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Engineering Dictionary:
Barium: What to do with someone when they die
Tablet: A little table
Node: I knew it but now I don’t.
MESH ANALYSIS
Mesh analysis is another general procedure for analyzing circuits, using
mesh currents as the circuit variables. Using mesh currents instead of
element currents as circuit variables reduces the number of equations
that must be solved simultaneously.
Recall that a loop is a closed path with no node passed more than once.
A mesh is a loop that does not contain any other loop within it.
Nodal analysis applies KCL to nd unknown voltages in a given circuit,
while mesh analysis applies KVL to nd unknown currents. Mesh
analysis is not quite as general as nodal analysis because it is only
applicable to a circuit that is planar.
Planar Circuit:
A planar circuit is one that can be drawn in a plane with no
branches crossing one another; otherwise it is
nonplanar. 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 circuit in shown in a has two crossing branches, but it
can be redrawn as in b. Hence, the circuit is planar.
The circuit next is nonplanar, because there is no way to redraw it and avoid the branches
crossing. Nonplanar circuits can be handled using nodal analysis, but they will not be
considered in this course.
A mesh is a loop which does not contain any other loops within it.
In the circuit, paths abefa and bcdeb are meshes, but
path abcdefa is not a mesh. The current through a mesh
is known as mesh current. In mesh analysis, we are
interested in applying KVL to nd the mesh currents in
a given circuit.
We will start by applying mesh analysis to planar
circuits that do not contain current sources.
In the mesh analysis of a circuit with n meshes, we take the following three steps.
1. Assign mesh currents i1, i2, . . . , in to the n meshes.
2. Apply KVL to each of the n meshes. Use Ohm’s law to express the voltages in terms
of the mesh currents.
3. Solve the resulting n simultaneous equations to get the mesh currents.
To illustrate the steps, consider the circuit shown.
The rst step requires that mesh currents i1 and i2
are assigned to meshes 1 and 2.
Although a mesh current may be assigned to each
mesh in an arbitrary direction, it is conventional
to assume that each mesh current flows clockwise.
As the second step, we apply KVL to each mesh.
Applying KVL to mesh 1, we obtain
−V1 + R1i1 + R3(i1 − i2) = 0
or
(R1 + R3)i1 − R3i2 = V1
For mesh 2, applying KVL gives
R2i2 + V2 + R3(i2 − i1) = 0
or
−R3i1 + (R2 + R3)i2 = −V2
Note that in the first set of equations the coefcient of i1 is the sum of the resistances in
the rst mesh, while the coefcient of i2 is the negative of the resistance common to
meshes 1 and 2. Now observe that the same is true in the second set of equations.
The third step is to solve for the mesh currents. Putting the equations in matrix form
yields
which can be solved to obtain the mesh currents i1 and i2. We can use any technique for
solving the simultaneous equations. Previously we learned that if a circuit has n nodes, b
branches, and l independent loops or meshes, then
l = b − n + 1.
Hence, l independent simultaneous equations are required to solve the circuit using mesh
analysis.
Notice that the branch currents are different from the mesh currents unless the mesh is
isolated. To distinguish between the two types of currents, we use i for a mesh current
and I for a branch current. The current elements I1, I2, and I3 are algebraic sums of the
mesh currents.
It is evident from the circuit above that
I1 = i1, I2 = i2, I3 = i1 − i2
Given the circuit below, solve for the loop currents i1 and i2 indicated using mesh
analysis.
Solution of Simultaneous Equations Using Cramer’s Rule
In circuit analysis, we often encounter a set of simultaneous equations having the form
where there are n unknown x1, x2, . . . , xn to be determined. These equations can be
written in matrix form as
This matrix equation can be put in a compact form as
AX =B
where
A is square (n x n) matrix while X and B are column matrices.
There are several methods for solving these equations. These include substitution,
Gaussian elimination, Cramer’s rule, and numerical analysis. In many cases, Cramer’s
rule can be used to solve the simultaneous equations we encounter in circuit analysis.
Cramer’s rule states that the solutions are
Where the Delta’s are the determinants given by
Notice that ∆ is the determinant of the matrix A and ∆k is the determinant of the matrix
formed by replacing the kth column of A by B. Cramer’s rule applies only when ∆ != 0.
When ∆ = 0, the set of equations has no unique solution, because the equations are
linearly dependent.
