NODAL ANALYSIS WITH VOLTAGE SOURCES
How do voltage sources affect nodal analysis?
Consider the following two possibilities.
1.
If a voltage source is connected between the reference node and a nonreference node, we simply set the voltage at the nonreference node equal to the voltage of the voltage source. In this case, v
1
= 10 V
Thus our analysis is somewhat simplified by this knowledge of the voltage at this node.
2 If the voltage source (dependent or independent) is connected between two nonreference nodes, the two nonreference nodes form a generalized node or supernode; we apply both KCL and KVL to determine the node voltages.
A supernode is formed by enclosing a (dependent or independent) voltage source connected between two nonreference nodes and any elements connected in parallel with it.
In the circuit, nodes 2 and 3 form a supernode. (We could have more than two nodes forming a single supernode.) We analyze a circuit with supernodes using the same three steps mentioned in the previous section except that the supernodes are treated differently.
Why?

Because an essential component of nodal analysis is applying KCL, which requires knowing the current through each element and there is no way of knowing the current through a voltage source in advance.
However, KCL must be satisfied at a supernode like any other node. Hence, at the supernode i
1
+ i
4
= i
2
+ i
3 or v
1
 v
2
 v
1
 v
3  v
2

0

2 4 8
To apply Kirchhoff’s voltage law to the supernode, we v
3 redraw the circuit as shown. Going around the loop in the
6

0 clockwise direction gives
−v
2
+ 5 + v
3
= 0 -> v
2
− v
3
= 5
We now have 3 equations and 3 unknowns so we can obtain the node voltages.
Find v
1
, v
2
, and v
3
.
Note the following properties of a supernode:
1. The voltage source inside the supernode provides a constraint equation needed to solve for the node voltages.
2. A supernode has no voltage of its own.
3. A supernode requires the application of both KCL and KVL.
Given the circuit below, solve for Vx.
+
+
PRESENT everything you know about the problem.
The goal of the problem is to find Vx . Letting the lower node be the reference node, we need to find the voltage to the right of the 10-V voltage source.

A supernode is formed by enclosing a (dependent or independent) voltage source connected between two non-reference nodes and any elements connected in parallel with it. Hence, it is clear that the two nodes on either side of the 10-V voltage source form a supernode.
A supermesh results when two meshes have a (dependent or independent) current source in common. Hence, the two meshes on the right half of the circuit create a supermesh, with the 9-A current source in common.
Establish a set of ALTERNATIVE solutions and determine the one that promises the greatest likelihood of success .
Because the goal of the problem is to find a voltage, the obvious choice is to use nodal analysis to find the voltages at each node of the circuit.
ATTEMPT a problem solution.
Begin the problem solution by identifying the nodes, including the supernode.

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 find unknown voltages in a given circuit, while mesh analysis applies KVL to find 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 find 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 i
1
, i
2
, . . . , i n
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 first step requires that mesh currents i
1
and i
2 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
−V
1
+ R
1 i
1
+ R
3
(i
1
− i
2
) = 0 or
For mesh 2, applying KVL gives
(R
1
+ R
3
)i
1
− R
3 i
2
= V
1
R
2 i
2
+ V
2
+ R
3
(i
2
− i
1
) = 0 or
−R
3 i
1
+ (R
2
+ R
3
)i
2
= −V
2

Note that in the first set of equations the coefficient of i
1
is the sum of the resistances in the first mesh, while the coefficient of i
2
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 I
1
, I
2
, and I
3
are algebraic sums of the mesh currents.
It is evident from the circuit above that
I
1
= i
1
, I
2
= i
2
, I
3
= i
1
− i
2
Given the circuit below, solve for the loop currents i
1
and i
2
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 first column:

= a
11
M
11
− a
21
M
21
+ a
31
M
31
+ · · · + (−1) n+1 a n1
M n1
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 first 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 x k
=
 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.
Do Nodal Analysis Lab