# Mesh Current Analysis with Example and Current Source

The Mesh Current Analysis provides a procedure for electric circuit analysis using mesh current as the circuit variable. The mesh analysis makes use of Kirchhoff’s Voltage Law is a basic key to analyzing 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 that 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.

## Mesh Current Analysis Method :

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 several meshes.
4. Use any method to solve these simultaneous equations for n mesh current.

## Mesh Current Analysis Examples:

Suppose we know the following parameter of the given circuit.

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

### Mesh Current Analysis Steps:

#### 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:

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

Note firstly that R3 is common in both meshes and is counted 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 the current I­­1 direction and subtracted current I­2 from current I1. For the mesh 2 equation, I followed the 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 finding a 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 the 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 with Current Source:

Mesh analysis theorem uses KVL, in the current source we don’t know the voltage of the source but rather the 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 a type of circuit is less complex than voltage source circuits because it reduces the number of equations. Suppose the current source is 5A.

#### Case 2 :

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

## Mesh Analysis Theorem Limitations:

• Mesh analysis theorem is limited to planner circuits
• Where planner circuits mean that no branch crosses another branch if the circuit is drawn over a plan.

### 1 thought on “Mesh Current Analysis with Example and Current Source”

1. Is there an algorithm to select R-L-C network tree such that R, L mesh matrices are non-singular? Or an algorithm to transform any given tree such same condition is met?