The value of the determinant ∆, for example, can be obtained by expanding along the first
row:
where the minor Mij is an (n − 1) × (n − 1) determinant of the matrix formed by striking
out the ith row and jth column. The value of ∆ may also be obtained by expanding along
the rst column:
∆ = a11M11 − a21M21 + a31M31 + · · · + (−1)n+1
an1Mn1
We now specifically develop the formulas for calculating the determinants of 2 ×2 and 3
×3 matrices, because of their frequent occurrence. For a 2 × 2 matrix,
For a 3 x 3 matrix:
An alternative method of obtaining the determinant of a 3 × 3 matrix is by repeating the
rst two rows and multiplying the terms diagonally as follows.
In summary:
The solution of linear simultaneous equations by Cramer’s rule boils down to finding
xk = ∆k / ∆, k = 1, 2, . . . , n
where ∆ is the determinant of matrix A and ∆k is the determinant of the matrix formed
by replacing the kth column of A by B.
Apply Mesh Analysis to find in the circuit below:
MESH ANALYSIS WITH CURRENT SOURCES
Applying mesh analysis to circuits containing current sources (dependent or independent)
may appear complicated. But it is actually much easier than what we encountered in the
previous section, because the presence of the current sources reduces the number of
equations. Consider the following two possible cases.
Case 1. When a current source exists only in one mesh:
Consider the circuit shown. We set i2 = −5 A and write a
mesh equation for the other mesh in the usual way, that is,
−10 + 4i1 + 6(i1 − i2) = 0 -> i1 = −2 A
Case 2: When a current source exists between two meshes:
Consider the circuit in (a), for example. We create a supermesh by excluding the current
source and any elements connected in series with it,
as shown in (b).
A supermesh results when two meshes have a (dependent or independent)
current source in common.
As shown in (b), we create a supermesh as the periphery of the two meshes and treat it
differently. (If a circuit has two or more supermeshes that intersect, they should be
combined to form a larger supermesh.) Why treat the supermesh differently? Because
mesh analysis applies KVL—which requires that we know the voltage across each
branch—and we do not know the voltage across a current source in advance. However, a
supermesh must satisfy KVL like any other mesh.
Therefore, applying KVL to the supermesh in (b) gives
−20 + 6i1 + 10i2 + 4i2 = 0
or
6i1 + 14i2 = 20
We apply KCL to a node in the branch where the two meshes intersect.
Applying KCL to node 0 in (a) gives
i2 = i1 + 6
Solving we get
i1 = −3.2 A, i2 = 2.8 A (3.20)
Note the following properties of a supermesh:
1. The current source in the supermesh is not completely ignored; it provides the
constraint equation necessary to solve for the mesh currents.
2. A supermesh has no current of its own.
3. A supermesh requires the application of both KVL and KCL.
Use mesh analysis to obtain i0 in the circuit below
1. Identify the supermesh
2. write a mesh equation for the super mesh.
3. Write mesh equations for any addition meshes
4. Write down any equations from KCL
5. Solve
Do Online Tutorial on Mesh Analysis
NODAL VS MESH ANALYSIS
Both nodal and mesh analyses provide a systematic way of analyzing a complex network.
Someone may ask: Given a network to be analyzed, how do we know which method is
better or more efficient?
Mesh analysis:
Networks that contain many series-connected elements, voltage sources, or supermeshes.
a circuit with fewer meshes than nodes is better analyzed using mesh analysis.
Nodal Analysis
Networks with parallel-connected elements, current sources, or supernodes are more
suitable for nodal analysis. A circuit with fewer nodes than meshes is better analyzed
using nodal analysis
The key is to select the method that results in the smaller number of equations.
The second factor is the information required. If node voltages are required, it may be
expedient to apply nodal analysis. If branch or mesh currents are required, it may be
better to use mesh analysis.
It is helpful to be familiar with both methods of analysis, for at least two reasons. First,
one method can be used to check the results from the other method, if possible. Second,
since each method has its limitations, only one method may be suitable for a particular
problem. For example, mesh analysis is the only method to use in analyzing transistor
circuits. Mesh analysis cannot easily be used to solve an op amp circuit, because there is
no direct way to obtain the voltage across the op amp itself.
For nonplanar networks, nodal analysis is the only option, because mesh analysis only
applies to planar networks. Also, nodal analysis is more amenable to solution by
computer, as it is easy to program. This allows one to analyze complicated circuits that
defy hand calculation.
