Simplifying Complex Resistor Circuits

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Discussion Overview

The discussion revolves around simplifying a complex resistor circuit presented in a homework problem. Participants explore various methods to analyze the circuit, including identifying series and parallel resistors, and addressing confusion regarding a specific resistor's role in the circuit.

Discussion Character

  • Homework-related
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant expresses confusion about the configuration of the resistors, particularly the role of a 10kΩ resistor that appears to connect back to itself.
  • Another participant suggests that the 10kΩ resistor may not be necessary for the circuit analysis and proposes removing it to simplify the circuit.
  • A different viewpoint posits that the 10kΩ resistor can be considered in parallel with a 0Ω resistor, implying it does not affect the overall resistance.
  • Several participants provide calculations for the equivalent resistance of the circuit, indicating a specific method to combine resistors in parallel and series.
  • Some participants remind others of the forum rules, emphasizing the importance of guiding rather than providing direct answers to homework questions.

Areas of Agreement / Disagreement

There is no consensus on how to handle the 10kΩ resistor, with some participants suggesting it can be ignored while others discuss its potential impact. The discussion remains unresolved regarding the best approach to simplify the circuit.

Contextual Notes

Participants express uncertainty about the configuration of the resistors and the implications of removing or considering the 10kΩ resistor in the analysis. There are also reminders about the forum's guidelines on providing assistance.

careless25
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Homework Statement



See attached image.

Homework Equations





The Attempt at a Solution



I cannot find any resistors in parallel or series in the circuit. I am confused by the lone branch going from the 10kOhm resistor back to the other end of the same resistor. How do I deal with that?
I tried specifying nodes, Node A being connected to the 10kOhm, 20kOhm and maybe the 30kOhm and 80kOhm resistors? Node B connected to the 8kOhm and the 80kOhm resistors. Node C connected to the 20kOhm, 30kOhm and the 8kOhm resistors.

This way I get 20kOhm and the 30kOhm resistors in parallel, but then I am confused about how to deal with the 10kOhm resistor. Is it in series? Parallel? with the other 2?
 

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The 10ohm resistor seems to have been put in just to see if you know what you are doing, since it isn't REALLY in the circuit. Take it out and redraw the circuit and you'll find parallel and then series resistors (after you reduce the parallels)
 
One way to look at it: the 10kΩ resistor is actually in parallel with a 0Ω resistor (the wire).
 
80Ω//( 8Ω series with (20Ω//30Ω))
So, equivalent resistor of cirrcuit is 80//(8+12)=16Ω
10 Ohm resistor is shorted cirrcuit.
 
hoangkyem said:
80Ω//( 8Ω series with (20Ω//30Ω))
So, equivalent resistor of cirrcuit is 80//(8+12)=16Ω
10 Ohm resistor is shorted cirrcuit.

Please read the forum rules. It is innappropriate to GIVE the answer. The point of this forum is not to give people the answers to homework problems but to help them learn how to THINK and get their own answers. Give HINTS, not answers.
 
phinds said:
Please read the forum rules. It is innappropriate to GIVE the answer. The point of this forum is not to give people the answers to homework problems but to help them learn how to THINK and get their own answers. Give HINTS, not answers.
you are right. sorry
 

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