

kVV

cba

cbbab

991

41

31

31

0943

=

+++−

=−

++−

kV

kV

kkkV

kVV

kV

kVV

sba

baasa

9331

61

91

0369

=−

++

=−

++−

Vs is a constant Vs = 12 v

031

91

91

39:@

=

+−

=−

kkV

kV

kV

kVVV

cb

ccbC

ELT221

12

NODAL ANALYSIS

Enzo Paterno

VVVVVVVVV

kVV

kV

kVV

5.16)6(124

02

012612

2

2

32212

32212

==−+=

=−++−

=−

++−

VVVV

612

3

1

−==

V1 =12 V V3 = -6 V Only one KCL needed

ELT221

13

NODAL ANALYSIS

Enzo Paterno

V1

V2 V3

ELT221

14

NODAL ANALYSIS

Enzo Paterno

V1

I1

I3 I2

723413:14433672

68412

:@

321

21311

21311

3211

=−−−+−=−

−+

−=

−+=

VVVEqVVVVV

kVV

kVV

kV

IIIV

V2 V3

ELT221

15

NODAL ANALYSIS

Enzo Paterno

V2

I3

I4 I5

V1

V3 06134:2

46644846

:@

321

23221

23221

5432

=+−+−=−

+−

=−

+=

VVVEqVVVVV

kV

kVV

kVV

IIIV

ELT221

16

NODAL ANALYSIS

Enzo Paterno

V1

V2 V3

I2

I4 I6

042:322

884

:@

321

33132

33132

6243

=−+=−+−

=−

+−

=+

VVVEqVVVVVk

VkVV

kVV

IIIV

ELT221 NODAL ANALYSIS

17 Enzo Paterno

=

−−

−−

0072

42161343413

3

2

1

VVV

−−

−−=

−

0072

42161343413 1

3

2

1

VVV

723413:1 321 =−− VVVEq

06134:2 321 =+− VVVEq

042:3 321 =−+ VVVEq

32 VVVo −=

ELT221 NODAL ANALYSIS

18 Enzo Paterno

vVvVvVvV

22.010.432.478.7

0

3

2

1

====

ELT221 NODAL ANALYSIS – Circuits with independent current sources

19 Enzo Paterno

Suppose that the network below has the following parameters: IA = 1 mA, IB = 4 mA, R1 = 12 kΩ, R2 = 6 kΩ, R3 = 6 kΩ. Determine all node voltages and branch currents.

22

121

2

21

1

1

211

111

:@

vR

vRR

I

Rvv

RvI

IIIV

A

A

A

−

+=

−+=

+=

232

12

232

21

32322

111

1:@

vRR

vR

I

vRR

vvI

IIIIIIV

B

B

BB

+−−=

−−

=

−=⇒+=

ELT221

20 Enzo Paterno

22

121

2

21

1

1

211

111

:@

vR

vRR

I

Rvv

RvI

IIIV

A

A

A

−

+=

−+=

+=

232

12

232

21

32322

111

1:@

vRR

vR

I

vRR

vvI

IIIIIIV

A

B

BB

+−−=

−−

=

−=⇒+=

321 101

61

61

121 −=−

+ xv

kv

kk3

21 10461

61

61 −−=

+−− xv

kkv

k

R1 = 12 kΩ, R2 = 6 kΩ, R3 = 6 kΩ

NODAL ANALYSIS – Circuits with independent current sources

ELT221

21 Enzo Paterno

−

−−

=

−

−−

−−

−−

3

31

33

33

2

1

104101

1033.1016.1016.1025.

xx

xxxx

VV

NODAL ANALYSIS – Circuits with independent current sources

ELT221

22 Enzo Paterno

Inverse

-6 -15

NODAL ANALYSIS – Circuits with independent current sources

ELT221

23 Enzo Paterno

Figure 3.5

NODAL ANALYSIS – Circuits with independent current sources

ELT221

24 Enzo Paterno

NODAL ANALYSIS – SUPERNODE

Consider the network below. To eliminate the problem of dealing with a current through a voltage source, we generate a dashed surface which here includes the 6v voltage source and is called a supernode, The KCL for the supernode is: 6mA = I1 + I2 + 4 mA

Eq1: 𝑉16𝑘

+ 𝑉212𝑘

= 2mA Eq2: V1 – V2 = 6v V1 = 6 + V2 Sub Eq3 into Eq1: V2 = 4v V1 = 10v

ELT221

25

MESH LOOP ANALYSIS

Enzo Paterno

In a loop analysis, the variables in the circuit are selected to be the branch currents and KVL equations are used so that the variables are the unknown currents. The number of LI KVL equations necessary to determine all the currents in a network with B branches, N nodes is: B – N + 1. For example the circuit below requires 7 – 6 + 1 = 2 LI KVL. We identify two independent mesh loops ABEF and BCDE. A mesh is a loop that does not contain any other loop within it.

ELT221

26 Enzo Paterno

( ) 1323211

132312111

213231111 0:@

S

S

S

vRiRRRivRiRiRiRi

RiRiRiRivABEF

−=+++−−=+−−−=−+−−

MESH LOOP ANALYSIS

( ) 2543231

232524231

313252422 0:@

S

S

S

vRRRiRivRiRiRiRiRiRiRiRiv

BCDE

=++−+=−−−+

=+−−−−

- +

( )( )

( )( )

−=

++−

++−

=++−+−=+++−

2

1

2

1

5433

3321

2543231

1323211

S

S

S

S

vv

ii

RRRRRRRR

vRRRiRivRiRRRi

+ -

ELT221

27 Enzo Paterno

MESH LOOP ANALYSIS

12612066612

21

211

=−=+−−

ikikikikik

+

+

39603663

21

212

=−=−+−−

ikikikikik

39612612

21

21

=−=−

ikikikik

6181212612

21

21

−=+−=−

ikikikik

612 2 =ikmA

ki 5.0

126

2 == mAk

i 25.11215

1 ==

Use Mesh analysis to solve for I1 and I2.

