Supermesh Analysis Easy Problem Solving

Supermesh analysis is the advanced method of mesh current analysis.

Have you learned about mesh analysis completely? Is it easy or hard to understand? If you think the previous post about mesh current method or mesh analysis is quite hard, try to read it one more time.

Like we stated in that post, we have limited the use of the current source in a circuit if we want to use mesh analysis. Applying mesh analysis to an electrical circuit containing current sources (dependent or independent) may appear complicated.

Is mesh analysis can’t analyze an electric circuit containing current sources?

Of course not, we only limited the current source to give you the basic principle with the easiest approach.

Supermesh Mesh Analysis

If the current source is found just in a single branch then it is easy to solve, we can use the mesh analysis directly.

We will need another version of mesh analysis if we found the current source in a branch where two meshes intersect.

Now we will use current sources in a circuit and analyze it with mesh analysis or we may call it Supermesh Analysis.

There are some understanding you need to follow:

  • A supermesh is formed by the presence of a current source between two meshes.
  • A supermesh is created in the condition where a current source exists between two meshes.
  • Supermesh analysis is used in case a current source is found between two meshes and we insist on using Mesh Analysis.

What is supermesh? In short, supermesh is where we make a mesh current path where there is no current source included.

Read also : 10k ohm resistor color code

Supermesh Analysis Problems with Solutions

Even if we need to modify the mesh current method analysis a little bit, let me be honest with you about the good thing about it.

The current sources will reduce the number of equations, making it easier to solve. Let us take a look at two possibilities.

CASE 1 – When a current source exists only in one mesh: Consider in Figure.(1) for example,

supermesh analysis 1

we set i2 = -5 A and write equation for the other mesh in the usual ways; that is,

(1)   \begin{align*}-10+4i_{1}+6(i_{1}-i_{2})&=0\\i_{1}&=-2A\end{align*}

CASE 2 – When a current source exists between two meshes: Consider the circuit in Figure.(2), for example.

supermesh analysis 2

We use the supermesh technique where we make a supermesh by excluding the current source and any elements connected in series with it (in the circle), as shown in Figure.(3).

supermesh analysis 3

Supermesh definition:

A supermesh results when two meshes have a (dependent or independent) current source in common.

As shown in Figure.(3), we create a supermesh as the periphery of the two meshes and treat it differently. (If a circuit has two or more supermeshes that intersect, they should be combined to form a larger supermesh).

Why treat supermesh differently? Because mesh analysis applies KVL – which requires that we know the voltage across each branch – and we do not know the voltage across the current source in advance.

However, a supermesh must satisfy KVL like any other mesh, this is also known as KVL supermesh. Therefore, applying KVL to the supermesh in Figure.(3) gives

(2)   \begin{align*}-20+6i_{1}+10i_{2}+4i_{2}&=0\\6i_{1}+14i_{2}&=20\end{align*}

We apply KCL to a node in the branch where the two meshes intersect. Applying KCL to node 0 in Figure.(2) gives

(3)   \begin{align*}i_{2}=i_{1}+6\end{align*}

Solving Equations.(2) and (3), we get

(4)   \begin{align*}i_{1}=-3.2A \qquad i_{2}=2.8A\end{align*}

Note the following properties of a supermesh :

  • The current source in the supermesh provides the constraint equation necessary to solve for the mesh currents.
  • A supermesh has no current of its own.
  • A supermesh requires the application of both KVL and KCL.

Procedure of Supermesh Current Analysis

This is the summary of supermesh analysis’ step-by-step

  • Make sure the circuit is planar.
  • Redraw the circuit if we can simplify it.
  • Make meshes in every loop you can find and assign the labels. It is easier to draw the meshes in the clockwise direction (it is up to you, though).
  • Form a supermesh circuit if you find a current source between two meshes.
  • Use KVL and maybe some KCL to the supermesh branch. One KCL is needed for each supermesh branch.
  • Solve all the math equations including the supermesh equation.

Supermesh Analysis Solved Problems

If there is a current source in the circuit, we have to treat it as a “supermesh”. In a supermesh, the analysis route will avoid the current source since the voltage across the current source is unknown.


