A wire of `20 Omega` resistance is gradually stretched to double its original length. It is then cut into two equal parts. These parts are then connected in parallel across a 4.0 volt battery. Find the current drawn from the battery.

Explanation

When any resistor is stretched to double its original length, the new resistance becomes four times of its original resistance.
Here, `R=20 Omega` and `V=4.0` volt
`:.` New resistance `=4R=4xx20=80 Omega`
Resistance of each part `=40 Omega`
`:. R_(1)=40 Omega, R_(2)=40 Omega`
Effective resistance in pararllel combination `R_(P)` is
`1/R_(P)=1/40+1/40=2/40=1/20" "[ :' 1/R_(P)=1/R_(1)+1/R_(2)]`
`:. R_(P)=20 Omega`
Current, `I=V/R_(P)=4.0/20=0.2 A`