# Solvability of a circuit

1. May 19, 2013

### Bipolarity

Let's say I have a circuit consisting only of a finite number of batteries and resistors, all ideal. Given an arbitrary shape of this circuit, will I always be able to "solve" this circuit, i.e. find the missing variables (current through any wire, voltage across any two points) by using the Kirchhoff's laws in conjunction with Ohm's laws?

If yes, can this be proven mathematically? Where would I find a proof? Surely it involves proving that a certain coefficient matrix of some linear system is invertible? Perhaps it will require some graph theory?

If not, under what conditions can the circuit be solved for?

One thing I noticed is that if you have a circuit with N nodes and write out equations for all nodes, 1 of the equations will be linearly dependent on the other (n-1) equations.

I am reading my introductory circuits text, but have a good linear algebra background, so I like to see a very rigorous justification of the methods employed by my text.

Thanks!

BiP

2. May 19, 2013

### Staff: Mentor

Yes. In a problem with N nodes there are N unknown voltages. As you noticed, this always results in a system of N-1 independent linear equations. Since you can arbitrarily set any one node as ground that gives you N equations in N unknowns.

I don't know about a rigorous proof.

3. May 21, 2013

### stevenb

I don't know about a rigorous proof either.

However, there will be pathological cases. For example, putting two batteries with unequal voltages in parallel is a no-no.

4. May 21, 2013

### Bipolarity

What happens if you did do that?

BiP

5. May 21, 2013

### CWatters

Booooommmmm

Calculate the current flow. Ohms law says..

V=IR
so
I = V/R
but
R=0
so
I=∞
Then
Power = Voltage * Current
so
Boooommmm

Some batteries really do go bang. Depends how well they approximate an ideal voltage source. Some relatively ordinary rechargable batteries can manage several hundred amps (but not for long).

Other things to avoid include disconnecting (open circuiting) an ideal current source.

Last edited: May 21, 2013
6. May 21, 2013

### CWatters

Lithium battery short circuited....

Last edited by a moderator: Sep 25, 2014
7. May 21, 2013

### stevenb

Well, above you see the practical concerns, but I was thinking more about the mathematics. Just write out the equations to see it.

V1=V2 Kirchhoff's Voltage Law
V1=10V Ideal voltage law for battery # 1
V2=20V Ideal voltage law for battery # 2

So, two batteries in parallel can't obey Kirchhoff's Voltage Law, or if they do, then at least one of them can't obey it's own ideal voltage law.

It's like an irresistible force pushing an unmovable object. It just can't be resolved logically.

Hence, any rigorous proof would isolate these types of pathological cases first, and then proceed to prove that the other cases are all solvable. Or, perhaps you already have your answer now because you set out to prove if all cases are solvable, but by showing even one case that is not solvable, it's clear that it can't be done.

As a practical matter, when solving real problems, these issues tend not to come up unless you make a mistake, and then the mistake (or poor assumption) becomes obvious when you can't get a solution.

8. May 21, 2013

### AlephZero

Real circuits haven't read your textbook, and they don't care if your over-simplified models of idealised components don't have solutions. As I said in another thread recently, physics is not a sub-branch of mathematics!

To answer the OP's question - try working backwards. Write down a matrix equation that has no solutions, or multiple solutions (obviously the matrix must be singular). Then invent a circuit diagram that corresponds to the matrix equations.

9. May 21, 2013

### Staff: Mentor

You can add a tiny resistance in parallel to every voltage source to get a physical solution for all scenarios.
With batteries and resistors only, every connection between two vertices can be expressed as series connection of a voltage source (can be 0V) and a single resistor with a positive resistance.

I do not know if the star-mesh transform is possible with voltage sources inside. If it is, every circuit can be reduced to trivial circuits.

There are iterative ways to solve the equation system - it would be sufficient to show their convergence.

10. May 21, 2013

### Staff: Mentor

Well, mathematically you can always take a basic circuit like this and put it into a system of linear equations. Then you can use standard linear algebra techniques to solve it or to determine that it doesn't have a solution (i.e. determinant not equal to 0).

11. May 22, 2013

### stevenb

Well, yes, of course. The OP mentioned "invertibility" of the linear system, which I think amounts to the same thing. And, there are other ways to check it too. This is one point I was trying to make; that, as a practical matter, we can check it on a case by case basis as needed. However, if the OP truly wants a rigorous (and I assume that implies "general" too) proof, a case by case checking doesn't satisfy that. Another point I am making is that there are clearly some cases that are not solvable "given an arbitrary shape of this circuit". So, a general/rigorous proof would identify the unrealizable/pathological cases and (presumably) would show that the interesting real-world practical cases are all solvable.

This is just a guess on my part, but I think that requiring all ideal voltage sources to have a nonzero series resistor with it will give circuits that are practical and always have a solution. If this is correct, then this might be the starting point for the rigorous proof he wants. Using either graph theory or matrix formulations could allow him to show that a general system always has a nonzero determinant. But this does leave out some solvable cases that are not physical (let's ignore superconductivity for this discussion).

Last edited: May 22, 2013