Calculating combined resistance of wires

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SUMMARY

The discussion focuses on calculating the combined resistance of six copper wires and a steel core, emphasizing the importance of understanding parallel resistor configurations. The correct approach involves using the formula for resistors in parallel, specifically 1/R = 6(1/10) + 1/100, leading to a total resistance of 1.6Ω. The steel core's resistance is not considered in this scenario due to its negligible impact compared to the copper wires. Participants confirm that the combined resistance of multiple copper wires is lower than that of a single copper wire of the same cross-sectional area.

PREREQUISITES
  • Understanding of Ohm's Law and resistance calculations
  • Familiarity with the concept of resistors in parallel
  • Knowledge of material properties affecting resistance (e.g., copper vs. steel)
  • Basic algebra for manipulating resistance equations
NEXT STEPS
  • Study the principles of resistors in parallel and series configurations
  • Learn about the specific resistivity values of different materials, such as copper and steel
  • Explore practical applications of combined resistance in electrical engineering
  • Review advanced topics in circuit analysis, including Thevenin's and Norton's theorems
USEFUL FOR

Students in electrical engineering, physics enthusiasts, and anyone involved in circuit design or analysis will benefit from this discussion.

toforfiltum
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Homework Statement



upload_2015-5-22_8-19-31.png

Homework Equations


  • R ∝ 1/A

The Attempt at a Solution


Since there are six copper wires and a steel core, I added the resistance of all the wires and the core and divided them by 7, though I don't think my approach is right since the resistance of the copper and steel wires are different. As such my answer is about 22.9Ω, which is wrong. The answer is B, 1.6Ω, which is just the quotient of the the total resistance of the six copper wires. Why is the resistance of steel core not taken into account in this case?
 

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Do you know how to calculate the net resistance of resistors in parallel?
 
Ermm...yes. But why is the steel not taken into account?
 
toforfiltum said:
Ermm...yes. But why is the steel not taken into account?
Why are you so sure it isn't? Have you done the calculation taking the steel into account?
 
Oh, so is the working 6(1/10) + 1/100 correct for 1/R?
 
toforfiltum said:
Oh, so is the working 6(1/10) + 1/100 correct for 1/R?
Yes.
 
Thanks. So is this combined resistance of six different copper wires and steel less than a power cable just made up of one copper wire of the same cross-sectional area?
 
toforfiltum said:
Thanks. So is this combined resistance of six different copper wires and steel less than a power cable just made up of one copper wire of the same cross-sectional area?
What do you think?
 
I think the resistance of six different copper wires and the steel would be lesser. However if it's just the six copper wires without the steel, then they would be the same.
 
  • #10
toforfiltum said:
I think the resistance of six different copper wires and the steel would be lesser. However if it's just the six copper wires without the steel, then they would be the same.
Quite so.
 
  • #11
Why quite so? Is it not really correct?:smile:
 
  • #12
toforfiltum said:
Why quite so? Is it not really correct?:smile:
"Quite so" means, yes, that is exactly correct. (British English idiom?)
 
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  • #13
Oh I see.
 
  • #14
Since resistance values are given we need think about type of wire.
From situation they are in parallel connection.
Please see the attachment
 

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  • #15
Neelima said:
Since resistance values are given we need think about type of wire.
From situation they are in parallel connection.
Please see the attachment
Yes, the OP got this in post #5. Five years ago.
 
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