Question

A wire of resistance \( R \), connected to an ideal battery, consumes a power \( P \). If the wire is gradually stretched to double its initial length, and connected across the same battery, the power consumed will be:

• A. \( \frac{P}{4} \)
• B. \( \frac{P}{2} \)
• C. \( P \)
• D. \( 2P \)

Hint:

When a wire is stretched, its length increases, causing an increase in resistance. The power consumed by the wire decreases as the resistance increases, and the relationship between power and resistance is inversely proportional.

Answer 1

The Correct Option is A

Let the initial resistance of the wire be \( R \). According to Joule’s law, the power consumed by the wire is given by: \[ P = \frac{V^2}{R} \] where \( V \) is the voltage of the ideal battery. When the wire is stretched to double its initial length, the resistance changes because the resistance of a wire is given by: \[ R = \rho \frac{L}{A} \] where \( \rho \) is the resistivity of the material, \( L \) is the length of the wire, and \( A \) is the cross-sectional area. When the length \( L \) of the wire is doubled, the new resistance \( R' \) becomes: \[ R' = 2 \times R \] because resistance is directly proportional to length. Now, the power consumed when the wire is stretched is given by: \[ P' = \frac{V^2}{R'} \] Since \( R' = 2R \), we have: \[ P' = \frac{V^2}{2R} = \frac{P}{2} \] Thus, the power consumed after stretching the wire is \( \frac{P}{2} \). However, when the wire is stretched, its cross-sectional area decreases in proportion to the increase in length (assuming the volume of the wire remains constant). This causes the resistance to increase further. In this case, when the wire is stretched to double its length, the final power consumed will be \( \frac{P}{4} \). Thus, the correct answer is \({\frac{P}{4}} \).