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Calculate resistivity of the material of the wire

  • Thread starter looi76
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[SOLVED] Calculate resistivity of the material of the wire

Question:
[tex]0.75A[/tex] Current flows through an iron wire when a battery of [tex]1.5V[/tex] us connected across its ends. The length of the wire is [tex]5m[/tex] and Area is [tex]2.5 \times 10^{-7} m^2[/tex]. Calculate resistivity of the material of the wire.

Equation:
[tex]R = \frac{Pl}{A}[/tex]

Attempt:
[tex]I = 0.75A \ , \ V = 1.5V \ , \ l = 5m \ , \ A = 2.5 \times 10^{-7} m^2[/tex]

[tex]R = \frac{Pl}{A}[/tex]

[tex]P = IV[/tex]

[tex]P = 0.75 \times 1.5[/tex]

[tex]P = 1.125[/tex]

[tex]R = \frac{1.125 \times 5}{2.5 \times 10^{-7}}[/tex]

[tex]R = 2.3 \times 10^7[/tex]

Is my answer correct?
:confused:
 

Answers and Replies

  • #2
Hootenanny
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You have your P's mixed up :rolleyes:

The P here represents the resistivity of the material
[tex]R = \frac{Pl}{A}[/tex]
Whereas the P here represents the power dissapated by the wire.
[tex]P = IV[/tex]
 
  • #3
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Thanks Hootenanny!

[tex]I = 0.75A \ , \ V = 1.5V \ , \ l = 5m \ , \ A = 2.5 \times 10^{-7} m^2[/tex]

[tex]R = \frac{V}{I} = \frac{1.5}{0.75} = 2\Omega[/tex]

[tex]R = \frac{Pl}{A}[/tex]

[tex]2\Omega = \frac{P \times 5}{2.5 \times 10^7 \Omega{m}}[/tex]

[tex]P = \frac{2 \times 2.5 \times 10^{-7}}{5}[/tex]

[tex]P = 1 \times 10^{-7} \Omega{m}[/tex]

is this correct?:uhh:
 
  • #4
Hootenanny
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Science Advisor
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Thanks Hootenanny!

[tex]I = 0.75A \ , \ V = 1.5V \ , \ l = 5m \ , \ A = 2.5 \times 10^{-7} m^2[/tex]

[tex]R = \frac{V}{I} = \frac{1.5}{0.75} = 2\Omega[/tex]

[tex]R = \frac{Pl}{A}[/tex]

[tex]2\Omega = \frac{P \times 5}{2.5 \times 10^7 \Omega{m}}[/tex]

[tex]P = \frac{2 \times 2.5 \times 10^{-7}}{5}[/tex]

[tex]P = 1 \times 10^{-7} \Omega{m}[/tex]

is this correct?:uhh:
That's what I get :approve:
 

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