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 \( R \). The power consumed, according to Joule's law, is \( P = \frac{V^2}{R} \), where \( V \) is the voltage. When the wire is stretched to double its initial length, its resistance changes. The resistance of a wire is given by \( R = \rho \frac{L}{A} \), where \( \rho \) is resistivity, \( L \) is length, and \( A \) is cross-sectional area. If the length \( L \) doubles, the new resistance \( R' \) becomes \( R' = 2R \) because resistance is directly proportional to length. The power consumed with the stretched wire is \( P' = \frac{V^2}{R'} \). Substituting \( R' = 2R \), we get \( P' = \frac{V^2}{2R} = \frac{P}{2} \). This implies the power consumed is halved. However, stretching the wire reduces its cross-sectional area proportionally while keeping the volume constant. This further increases resistance. When the length doubles, the area halves, leading to a fourfold increase in resistance (\( R'' = \rho \frac{2L}{A/2} = 4 \rho \frac{L}{A} = 4R \)). Consequently, the final power consumed after stretching is \( P'' = \frac{V^2}{R''} = \frac{V^2}{4R} = \frac{P}{4} \). The correct answer is \( \frac{P}{4} \).