Why does the Mesh Currents Method work?

Question

When there is an n-mesh circuit, we define n independent mesh equations for the unknown n currents. While doing this, we use an assumption that each mesh has a separate current flowing around it.

I don't understand this assumption. How do we do it? What is the proof that an assumption like this will work? And, you should know that a statement like "I tried this a billion times and it worked every time." is not a solid mathematical proof. There must be an untold background of this theorem. Is it related to the Stokes' Theorem by any chance?

Answer 1

None of the given answers really address the question. The mesh analysis approach allows us to reduce the number of degrees of freedom significantly. In the example, mesh analysis provides 3 variables while there are 10 total edges in the circuit. So clearly, we could get and solve 10 equations and 10 variables, but mesh analysis tells us we only need to solve 3.

Two facts are true to allow this reduction.

1. Using the mesh currents approach will satisfy the node and loop rules
2. A set of currents and voltages that obey both rules is unique.

The first is easy to see, but the second explains why mesh analysis works. Since the mesh currents satisfy the node and loop rules, it must be the only solution.

It is not straightforward to see why statement 2 must be true. The proof by contradiction assumes that two such current-voltage arrangements exist satisfying the node and loop rules. It then shows that you can find a loop where the sum of voltage drops is non-zero. https://math.stackexchange.com/questions/1742680/unique-solution-for-circuits-in-linear-algebra

Answer 2

I1, I2, I3 the so-called "mesh currents" are, in fact, the real current only at the elements of the mesh that are not shared with any other mesh. That's it I1 is the real current of R1 and V1, I2 is the real current of R3 and I3 is the real current in R3, R6, and V2.

Then by first law of Kirchhoff the current through R4 is (I1-I2), the current through R5 is (I2-I3). Then you simply apply the second law of Kirchhoff (summatory of voltages equal zero) at every mesh.

Answer 3

Imagine you start at the front door of a house, you go on a hike through curvy terrain, and then you come back to the same door where you started. As you went on that walk-about and came back to the same point, all the ups and downs summed up to zero. You are at the same altitude as when you started. Here you have a simple reasoning for Kirchoff's Voltage Law, and why the voltages sum up to zero.

Now imagine you and a friend start at work, but before lunch, you need to go the bank, and your friend needs to go the post office. You each take different routes, but for lunch you meet at the cafe, and after lunch you go back together. You have both completed your circuit, and are back where you started. Now imagine that these two trips took place in two parallel worlds where you are each all alone. Would the total amount of people coming back from the cafe be any different ?

How the mesh current method works is basically like that. Where the mesh currents meet you sum the contributions.

In linear circuit analysis mesh analysis (and nodal analysis) almost always works as a method to provide a single solution. In part I think it is helpful to look at the superposition principle. The end-game is anyway that for a linear circuit mesh analysis provides N equations for N mesh currents, with no unknown variables (all mesh currents were included). From linear algebra we know that N equations is exactly what we need to solve N variables.

However a couple of times mesh analysis can fail. Like in linear algebra, if our set of equation has a zero determinant, then there may be many (infinite) solutions. A simple way of achieving a circuit like this is to hook up two equal voltage sources in parallel. How much current goes from one source into the other ? In theory it could be anything.

Another case that gets simple mesh analysis into trouble is when there is no way to "flatten" the circuit into a 2D planar circuit. E.g. consider adding a resistor between A and B in the drawing below. How would you describe the mesh current in the other elements between A and B? For this there is loop analysis.

However a linear circuit is only an approximation and it can meet it's limits. Say that voltage over a capacitor far exceeds it's voltage rating - it would break down - it might even explode - and thus completely change your circuit. That's a bit like you and your friend crossing a bridge on your way back to work, and 10 million other people happened to be on the same bridge (it breaks down!).

Non-linear elements also quickly make trouble for mesh analysis as a general method. By this I mean that mesh analysis may not be able to provide you with a solution anymore, although the mesh equations it provides are not wrong per se. With non-linear elements the set of mesh equations can fairly quickly become analytically unsolvable, there could be multiple solutions, unstable solutions, or no solutions at all...

Answer 4

Ultimately you're solving ohm's law for simultaneous linear equations. The "assumption" that there is a current doesn't imply that current cannot be zero! It is just a variable when solving the equations.

As far as the reasoning behind why this works (at least mathematically), I would have a look at this entry on solving linear equations under General Behavior. http://en.wikipedia.org/wiki/System_of_linear_equations

Note that for a system with the same number of equations as unknowns has a single unique solution, which is exactly the case here (three mesh equations).

Hopefully this helps

Answer 5

If we find a solution to mesh current equations, then currents we find should satisfy Kirchhoff voltage law. But does these currents satisfy Kirchhoff current law? Let us do the algebra. For any node, if there are n meshes flowing around that node, then we can name these mesh I1, I2,...,In. Then current we find on the 1st edge out that node is I2-I1. The current we find on the 2nd edge out that node is I3-I2. So on. The current we find on the last edge out that node is I1-In. If we add all currents together, then we have

I2-I1+I3-I2+I4-I3+...+In-I(n-1)+I1-In=0.

So the currents we find satisfy Kirchhoff current law. So mesh current method works. The above proof is for a node with whose edge number is greater or equal to 3. If a node has 2 edges, then the proof is trivia. We only has 1 mesh for that node. We name this mesh I1. So there are 2 currents for this node. One is I1. The other is -I1. If we add all currents together, then we have I1+(-I1)=0. So the currents we find satisfy Kirchhoff current law thus complete the proof.