Question

Using aluminum wire(2.82x10^-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?

Answer 1

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