A wire of `15 Omega` resistance is gradually stretched to double in original length. it is then cut into two equal parts .These parts are then connected in parallel across a `3.0` volt battery. Find the current draw from the battery .

Explanation

When any resistor is streched to double is original length. The new resistance becomes four time of its original resistance.
Here, `R=15 Omega and V=3.0` Volt
`:.` New resistance `=4R=4xx15=60 Omega`
Resistance of each part `=30 Omega`
`:. R_(1)=30 Omega, R_(2)=30 Omega`
Effective resistance in parallel combination `R_(P)` is
`1/R_(P)=1/30+1/30=2/30=1/15" "[ :' 1/R_(P)=1/R_(1)+1/R_(2)]`
`:. R_(P)=15 Omega`
Current, `I=V/R=3.0/15 =0.2 A`.