1 METHODS OF CIRCUIT ANALYSIS. 2 Methods of Circuit Analysis Mesh Analysis Nodal Analysis.
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1
METHODS OF CIRCUIT ANALYSIS
2
Methods of Circuit Analysis
Mesh Analysis Nodal Analysis
3
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)
4
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,
5
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)
8
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)
12
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
19
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
21
Solution
22
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
39
Solution
Convert into frequency domain
40
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)
41
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
42
Example 10.2Compute V1 and V2 in the circuit
43
Solution
44
SolutionNodes 1 and 2 form a supernode.
12
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
46
Practice Problem 10.2Calculate V1 and V2 in the circuit using nodal analysis
47
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
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