Using aluminum wire(2.82×10^-8 ohm) of diameter 1.60mm, you want to wind a resistor that will dissipate 25.0 W when 25.0 V is applied across

Question

Using aluminum wire(2.82×10^-8 ohm) of diameter 1.60mm, you want to wind a resistor that will dissipate 25.0 W when 25.0 V is applied across it . What length( in m) of wire is needed?

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Calantha 3 years 2021-08-06T15:57:57+00:00 1 Answers 6 views 0

Answers ( )

  1. Dau
    0
    2021-08-06T15:59:17+00:00
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    Answer:

    4565 meters

    Explanation:

    W = V²/R

    25 = (25)²/R

    R = 25 ohm

    R = (pl)/A

    R = \frac{pl}{2\pi( \frac{d}{2} )}

    25 = \frac{( 2.82\times {10}^{ - 8} \times l)}{2\pi (\frac{1.60 \times {10}^{ - 6} }{2}) }

    l = 4565 meters approximately

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