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Current in wire - Please help

  • Thread starter jtw2e
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  • #1
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Current in wire -- Please help

Homework Statement



A 1.63-m length of wire is made by welding the end of a 100-cm long silver wire to the end of a 63-cm long copper wire. Each piece of wire is 1.1 mm in diameter. A potential difference of 5.0 mV is maintained between the ends of the 1.63-m composite wire.
ρsilver = 1.47e-8Ωm
ρcopper = 1.72e-8Ωm


Homework Equations



What is the current in the silver section? (mA)
What is the current in the copper section? (mA)
What is the potential difference (voltage drop) between the ends of the silver section of wire? (mV)
What is the magnitude of the electric field in the silver? (N/C)

The Attempt at a Solution



Part A:
I assumed we had to find the resistance in each part, sum them, then find the current. The current from each section should be the same, correct? The value I have below (148 milliAmps) is wrong according to the online homework system.

Rs = ρL/A
= [(1.47e-8Ωm)(1.0m)]/[π(.0011m/2)^2]
= 0.015468Ω

Rc = ρL/A
= [(1.72e-8Ωm)(.63m)]/[π(.0011m/2)^2]
= 0.018098Ω

Rtotal = Rs + Rc = 0.033567Ω

V = IR
I = V/R
= (0.005V)/(0.033567Ω)
= 0.14895 Amp
= 148.954 milliAmps
 
Last edited:

Answers and Replies

  • #2
798
1


Hello,
As long as you do not have any arithmetic errors in your calculations then you should be good. You have the right idea.
 
  • #3
27
0


Hello,
As long as you do not have any arithmetic errors in your calculations then you should be good. You have the right idea.
Cannot find any math errors. Been on this for 5 hours and can't move on until I get it. :/
 
  • #4
798
1


Okay, well the current will be the same in both instances as they are in series with one another. Since you know the resistance of each portion, you can apply a voltage divider to determine the voltage across each portion. For example,
[tex]V_{silver}=V_{total}\frac{R_{silver}}{R_{total}}[/tex]

I'm not entirely sure if there is something in specific you are having troubles with?
 
  • #5
798
1


Check your calculation for the resistance of the copper portion. I get 0.0114023 Ohms.
 
  • #6
27
0


Okay, well the current will be the same in both instances as they are in series with one another. Since you know the resistance of each portion, you can apply a voltage divider to determine the voltage across each portion. For example,
[tex]V_{silver}=V_{total}\frac{R_{silver}}{R_{total}}[/tex]

I'm not entirely sure if there is something in specific you are having troubles with?
FOUND IT, finally... somehow the value I got for Resistance through the copper is wrong.

Thanks for telling me I was on the right track so I knew it had to be a math error (or a programming error on their part).
 
  • #7
798
1


Hehe, no problem.
 

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