A wire of 15Ω resistance is gradually stretched to double its origin

1.	A wire of 15Ω 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 3.0 volt battery. Find the current drawn from the battery.
Answer» When the wire of 15Ω resistance is stretched to double its original length, then its resistance becomes R’ = n² × 15 = 2² × 15 = 60Ω When it cut into two equal parts, then resistance of each part becomes R' = R'/2 = 60/2 = 30Ω These parts are connected in parallel, then net resistance of their combination is R = R'/2 = 30/2 = 15Ω So, the current drawn from the battery l = V/R = 3/15 = 1/5 or l = 0.2 A.

A wire of 15Ω 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 3.0 volt battery. Find the current drawn from the battery.

Answer»

When the wire of 15Ω resistance is stretched to double its original length, then its resistance becomes

R’ = n² × 15 = 2² × 15 = 60Ω

When it cut into two equal parts, then resistance of each part becomes

R' = R'/2 = 60/2 = 30Ω

These parts are connected in parallel, then net resistance of their combination is

R = R'/2 = 30/2 = 15Ω

So, the current drawn from the battery

l = V/R = 3/15 = 1/5

or l = 0.2 A.