Week Seven: "Mesh Analysis"

What is a Mesh Analysis?

Mesh analysis provides another general procedure for analyzing circuits, using mesh currents as the circuit variables. Using mesh currents instead of element currents as circuit variables is convenient and 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. 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.

To understand mesh analysis, we should first explain more about what we
mean by a mesh.

In Figure below, for example, 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.

In this section, we will apply mesh analysis to planar circuits that do not
contain current sources. In the next sections, we will consider circuits with 
current sources. In the mesh analysis of a circuit with n meshes, we take the 
following three steps.

Steps to Determine Mesh Currents:

1. Assign mesh currents i1,i2,...,into the n meshes.

1. Apply KVL to each of then meshes. Use Ohm’s law to express the
   voltages in terms of the mesh currents.

1. Solve the resulting n simultaneous equations to get the mesh currents.
   
   

Mesh Analysis Without Current Sources:

Example:

 

Now write KVL equations for each loop.

@Mesh 1:

-18V + 5(i1-i2) + 4(i1-i3) + 1(i1) = 0

then gather terms:

10i1 - 5i2 - 4 i3 -18V = 0

Note that the i1 term is positive, and all other current terms are negative (because 

they are all clockwise, all other panes will contribute a negative term). Let's do the 

other two panes with terms gathered up directly (write the total resistance of the 

loop multiplied by the mesh current that goes through that total resistance):

@Mesh 2:

-5i1 + 10i2 - 3i3 - 12 = 0

@ Mesh 3:

-4i1 -3i2 +9i3 = 0

Now solve the three equations in three unknowns:

i1 = 7.02A
i2 = 6.28A
i3 = 5.21A

The current through the 5 ohm resistor is just i1 - i2, or 0.74A. The current 

through the 18V is i1, or 7.02A. All the branch currents are shown below from a 

Pspice simulation:

 

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. We set

mesh current i equal to that current source.

Case 2:

When a current source exists between two meshes. We create a supermesh 
by excluding the current source and any elements connected in series with 
it.

Examples:

Case1:

First, define a mesh current arround each mesh (window pane) of the circuit. 
Define each one in a clockwise direction.

Now write KVL equations for each loop. On the first loop, we run into a problem. 
We do not know the voltage across the 5A source. (This is one of the difficulties 
you can run into when using mesh current.) We add a new unknown to handle this problem. We run into the same problem with the 2A source, so two new unknowns 
 (V5 and V2) are labeled:

Now we are ready to write the KVL equations.

@Mesh 1:

-V5 + 5(i1-i2) = 0

@ Mesh 2:

5(i2-i1)+10(i2-i4)+2(i2-i3) = 0

or

-5i1 + 17i2 - 2i3 = 0

@Mesh 3:

2(i3-i2) + V2 = 0

@Mesh 4:

10i4 + 4V + 10(i4-i2) = 0

or

-10i2 + 20i4 = -4

We have two extra unknowns, so we need two more equations. These can be 

found by examining the loops with the sources:

i1= 5A

i3 = -2A

We can then solve for the remaining unknowns:

i2 = 1.58A
i4 = 0.59A

We can then find the current through the 5 ohm resistor as:

I = i1 - i2 = 3.42A

The current through the 4V source is simply i4, or 0.59A.

Case 2:

Determine the values of the mesh currents, i1and i2, for this circuit.

Solution:

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

(KCL)

 i1 - i2 = 1.5  or  i1 = 1.5 + i2

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 to

the current source. Shown below in

blue, this supermesh is the perimeter of the two meshes that each contain the
current source.

 

 @ Supermesh:

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

Solving the equations gives:

i1 = 1.4167A   and   i1 = -83.3mA

Some Learnings:

• Mesh Analysis uses KVL to obtain unknown mesh currents, while Nodal Analysis uses KCL to obtain unknown nodal voltages.
• Mesh Analysis is only applicable to planar circuits. Planar circuits are circuits that can be drawn on a plane surface, with no wires crossing each other.
• Mesh Analysis is different from Loop Analysis. Loop Analysis is applicable to both planar or not. Both Mesh & Loop Analysis make use of Kirchhoff's 
• Voltage Law or KVL.
• Mesh Analysis is usually easier to use when the circuit is planar, compared to Loop Analysis.

Video:

       For more information, watch the video below:

 

That's all. Thank You for visiting my blog.

GOD Bless! :)

By:

AYALA, ARNY  S.   BSECE -3

ECE 311

Professor:

ENGR. JAY S. VILLAN, MEP - EE