ELT221

28 Enzo Paterno

MESH LOOP ANALYSIS

+

+

39612612

21

21

=−=−

ikikikik

Use Mesh analysis to solve for I1 and I2.

=

−−

312

96612

2

1

ii

kkkk

mAk

ik

i

=

=

−− −

50.025.1

312

96612

2

11

ELT221

29 Enzo Paterno

MESH LOOP ANALYSIS

+

+

Use Mesh analysis to solve for I1 and I2.

mAk

ki 25.1

7209.0

96612

9003.06012.0

1 =−−

=

−−−−

= mAk

ki 5.0

96612

003.06012.012

2 =

−−

=

39612612

21

21

=−=−

ikikikik

ELT221

30 Enzo Paterno

MESH LOOP ANALYSIS

PSPICE Simulation

ELT221

31 Enzo Paterno

MESH LOOP ANALYSIS

i1 i2

Use Mesh analysis to solve for I1 and I2.

+

+

+

+

+

Mesh1 Mesh2

V + -

10055.120555.7100

21

211

=−+=+−−

ikikikikik

010505523

21

1222

=−+=+−−−

ikikikikikik

Mesh1: Mesh2:

ELT221

32 Enzo Paterno

MESH LOOP ANALYSIS

010510055.12

21

21

=−+=−+

ikikikik

mA

kkkkkk

i 10

10555.12

1005100

1 =

−−

−−

=

We use Cramer’s rule to find the currents i1 and i2:

Remark: This step could have been solved using Elimination by addition or Elimination by substitution techniques

bcaddcba

−=

2x2 determinant

mA

kkkk

kk

i 5

10555.1205

1005.12

2 =

−−

=

ELT221

33 Enzo Paterno

MESH LOOP ANALYSIS

i1

i2

Use Ohm’s law to solve for I1 and I2.

+

+

+

+

+ V + -

i2

RT = 7.5k + [ 5k // (3k + 2k) ] = 10k I1 = 100v / 10k = 10 mA I2 = 10 mA / 2 = 5 mA

ELT221

34 Enzo Paterno

MESH LOOP ANALYSIS

i1 i2 VOUT +

-

Mesh1 Mesh2

+

+

+

+

Use Mesh analysis to solve for VOUT.

V1

121

2111

20)1()1()1(

ViiiiiV

=−+=+−−

V2

Mesh1:

221

1222

30)1()1()2(

ViiiiiV

=+−=+−−

Mesh2:

ELT221

35 Enzo Paterno

MESH LOOP ANALYSIS

212

1

252

51

3112

12

VVVV

i +=

−−

−=

VOUT = 2 i2. We use Cramer’s rule to find the current i2:

2123254

522 VViRiVOUT +===∴

221

121

32

ViiVii

=+−=−+

If V1 = V2 = 10 v then VOUT = 4 + 8 = 12 V

ELT221

36 Enzo Paterno

MESH LOOP ANALYSIS

Use Mesh analysis to solve for I1, I2, and I3.

661006646

31

311

=+−=+−−−

ikikikikik

+

+

+ +

631209336

32

232

=−+=−+−+

ikikikikik

021360331266

321

23313

=−++=+−−+−

ikikikikikikikik

=

−−

−

066

21363120

6010

3

2

1

iii

kkkkk

kk

−−

−=

−

0

6

6

21363120

6010 1

3

2

1

k

k

iii

1:

2:

3:

ELT221

37 Enzo Paterno

MESH LOOP ANALYSIS

−−

−=

−

0

6

6

21363120

6010 1

3

2

1

k

k

iii

ELT221

38 Enzo Paterno

MESH LOOP ANALYSIS

−−

−=

−

0

6

6

21363120

6010 1

3

2

1

k

k

iii i1 = -0.678 mA

i2 = 0.468 mA i3 = -0.126 mA

ELT221

39 Enzo Paterno

MESH LOOP ANALYSIS

Use Mesh analysis to solve Vo.

There are two mesh currents. I1 is known & only one mesh equation is needed for I2.

mAk

i

ik

ikikikikik

75.086

284

28202262

2

2

21

122

==

=+−

=+−=+−−+

+

+

I1 = 2 mA

2:

Vo = 0.75 mA (6kΩ) = 4.5 v

ELT221

40 Enzo Paterno

MESH LOOP ANALYSIS

There are three mesh currents. I1 and I2 are known. Only one mesh equation is needed for I3.

I1 = 4 mA I2 = -2 mA

mAk

i

ik

ikikikikikikikik

25.012

331288

312420622443

3

3

321

31323

==

=++−

=+−−=−+−+−+

+

+

+

+

Use Mesh analysis to solve Vo.

Vo = - [0.25 mA (6kΩ) ] + 3 = 1.5 v

ELT221

41 Enzo Paterno

MESH LOOP ANALYSIS

Use Mesh analysis to solve for I1, I2, and I3.

12

122

13

133

240222

12601116

ikikVikikikV

ikikVikikVik

x

x

x

x

−==+−−

+−+==+−−−+

+ +

kikiki

ikikikikikikVikik x

2634

6124224126

123

1123

1213

−−−

=

−−−=−−==+−+

+

+

@3:

@2:

I2 = (4 - 2/3) = 3.33 mA

One could have used SuperMesh (take out the 4 mA )

Eq1:

Eq2:

Eq1 Eq2

I1 = 2 mA We see that 4mA = I2 – I3

mAik

iki

kimAki

32

244

266)4(4

33

3

33

−=⇒

−+

=

−−−+

=