Question

A large spool in an electrician's workshop has 69 m of insulation-coated wire coiled around it. When the electrician connects a battery to the ends of the spooled wire, the resulting current is 2.5 A. Some weeks later, after cutting off various lengths of wire for use in repairs, the electrician finds that the spooled wire carries a 3.7-A current when the same battery is connected to it. What is the length of wire remaining on the spool

Answer 1

Answer:

46.62m

Explanation:

Given the length of the wire as 69m, current as 2.5A, and as 3.7A after 2 weeks;

Let the initial and final lengths of the wire be expressed as:

$R=p(\frac{L}{A})->L=\frac{RA}{p}\\\\L_o=\frac{R_oA}{p}\\\\L_f=\frac{R_fA}{p} \ \ \ \ \ \....i$

Ro is the initial resistance and Rf the final resistance.

$\frac{L_f}{L_o}=\frac{R_f}{R_o}$

From $R=\frac{V}{I}$ equation, Ro and Rf of the spooled wire are expressed as:

$R_o=\frac{V}{I_o}, \ \ \ \ R_f=\frac{V}{I_f}\\\\#From \ above\\\\L_f=\frac{(V/I_f)}{V/I_o}L_o\\\\=\frac{I_o}{I_f}L_o\\\\=\frac{2.5A}{3.7A}\times 69m\\\\=46.62m$

Hence, the  length of wire remaining on the spool is 46.62m

Answer 2

Answer:

Explanation:

initial length, L = 69 m

initial current, i = 2.5 A

final current, i' = 3.7 A

Let the final length is L'.

As the resistance is directly proportional to the length of wire and the voltage remains same so

current is inversely proportional to the resistance and hence the current is inversely proportional to the length of the wire.

So,

$\frac{i}{i'}=\frac{L'}{L}$

$\frac{2.5}{3.7}=\frac{L'}{69}$

L' = 46.6 m