A battery is used in an experiment and is connected by a 37.5 cm long gold wire with a diameter of 0.590 mm. If the battery has a voltage of 1.70 V and the resistivity of the wire is 2.4×10-8 Ω m, what is the current in the wire?

Answer 1

L = 0.375 m.
p = 2.4*10^-8 Ohm m.
r = 0.590/2 = 0.295 mm. = 2.95*10^-4 m.
A = pi*r^2 = 3.14*(2.95*10^-4)^2 = 2.7*10^-7 m^2.

R = pL/A = 2.4*10^-8*0.375/2.7*10^-7 = 0.033 Ohms.

I = V/R = 1.7/0.033 = 51A.