Question

(a) How long must a 0.60 mm diameter aluminum wire be to have a0.54A current when connected to the terminals of a 1.5 V flashlight battery?
m

(b) What is the current if the wire is half this length?
A

Answer 1

resistivity=2.8*10^-8

so

resistance=1.5/0.54

=2.778 ohm.so,

2.778=(2.8*10^-8*l)/(0.25*3.142*(6*10^-4)^2)

=28.0558 m

b)resistace will become half=1.389 ohm

so current=1.5/1.389

=1.08 A