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METHODS OF CIRCUIT ANALYSIS

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Methods of Circuit Analysis

Mesh Analysis Nodal Analysis

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Mesh Analysis Kirchhoff’s Voltage Law (KVL) forms the

basis of mesh analysis. This technique is applicable to

Basic circuit Circuit with dependent source Circuit with current source

Case 1: Current source at the outer most boundary (known as mesh current)

Case 2: Current source in between two loops (known as supermesh)

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Step to determine Mesh Current

Assign mesh currents I1, I2…, In to the n meshes

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

Solve the resulting n simultaneous equation to get the mesh current,

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Example 10.3For the circuit below, find Io using mesh analysis

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Solution

0j10Ij2)I(j2)Ij10(8 321

09020j2)I(j2)I(j2)Ij2(4 0312

j50j2Ij8)I(8 21

j10j20j4)I(4j2I 21

Applying KVL to Mesh 1

Mesh 2

Substitute (I3=5) into meshes (1) and (2)

…(1)

…(2)

…(3)

…(4)

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Solution

j30

j50

I

I

j44j2

j2j88

2

1

684j)j)(132(1j44j2

j2j88Δ

02 35.22416.17j240340

j30j2

j50j88Δ

Put equation (3) and (4) in matrix form

Find determinant for the matrix (Cramer’s Rule)

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Solution

oo

22 35.226.12

68

35.22416.17

Δ

ΔI

o78.14412.6 Io = (-I2) =

Use Cramer’s rule to solve for I2

Hence

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Practice Problem 10.3For the circuit below, find Io using mesh analysis

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Solution

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Solution

0j4Ij4)Ij2(8 21

21 j4Ij2)I(8

030106Ij4Ij4)I(6 o312

2I3

Mesh 1

Mesh 2

Mesh 3

Insert Mesh 3 into Mesh 2

o12 301012j4Ij4)I(6

…(1)

…(2)

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Solution

112 j2)I(0.5Ij4

j28I

o11 301012j4Ij2)Ij4)(0.5(6

j5)(20.66j14)I(11 1

Simplify Equation (1)

Substitute equation (3) into (2)

j1411

j5)(20.66I1

…(3)

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Solution

o

o

1o51.8417.8

13.621.256

j1411

j520.66II

oo 65.441.194I

Hence

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Example 10.4For the circuit below, find Vo using mesh analysis

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Solution

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Solution

Mesh 1

08Ij2)I(j2)I(810 321 108Ij2)I(j2)I(8 321

3I2

0j5Ij5)I(68Ij4)I(8 2413

4II 34

Mesh 2

Supermesh

Due to current source between meshes 3 and 4 at node A

…(1)

…(2)

…(3)

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Solution

j6108Ij2)I(8 31

j3524j)I(148I 31

j3524

j610

I

I

j148

8j28

3

1

Combine I2 = -3 into equation (1)

Combine I2 = -3 into equation (2) and (3)

…(4)

…(5)

Put equation (4) and (5) into matrix

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Solution

j2050642j28j8112j148

8j28Δ

j2801926j84j10140j14j3524

8j610Δ1

j18658

o11 274.53.618

j2050

j18658

Δ

ΔI

Use Cramer’s Rule to solve for I1

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Solution

3)274.5j2(3.618)Ij2(IV o21o

o222.329.756j6.5687.2134

Solve for Vo

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Practice Problem 10.4

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Solution

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Solution

Mesh 1

05Ij4)I(j4)I(1550 321

505Ij4Ij4)I(15 321

0j4)I(5j6)I(5j4)I(j8 132

2II 23

Supermesh

Also the current source between meshes 2 and 3

…(1)

…(2)

…(3)

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Solution

60j4)I5(j4)I(15 21

j1210j2)I(5j4)I5( 21

j1210

60

I

I

j25j45

j45j415

2

1

Eliminating I3 from equation (1) and (2)

…(4)

…(5)

Put equation (4) and (5) into matrix

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Solution

o9.7858.86j1050j25j45

j45j415Δ

o1 3.84298.67j20298

j25j1210

j4560Δ

A5.945.074j20298

j1058

Δ

ΔII o11o

Use Cramer’s Rule to solve for I1 and then Io

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Exercise III (Problem 10.38)

Using mesh analysis, find Io

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Solution

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Solution

0j10j4I2(2)j4)I(2 42

Mesh 1

2I1

09010j4)I(2Ij4)I(2 o412

Mesh 2

Substitute (1) into (2)

…(1)

…(2)

…(3)j10-4j4Ij4)I(2 42

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Solution

0j4Ij2(2)-j4)I-(14)j2)(I(1 244

Supermesh

0j4)I(j2Ij4)I(1j2)I(1 2143

4II 43

Substitute (1) and (5) into (4)

…(4)

…(5)

…(6)j124j2)I(2j4I 42

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Solution

A5.713.35j12)(12

j44)(36

Δ

ΔI o1

2

j124

j104

I

I

2j2j4

j4j42

4

2

Put equation (3) and (6) into matrix

Use Cramer’s Rule to solve for I2

A174.33.35II o2o

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Nodal Analysis

The basis of nodal analysis is Kirchhoff’s Current Law (KCL).

