Understanding Copper Wire Resistance Change

This problem involves understanding how the physical dimensions of a conductor, specifically a copper wire, affect its electrical resistance. We are given the initial resistance of the wire and the factor by which its length is increased due to stretching. The goal is to determine the resulting resistance.

Physics Principles of Resistance

The electrical resistance ($R$) of a wire is determined by its material's resistivity ($\rho$), its length ($L$), and its cross-sectional area ($A$). The fundamental formula relating these quantities is:

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

From this formula, we can observe:

• Resistance ($R$) is directly proportional to the length ($L$) of the wire. If the length increases, resistance increases, assuming other factors remain constant.
• Resistance ($R$) is inversely proportional to the cross-sectional area ($A$) of the wire. If the area increases, resistance decreases, and vice versa.

Analyzing the Effect of Stretching

Let the original length, cross-sectional area, and resistance of the copper wire be $L_1$, $A_1$, and $R_1$, respectively. We are given that the initial resistance is $R_1 = 10 \, \Omega$.

The initial resistance can be expressed as:

$$R_1 = \rho \frac{L_1}{A_1} = 10 \, \Omega$$

When the wire is stretched to double its original length, the new length ($L_2$) becomes:

$$L_2 = 2L_1$$

A crucial aspect of stretching a wire is that its volume ($V$) remains constant, assuming no material is lost. The volume ($V$) of the wire can be calculated as the product of its length and cross-sectional area:

$$V = L_1 A_1$$

Since the volume remains constant after stretching:

$$V = L_2 A_2$$

Therefore, we have the relationship:

$$L_1 A_1 = L_2 A_2$$

Now, substitute the new length $L_2 = 2L_1$ into the volume conservation equation:

$$L_1 A_1 = (2L_1) A_2$$

To find the new cross-sectional area ($A_2$), we rearrange the equation:

$$A_2 = \frac{L_1 A_1}{2L_1}$$

$$A_2 = \frac{A_1}{2}$$

This calculation shows that when the length of the wire is doubled, its cross-sectional area is reduced to half of its original value.

Calculating the New Resistance Value

The new resistance ($R_2$) can be calculated using the new dimensions ($L_2$ and $A_2$) and the same resistivity ($\rho$):

$$R_2 = \rho \frac{L_2}{A_2}$$

Substitute the expressions for $L_2$ and $A_2$:

$$R_2 = \rho \frac{(2L_1)}{(A_1/2)}$$

Simplify the fraction:

$$R_2 = \rho \frac{2L_1 \times 2}{A_1}$$

$$R_2 = 4 \left( \rho \frac{L_1}{A_1} \right)$$

Recognize that the term $\rho \frac{L_1}{A_1}$ is the original resistance $R_1$. Substituting $R_1$ back into the equation:

$$R_2 = 4 R_1$$

Given the original resistance $R_1 = 10 \, \Omega$, we can find the new resistance:

$$R_2 = 4 \times 10 \, \Omega$$

$$R_2 = 40 \, \Omega$$

Final Result

The new resistance of the copper wire after being stretched to double its original length is $40 \, \Omega$. This demonstrates that resistance increases quadratically with the length when stretching, due to the simultaneous decrease in cross-sectional area.