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mesh_analysis.md

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Mesh Analysis

In the Mesh analysis the unknown parameters are mesh currents instead of the node voltages.

A mesh is a loop which does not contain any other loops within it.

In the Nodal analysis we have used Kirchhoff's Current Law, in the Mesh Analysis Kirchoof's Voltage Law will be used.

Nodal Analysis Mesh Analysis
Unknowns Node Voltages Mesh Current
Method Kirchoof's Current Law Kirchoof's Voltage Law
Supernode/ Supermesh Voltage Sources Current Sources

Mesh analysis can only be applied to planar circuit. A planar circuit can be drawn with no branches crossing one another.

Method includes the following steps:

  • Label each mesh in the circuit.
  • Write KVL equations for each mesh.
  • Solve equations for the mesh currents.

Things to be careful:

  • Although, the direction of mesh currents (clockwise or counterclockwise) is arbitrary, and the equations are valid for both directions, it is conventional to use clockwise current direction.
  • Use - sign if you encounter the negative polarity of the voltage source first, use + sign otherwise.

Example 1:

Find the branch currents.

Write the KVL equations for each mesh:

Mesh1: $$-15 + 5i_1 + 10 (i_1 - i_2) + 10 =0$$ $$15i_1 - 10 i_2 = 5$$ which is equal to: $$3i_1 - 2i_2 =1$$

Mesh2: $$6i_2 + 4i_2 -10 + 10 (i_2 - i_1) =0$$ $$2i_2-i_1=1$$

Using the substitution method: $$i_1 = 2i_2 -1$$ $$i_1 = 1 A \quad i_2 = 1 A$$

The branch currents are: $$I_1 = i_1$$ $$I_2 = i_2$$ $$I_3 = i_1-i_2$$ Thus: $$I_1 = 1 A$$ $$I_2 = 1 A$$ $$I_3 = 0$$

Example 2:

Find the branch currents

Mesh1: $$-45 + 2i_1 + 12 (i_1 - i_2) + 4i_1 = 0$$ $$6i_1-4i_2=15$$

Mesh2: $$12(i_2-i_1)+9i_2 +30 + 3i_2 =0$$ $$4i_2 - 2i_1 = -5$$

$$i_1 = 2.5 A, i2=0 A$$

Example 3:

Use mesh analysis to find Io.

Write KVL equation for each mesh:

Mesh 1: $$-24 + 10(i_1 - i_2) + 12 (i_1 - i_3) =0$$ $$11i_1 - 5i_2-6i_3 = 12$$

Mesh 2: $$24i_2 + 4(i_2-i_3) + 10(i_2 -i_1)=0$$ $$-5i_1 + 19i_2 - 2i_3 = 0 $$

Mesh3: $$12(i_3-i_1) + 4 (i_3 -i_2) + 4I_o =0$$ Io equals to: $$I_o = i_1 - i_2$$ Thus: $$12(i_3-i_1) + 4 (i_3 -i_2) + 4(i_1-i_2) = 0 $$ which is equal to: $$-8i_1-8i_2+16i_3 =0$$ $$-i_1-i_2+2i_3 =0$$ Thus: $$i_1 = 2.25 A \quad i_2 = 0.75 A \quad i_3 = 1.5 A$$

We need to find Io, which is equal to: $$I_o = i_1 -i_2 = 1.5 A$$

Try to solve the same problem with node voltage analysis.

Example 4:

Use mesh analysis to find Io.

Mesh 1: $$-16+4(i_1-i_3)+2(i_1-i_2)=0$$ $$6i_1 - 2 i_2 - 4i_3 =16$$ $$3i_1 - i_2 - 2i_3 = 8$$

Mesh 2: $$2(i_2 - i_1) + 8(i_2-i_3) -10 I_o =0$$ $$I_o = i_3$$ $$10i_2 -2i_1 - 8i_3 -10i_3 =0$$ $$-i_1+5i_2-9i_3=0$$

Mesh 3: $$6i_3 + 8(i_3-i_2) + 4(i_3-i_1) =0$$ $$18i_3-8i_2-4i_1=0$$ $$-2i_1-4i_2+9i_3=0$$

$$\begin{bmatrix} 3 & -1 & -2 \ -1 & 5 & -9 \ -2 & -4 & 9 \end{bmatrix} \begin{bmatrix} i_1 \ i_2 \ i_3 \ \end{bmatrix}

\begin{bmatrix} 8\ 0 \ 0 \ \end{bmatrix} $$

$$ \begin{bmatrix} i_1 \ i_2 \ i_3 \ \end{bmatrix}

\begin{bmatrix} -2.57\ -7.71 \ -4 \ \end{bmatrix} $$

$$I_o = i_3 = -4 A$$