La resistividad de cierto alambre es 1.72x10 -8 Ωm a 20 oC .Una batería de 6v se conecta a una bobina de 20 m de este alambre, con un diámetro de 0.8 mm. ¿Cuál es la corriente en el alambre?

   i = 1.09 10⁴ A

Explanation:

For this exercise we will look for the resistance of the wire

          R = ρ L / a

the area of ​​the wire is

          a =ππ r² = π πd² / 4

we substitute

         R = ρ L 4 / π d²

          R = 1.72 10⁻⁸   20  4/π 8 10⁻⁴

         R = 5.47 10⁻⁴ Ω

to calculate the current we use ohm's law

           V = R i

            i = V / R

            i = 6 / 5.47 10⁻⁴

            i = 1.09 10⁴ A

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