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

[skiboka33]

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?

 

[berkeman]

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.

 

[m2003]

I would first start by carefully analyzing the given diagram and understanding the connections between the resistors. From the diagram, we can see that there are resistors in series and in parallel.

To find the equivalent resistance between points A and B, we can use the formula 1/Req = 1/R1 + 1/R2, where R1 and R2 represent the resistors in series. In this case, R1 = 1+3+5 = 9 and R2 = 1+1+5 = 7. Therefore, 1/Req = 1/9 + 1/7 = 16/63. Taking the reciprocal, we get Req = 63/16.

However, this is not the final answer as there are still resistors in parallel. To account for this, we can use the formula Req = R1 + R2, where R1 and R2 represent the equivalent resistances of the parallel resistors. In this case, R1 = 1/(1/9 + 1/7) = 63/16 and R2 = 1/(1/5 + 1/5) = 5/2. Therefore, Req = 63/16 + 5/2 = 27/17, which is the desired answer.

In conclusion, by carefully analyzing the given diagram and using the appropriate formulas, we can find the equivalent resistance between points A and B to be 27/17.