# Special Case of KVL - Supermesh Analysis

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If you’ve read our tutorial on Kirchhoff’s Voltage Law (KVL), you already know how useful and important KVL is for the analysis of any electrical network. The knowledge of KVL is also really useful if you are studying subjects like ‘Analog Electronics’ and ‘Control systems’. So getting a good command over KVL is extremely important. In this tutorial, we’ll be looking at a special case of KVL. What is that? Let’s find out.

Before we start with our example, here’s a quick recap of KVL: It states that adding up the voltage sources and the voltage drop across network elements in a closed-loop results in a zero.

Consider the below network,

We are supposed to find the mesh currents i_{1} and i_{2}. Take a moment and try to recall how you can apply KVL in a closed loop of a network.

So let’s begin with reviewing the circuit. We can see that we have two resistors and two voltage sources with their values provided to us already. We also have a current source in the middle branch which is providing 1A current in the downward direction. The assigned mesh currents i_{1} and i_{2} are flowing in the clockwise directions. It is always a good practice to remember the directions of the mesh currents while solving such problems.

In mesh 1, we can see that there’s a 21V voltage source, 6Ω resistor, and 1A current source.

So, per KVL, the addition of the voltage drop across the resistor and the current source should be equal to 21 Volts. We can easily figure out the voltage drop across the resistor using Ohm’s law, but what about the current source? Can you see the problem now?

We cannot find the voltage drop across the current source because we don’t know its resistance. It could be anything. Right now, we don’t have enough information to figure it out. So we need a workaround.

Let’s start by assuming that the voltage drop across the current source will be some value V_{X}.

Now we can apply KVL at mesh 1 considering the voltage drop across the current source as Vx and this is what we get

Rearranging for V_{x} we’ll get,

Similarly, we will apply KVL at mesh 2. But always make sure to double-check the polarity of the voltage source before writing the KVL equation.

If you notice, we have the term Vx in both the mesh equations. So we can equate the two equations to get a common mesh equation.

Rearranging, we get our common mesh equation as

We’ve still got two unknowns i_{1} and i_{2} in our equation, so we’ll be needing another equation to solve it. This is where the current source branch comes into our help.

You can see from the figure that the mesh current I _{1} flows in the same direction as the 1A current from the current source, but I _{2} opposes the direction of the current source. We can form a mathematical relation expressing this condition and that is:

Now we have two equations for the two unknowns and the math after it is quite easy if you haven’t flunked your linear algebra classes. We can easily get the final answer by solving the two equations. This is one of the approaches to solve this problem. But we would like to introduce you to another method of solving this problem, and a relatively shorter one. It’s known as the **‘super-mesh’** method.

Let’s redo this question using the ‘super-mesh’ method. And don’t be worried, this time it’ll be much quicker.

In the ‘super-mesh’ method, we start off by** removing the branch containing the current source which is common between any two meshes.** For us, it’s the middle branch common between mesh 1 and mesh 2. So after removing it, we are left with a bigger loop containing both the mesh currents. This is called a ‘super-mesh’. Now let’s write a KVL equation for this super-mesh.

Aha! As you can see, we’ve obtained the same common mesh equation once again, with fewer steps and without any hassle. This is the advantage of using the ‘super-mesh’ method. Now we can move forward to find the mesh currents.

Always remember that the current source branch which we remove for making the ‘super-mesh’ has an important role and that is to provide us with a relation between mesh currents which is necessary to solve the problem.

Now we can plug this value of i_{1} into the common mesh equation which we’ve obtained using the ‘super-mesh’ method to find the required values:

We can now substitute this value of i_{2} in the equation showing a relation between the two mesh currents

And we’ve finally obtained the mesh currents as:

Awesome! Now we have a shorter method to solve this problem. But still, this was a pretty basic example. Let’s try one which is a little bit more complicated.

Consider the network provided below. Review it and try to find the best approach to solve for the mesh currents i_{1}, i_{2}, and i_{3}.

This time we are given a network having 3 meshes. Other than the resistors and voltage sources that are given in the network, there’s also a current source that is getting shared between mesh 1 and mesh 3. A shared current source means it is time for the ‘super-mesh’ method once again.

We’ll remove the branch containing the current source to get our ‘super-mesh’ consisting of mesh 1 and mesh 3.

Mesh 2 with mesh current i_{2} stays as it is because it does not share any common branch with the current source. Now we can apply KVL in the super-mesh to get the following common mesh equation:

Applying KVL in mesh 2 with mesh current i_{2}

The branch we’ve removed will give us the third and most important equation, the relation between the two mesh currents. Because i_{1} supports the direction of 3A current due to the current source and i3 opposes it, we can express it as:

Awesome! Now we have 3 equations and 3 unknowns.

We don’t want to bore you to death by solving these equations step by step. That would be quite painful. Here’s a handy tool you can use to solve these linear equations:

So finally, we’ve got the values of our mesh currents as,

Now that you have the mesh currents, you can do all sorts of things like finding out the currents through each element of the network, and then you can make use of Ohm's law to find out the voltage drop across the resistors.

With this, you’ve added one more technique to your network analysis toolbox. We hope that now you can see how useful and quick the super-mesh method can prove to be when dealing with such complicated networks. So whenever you encounter a problem with a current source getting shared between two meshes, use the ‘super-mesh’ method and make your life a little bit easier.

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