A wire of resistance 4Ω is stretched to twice its original length. The resistance of stretched wire would be

Question

A battery of \( 6 \, \text{V} \) is connected to the circuit as shown below. The current \( I \) drawn from the battery is:

Answer 1

To solve the problem of finding the new resistance of a wire that is stretched to twice its original length, we need to understand the relationship between resistance, length, and area.

The resistance R of a wire is given by the formula:

R = \frac{\rho L}{A}

Where:

R is the resistance.
\rho is the resistivity of the material.
L is the length of the wire.
A is the cross-sectional area of the wire.

Given that the original resistance of the wire is 4 Ω and it is stretched to double its length, let's denote the original length of the wire as L and its new length after stretching as 2L.

When the wire is stretched, its volume remains constant. Therefore, the relationship between the original and the new cross-sectional area is given by:

A_1 \cdot L = A_2 \cdot 2L \Rightarrow A_2 = \frac{A_1}{2}

Substituting the new length and area into the resistance formula, the new resistance R_{new} becomes:

R_{new} = \frac{\rho \cdot 2L}{A_2} = \frac{\rho \cdot 2L}{A_1/2} = \frac{\rho \cdot 2L \cdot 2}{A_1} = 4 \cdot R

Substituting the original resistance of 4 Ω:

R_{new} = 4 \times 4 = 16 \, \Omega

Therefore, the resistance of the stretched wire is 16 Ω. This is confirmed as the correct answer.

Answer 2