Mesh Current Analysis with Example: Circuit Analysis

Mesh current Analysis provide a procedure for electric circuit analysis using mesh current as the circuit variable. The mesh analysis makes use of Kirchhoff’s Voltage Law as a basic key to analyze the circuit. In contrast to Nodal analysis, it uses loop current as a variable rather than element current, so it reduces the number of equations and complexity.

Mesh is a loop which does not contain any other loop. Referring to the figure below; ABEF and BCDE both are mesh but ACDF is not a mesh because it does contain another loop inside it.

Steps for analysis :

  1. Assign a name to each mesh current like i1, i2, and i3
  2. Apply KVL to each mesh and use ohm’s law to express the voltage drop in each circuit element.
  3. The second step will give you n number of simultaneous equations, where n is a number of meshes.
  4. Use any method to solve these simultaneous equations for n mesh current.

Example:

Suppose we know the following parameter of the given circuit.

V­­1= 12v,      V­­2= 8v
R1= 5Ω,       R­­2= 6Ω
R­­3= 10Ω

Mesh Current Analysis Example

Solution:

Assign Mesh Currents:

First, I start by:

  • Assigning the name of I­1 to the current in mesh ABEF in the direction specified in the following diagram.
  • And I2 to the current in mesh BCDE in the direction specified in the following diagram.

.

Write Mesh Equations :

In step 2: I start by writing an equation for each mesh using KVL

For mesh 1:$V_{1}= R_{1}I_{1}+ R_{3} (I_{1}- I_{2}) – (1) $

For mesh 2:
$V_{2}= R_{2}I_{2}+ R_{3} (I_{2}-I_{1}) – (2)$

Mesh Current Analysis Example

Note firstly that R3 is common in both meshes and is counted for twice (for each equation). Secondly, in resistance R3, the current I1 is flowing downward and current I2 is flowing upward (opposite of each other). For the mesh 1 equation, I followed current I­­1 direction and subtracted current I­2 from current I1. For mesh 2 equation, I followed current I­­2 direction and subtracted current I­1 from current I2.

Note: Remember a small mistake at this level will lead to a huge problem, so before proceeding to the next step make sure your equations are correct.

Solve Equations :

We had two mesh, that’s why we got two simultaneous equations. If you have written equations correctly, you are done. At this stage, you can use various software to calculate these mesh currents for you. Here I am converting these simultaneous equations to the matrix and find reduced echelon form of it.

$ (R_{1}+ R_{3})I_{1}- R_{3} I_{2}=V_{1}$

$-R_{3}I_{1}+(R_{2}+R_{3}) I_{2}=V_{2}$

Putting the appropriate values will get us:

$(5+ 10)I_{1}- 10 I_{2}=12$

$-10I_{1}+(6+10) I_{2}=8$

Converting to the augmented matrix for further operation:

$\lbrack \begin{matrix}
15 & -10 & \\
-10 & 16 & \\
\end{matrix}
\rbrack \lbrack \begin{matrix}
I_{1} & \\
I_{2} & \\
\end{matrix}
\rbrack = \lbrack \begin{matrix}
12 & \\
8 & \\
\end{matrix}
\rbrack $

Dividing Row 1 by 3:

$\lbrack \begin{matrix}
5 & -\frac{10}{3} & \\
-10 & 16 & \\
\end{matrix}
\rbrack \lbrack \begin{matrix}
I_{1} & \\
I_{2} & \\
\end{matrix}
\rbrack = \lbrack \begin{matrix}
4 & \\
8 & \\
\end{matrix}
\rbrack $

R2 + 2R1:

$\lbrack \begin{matrix}
5 & -\frac{10}{3} & \\
0 & \frac{68}{3} & \\
\end{matrix}
\rbrack \lbrack \begin{matrix}
I_{1} & \\
I_{2} & \\
\end{matrix}
\rbrack = \lbrack \begin{matrix}
4 & \\
16 & \\
\end{matrix}
\rbrack $

This means that:

$\frac{68}{3}I_{2}=16 $

OR

$I_{2}=\frac{48}{68}=0.7 A$

Now using the value of I$_{2}$:

$5I_{1}-\frac{10}{3}I_{2}=4$

OR

$5I_{1}-\frac{10}{3} (0.7)=4$

OR

$5I_{1}=4+\frac{7}{3}=\frac{19}{3}$

OR

$I_{1}=\frac{19}{15}=1.26 A$

Finally, we got both values of I1 and I2. Now, we can know the current of any resistor in the circuit and use Ohm’s law we can find the voltage drop across any resistor. Suppose I want to know the current in R3, and I take the direction of I2 as a reference means downward.

$I_{2}-I_{1}=0.7-1.26= -0.56A Downward$

If I take the direction of I$_{1}$ as a reference, means upward, so equation will become:

$I_{1}-I_{2}=1.26 -0.7= 0.56A upward$

In both cases, the expression means that the current is flowing upward with the magnitude of 0.56 A.

Mesh analysis having a current source :

Mesh analysis uses KVL, in the current source we don’t know the voltage of the source rather current. There are two possible cases for current sources.

Case 1 :

When the current source and voltage source are in separate mesh, as shown in the figure. Such type of circuit is less complex than voltage source circuits because it reduces the number of equations. Suppose if current source is 5A.

Mesh current Analysis

Case 2 :

When voltage source and current source both are in the same mesh, as shown in the figure. Write KCL equation for node near the current source and replace the current source with an open circuit which leads to super-mesh. Now write mesh equation.

Mesh Current Analysis

Limitation of Mesh Analysis :

Mesh analysis is limited to planner circuits, where planner circuits mean that no branch cross another branch if the circuit is drawn over a plan.

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