Question

A copper wire of diameter 0.5mm and resistivity of 1.6×10-8ohm-m. What will be the length of the copper wire required to make a resistance of 21ohm.

Answer 1

Answer:

R=10 Ω

ρ=1.6 x 10^(-8) Ω⋅m

d=0.05 mm

=0.05 x 10^(-3) m

=5 x 10^(-5)m

So r = 2.5 x 10^(-5) m

l=? m

now using,

R=ρl/A,

or R=ρl/π(r^2)

10= {[1.6 x 10^(-8)] x l}/ π x [(2.5 x 10^(-5)) ^2]

10= {[1.6 x 10^(-8)] x l}/[2 x 10^(-9)]

now find the value of l

Answer 2

Explanation:

$Answer: \\ </p><p>R=10 Ω \\ </p><p>ρ=1.6 x 10^(-8) Ω⋅m \\ </p><p>d=0.05 mm \\ </p><p>=0.05 x 10^(-3) m \\ </p><p>=5 x 10^(-5)m \\ </p><p>So \: r = 2.5 x 10^(-5) m \\ </p><p>l=? m \\ </p><p>now \: using, \\ </p><p>R=ρl/A, \\ </p><p>or R=ρl/π(r^2) \\ </p><p>10= {[1.6 x 10^(-8)] x l}/ π x [(2.5 x 10^(-5)) ^2] \\ </p><p>10= {[1.6 x 10^(-8)] x l}/[2 x 10^(-9)] \\ </p><p>now \: find \: the \: value \: of \: l \:$