CBSE 2025, Set 2

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). $2P$

Edvaya · Accepted Answer

The correct answer is $\frac{P}{4}$. The power consumed by a wire connected to an ideal battery (constant voltage $V$) is given by $P = \frac{V^2}{R}$.

When a wire is stretched, its volume remains constant. 
Let the initial length be $l$ and area be $A$. Volume $V_{ol} = A 	imes l$.
If the new length $l' = 2l$, the new area $A'$ must be $\frac{A}{2}$ to keep the volume constant.

The new resistance $R'$ is:
$$R' = ho \frac{l'}{A'} = ho \frac{2l}{A/2} = 4 \left( ho \frac{l}{A} ight) = 4R$$

The new power consumed $P'$ when connected to the same battery is:
$$P' = \frac{V^2}{R'} = \frac{V^2}{4R} = \frac{1}{4} \left( \frac{V^2}{R} ight) = \boxed{\frac{P}{4}}$$

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