1

2.1 Branch-Current Analysis

2.2 Mesh Analysis

2.3 Nodal Analysis

2.4 Mesh with Current Sources

2.5 Nodal with Voltage Sources

2.6 Bridge Network

Chapter 2: Circuit Analysis

2

Introduction

Consider this network

• Circuit simplification is difficult because there are

2 or more sources, and they are neither in series nor parallel.

• There will be an interaction between sources that

will not permit the reduction techniques that is used to find quantities such as the effective

resistance.

3

Steps required for Branch-Current Analysis: -

1. Assign a distinct current of arbitrary direction to

each branch of the circuit.

2. Add the polarities for each voltage drop across resistor.

3. Apply KVL for each mesh.

4. Apply KCL to a node that includes all the branch currents.

5. Solve the equations for branch currents.

2.1 Branch-Current Analysis

4

Example 1 :

Determine the current

in each branch of the network using branch-

current analysis

5

Solution:

• Assign distinct current of arbitrary direction to each branch of the network (one branch one current)

• Add the polarities for each R

6

Solution(continue…):

Kirchhoff’s Voltage Law:

Loop 1:

- E1 + V1 + V3 = 0

I1R1 + I3R3 = E1

Loop 2:

- V3 – V2 + E2 = 0

I2R2 + I3R3 = E2

7

Solution(continue…):

Kirchhoff’s Current Laws

I1 + I2 = I3

Solve the equations

2I1 + 0I2 + 4I3 = 2 ---(1)

0I1 + I2 + 4I3 = 6 ---(2)

I1 + I2 - I3 = 0 ---(3)

Solve by using elimination method or Cramer’s rule,

I1 = -1A, I2 = 2A, I3 = 1A

Observation: 3 branches, 3 equations

8

• Mesh Analysis – defines a unique array of currents (Mesh or Loop current) to the network.

• Steps required for Mesh Analysis: -

1. Assign current in clockwise direction to each closed loop of network.

2. Insert polarities for each resistor.

3. Apply KVL to each closed loop.

4. Solve the resulting equations.

Note: if a resistor has two or more current passing through it, the netcurrent = the mesh current of the closed loop + mesh currents from other loops in same direction - mesh currents from other loops in opposite direction.

2.2 Mesh Analysis

9

Example 2 : find the branch currents I1, I2, I3.

1.Assign current in clockwise direction

( )

123

51015

01010515

0

:1mesh For

21

21

211

2311

=−⇒

=−⇒

=+−++−

=+++−

ii

ii

iii

EVVE

( )

12

102010

0461010

0

:2mesh For

21

21

2212

4232

=+−⇒

=+−⇒

=++−+−

=+++−

ii

ii

iiii

VVVE

Solution:

+ -

+

-

+ -

+

-

2. Insert polarities for each R according to the direction of current.

3. Apply KVL to each closed loop in clockwise direction.

-

+

E1

V1 V2

V4

V3

E2

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Example 2 : find the branch currents I1, I2, I3.

AiAi 1,1 21 ==

Solution:

+ -

+

-

+ -

+

-

-

+

E1

V1 V2

V4

V3

E2

Solve the equations3i1 – 2i2 = 1 ---(1)

- i1+ 2i2 = 1 ---(2)

AiiIAiIAiI 0,1,1 2132211 =−=====∴

Observation: 2 meshes, 2 equations.

11

Example 3 : Use mesh analysis to find the current I0.

Solution:

+ -+

-

+

-

-

+

E1

V1

V2

V4

V3

E2

+

-

-

+- +

( )

126511

24121022

012)(1024

0

:1mesh For

321

321

3121

211

=−−⇒

=−−⇒

=−+−+−

=++−

iii

iii

iiii

VVE

( )

02195

043810

0424)(10

0

:2mesh For

321

321

32212

431

=−+−⇒

=−+−⇒

=−++−

=++

iii

iii

iiiii

VVV ( )

( )

02

01688

04)(12)(4

04)(124

0

:2mesh For

321

321

231321

23130

422

=+−−⇒

=+−−⇒

=−+−+−

=−+−+

=++

iii

iii

iiiiii

iiiiI

VVE

210 iiI −=

12

Example 3 : Use mesh analysis to find the current I0.

Solution:

+ -+

-

+

-

-

+

E1

V1

V2

V4

V3

E2

+

-

-

+- +

Solve the equations11i1 - 5i2 - 6i3 = 12 ---(1)-5i1 + 19i2 - 2i3 = 0 ---(2)

-i1 - i2 + 2i3 = 0 ---(3)

AiAiAi 5.1,75.0,25.2 321 ===

AiiI 5.175.025.2210 =−=−=∴

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2.3 Nodal Analysis• Nodal analysis - provides nodal voltages by using KCL.

• Steps required for Nodal Analysis: -

1. Determine the no. of nodes (junction of 2 or more branches).

2. Select a reference node (Ground), and label all other nodes.

3. Apply KCL at each node (except the reference node).

4. Solve the resulting equations.

Note: A network of N nodes require (N-1) equations to find (N-1) nodal voltages where by the Reference node is eliminated.

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Example 4 : Calculate the node voltages.

1. Determine the no. of nodes.2. Select a reference node (Ground),

and label all other nodes.

3. Apply KCL at each node except the reference node.

4. Solve the resulting equations.

203

52

2

0

45

:1

21

121

121

321

=−⇒

=+−⇒

−+

−=

+=

vv

vvv

vvv

iii

nodeFor

Solution:

6053

26012033

6

0510

4

:2

21

221

221

5142

=+−⇒

+=+−⇒

−+=+

−

+=+

vv

vvv

vvv

iiii

nodeFor

Observation: (3-1) nodes, 2 equations.

