Upload
preston-lindsey
View
266
Download
60
Embed Size (px)
DESCRIPTION
Nodal and Mesh Analysis. Nodal Analysis KCL Consider Node current Node voltage. Example 1. Given => node current Determine => node voltage =>Find voltage at node 1,2 and 3. Steps of Nodal Analysis. - PowerPoint PPT Presentation
Citation preview
Nodal and Mesh Analysis
Nodal Analysis– KCL– Consider
• Node current• Node voltage
Need to assign one node as a reference node.Normally it will be a ground node
R1, R2, R3, Vs, Is => electric elements
R1 R2
Vs Is
R1, R2, R3, Vs, Is => electric elements
R1 R2
Vs Is
Vi Vj
i j
vi & vj are node voltage
Branch voltage vij=vi-vj
Vi Vj
i j
vi & vj are node voltage
Branch voltage vij=vi-vj
Example 1
Given => node current
Determine => node voltage =>Find voltage at node 1,2 and 3
-8A. 1Ω 5Ω
7Ω
3Ω
4Ω
-25A.
-3A.
12
3
4
Steps of Nodal Analysis
5Ω
4Ω
-3A.
7Ω
-25A.
V1V2 V3
1Ω
-8A.
3Ω
1. Assign a reference node => a node having the most electric elements connecting to it or a ground node if it is given
2. Apply KCL to each node except the reference node.
3. Solving the simultaneous equations.
constant,,,,
constant,,
33131211
321
3333232131
2323222121
1313212111
aaaa
bbb
bxaxaxa
bxaxaxa
bxaxaxa
BAX
KCL@Node 1
V1 V23Ω
-3A.
4Ω
V3
-8A.
i1
i2 )2(1134
)1(08334
3
44
083
2131
2131
212
31131
21
vvvv
vvvv
vvi
vvvi
ii
KCL@Node 2
-3A.
3Ω 7Ω
1Ω
V1V2 V3
i2
i3
i4
)3(373
3173
03
23212
23221
432
vvvvv
vvvvv
iii
KCL@Node 3
From (2), (3) and (4) we get
v1 = 5.412 V , v2 = 7.736 V , v3 = 46.32 V
4Ω
7Ω
5Ω
-25A.
V1
V2 V3 )4(2554731323
vvvvv
Example 2
4Ω
2Ω 8Ω
4Ω2ix3A
.
(1)(2)
(3)
ix
Determine v1, v2 and v3
KCL @ node 1
)1(324
2131
vvvv
KCL @ node 2
)2(048223212
vvvvv
KCL @ node 3
)3(04
284
22313
vvvvv
From (1), (2) and (3) we getv1 = 4.8 V , v2 = 2.4 V , v3 = -2.4 V
Example 3
4Ω
3Ω
1Ω-8A.
-25A.
-3A.
22V.V3
V1V2
5Ω
Supernode
Determine v1, v2 and v3
Supernode internal relationship)1(2223 vv
KCL @ node 1
)2(1134
2131
vvvv
KCL @ supernode
)3(02554
313
313212
vvvvvv
From (1), (2) and (3) we getv1 , v2 , and v3
Example 4 (1)
0.5Ω 2Ω
0.2vy2.5Ω
12V.
14A.
0.5vx
1Ωvy
vx
V2
V3
V4
Ref.
+
-
-
+
Supernode
(2) 2.0
)1(V12
43
1
yvvv
v
Determine the voltage of the unknown node to reference voltages.
We know that
KCL @ node 1 Because it connects to reference node, the node equation has not to determine.
KCL @ node 2
)3(02
145.0
3212
vvvv
Example 4 (2)
0.5Ω 2Ω
0.2vy2.5Ω
12V.
14A.
0.5vx
1Ωvy
vx
V2
V3
V4
Ref.
+
-
-
+
Supernode
Determine the voltage of the unknown node to reference voltages.