A computer software package based on nodal analysis will be introduced next class.

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Day 6-notes-mesh-analysis

  • 1. Engineering Dictionary: Barium: What to do with someone when they die Tablet: A little table Node: I knew it but now I don’t. MESH ANALYSIS Mesh analysis is another general procedure for analyzing circuits, using mesh currents as the circuit variables. Using mesh currents instead of element currents as circuit variables reduces the number of equations that must be solved simultaneously. Recall that a loop is a closed path with no node passed more than once. A mesh is a loop that does not contain any other loop within it. Nodal analysis applies KCL to nd unknown voltages in a given circuit, while mesh analysis applies KVL to nd unknown currents. Mesh analysis is not quite as general as nodal analysis because it is only applicable to a circuit that is planar. Planar Circuit: A planar circuit is one that can be drawn in a plane with no branches crossing one another; otherwise it is nonplanar. 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 circuit in shown in a has two crossing branches, but it can be redrawn as in b. Hence, the circuit is planar.
  • 2. The circuit next is nonplanar, because there is no way to redraw it and avoid the branches crossing. Nonplanar circuits can be handled using nodal analysis, but they will not be considered in this course. A mesh is a loop which does not contain any other loops within it. In the circuit, paths abefa and bcdeb are meshes, but path abcdefa is not a mesh. The current through a mesh is known as mesh current. In mesh analysis, we are interested in applying KVL to nd the mesh currents in a given circuit. We will start by applying mesh analysis to planar circuits that do not contain current sources. In the mesh analysis of a circuit with n meshes, we take the following three steps. 1. Assign mesh currents i1, i2, . . . , in to the n meshes. 2. Apply KVL to each of the n meshes. Use Ohm’s law to express the voltages in terms of the mesh currents. 3. Solve the resulting n simultaneous equations to get the mesh currents. To illustrate the steps, consider the circuit shown. The rst step requires that mesh currents i1 and i2 are assigned to meshes 1 and 2. Although a mesh current may be assigned to each mesh in an arbitrary direction, it is conventional to assume that each mesh current flows clockwise. As the second step, we apply KVL to each mesh. Applying KVL to mesh 1, we obtain −V1 + R1i1 + R3(i1 − i2) = 0 or (R1 + R3)i1 − R3i2 = V1 For mesh 2, applying KVL gives
  • 3. R2i2 + V2 + R3(i2 − i1) = 0 or −R3i1 + (R2 + R3)i2 = −V2 Note that in the first set of equations the coefcient of i1 is the sum of the resistances in the rst mesh, while the coefcient of i2 is the negative of the resistance common to meshes 1 and 2. Now observe that the same is true in the second set of equations. The third step is to solve for the mesh currents. Putting the equations in matrix form yields which can be solved to obtain the mesh currents i1 and i2. We can use any technique for solving the simultaneous equations. Previously we learned that if a circuit has n nodes, b branches, and l independent loops or meshes, then l = b − n + 1. Hence, l independent simultaneous equations are required to solve the circuit using mesh analysis. Notice that the branch currents are different from the mesh currents unless the mesh is isolated. To distinguish between the two types of currents, we use i for a mesh current and I for a branch current. The current elements I1, I2, and I3 are algebraic sums of the mesh currents. It is evident from the circuit above that I1 = i1, I2 = i2, I3 = i1 − i2 Given the circuit below, solve for the loop currents i1 and i2 indicated using mesh analysis.
