Question

A large spool in an electrician's workshop has 60 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 2.0 A. Some weeks later, after cutting off various lengths of wire for use in repairs, the electrician finds that the spooled wire carries a 2.6-A current when the same battery is connected to it. What is the length of wire remaining on the spool?

Answer 1

The length of wire remaining on the spool is 46.15 meters.

Step-by-step explanation:

It is given that,

Length of the wire, L = 60 m

Current, I = 2 A

Some weeks later, after cutting off various lengths of wire for use in repairs,

New current, I' = 2.6 A

Let l' is the length of wire remaining on the spool. We know that he resistance of wire in terms of length and area is given by :

`R=(\rho L)/(A)`

If V is the voltage, then, V = IR

Current,
`I=(AV)/(\rho L)`

So, current is inversely proportional to the length of the wire. So,

`(I)/(I')=(L')/(L)`

L' is the length of wire remaining on the spool

`(IL)/(I')=L'`

`(2* 60)/(2.6)=L'`

L' = 46.15 meters

So, the length of wire remaining on the spool is 46.15 meters.