Question

A large spool in an electrician's workshop has 65 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 1.8 A. Some weeks later, after cutting off various lengths of wire for use in repairs, the electrician finds that the spooled wire carries a 2.9-A current when the same battery is connected to it. What is the length of wire remaining on the spool

Answer 1

`40.34\ \text{m}`

Step-by-step explanation:

`L_1` = Length of wire = 65 m

`I_1` = Initial current = 1.8 A

`I_2` = Final current = 2.9 A

We know

`R\propto (1)/(I)`

and

`R\propto L`

`(V)/(I)\propto L\\\Rightarrow L\propto (1)/(I)`

so

`(L_2)/(L_1)=(I_1)/(I_2)\\\Rightarrow L_2=(I_1)/(I_2)L_1\\\Rightarrow L_2=(1.8)/(2.9)* 65\\\Rightarrow L_2=40.34\ \text{m}`

The length of the wire remaining on the spool is
`40.34\ \text{m}`.