Question

1. seuutas kabel tembaga panjangnya 10 m berpenampang 2 mm² . berapakah hambatan kabel apabila hambatan jenis kawat 1,68 ×10 pangkat negatif 8 ohm meter ?

Answer 1

Diket:
l = 10 m
A = 2 mm² = 2x10^-6 m²
ρ = 1,68x10^-8 ohmmeter
Dit: R.....?
R = ρl/A
R = (1,68x10^-8 x 10)/2x10^-6 m²
R = 1,68x10^-7 / 2x10^-6
R = 0,084 ohm

Answer 2

1,68x10^-8x10/2x10^-6
=8,4x10^-2