How Do You Calculate the Equivalent Resistance in Complex Resistor Networks?

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SUMMARY

The discussion focuses on calculating the equivalent resistance in complex resistor networks, specifically addressing a circuit with resistors valued at 3, 1, and 5 ohms. The solution for the equivalent resistance R[ab] is stated to be 27/17 ohms. Participants emphasize the importance of understanding series and parallel resistor relationships, as well as applying Kirchhoff's Current Law (KCL) for solving the circuit. The original poster struggles to reach the provided solution despite using KCL and expresses a desire for a simpler method.

PREREQUISITES
  • Understanding of series and parallel resistor combinations
  • Familiarity with Kirchhoff's Current Law (KCL)
  • Basic circuit analysis techniques
  • Ability to manipulate circuit diagrams for clarity
NEXT STEPS
  • Study advanced techniques for simplifying resistor networks
  • Learn about mesh and nodal analysis in circuit theory
  • Explore software tools for circuit simulation, such as LTspice
  • Investigate the application of Thevenin's and Norton's theorems
USEFUL FOR

Electrical engineering students, circuit designers, and anyone involved in analyzing complex resistor networks will benefit from this discussion.

skiboka33
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Homework Statement



This problem involves a sketch of a system of resistors, so I will do my best.

A--xx1xx---|---xx1xx--B
.\.....|..../
...x...x...x
...x...x...x
...3...1...5
...x...x...x
...x...x...x
...\...|.../
.....\.|./
...\/

The numbers between the x's represent the resistors and their vaules.

The question asks to find the equivalent resistance R[ab].

Homework Equations



I am aware of the series/parallel relation ships between resistors, but this question seems ambiguous.

1/Req = 1/R1 + 1/R2
Req = R1 + R2

The Attempt at a Solution



The question states that the solution should be 27/17, but anyway I work this problem I dno't seem to get this answer, although I have gotten very close.

Any help?
 
Last edited:
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I don't see a simplification right off the bat (there may be one I'm missing), so what I would do is just solve the circuit using the KCL technique, and convert the answer into an overall resistance.

Like, turn the circuit 90 degrees clockwise, so A is on the top, and B is on the bottom. Ground B and put 1V in at A. Now write the two KCL equations for the voltages at the unknown nodes and solve for those voltages. That will give you the currents and voltages, so sum the currents through the two resistors leading out of A, and that along with the one 1V in will give you the overall resistance.

I often lose patience with these kind of problems early, and just do it the brute force way. BTW, if you find out that there's a simpler way to do it, please post it so I can learn it.
 

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