MESH ANALYSIS

The Mesh Current Method, also know as the Loop Current Method, is quite similar to the Branch current method in that it uses simultaneous equations, Kirchhoff Voltage Law, and Ohms Law to determine unknown currents in a network. It differs from the Branch current method in that it does not use Kirchhoff  Current Law, and it is usually able to solve a circuit with less unknown variables and less simultaneous equation, which is especially nice if you're forced to solve without a calculator

                The first step in the Mesh Current method is to identify “loops” within the circuit encompassing all components. In our example circuit, the loop formed by B1, R1, and R2 will be the first while the loop formed by B2, R2, and R3 will be the second. The strangest part of the Mesh Current method is envisioning circulating currents in each of the loops. In fact, this method gets its name from the idea of these currents meshing together between loops like sets of spinning gears:

The choice of each current's direction is entirely arbitrary, just as in the Branch Current method, but the resulting equations are easier to solve if the currents are going the same direction through intersecting components (note how currents I1 and I2 are both going “up” through resistor R2, where they “mesh,” or intersect). If the assumed direction of a mesh current is wrong, the answer for that current will have a negative value.

The next step is to label all voltage drop polarities across resistors according to the assumed directions of the mesh currents. Remember that the “upstream” end of a resistor will always be negative, and the “downstream” end of a resistor positive with respect to each other, since electrons are negatively charged. The battery polarities, of course, are dictated by their symbol orientations in the diagram, and may or may not “agree” with the resistor polarities (assumed current directions):

 

Using Kirchhoff's Voltage Law, we can now step around each of these loops, generating equations representative of the component voltage drops and polarities. As with the Branch Current method, we will denote a resistor's voltage drop as the product of the resistance (in ohms) and its respective mesh current (that quantity being unknown at this point). Where two currents mesh together, we will write that term in the equation with resistor current being the sum of the two meshing currents.

Tracing the left loop of the circuit, starting from the upper-left corner and moving counter-clockwise (the choice of starting points and directions is ultimately irrelevant), counting polarity as if we had a voltmeter in hand, red lead on the point ahead and black lead on the point behind, we get this equation:

 -28 + 2 (I1 - I2) + 4I1 = 0

Notice that the middle term of the equation uses the sum of mesh currents I1 and I2 as the current through resistor R2. This is because mesh currents I1 and I2 are going the same direction through R2, and thus complement each other. Distributing the coefficient of 2 to the I1 and I2 terms, and then combining I1 terms in the equation, we can simplify as such:

-28 + 2 (I1 - I2) + 4I1 = 0                  Original Form of Equation

..Distributing to terms within parenthesis..
-28 + 2I1+ 2I2 + 4I1 = 0

..Combining like terms..

-28 + 6I1 + 2I2 = 0                          Simplified Form of Equation

At this time we have one equation with two unknowns. To be able to solve for two unknown mesh currents, we must have two equations. If we trace the other loop of the circuit, we can obtain another KVL equation and have enough data to solve for the two currents. Creature of habit that I am, I'll start at the upper-left hand corner of the right loop and trace counter-clockwise:

-2 (I1 +I2) +7 - 1I2 = 0

Simplifying the equation as before, we end up with:

-2I1 - 3I2 +7 = 0

Now, with two equations, we can use one of several methods to mathematically solve for the unknown currents I1 and I2:

-28 + 6I1 + 2I2 = 0

-2I1 - 3I2 + 7 = 0

..rearranging equation for easier equation..

6I1 + 2I2 = -28

-2I1 - 3I2 = -7

Solutions:

I1 = 5 A

I2 = -1 A


CASES TO BE CONSIDERED FOR MESH ANALYSIS

CASE 1
  A current source exists only in one mesh

 

CASE 2

A current source exist between two meshes

 

We can write one mesh equation by considering the current source. The current source  is related to the mesh currents at by

i1=i2+1.5

In order to write the second mesh equation, we must decide what to do about the current source voltage. Notice that there is no easy way to express the current source voltage in terms of the mesh currents. In this example, illustrate two methods of writing the second mesh equation.

Apply KVL to the supermesh corresponding the current source. Shown in blue, this supermesh is the perimeter of the two meshes that each contain the current source. 

 

Apply KVL to the supermesh to get 

9i1+3i2+6i2-12=0 or 9i1+9i2=12

This is the same equation that was obtained using method 1. Applying KVL to the supermesh is a shortcut for doing three things.

1. labeling the current source voltage as v

2. applying KVL to both meshes that contain the current source

3. eliminating v from the KVL equations

In summary, the mesh equation are 

i1=i2+1.5

and 

9i1+9i2=12

by simplying this, the equation results to 

i1=1.4167 A          and      i2=-83.3 mA

Reflection:

In this topic : Mesh analysis , i have learned that it is use to solve planar circuits for the currents at any place in the circuit. this means that there no wires crossing to each other. The difference between a mesh and a loop is that, the mesh is a loop which does not contain any other loops within it while a loop is any continuity path that is available to a current flow from a given circuit. But so far in answering the mesh analysis, it is still challenging because it is hard for me to get equations in a given circuit especially in using matrix solving. I am troubling in using matrix because of lack of practice. That's why I have practice in order have correct answer to the problem.