Vv

Vv

20

333.13

2

1

=

=

1

2

15

Example 5:

For the circuit below, obtain v1 and v2.

Observation: 4 meshes, (3-1) nodes

i1 i2

i4

i5 i6

i3

1 2

16

At node 1,

At node 2,

)1(6058

65102

21

1112

3214

−−=+−

++=−

++=

vv

vvvv

iiii

Solution:

i1 i2

i4

i5 i6

i3

)2(3632

4236

21

212

5463

−−=+−

+−

=+

+=+

vv

vvv

iiii

Solving (1) and (2),

v1 = 0 V, v2 = 12 V

1 2

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2.4 Mesh with Current Sources

Case 1: If the current source exists only in one of the mesh, find the current directly.

Case 2: If the current source exists between two or more meshes, it forms a supermesh. Exclude the current source (replace with an open circuit) and the elements that connected in series with it. Apply KVL in the supermesh and KCL in the nodes of the excluded elements.

Consider the following case 1

Applying Mesh analysis in usual wayLoop 1:

( ) 06410 211 =−++− iii

Loop 2:

A 52 −=i A 21 −=∴ i

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Supermesh is formed when two meshes have a (dependent or independent) current source

in common.

Consider the following case 2

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Apply KVL on the supermesh loop figure (b):

20146

0410620

21

221

=+⇒

=+++−

ii

iii

Applying KCL:612 += ii

Solving Equation:

A 8.2

A 2.3

2

1

=∴

−=∴

i

i

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Example 6:

Find i1 to i4 using mesh analysis.

• Note that meshes 1 and 2 form a supermesh since theyhave an independent current source in common.

• Also, meshes 2 and 3 form another supermesh because

they have a dependent current source in common.• The two supermeshes intersect and form a larger supermesh as

shown.

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Applying KVL to the larger supermesh,

)1(0463

0)(8462

4321

43321

−−=−++

=−+++

iiii

iiiii

At node P,

)2(5

5

21

12

−−=+−

+=

ii

ii

At node Q,

)3(03

)(

)(3

3

432

40

243

203

−−=−−

−=

=−+

=+

iii

iI

iii

iIi

Q

For mesh 4,

)4(554

0102)(8

43

434

−−−=+−

=++−

ii

iii

22

)4(554

)3(03

)2(5

)1(0463

43

432

21

4321

−−−=+−

−−=−−

−−=+−

−−=−++

ii

iii

ii

iiii

Solve the equations

Ai

Ai

Ai

Ai

143.2

93.3

5.2

5.7

4

3

2

1

=

=

−=

−=

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Consider the following circuit

Find the nodal voltages forthe given circuit.

2.5 Nodal with Voltage Sources

Case 1: If the voltage source is connected to a reference node, set the voltage equal to the voltage source.

Case 2: If the voltage source is connected between two non-reference nodes, it forms a supernode. Exclude the voltage source and replace with a short circuit. Apply KCL in the supernode and KVL

to determine the nodes voltages.

1 21 2

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• Define the node voltage

• Replace voltage source with short circuit

• Apply KCL to the new network

1 21 2

25

244-6

:SupernodetheFor

21

2121

232131

VV

IIII

IIIIII

SS

SS

+=⇒

+=−⇒

+++=+

2112

:circuit original theRefer to

VV −=

12

250250

:equations Resulting

21

21

=−

=+

V V

V.V.

*challenge: use source conversion or mesh analysis

12V1 2

1 2

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Example 7:

Find the node voltages in the circuit shown below.

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At supernode 1-2

)1(6025

23610

10

4321

14123

213

−−=−−+

+−

=−

+

+=+

vvvv

vvvvv

iii

At supernode 3-4

)2(016524

4163

4321

342341

5431

−−=−−+

++−

=−

++=

vvvv

vvvvvv

iiii

28

For mesh 1

)3(2021 −−=− vvFor mesh 3

)5(2022

0)()(220

06220

06320

4321

2341

3

3

−−=++−−

=−+−−−

=+−−

=+−+−

vvvv

vvvv

iv

ivv

x

xx

For mesh 2

)4(023

)(3

3

431

4143

43

−−=−−

−=−

=−

vvv

vvvv

vvv x

233

41

6 vvi

vvvx

−=

−=

29

)5( 202 2

)4( 02 3

)3( 20

)2( 016524

)1( 602 5

4321

431

21

4321

4321

−−=++−−

−−=−−

−−=−

−−=−−+

−−=−−+

vvvv

vvv

vv

vvvv

vvvv

Solve the equations

Vv

Vv

Vv

Vv

67.46

33.173

667.6

67.26

4

3

2

1

−=

=

=

=

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Summary: Mesh-Nodal Analysis

(i) Mesh

• To find mesh currents

• Suitable to be used with

circuit with many series-

connected elements,

voltage sources &

Supermesh

Supermesh

• Replace current sources

with an open circuits

(ii) Nodal

• To find node voltages

• Suitable to be used with

circuit with many parallel-

connected elements,

current sources &

Supernode

Supernode

• Replace voltage sources

with a short circuits

Mesh Analysis vs. Nodal Analysis

31

2.6 Bridge Network

• Belong to the family of complex networks because

the elements are neither in series nor in parallel.

• Bridge configuration may be analyzed by using either

mesh or nodal analysis.

32

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( ) 052152

054254

2024243

213

312

321

=−−++

=−−++

=−−++

III

III

III

0852

05114

20249

321

321

321

=+−−

=−+−

=−−

III

III

III

A 667.2

A 667.2

A 4

3

2

1

=∴

=∴

=∴

I

I

I

Note that:A 0325 =−= III R

A bridge network is said to

be balanced, if the current

or voltage through the bridge

arm is 0A or 0V.