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For the combination of the resistors, shown in the figure. Calculate the equivalent resistance between A and B, please help me to find the answer of this complicated question this is a question from an easy part but, I don't know how to solve this.

For the combination of the resistors, shown in the figure. Calculate the equivalent resistance between A and B, please help me to find the answer of this complicated question this is a question from an easy part but, I don't KNOW TO SOLVE THIS

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closed as off-topic by Emilio Pisanty, Qmechanic Sep 5 '13 at 16:27

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Emilio Pisanty, Qmechanic
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Hi. Please note that Homework questions are supposed to supposed to show more effort. For more details, please see the Homework Policy. –  Dimensio1n0 Sep 5 '13 at 15:57
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Start from the B end, and try to replace two resistors by an equivalent one. Then repeat until finished. It should not be hard: "resistance is futile". –  babou Sep 5 '13 at 16:03
    
It is not always that easy. Sometimes you have to write equations, because you cannot isolate pairs of resistors. –  babou Sep 5 '13 at 16:18
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1 Answer 1

This is a homework(-like) question so I may not give a full answer.

Let $\alpha$ be the bottom resistor with 100 omega resistance.
Let $\beta$ be the middle resistor with 100 omega resistance.
Let $\gamma$ be the top resistor with 100 omega resistance.
Let $\delta$ be the resistor with 25 omega resistance.
Let $\varepsilon$ be the resistor with 120 omega resistance.
Let $\zeta$ be the resistor with 40 omega resistance.

If you see the resistors $\alpha$ and $\delta$, they are in parallel, so they add up to a resistance given by $\frac1R=\frac1{R_1}+\frac1{R_2}$. Let's call this "parallel addition".

This added up resistance is in series, with resistor $\beta$, so they're total resistance is merely their sum. Let's call this "serial addition".

This resistance is to be parallelly added to the resistor $\varepsilon$, which can be serially added to the resistor $\gamma$ and so on.

The formulas I mentioned can be obtained from any standard introductory physics or such textbook. My personal favourite is Jewett and Serway Physics for Scientists and Engineers with Modern Physics.

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@GeorgeE.Smith: Yes, but the above is what the OP asks for. –  Dimensio1n0 Sep 6 '13 at 1:51
    
@GeorgeE.Smith: "OP" is "Original poster", "29231". "above" is my answer. You're comment is a full solution to a hw problem, which is not allowed. –  Dimensio1n0 Sep 6 '13 at 11:27
    
@GeorgeE.Smith: There's no use ranting to me. Go to Physics Meta and post a feature request. "Hardware" problems are not allowed because they're engineering, not physics. Give me one question asked in another language, please. I agree there are some arbitrary rules, such as for book questions, list questions, etc.; but those rules may be revised soon. –  Dimensio1n0 Sep 7 '13 at 3:32
    
If answering hardware problems isn't allowed; why did YOU answer it. You give an answer that is specific to this network; not all networks. And it is obvious, from the specific numbers in the question, that the author's full intent was for it to be solved, by inspection exactly as I explained to the "OP", as you call him/er. If it was my question on an exam paper, I would dock the student points for using the laborious method you describe; in fact I have done precisely that, for a similar network problem that had an obvious answer by inspection, that I asked on an exam paper. –  user26165 Sep 8 '13 at 5:11
    
@GeorgeE.Smith: (1) Becausew this isn't a hardware question. By "hw", I meant "homework". Homework questions are allowed here, but they need to show effort, and have to ask about a conceptual problem. Full solutions aren't allowed. That's the fact. Please. Stop. Ranting. To. Me, . Thank. You. c –  Dimensio1n0 Sep 9 '13 at 8:09
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