Is the Book's Answer Correct for This Circuit Problem?

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Homework Help Overview

This discussion revolves around a circuit problem involving resistors, specifically focusing on calculating the equivalent resistance in a complex circuit setup. The original poster references a textbook answer of 4.7 ohms and presents their own calculation of 4.33 ohms, raising questions about the accuracy of both their work and the book's answer.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the accuracy of the diagram provided and the number of resistors accounted for in the calculations. There are questions about the method of calculating equivalent resistance, with some participants suggesting different approaches and interpretations of the circuit layout.

Discussion Status

Multiple interpretations of the circuit are being explored, with participants providing guidance on how to approach the problem. Some have expressed confusion regarding the original poster's calculations and the diagram, while others have shared their own results, indicating a lack of consensus on the correct answer.

Contextual Notes

There is mention of a discrepancy between the number of resistors in the diagram and those used in calculations, as well as potential rounding differences in the textbook answer. Participants are encouraged to clarify their assumptions and redraw the circuit for better understanding.

KingNothing
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Hey everyone, this is a problem straight from my book. It's an odd number, so the book has the answer in the back: 4.7 ohms

The setup is near impossible to explain, so I've attached a diagram of the setup. Basically, it is a circuit with some resistors (represented by squiggly lines) each with R=2.8 kilo-ohms.

I came up with:
[tex]R=(\frac{1}{3*2.8}+\frac{1}{2*2.8}+\frac{1}{2.8})^-1+2.8=4.33\Omega[/tex]
The -1 right after the frist parentheses is a ^-1, it's kind of hard to tell.
Did I do it wrong? Is the book wrong? Is it just a case where they rounded and I didnt?
 

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Are you sure the diagram you posted is correct? Your equation has 7 resistors but I see only 6 in the diagram.
 
Yes, I'm fully aware of that...you see, I figured that if it exits the smaller line there on the bottom, and travels up to the intersection, that if it took either of the two uppermost paths (speaking literally according to diagram) that it would have to pass through the upper resistor on the right side, so I accounted for that twice. What is the proper way to do this?
 
The two in the top left are in series, and are in parallel with the diagonal. That grouping is in series with the top one on the right. That whole grouping is in parallel with the middle one. Then that whole thing is in series with the bottom one on the right.
 
It certainly doesn't go through any resistor twice.

First I'd re-draw the diagram to make it less confusing. Then I think the solution is pretty easy. But I don't get your book's answer either. I get 4.55 [tex]\Omega[/tex]
 

Attachments

  • circuit.png
    circuit.png
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I get 4.55

I broke it into 3 parts.
1. solve for 4 top resistors (as a parallel circuit)
2. solve for the middle resistor in parallel with the top 4
3. add the 1 resistor in series

Here is what the top 4 resistors are like, I think.

[tex]2.8 + [2.8^{-1} + [(2)(2.8)]^{-1}]^{-1}[/tex]

Now write that as A in your calculator. Trying to sub that into an even more complicated equation will just cause problems.

Now here is what the parallel portion of the entire circuit looks like

[tex][2.8^{-1} + A^{-1}]^{-1}[/tex]

Now add the series portion which is just 2.8

[tex]R = 2.8 + [2.8^{-1} + A^{-1}]^{-1}[/tex]

[tex]R = 4.55[/tex]
 
Thank you everyone...I can't tell you how relieving it is to see exactly what I did wrong. It's such a great feeling! Thank you a ton Shawn and enigma and gnome. I got 4.55 as well.
 

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