  • 4. Solution of Simultaneous Equations Using Cramer’s Rule In circuit analysis, we often encounter a set of simultaneous equations having the form where there are n unknown x1, x2, . . . , xn to be determined. These equations can be written in matrix form as This matrix equation can be put in a compact form as AX =B where A is square (n x n) matrix while X and B are column matrices. There are several methods for solving these equations. These include substitution, Gaussian elimination, Cramer’s rule, and numerical analysis. In many cases, Cramer’s rule can be used to solve the simultaneous equations we encounter in circuit analysis. Cramer’s rule states that the solutions are
  • 5. Where the Delta’s are the determinants given by Notice that ∆ is the determinant of the matrix A and ∆k is the determinant of the matrix formed by replacing the kth column of A by B. Cramer’s rule applies only when ∆ != 0. When ∆ = 0, the set of equations has no unique solution, because the equations are linearly dependent. The value of the determinant ∆, for example, can be obtained by expanding along the rst row: where the minor Mij is an (n − 1) × (n − 1) determinant of the matrix formed by striking out the ith row and jth column. The value of ∆ may also be obtained by expanding along the rst column: ∆ = a11M11 − a21M21 + a31M31 + ¡ ¡ ¡ + (−1)n+1 an1Mn1 We now specically develop the formulas for calculating the determinants of 2 ×2 and 3 ×3 matrices, because of their frequent occurrence. For a 2 × 2 matrix,
  • 6. For a 3 x 3 matrix: An alternative method of obtaining the determinant of a 3 × 3 matrix is by repeating the rst two rows and multiplying the terms diagonally as follows. In summary: The solution of linear simultaneous equations by Cramer’s rule boils down to nding xk = ∆k / ∆, k = 1, 2, . . . , n where ∆ is the determinant of matrix A and ∆k is the determinant of the matrix formed by replacing the kth column of A by B. Apply Mesh Analysis to find in the circuit below:
  • 7. MESH ANALYSIS WITH CURRENT SOURCES Applying mesh analysis to circuits containing current sources (dependent or independent) may appear complicated. But it is actually much easier than what we encountered in the previous section, because the presence of the current sources reduces the number of equations. Consider the following two possible cases. Case 1. When a current source exists only in one mesh: Consider the circuit shown. We set i2 = −5 A and write a mesh equation for the other mesh in the usual way, that is, −10 + 4i1 + 6(i1 − i2) = 0 -> i1 = −2 A Case 2: When a current source exists between two meshes: Consider the circuit in (a), for example. We create a supermesh by excluding the current source and any elements connected in series with it, as shown in (b).
  • 8. A supermesh results when two meshes have a (dependent or independent) current source in common. As shown in (b), we create a supermesh as the periphery of the two meshes and treat it differently. (If a circuit has two or more supermeshes that intersect, they should be combined to form a larger supermesh.) Why treat the supermesh differently? Because mesh analysis applies KVL—which requires that we know the voltage across each branch—and we do not know the voltage across a current source in advance. However, a supermesh must satisfy KVL like any other mesh. Therefore, applying KVL to the supermesh in (b) gives −20 + 6i1 + 10i2 + 4i2 = 0 or 6i1 + 14i2 = 20 We apply KCL to a node in the branch where the two meshes intersect. Applying KCL to node 0 in (a) gives i2 = i1 + 6 Solving we get i1 = −3.2 A, i2 = 2.8 A (3.20) Note the following properties of a supermesh: 1. The current source in the supermesh is not completely ignored; it provides the constraint equation necessary to solve for the mesh currents. 2. A supermesh has no current of its own. 3. A supermesh requires the application of both KVL and KCL. Use mesh analysis to obtain i0 in the circuit below
  • 9. 1. Identify the supermesh 2. write a mesh equation for the super mesh. 3. Write mesh equations for any addition meshes 4. Write down any equations from KCL 5. Solve
  • 10.
  • 11. Do Online Tutorial on Mesh Analysis NODAL VS MESH ANALYSIS Both nodal and mesh analyses provide a systematic way of analyzing a complex network. Someone may ask: Given a network to be analyzed, how do we know which method is better or more efficient? Mesh analysis: Networks that contain many series-connected elements, voltage sources, or supermeshes. a circuit with fewer meshes than nodes is better analyzed using mesh analysis. Nodal Analysis Networks with parallel-connected elements, current sources, or supernodes are more suitable for nodal analysis. A circuit with fewer nodes than meshes is better analyzed using nodal analysis The key is to select the method that results in the smaller number of equations. The second factor is the information required. If node voltages are required, it may be expedient to apply nodal analysis. If branch or mesh currents are required, it may be better to use mesh analysis. It is helpful to be familiar with both methods of analysis, for at least two reasons. First, one method can be used to check the results from the other method, if possible. Second, since each method has its limitations, only one method may be suitable for a particular problem. For example, mesh analysis is the only method to use in analyzing transistor circuits. Mesh analysis cannot easily be used to solve an op amp circuit, because there is no direct way to obtain the voltage across the op amp itself. For nonplanar networks, nodal analysis is the only option, because mesh analysis only applies to planar networks. Also, nodal analysis is more amenable to solution by computer, as it is easy to program. This allows one to analyze complicated circuits that defy hand calculation. A computer software package based on nodal analysis will be introduced next class.