A wire of 15Ω resistance is gradually stretched to double its original length.
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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When length of a given wire is made n-times by strecting it, its resistance becomes n2 times
i.e., R' =n2R = (2)2 x 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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