This technique is applicable to Basic Circuit Circuit with dependent source Circuit with voltage source

Case 1: Voltage source in between reference node and essential node

Case 2: voltage source in between two nodes

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Step to determine Node Voltages

Select a node as the reference node. Assign voltages V1,V2…,Vn-1 to the

remaining n-1 nodes. Apply KCL to each of the n-1 nonreference

node. Use Ohm’s Law to express the branch currents in term of node voltages.

Solve the resulting simultaneous equation to obtain the unknown node voltage.

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Example 10.1Find Ix in the circuit using nodal analysis

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Solution

Convert the circuit into frequency domain

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Solution

j4

VV

j2.5

V

10

V20 2111

20j2.5Vj1.5)V(1 21

Applying KCL at node 1

Iin = Ix + I2

…(1)

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SolutionApplying KCL at node 2

j2

V

j4

VV2I 221

X

j2.5

VI 1

X

j2

V

j4

VV

j2.5

2V 2211

015V11V 21

Ix + I2 = I3

Hence

But

…(2)

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Solution

0

20

V

V

1511

j2.5j1.51

2

1

j5151511

j2.5j1.51Δ

300150

j2.520Δ1 220

011

20j1.51Δ2

Put equation (1) and (2) into matrix

Find determinant

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Solution

o11 18.4318.97

j515

300

Δ

ΔV

o22 198.313.91

j515

220

Δ

ΔV

Solve for V1 and V2 using Cramer’s Rule

Solve for Ix

oo

o1

X 108.47.59902.5

18.4318.97

j2.5

VI

)108.47.59cos(4ti oX

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Practice Problem 10.1Find V1 and V2 usind nodal analysis

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Solution

Convert into frequency domain

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Solution

4

V3V

j2.5

VV

j4

V 2x212

)V2.5(3V)Vj4(Vj2.5V 21212

0j1.5)V(2.5j4)V.57(- 21

j2.5

VV

2

V10 211

21 j4Vj4)V5(100

At node 1

…(1)

At node 2

where 1x VV

…(2)

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Solution

0

100

V

V

j1.55.2j4)7.5(

4jj45

2

1

V60.0111.32V o1

V57.1233.02V o2

Put equation (1) and (2) into matrix

Solving for V1 and V2 using Cramer’s Rule

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Example 10.2Compute V1 and V2 in the circuit

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Solution

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SolutionNodes 1 and 2 form a supernode.

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V

j6

V

j3

V3 221

21 j2)V(1j4V36

o21 4510VV

But a voltage source is connected between nodes 1 and 2

Applying KCL to the supernode gives

…(2)

…(1)

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Solution

2o j2)V(11354036

o2 87.1831.41V

oo1 70.4825.784510VV 2

Substitute equation (2) in (1) result in

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Practice Problem 10.2Calculate V1 and V2 in the circuit using nodal analysis

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Solution

2

V

j

V

j4

V

4

V15 2211

21 j4)V(2j)V(115 2211 2Vj4VjVV15

The only non-reference node is supernode

o21 6020VV

The supernode gives

…(1)

…(2)

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Solution

o1 69.6719.36V

oo

oo

2 165.73.376454.243

210.7214.327

3j3

)60j)(20(115V

j17.32)(10j0.8327)3.272(6020VV o21

Substitute (2) into (1) gives

Therefore

j18.1546.728V1

2o j3)V(360j)20(115

o2 165.73.376V

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Exercise III (Problem 10.9)

Find Vo in the circuit using nodal analysis

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Solution

Convert into frequency domain

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Solution

j1030

V4I

j20-

VV 2o

21

j1030

V

20

4V

j20-

VV 2121

0j200)V600(j2600)V200( 21

j20

VV

20

V

20

V10 2111

j4000400Vj800)V(400 21

Node 1

…(1)

Node 2

Substitute20

VI 1

o

…(2)

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Solution

0

j40

V

V

j26j262

4j84

2

1

j160)16(j26j262

4j84Δ

j80)1040(0j262

j40j84Δ2

Divide both equation (1) and (2) with 100 to simplify the equations and put into matrix

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Solution

V88.696.487j16016

j801040

Δ

ΔV o2

2

Solve for V2 using Cramer’s Rule

Solve for Vo by using voltage divider rule

V70.266.154Vj1030

30V o

2o

)V70.26t06.154cos(1(t)v o3o