Question

A large spool in an electrician's workshop has 70 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 2.7 A. Some weeks later, after cutting off various lengths of wire for use in repairs, the electrician finds that the spooled wire carries a 3.5-A current when the same battery is connected to it. What is the length of wire remaining on the spool

Answer 1

Answer:

54.0m

Explanation:

#First we solve for

$R=p(L/A)\\\\L=RA/p$

let's denote the initial and final length of the rope as $L_o, L_f$ respectively and given as:

$L_o=\frac{R_oA}{p}, \ \ \ \ L_f=\frac{R_fA}{p}\ \ \ \ \ .....eqtn1$

$R_o$ is the initial resistance and Rf the final of the wire.

$\frac{L_f}{L_o}=\frac{R_fA/P}{R_oA/P}=R_f/R_o\\\\\\or \ L_f=(R_f/R_o)L_o\ \ \ \ \ \ ...eqtn2$

From $R=V/I,$ the initial resistance $R_o, \ and\ R_f$ of the spooled wire are:

$R_o=V/I_o,\ \ \ \ \ \ R_f=V/I_f \ \ \ \ \ \ \....eqtn3$

#Substituting eqtn 3 in 2, we get

$L_f=(V/I_f)/(V/I_o)L_o\\\\=\frac{2.7A}{3.5A}\times 70m\\\\=54m$

#the length of wire remaining on the spool is 54.0m