Mesh analysis is a method based Kirchhoff's voltage law that can be used to compute the currents flowing in each branch of a circuit. The technique is based on the fact that, if a circuit has $n$ minor meshes, the $n$ mesh currents can be found by solving a system of $n$ independent equations that can be obtained by applying KVL, in which the voltage across each component is computed based on its voltage-current characteristic. If the circuit contains $m$ current sources (notice that in the case of current sources the voltage vs. current is a multivalue function), $m$ equations will be the current constrained equations of each current source, as described below.

Assume we have a circuit with $n$ meshes (excluding the outer mesh), $m$ current sources and $c$ control variables.

Step 1. Identify the meshes in the circuit. The outer mesh is selected as a reference mesh (i.e. the mesh current of the outer loop is equal to 0).

Step 2. Label the currents at each of the $n$ meshes with $i_1$, $i_2$, ..., $i_n$.

Step 3. Write the system of mesh analysis equations, which will have $n+c$ equations ($m$ current constrained equations, $n-m$ KVL equations, and $c$ equations for the control variables). It is a good practice to write the $n+c$ equations in the order specified below:

Step 4. Solve the system on mesh analysis equations to compute the $n$ mesh currents and $c$ control variables.

Step 5. Compute the sought variables.

• A supermesh is a mesh formed by combining two or more meshes that include one or more current sources inside the combined mesh; the combined mesh cannot include any current source on its branches. Usually, we need to write equations for supermeshes whenever we have current sources that are not part of the outer mesh and, by the combined mesh does not contain any current source on it borders. Since a supermesh contains current sources inside itself, we cannot assign a single current to the supermesh.

  Fig. 1 shows an example of a circuit with one supermesh. Current source $I_2$ results in a supermesh formed by combining $i_4$ and $i_5$. Notice that the combined supermesh is shown in red and does not contain any current sources on its branches. Current source $I_1$ does not result in a supermesh because the supermesh that it had created made by combining $i_1$ and $i_2$ would contain current source $2I_x$ on its branches. If the $2I_x$ source was a resistor, current source $I_1$ would result in another supermesh. The KVL equation for the supermesh shown in Fig. 1 is $$\begin{equation}V_1+i_5 R_5+(i_4-i_2)R_4=0\end{equation}$$

• The number of KVL equations that we write at steps 3.C and 3.D should be $n-m$.
• Sometimes, it is convenient to add the equations for the sought variables to the $n+c$ mesh analysis equations. In this case, the solution of the system will also contain the values of the sought variables.
• It is not mandatory to chose the outer mesh as a reference mesh. In principle, one could chose any mesh as the reference mesh and include the outer mesh as a regular mesh. However, this is almost never done in practice when solving mesh analysis problems.
• Mesh analysis can be used to analyze both linear and nonlinear circuits, DC and AC circuits, as well as time-dependent circuits. In the case of DC circuits we obtain a system of equations with real coefficients; in the case of AC circuits we obtain a system of equations with complex coefficients; in the case or time-dependent problems we obtain a system of integro-differential equations (which can usually be reduced to a system of ordinary differential equations).

The sought variables can usually be expressed directly in terms of the mesh currents.

• The current flowing through a branch of the circuit can be expressed as the difference of the neighboring two mesh currents. For instance, current $I_x$ in Fig. 1 is $I_x=i_4-i_3$.
• The voltage across a resistor can be computed using Ohm's law as a function of the current going through the resistor and the resistance. For instance, the voltage from left to right across resistor $R_1$ in Fig. 1 is $V_{R_1}=R_1 i_3$.
• The voltage across a current source can be computed indirectly, by writing KVL for a loop which contains the current source. In this way, we compute the sum of all the voltage drops around the loop, and by isolating the term associated with the current source we obtain its voltage. For instance, to compute voltage $V_0$ across current source $I_2$ in Fig. 1, we can write that $$\begin{equation}V_0 = I_5 R_5\end{equation}$$ Or, we can also write $$\begin{equation}V_0 = -V_1+(i_2-i_4)R_4\end{equation}$$

• The power dissipated by a resistor can be found by computing the current going through the resistor, say $I_R$, and using $P=R I_R^2$. For instance, the power dissipated by resistor $R_3$ in Fig. 1 is $(i_1-i_3)^2/R_3$.

• The power generated by a current source can be computed by multiplying the the current of the source by the voltage from node where the current goes out of the source to the node where current gets in the source. For instance, the power generated by source $I_2$ in Fig. 1 is $P_g=I_2\cdot(-V_0)$ (where $I_0$ was computed above).

• The power generated by a voltage source can be computed by multiplying the the voltage of the source by the current going from the negative to the positive terminal of the source. For instance, the power generated by source $V_1$ in Fig. 1 is $P_g=V_1(i_3-i_4)$.