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NEETPhysicsCurrent Electricity

A uniform metal wire of resistance 10 is stretched to double its original length. The stretched wire is then cut into two equal halves. These two halves are connected in parallel across a battery of EMF 12 V and internal resistance 2 . The current drawn from the battery is

Options

  1. A1 A
  2. B1.2 A
  3. C12 7 A
  4. D8 3 A

Correct answer

A. 1 A

Step-by-step solution

Initial resistance of the wire, R₀ = 10 . When a wire is stretched to n times its original length, its volume remains constant. The new resistance becomes R_ new = n^2 R₀ . Here, n = 2 , so R_ new = (2)^2 10 = 40 . The wire is then cut into two equal halves. The resistance of each half is proportional to its length, so each half has a resistance of 40 2 = 20 . These two halves are connected in parallel. The equivalent external resistance R_ ext is: R_ ext = 20 20 20 + 20 = 10 . The total resistance of the circuit i

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