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alfers101

  • 3 years ago

A silverwire has a resistance of 1.5Ω at 0°C and a temperature coefficient of resistance of 0.00375/°C. To what temperature must be the raised to double the resistance?

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  1. JustHere
    • 3 years ago
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    I had a smiliar problem expect I was asked to find the R. I used R= (Delta V ^2 x time)/mc(T2-T1)

  2. alfers101
    • 3 years ago
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    so that is the formula?

  3. JustHere
    • 3 years ago
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    My bad, that's not the correct formula for this type of problem. You aren't given the potential difference.

  4. alfers101
    • 3 years ago
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    :( actually i can't understand the problem

  5. kropot72
    • 3 years ago
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    \[R _{t}=R _{0}+\alpha t\] \[R _{t}-R _{0}=\alpha t\] 3 - 1.5 = 0.00375 * t Now solve for t

  6. alfers101
    • 3 years ago
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    why 3?

  7. kropot72
    • 3 years ago
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    Raising the temperature needs to double the resistance. 1.5 ohms * 2 = 3 ohms

  8. alfers101
    • 3 years ago
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    t= 3-15 / 0.0037 ??

  9. kropot72
    • 3 years ago
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    \[t=\frac{3-1.5}{0.00375}\]

  10. alfers101
    • 3 years ago
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    thanks. so the unit must be degree celsius ?

  11. kropot72
    • 3 years ago
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    Yes. The unit for the answer is degrees Celsius :)

  12. alfers101
    • 3 years ago
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    wow thanks a lot. :D

  13. kropot72
    • 3 years ago
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    You're welcome :)

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