Mesh Analysis - confusion on writing equations

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SUMMARY

This discussion focuses on mesh analysis in electrical circuits, specifically applying Kirchhoff's Voltage Law (KVL) to loops containing resistors. The correct KVL equation for a loop with a 3-ohm and 2-ohm resistor is established as 0 = 3I + 2I, leading to I = 0 due to the absence of sources. The importance of the passive sign convention is emphasized, clarifying that voltage drops are negative and voltage gains are positive in KVL equations. An example equation for a second loop is provided: 0 = 3000(I1 - I2) - 2000I2 - 2000I2.

PREREQUISITES
  • Understanding of Kirchhoff's Voltage Law (KVL)
  • Knowledge of passive sign convention in circuit analysis
  • Familiarity with basic resistor calculations (Ohm's Law)
  • Experience with writing and solving linear equations
NEXT STEPS
  • Study the application of KVL in complex circuits with multiple loops
  • Learn about the passive sign convention in detail
  • Explore circuit simulation tools like LTspice for practical mesh analysis
  • Investigate advanced topics such as nodal analysis for circuit analysis
USEFUL FOR

Electrical engineering students, circuit designers, and anyone seeking to deepen their understanding of mesh analysis and KVL in electrical circuits.

KingNothing
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I'm quite confused by mesh analysis. I understand that the idea is basically to apply KVL to each mesh. For example, if you have a loop that contains, say, a 3ohm and 2ohm resistor, and that's it, what is the correct KVL equation? Is it simply 0=3I+2I?

As another example, check out this page: http://en.wikibooks.org/wiki/Electronics/Mesh_Analysis
At the bottom, I can understand the first equation: you have a 9V source, and the resistors must cause a 9V voltage drop. But the second equation, I don't understand. Why is the first term positive?
 
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On your first question, yes, you would write 0=3I+2I. And you would solve it and get I=0, because there are no sources in your example.

In the second loop, the first term is positive because of the +/- sign convention on that resistor. In the first loop, the voltage across that resistor is a *voltage drop* following the direction of the first loop current as shown. But it is a *voltage gain* in the direction of the second current loop as shown. Make sense?

You just have to be careful and consistent in KVL loops to write drops as - voltage changes, and gains as + voltage changes.
 
Loop2: 0 = 3000(I1 - I2) - 2000I2 - 2000I2
yes it has to do with the passive sign convention. This equation can also
be written as follows
0 = -3000(I2 - I1) - 2000I2 - 2000I2.
I2 and I1 goes through 3Kohms, but in opposite directions. in the second loop just follow I2 and write the equation i wrote :) this avoids confusion
cheers
 

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