Mesh Analysis

As-Salaam-Alaikum,Welcome to my blogger "World of Knowledge".Here we discuss the topic mesh analysis.

Points to be discuss in this Article are given below:

  • Mesh Analysis
  • Steps to Analyse the mesh analysis technique
  • Super Mesh Analysis
  • Super Mesh with examples



Mesh Analysis:

                                   Mesh Current Analysis is a technique used to find the currents circulating around a loop or mesh with in any closed path of a circuit.

 Example:



One simple method of reducing the amount of math’s involved is to analyse the circuit using Kirchhoff’s Current Law equations to determine the currents, I1 and I2 flowing in the two resistors. Then there is no need to calculate the current I3 as its just the sum of I1 and I2. So Kirchhoff’s second voltage law simply becomes,
                                             10 =  50I1 + 40I2                                                                                                                                                 
                                             20 =  40I1 + 60I2

We get these two equation by applying mesh.

Steps to Analyse the mesh analysis technique:
  • Check whether there is a possibility to transform all current sources in the given circuit to voltage sources.
  • Assign the current directions to each mesh in a given circuit and follow the same direction for each mesh.
  • Apply KVL to each mesh and simplify the KVL equations.
  • Solve the simultaneous equations of various meshes to get the mesh currents and these equations are exactly equal to the number of meshes present in the network. 

Super Mesh Analysis:

A super mesh is formed when two adjacent meshes share a common current source and none of these (adjacent) meshes contains a current source in the outer loop.

Example of Super Mesh Analysis:

 The current source is common to the meshes 1 and 2 and hence it must be analysed independently. To achieve this, assume the branch that contains current source is open circuited and create a new mesh called super mesh 


Writing KVL to the super mesh we get,
V = I1R1 + (I2 – I3) R3
 V = I1R1 + I2R3 – I3R3
Applying KVL to the Mesh 3 we get,,
(I3 – I2) R3 + I3R4 = 0
And the difference between the two mesh currents gives the current from the current source. Here the current source direction is in the loop current direction I1. Hence I1 is more than I2, then
I = I1 – I2
Thus, by using these three mesh equations we can easily find the three unknown currents in the network
Another example regarding mesh analysis:
By applying the KVL to the mesh 1 we get
1I1 + 10 (I1 – I2) = 2
11I1 – 10 I2 = 2
The meshes 2 and 3 consist of 4A current source and hence form a super mesh. The current from 4A current source is in the direction of I3 and thus the super mesh current is given as
I = I3 – I2
I3 – I2 = 4
By applying KVL to the outer loop of the super mesh we get,
– 10 (I2 – I1) – 5I2 – 15I3 = 0
10I1 – 15I2 – 15I3 = 0
By solving 1, 2 and 3 equations, we get
I1 = –2.35 A
I2 = –2.78 A
I3 = 1.22 A
Hence the current through the 10 ohms resistor is
                                            I1 – I2 = –2.35 + 2.78 A = 0.43 A

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