Mesh currents are analysis variables that are useful in circuits containing many elements
connected in series. Mesh analysis provides another general procedure for analyzing circuits, using mesh currents as the circuit variables. Using mesh currents instead
of element currents as circuit variables is convenient and reduces the
number of equations that must be solved simultaneously.
#### What is loop in circuit analysis?

#### What is mesh in circuit analysis?

To understand mesh analysis, we should first explain more about
what we mean by a mesh.
**Fig.no.1: **A circuit with two meshes.
In Fig.no.1, for example, paths *abefa* and *bcdeb* are meshes, but path
*abcdefa* is not a mesh. The current through a mesh is known as mesh
current. In mesh analysis, we are interested in applying KVL to find the
mesh currents in a given circuit. Mesh analysis is not quite as general as nodal analysis because it is only applicable to a circuit that is planar.
#### What is planar circuit?

A planar circuit is one that can be drawn in a plane with no branches crossing one another; otherwise it is nonplanar. A circuit may have crossing branches and still be planar if it can be redrawn such that it has no crossing branches.
**Fig.no.2: **(a) A planar circuit with crossing
branches, (b) the same circuit redrawn with no
crossing branches.
For example, the circuit in Fig.no.2(a) has two crossing branches, but it can be redrawn
as in Fig.no.2(b). Hence, the circuit in Fig.no.2(b) is planar. However,
the circuit in Fig.no.3 is nonplanar, because there is no way to redraw
it and avoid the branches crossing. Nonplanar circuits can be handled
using nodal analysis.
**Fig.no.3: **A nonplanar circuit.
#### Mesh Analysis Procedure

**Example 1: **
For the circuit in Fig.no.4, find the branch currents $I_1$, $I_2$, and $I_3$ using
mesh analysis.
**Fig.no.4: **For Example 1.
**Solution:**

We first obtain the mesh currents using KVL. For mesh 1, $$-15 + 5i_1 + 10(i_1 - i_2) + 10 = 0$$ or $$3i_1 - 2i_2 = 1 \text{ ......... eq. (1)} $$ For mesh 2, $$6i_2 + 4i_2 + 10(i_2 - i_1) - 10 = 0$$ or $$i_1 = 2i_2 - 1 \text{ ......... eq. (2)} $$ METHOD 1: Using the substitution method, we substitute Eq. (2) into Eq. (1), and write $$3(2i_2 - 1) - 2i_2 = 1$$ $$ 6i_2 - 3 - 2i_2 = 1$$ $$ 4i_2 = 4, i_2 = 1 A$$ From Eq. (2), $$i_1 = 2i_2 - 1 = 2(1) - 1 = 1 A.$$ Thus, $I_1 = i_1 = 1 A$, $I_2 = i_2 = 1 A$ , $I_3 = i_1 - i_2 = 0$

METHOD 2: To use Cramer's rule, we cast Eqs. (1) and Eqs. (2) in matrix form as \begin{gather} \begin{bmatrix} 3 & -2 \\ -1 & 2 \\ \notag \end{bmatrix} \begin{bmatrix} i_1 \\ i_2 \\ \notag \end{bmatrix}= \begin{bmatrix} 1 \\ 1 \\ \notag \end{bmatrix} \end{gather} We obtain the determinants \begin{gather} \Delta = \begin{vmatrix} 3 & -2 \\ -1 & 2 \\ \notag \end{vmatrix} = (6-2) = 4 \end{gather} \begin{gather} \Delta_1 = \begin{vmatrix} 1 & -2 \\ 1 & 2 \\ \notag \end{vmatrix} = (2+2) = 4 \end{gather} \begin{gather} \Delta_2 = \begin{vmatrix} 3 & 1 \\ -1 & 1 \\ \notag \end{vmatrix} = (3+1) = 4 \end{gather} $$i_1 = {\Delta_1 \over \Delta}={4 \over 4} = 1A$$ $$i_2 = {\Delta_2 \over \Delta}={4 \over 4} = 1A$$

- Assign a distinct current in the clockwise direction to each independent, closed loop of the network. It is not absolutely necessary to choose the clockwise direction for each loop current. In fact, any direction can be chosen for each loop current with no loss in accuracy. However, by choosing the clockwise direction as a standard, we can develop a shorthand method for writing the required equations that will save time and possibly prevent some common errors.
- Indicate the polarities within each loop for each resistor as determined by the assumed direction of loop current for that loop.
- Apply Kirchhoff's voltage law around each closed loop in the clockwise direction. Again, the clockwise direction was chosen to establish uniformity.
- The polarity of a voltage source is unaffected by the direction of the assigned loop currents.
- Solve the resulting simultaneous linear equations for the assumed loop currents.

We first obtain the mesh currents using KVL. For mesh 1, $$-15 + 5i_1 + 10(i_1 - i_2) + 10 = 0$$ or $$3i_1 - 2i_2 = 1 \text{ ......... eq. (1)} $$ For mesh 2, $$6i_2 + 4i_2 + 10(i_2 - i_1) - 10 = 0$$ or $$i_1 = 2i_2 - 1 \text{ ......... eq. (2)} $$ METHOD 1: Using the substitution method, we substitute Eq. (2) into Eq. (1), and write $$3(2i_2 - 1) - 2i_2 = 1$$ $$ 6i_2 - 3 - 2i_2 = 1$$ $$ 4i_2 = 4, i_2 = 1 A$$ From Eq. (2), $$i_1 = 2i_2 - 1 = 2(1) - 1 = 1 A.$$ Thus, $I_1 = i_1 = 1 A$, $I_2 = i_2 = 1 A$ , $I_3 = i_1 - i_2 = 0$

METHOD 2: To use Cramer's rule, we cast Eqs. (1) and Eqs. (2) in matrix form as \begin{gather} \begin{bmatrix} 3 & -2 \\ -1 & 2 \\ \notag \end{bmatrix} \begin{bmatrix} i_1 \\ i_2 \\ \notag \end{bmatrix}= \begin{bmatrix} 1 \\ 1 \\ \notag \end{bmatrix} \end{gather} We obtain the determinants \begin{gather} \Delta = \begin{vmatrix} 3 & -2 \\ -1 & 2 \\ \notag \end{vmatrix} = (6-2) = 4 \end{gather} \begin{gather} \Delta_1 = \begin{vmatrix} 1 & -2 \\ 1 & 2 \\ \notag \end{vmatrix} = (2+2) = 4 \end{gather} \begin{gather} \Delta_2 = \begin{vmatrix} 3 & 1 \\ -1 & 1 \\ \notag \end{vmatrix} = (3+1) = 4 \end{gather} $$i_1 = {\Delta_1 \over \Delta}={4 \over 4} = 1A$$ $$i_2 = {\Delta_2 \over \Delta}={4 \over 4} = 1A$$