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

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 3.0volt battery. Find the current drawn from the battery.
Answer» When length of a given wire is made n-times by strecting it, its resistance becomes n² times i.e., R' =n²R = (2)²x 15 = 60Ω Resistance of each half part 60/2 = 30Ω When both parts are connected in parallel, final resistance = 30/2 = 15Ω Current drawn from battery, I = V/R = 3.0/15 = 0.2A

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 3.0volt battery. Find the current drawn from the battery.

Answer»

When length of a given wire is made n-times by strecting it, its resistance becomes n² times

i.e., R' =n²R = (2)²x 15 = 60Ω

Resistance of each half part 60/2 = 30Ω

When both parts are connected in parallel, final resistance = 30/2 = 15Ω

Current drawn from battery,

I = V/R

= 3.0/15 = 0.2A

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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 3.0volt battery. Find the current drawn from the battery.

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