KCL @ supernode (3,4)
)6(
)5(
)4(05.21
5.02
14
12
14423
vvv
vvv
vvvv
vv
y
x
x
From (1) - (6) we getv1 = -12 V, v2 = -4 V, v3 = 0 V and v4 = -2 V
Summarize of Nodal Analysis
1. Select the node to which the highest number of branches is connected as the referenced node.
2. Set up KCL equations for other nodes by expressing the unknown currents as a function of the node voltages measured with respect to the referenced node.
3. If the given circuit contains voltage sources, KCL equations of those two nodes connected by a voltage source are combined to eliminate the redundancy of KCL equations since the additional information is available through the node voltage.
4. Solve KCL equations to determine the node voltages.
Mesh Analysis
Mesh = a smallest loop
(a loop does not contain other loops inside)
Mesh analysis can be used only with planar-network.
Figure 3.5 Examples of planar and nonplanar networks. (a) and (c) are planar networks whereas (b) is a nonplanar network.
Figure 3.5 Examples of planar and nonplanar networks. (a) and (c) are planar networks whereas (b) is a nonplanar network.
Mesh Analysis Procedure
1. Assign all mesh currents in the same direction.
2. Set up KVL equation for each mesh. Use Ohm’s law to express the voltages in term of the mesh currents
3. If the given circuit contains current sources on the perimeter of any meshes. That is, two meshes share current sources in common. Such meshes form a supermesh to eliminate the redundancy of KVL equations.
4. Solve KVL equations to determine the mesh currents
Example 5
3Ωi1 i242V. 10V.
6Ω 4Ω)1(42)(36 211 iii
)2(10)(34 122 iii
Determine the current flowing through 3 Ω resistor using mesh analysis.
KVL @ mesh 1
A4
A6
2
1
i
i
KVL @ mesh 2
From (1) + (2),
current flowing through 3 Ω = 6 – 4 = 2 A. from top to bottom.
Example 6
2Ω
1Ω
6V.
2Ω
7V.
1Ω
3Ωi1
i2
i3
)1(123
122
7)(26)(
321
3121
3121
iii
iiii
iiii
)2(036
0)()(32
321
12322
iii
iiiii
Find i1, i2 and i3
KVL @ mesh 1
KVL @ mesh 2
KVL @ mesh 3
)3(6632
6)(2)(3
321
13323
iii
iiiii
A3 and,A2,A3 321 iii
Example 6
3Ωi1 i210V. 5A.
6Ω 4Ω
ix
9
40
)5(9
59
59
)5(310
9
310, Substitute
10)(36
5
21
212
211
2
iii
iii
Viii
Ai
xFind ix using mesh analysis
Because 5A source is at perimeter,The value i2 is already know
Example 7
A8.2andA2.3
(2), and (1) From
)2(20146
eshKVL@superm
)1(6
iprelationsh internalsupermesh
21
21
12
ii
ii
ii
20V.
6A.
4Ω
10Ω6Ω
2Ωi1 i2
20V.
10Ω6Ω
i1 i2
Find i1 , i2
Supermesh
Example 8 (1)
i1
i2i3
9V.
2.2kΩ
10kΩ
3.3kΩ4.7kΩ
2.2kΩ
1mA.
3mA.
Find current flowing through 4.7 k Ω and 2.2 k Ω.
i1
i2i3
10kΩ
3.3kΩ4.7kΩ
2.2kΩ
3mA.
mA3
mA1
32
1
ii
i
02.210)(3.37.4 32123 kikiiikki
Mesh analysis
Example 8 (2)
i1
i2i3
9V.
2.2kΩ
10kΩ
3.3kΩ4.7kΩ
2.2kΩ
1mA.
3mA.
Find current flowing through 4.7 k Ω and 2.2 k Ω.
9V.
2.2kΩ
10kΩ
3.3kΩ4.7kΩ
2.2kΩ
1mA.
3mA.
V2 V1
Node analysis