Simple Mesh Analysis Homework: Solving for Current in a 0.25 Ohm Resistor

[fogel1497]

Homework Statement

Please see the attached file for the problem statement. I am attempting to find the current across the .25 Ohm resistor.

Homework Equations

Using kirchhoffs voltage law I write the following equations:

(1 Ohm)(2 amps) + (.25 Ohm)(2 - M₁) = 0

The Attempt at a Solution

Solving this equation for M₁, I get that the current through the .25 Ohm resistor is 10 Amps, but I do not believe this to be correct.

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Can someone please explain to me where I am going wrong?

 

[MisterX]

The 1 Ω resistor is in two loops. Also, current sources may have a voltage across them.

 

[gneill]

You have only current sources, so it might be easier to write node voltage equations rather than mesh current equations.

 

[fogel1497]

Thanks for your replies, mesh analysis is mandated for this problem. If i annotate my solution such that the 1 ohm resistor has the differences of the two currents crossing it rather then just the single current, is the remainder of my analysis and procedure accurate?

 

[gneill]

The proof is in the pudding, as they say. Do you get correct results from your analysis?

 

[MisterX]

Thanks for your replies, mesh analysis is mandated for this problem. If i annotate my solution

http://www.merriam-webster.com/dictionary/annotate

such that the 1 ohm resistor has the differences of the two currents crossing it rather then just the single current, is the remainder of my analysis and procedure accurate?

As I already wrote, current sources may have a voltage across them. So to apply Kirchoffs voltage law properly there you must include the current sources's voltage in the sum of all the voltages in that loop.

 

[jegues]

You have only current sources, so it might be easier to write node voltage equations rather than mesh current equations.

Wait, what?

When I see current sources, it tempts me to use mesh analysis as oppose to nodal analysis because it often allows you solve some of the mesh current by inspection.

Am I missing something?

 

[gneill]

Wait, what?

When I see current sources, it tempts me to use mesh analysis as oppose to nodal analysis because it often allows you solve some of the mesh current by inspection.

Am I missing something?

Mesh equations with voltage sources can be speedily written in matrix form using a simple shortcut method that requires almost no thought. Mesh equations with current sources requires you to introduce more variables for the voltages across those current sources, and write additional equations relating the mesh currents to the constraints of the sources. I'm not aware of any simple method of translating that directly to the minimal equation set (one per mesh).

 

[jegues]

Mesh equations with current sources requires you to introduce more variables for the voltages across those current sources, and write additional equations relating the mesh currents to the constraints of the sources.

They either greatly simplify the problem by allowing you to solve the mesh current by inspection, or it results in a simple super-mesh.

Writing the equation for a super-mesh is no more difficult than writing an equation for a regular mesh(or it shouldn't be), you just have to pay attention to which mesh current is flowing where.

For this problem, one can solve the mesh current in M2 by inspection.

Write one mesh equation for M2(or better yet a simple KCL), one super-mesh equation and you're done.

With nodal analysis certainly you'd have to write more than 2 equations.

 

[gneill]

They either greatly simplify the problem by allowing you to solve the mesh current by inspection, or it results in a simple super-mesh.

Writing the equation for a super-mesh is no more difficult than writing an equation for a regular mesh(or it shouldn't be), you just have to pay attention to which mesh current is flowing where.

Sure. But having to "pay attention to which mesh current is flowing where" is not required when doing the voltage source only analysis. And you don't have to seek out super-meshes and the cases that require them.

If there are three loops, create a 3 x 3 matrix. Write the sum of the loop impedances on the diagonal entries, write the negative of the impedances that straddle loops in the off-diagonal entries, and stick the sum of the loop voltage sources into a separate vector. Throw the lot at a solver and out pops the results for the loop currents. Done and dusted. Almost no thinking required!

 

[jegues]

Sure. But having to "pay attention to which mesh current is flowing where" is not required when doing the voltage source only analysis. And you don't have to seek out super-meshes and the cases that require them.

If there are three loops, create a 3 x 3 matrix. Write the sum of the loop impedances on the diagonal entries, write the negative of the impedances that straddle loops in the off-diagonal entries, and stick the sum of the loop voltage sources into a separate vector. Throw the lot at a solver and out pops the results for the loop currents. Done and dusted. Almost no thinking required!

If I'm going to go through all that trouble, why wouldn't I just simulate the circuit and simply record the measured voltages and currents? That requires even less thought.

I'd imagine the student is being forced to use mesh analysis to solve the problem while using a regular calculator, one that can't compute matrices and do the algebra for him. (I know I didn't have access to one when I took my circuits courses) In fact, algebra is where most students mistakes lie, so it'd be good for him to keep his algebra in check.

Irregardless, I'd take an intuitive, understanding based approach over an "algorithm" based approach any day.

If you understand what you're doing, it's no more difficult.

 

[gneill]

If I'm going to go through all that trouble, why wouldn't I just simulate the circuit and simply record the measured voltages and currents? That requires even less thought.

I'd imagine the student is being forced to use mesh analysis to solve the problem while using a regular calculator, one that can't compute matrices and do the algebra for him. (I know I didn't have access to one when I took my circuits courses) In fact, algebra is where most students mistakes lie, so it'd be good for him to keep his algebra in check.

I agree. Minimizing mistakes due to algebra slip-ups is of course one of the benefits of having an essentially foolproof algorithm for obtaining the equations by inspection. The matrix form for the equations is simply the usual loop equations, already in simplified, in a compact form. The individual equations are there to be read in the matrix.

Irregardless, I'd take an intuitive, understanding based approach over an "algorithm" based approach any day.

If you understand what you're doing, it's no more difficult.

While I agree that it is important for students to understand where the equations come from, what motivates them, and how to solve them by any of a number of methods (matrix methods just being another tool -- back substitution is no different), once the theory is understood and intuition is ingrained, time saving and error saving methods become desirable.

While I understand that clothes can be washed by beating them on rocks by the river, and I can appreciate the bold simplicity of the method, I still prefer throwing them in the washing machine.

In the case of the present problem, I can appreciate the learning experience that goes along with writing and solving the mesh equations when current sources abound in the circuit. It was not clear (to me at least) at the outset that the mesh method was the only one permitted for the problem. From experience, I simply pointed out that the circuit admitted another, possibly less painful route to the solution.

Anyway, at this juncture I suggest that we might get back to helping fogel1497 with his problem. He seems to have grown quiet...