SUPER MESH ANALYSIS ~ ECE

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Sunday, 24 February 2013

SUPER MESH ANALYSIS

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SUPERMESH:

When a current source is common to two meshes we use the concept of super mesh. A supermesh is created from two meshes that have a current source in common

EXAMPLE

SOLUTION

At node A current i₁ is entering,i₂ and 1.5A are leaving

So by appling KCL at node A

-i₁ + i₂ + 1.5 = 0

i₁ = i₂ + 1.5 -------------------------------------- 1

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.

METHOD 1:

Assign a name to the current source voltage. Apply KVL to both of the meshes. Eliminate the current source voltage from the KVL equations.

Here’s the circuit after labeling the current source voltage V.

The KVL equation for mesh 1 is

12 – 9i₁ –v = 0

The KVL equation for mesh 2 is

-3i₂ -6i₂ + v =0

Solving above equations we get

9i₁+9i₂ = 12 ------------------ 2

Method 2:

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.

Apply KVL to the supermesh to get

12 -9i₁ -3i₂ -6i₂ = 0

9i₁+9i₂ = 12 ----------------------------2

Solving equation 1 & 2 we get

i1 = 17/2

i2 = -1/12

8 Responses to “SUPER MESH ANALYSIS”

Anonymous said... 28 March 2014 at 11:54 you did it wrong
Senthil said... 27 August 2014 at 05:38 thanks man
Anonymous said... 22 December 2014 at 13:55 this is wrong method or right?
Right Supermesh circuit Analysis said... 29 January 2015 at 00:49 Ya sure... let us check in detail for mesh analysis
Altin Guberi said... 31 January 2015 at 03:08 second way with the blue mesh is right
Anonymous said... 6 October 2017 at 04:46 this actually correct but he did a extra method which is not needed that is applying kvl to both mesh's just apply supermesh you will get the answer
Unknown said... 20 July 2019 at 11:28 Second one is rgt
Kenneth said... 25 September 2019 at 23:16 Method right though, i1 = 17/12. Not 17/2