Mesh Analysis / Sign Convention - Conceptual Problem

In summary, the author is having trouble using mesh analysis and is trying to stick to the passive sign convention. They add mesh currents in orange and write equations for loops 1 and 3 that are incorrect because the voltage source has different polarities simultaneously. They eventually figure out how to assign the polarity to the voltage source and the problem is solved.
  • #1
Zatman
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Homework Statement



I am having a little trouble using mesh analysis. I'm trying to stick rigorously to the passive sign convention, as I'm told that this is sensible.

Consider the circuit diagram attached, where I have added the mesh currents in orange. In red are the polarities for the components where it is obvious - of course, using the passive sign convention.

attachment.php?attachmentid=64166&d=1385058228.jpg


How do I assign the polarities for the other components, i.e. the 6V voltage source and the 5 and 1 ohm resistors? For example, the 6V source cannot be assigned a polarity that satisfies the passive sign convention for both mesh 1 and mesh 3.

As it turns out, assigning positive to the top and negative to the bottom gives the correct answer (after the mesh analysis), but I'm not sure how you can tell.
 

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  • #2
Potential drops in the direction of the current flow. Since mesh currents are treated as separate currents in shared components (like the 5 and 1 ohm resistors in your example), there's a separate potential drop associated with each current.

attachment.php?attachmentid=64170&stc=1&d=1385059766.gif


You need to account for both changes in potential when you write the equation for a loop.
 

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  • #3
Thank you gneill, that makes sense. However, I still get the incorrect equations when I try to analyse my circuit, so I guess I'm still missing something.

attachment.php?attachmentid=64173&d=1385063770.jpg


I get, for mesh 1, 2 and 3 respectively,

[itex]2i_1+5(i_1-i_2)+6=0[/itex]
[itex]\Rightarrow7i_1-5i_2=-6[/itex]

[itex]12+(i_2-i_3)+5(i_2-i_1)=0[/itex]
[itex]\Rightarrow 5i_1-6i_2+i_3=12[/itex]

[itex]4i_3+6+(i_3-i_2)=0[/itex]
[itex]\Rightarrow -i_2+5i_3=-6[/itex]

Not sure what I'm doing wrong. :(
 

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  • #4
Well, for one thing, in loops 1 and 3, the 6V source has different polarities simultaneously. This can only lead to confusion.

http://en.wikipedia.org/wiki/Mesh_analysis

It's best to pick one polarity and adjust the sign of the voltage according to the assumed direction of the current when writing your mesh equations. There is no rule that says voltages must always be positive.
 
  • #5
Okay, but that raises the question: how do I know which way to assign the polarity to the 6V source? I could either assign it to match up with the mesh 1 current (which gives the correct answer) or the mesh 3 current (which gives the incorrect answer).
 
  • #6
Zatman said:
Okay, but that raises the question: how do I know which way to assign the polarity to the 6V source? I could either assign it to match up with the mesh 1 current (which gives the correct answer) or the mesh 3 current (which gives the incorrect answer).

Voltage sources are not passive components like resistors. They have a fixed polarity which you must use. A voltage source always produces the same potential difference across its terminals regardless of the current flowing though it.
 
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  • #7
gneill said:
Voltage sources are not passive components like resistors. They have a fixed polarity which you must use. A voltage source always produces the same potential difference across its terminals regardless of the current flowing though it.
Got it, thanks!

For the record, the book I'm using doesn't use the passive sign convention. So I've got into the habit of ignoring their labelling and redoing it. So in this example I changed the labelling of the voltage sources!
 

Related to Mesh Analysis / Sign Convention - Conceptual Problem

1. What is mesh analysis and why is it useful in circuit analysis?

Mesh analysis is a method used to analyze electric circuits by applying Kirchhoff's Voltage Law (KVL) to find unknown currents in a circuit. It is useful because it simplifies the analysis process and reduces the number of equations needed to solve a circuit.

2. What is the sign convention used in mesh analysis?

The sign convention used in mesh analysis states that the voltage drop across a resistor is considered positive if the current flows in the direction of the voltage drop, and negative if the current flows in the opposite direction. This convention is necessary to ensure accurate results in the analysis.

3. Can the sign convention in mesh analysis be applied to all circuit elements?

No, the sign convention in mesh analysis can only be applied to passive circuit elements, such as resistors, capacitors, and inductors. It cannot be applied to active elements like voltage and current sources.

4. What are the steps involved in solving a circuit using mesh analysis?

The steps involved in solving a circuit using mesh analysis are as follows:
1. Identify the meshes in the circuit
2. Assign a current variable for each mesh
3. Apply KVL to each mesh and write equations
4. Solve the equations simultaneously to find the unknown mesh currents
5. Use Ohm's Law to find the voltage drops across resistors using the calculated mesh currents.

5. What are the limitations of mesh analysis?

Mesh analysis is limited to circuits with a maximum of three meshes. It also assumes that all the circuit elements are linear and can be represented by resistors, capacitors, and inductors. Additionally, it cannot be used to analyze circuits with dependent sources or time-varying elements.

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