1. Find the value of i with supermesh analysis!

supermesh analysis 4


Circuit above is a supermesh with current source.

supermesh analysis 5

Observe loop I1:


Observe loop I2 and I3:

(1)   \begin{align*}I_{3}-I_{2}&=3A\\I_{3}&=3+I_{2}\end{align*}

Observe supermesh path:

supermesh analysis 6

(2)   \begin{align*}\Sigma v=0\\8(I_{2}-I_{1})+16I_{2}+12I_{3}=0\end{align*}

Substitute Equation.(1) and (2):

    \begin{align*}8(I_{2}-9)+16I_{2}+12(3+I_{2})=0\\8I_{2}-72+16I_{2}+36+12I_{2}=0\\36I_{2}=36 \quad\rightarrow\quad I_{2}=\frac{36}{36}=1A\end{align*}



2. Find the value of V with supermesh analysis!

supermesh analysis 7


Observe that the circuit above has a supermesh with a dependent voltage source, voltage source, and a current source.

supermesh analysis 8

Observe loop I1 and I2:

(1)   \begin{align*}I_{2}-I_{1}&=6A\\I_{1}&=I_{2}-6\end{align*}



Observe the supermesh with voltage source and resistor path:

(2)   \begin{align*}\Sigma v=0\\-12+1.I_{1}+2i+3I_{2}=0\\-12+I_{1}+2I_{1}+3I_{2}=0\\3I_{1}+3I_{2}=12\end{align*}

Substitute Equation.(1) and (2):

    \begin{align*}3(I_{2}-6)+3I_{2}=12\\3I_{2}-18+3I_{2}=12\\6I_{2}=30 \quad\rightarrow\quad I_{2}=\frac{30}{6}=5A\end{align*}



3. Find the value of i with supermesh analysis!

supermesh analysis 9


supermesh analysis 10

Observe loop I1:

(1)   \begin{align*}6I_{1}+12+12(I_{1}-I_{2})&=0\\18I_{1}-12I_{2}&=-12\end{align*}

Observe loop I2 and I3:

(2)   \begin{align*}I_{3}-I_{2}=3\end{align*}

Observe supermesh path:

supermesh analysis 11

(3)   \begin{align*}\Sigma v&=0\\4I_{2}+6I_{3}+12(I_{2}-I_{1})-12&=0\\16I_{2}-12I_{1}+6I_{3}&=12\end{align*}

Using Cramer’s method:

    \begin{align*}&\begin{bmatrix}18 & -12 & 0\\0 & -1 & 1\\-12 & 16 & 6\end{bmatrix}\begin{bmatrix}I_{1}\\I_{2}\\I_{3}\end{bmatrix}=\begin{bmatrix}-12\\3\\12\end{bmatrix}\\\\I_{3}&=\frac{\begin{vmatrix}18 & -12 & -12\\0 & -1 & 3\\-12 & 16 & 12\end{vmatrix}}{\begin{vmatrix}18 & -12 & 0\\0 & -1 & 1\\-12 & 16 & 6\end{vmatrix}}\\\\&=\frac{18\begin{vmatrix}-1 & 3\\16 & 12\\\end{vmatrix}+12\begin{vmatrix}0 & 3\\-12 & 12\\\end{vmatrix}-12\begin{vmatrix}0 & -1\\-12 & 16\\\end{vmatrix}}{18\begin{vmatrix}-1 & 1\\16 & 6\\\end{vmatrix}+12\begin{vmatrix}0 & 1\\-12 & 6\\\end{vmatrix}}\\&=2A\end{align*}



Supermesh Analysis Examples

For better understanding, let us review the example below :
1. Observe the circuit below and find i1 to i4 using supermesh analysis.

supermesh analysis 12

Solution :

We can see that the circuit above has a supermesh dependent current source.

Note that meshes 1 and 2 form a supermesh since they have an independent current source in common.

Also, meshes 2 and 3 form another supermesh because they have a dependent current source in common.

The two supermeshes intersect and form a larger supermesh as shown.

supermesh analysis 13

Applying KVL to the larger supermesh,

(1)   \begin{align*}2i_{1}+4i_{3}+8(i_{3}-i_{4})+6i_{2}=0\\i_{1}+3i_{2}+6i_{3}-4i_{4}=0\end{align*}

For the independent current source, we apply KCL to node P :

(2)   \begin{align*}i_{2}=i_{1}+5\end{align*}

For the dependent current source, we apply KCL to node Q :


But Io = – i4 , thus,

(3)   \begin{align*}i_{2}=i_{3}-3i_{4}\end{align*}

Applying KVL in mesh 4,

(4)   \begin{align*}2i_{4}+8(i_{4}-i_{3})+10&=0\\5i_{4}-4i_{3}&=-5\end{align*}

From (1) to (4),


4 thoughts on “Supermesh Analysis Easy Problem Solving”

  1. All these are well explained examples.
    kindly explain how you replaced io=-i4 in 1.3 equation.
    It’ll be a great favor.

    • If you observe the most right loop, especially on the 2 ohm resistor and 10 V voltage source, you will find the current flowing through them is i4. The variable Io we have on that loop has the same value with i4 but the direction is reversed, hence we give negative value (because the direction of the current is opposite to i4).

      TLDR; the value of Io = i4, because the direction is opposite each other then Io = – i4
      Hope it helps.


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