the mesh current method
TRANSCRIPT
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Dr.
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Dr.
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THE MESH CURRENT METHOD
In the mesh current method a current is assigned to each window of the network such that the currents complete a closed loop. They are sometimes referred to as loop currents. Each element and branch therefore will have an independent current . When a branch has two of the mesh currents, the actual current is given by their algebraic sum. The assigned mesh currents may have either clockwise or counterclockwise directions, although at the outset it is wise to assign to all of the mesh currents a clockwise direction. Once the currents are assigned , Kirchhoff’s voltage law is written for each loop to obtain the necessary simultaneous equations .
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When KVL is applied to the three-mesh network of the following Fig., the following three equations are obtained:
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If there are m independent meshes in any linear network, then the mesh equations can be written in the matrix form as under:
The above equation can be written in a more compact form as:
Ohms law in matrix
form
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The three loop circuit in generalized matrix form can be solved by Cramer’s rule, as follows:
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Mesh analysis applies KVL to find unknown currents. Mesh analysis is only applicable to a circuit that is planar. A planar circuit is one that can be drawn in a plane with no branches crossing one another; otherwise it is non planar. A circuit may have crossing branches and still be planar if it can be redrawn such that it has no crossing branches. For example, the
Planar Circuits Non-Planar Circuits
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Example: Solve the current I1 in the given circuit by the mesh current
method.
Applying KVL to each mesh results in
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Rearranging terms and putting the equations in matrix form,
Using Cramer’s rule to find I1,
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Example: Write the current matrix equation for the given network by
inspection, and solve for the currents.
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For loop 1
For loop 2
For loop 3
Example: Apply mesh analysis to find i in the given Fig.
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Example: Apply mesh analysis to find
V0 in the given Figure.
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7 -5
-5 9
i1
i2
=
30 - 20
20
7 -5
-5 9
i1
i2
=
10
20
7 10
-5 9
i2 = 7 -5
-5 9
= (140 + 50)
(63 - 25) =
190
38 = 5 A
V0 = i2 X 4 = 20 V
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Example: Use mesh analysis to find the current io in the circuit in the
following Fig.
(1)
(2)
(3)
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In matrix form, Eqs. (1) to (3) become
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Example: For the bridge network in the given Figure, find io using mesh analysis.
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12 -6 -4
-6 14 -2
-4 -2 10
i1
i2
i3
=
30
0
0
6 -3 -2
-3 7 -1
-2 -1 5
i1
i2
i3
15
0
0
=
15 -3 -2 0 7 -1 0 -1 5
6 -3 -2 -3 7 -1 -2 -1 5
= 15 (35 – 1) 6 (35 – 1) + 3 (-15 -2) -2 (3 + 14) = 4.286 A i1 =
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Example: Apply mesh analysis to find
V0 in the given Figure.