When a wire of resistivity r is stretched to double its length?
2 Answers
When a wire is stretched to double its length, several factors related to the wire's resistance change due to its new dimensions. Let’s analyze this step by step.
### 1. **Initial Resistance Formula**:
The resistance \( R \) of a wire is given by:
\[
R = \frac{\rho \cdot L}{A}
\]
Where:
- \( \rho \) is the resistivity of the material,
- \( L \) is the length of the wire,
- \( A \) is the cross-sectional area.
### 2. **Effect of Stretching**:
If the wire is stretched to double its original length, its new length becomes \( 2L \). However, since the volume of the wire remains constant, the cross-sectional area \( A \) will decrease as the length increases.
The volume of the wire \( V \) is:
\[
V = A \cdot L
\]
After stretching, the new cross-sectional area \( A_{\text{new}} \) becomes:
\[
A_{\text{new}} = \frac{A}{2}
\]
since the volume remains the same and the length has doubled.
### 3. **New Resistance**:
Now, let’s calculate the new resistance \( R_{\text{new}} \) using the new length \( 2L \) and the new cross-sectional area \( A_{\text{new}} = \frac{A}{2} \):
\[
R_{\text{new}} = \frac{\rho \cdot (2L)}{A_{\text{new}}}
\]
Substitute \( A_{\text{new}} = \frac{A}{2} \) into the equation:
\[
R_{\text{new}} = \frac{\rho \cdot (2L)}{\frac{A}{2}} = \frac{2 \cdot \rho \cdot L}{\frac{A}{2}} = 4 \cdot \frac{\rho \cdot L}{A}
\]
### 4. **Conclusion**:
Thus, the new resistance \( R_{\text{new}} \) is:
\[
R_{\text{new}} = 4R
\]
This means that **when a wire is stretched to double its length, its resistance increases by a factor of 4**.
### Key Points:
- The resistivity \( \rho \) of the wire remains unchanged because it is a property of the material.
- The cross-sectional area decreases as the wire is stretched.
- The overall resistance increases by 4 times due to the combination of increased length and reduced cross-sectional area.
### 1. **Initial Resistance Formula**:
The resistance \( R \) of a wire is given by:
\[
R = \frac{\rho \cdot L}{A}
\]
Where:
- \( \rho \) is the resistivity of the material,
- \( L \) is the length of the wire,
- \( A \) is the cross-sectional area.
### 2. **Effect of Stretching**:
If the wire is stretched to double its original length, its new length becomes \( 2L \). However, since the volume of the wire remains constant, the cross-sectional area \( A \) will decrease as the length increases.
The volume of the wire \( V \) is:
\[
V = A \cdot L
\]
After stretching, the new cross-sectional area \( A_{\text{new}} \) becomes:
\[
A_{\text{new}} = \frac{A}{2}
\]
since the volume remains the same and the length has doubled.
### 3. **New Resistance**:
Now, let’s calculate the new resistance \( R_{\text{new}} \) using the new length \( 2L \) and the new cross-sectional area \( A_{\text{new}} = \frac{A}{2} \):
\[
R_{\text{new}} = \frac{\rho \cdot (2L)}{A_{\text{new}}}
\]
Substitute \( A_{\text{new}} = \frac{A}{2} \) into the equation:
\[
R_{\text{new}} = \frac{\rho \cdot (2L)}{\frac{A}{2}} = \frac{2 \cdot \rho \cdot L}{\frac{A}{2}} = 4 \cdot \frac{\rho \cdot L}{A}
\]
### 4. **Conclusion**:
Thus, the new resistance \( R_{\text{new}} \) is:
\[
R_{\text{new}} = 4R
\]
This means that **when a wire is stretched to double its length, its resistance increases by a factor of 4**.
### Key Points:
- The resistivity \( \rho \) of the wire remains unchanged because it is a property of the material.
- The cross-sectional area decreases as the wire is stretched.
- The overall resistance increases by 4 times due to the combination of increased length and reduced cross-sectional area.
When a wire is stretched to double its original length, its resistance changes due to alterations in both its length and cross-sectional area. Let’s break down the effect step by step:
1. **Original Resistance**:
Suppose the wire has an original length \( L \), cross-sectional area \( A \), and resistivity \( \rho \). The resistance \( R \) of the wire can be calculated using the formula:
\[
R = \rho \frac{L}{A}
\]
2. **Stretching the Wire**:
When the wire is stretched to double its length, the new length \( L' \) becomes:
\[
L' = 2L
\]
The volume of the wire remains constant during stretching. If the original volume is \( V = L \times A \), then the new volume \( V' \) must also be \( V \). With the new length \( L' \), the new cross-sectional area \( A' \) can be found by setting the original volume equal to the new volume:
\[
L \times A = L' \times A'
\]
Substituting \( L' = 2L \):
\[
L \times A = 2L \times A'
\]
Solving for \( A' \):
\[
A' = \frac{A}{2}
\]
3. **New Resistance**:
With the new length \( L' \) and the new cross-sectional area \( A' \), the new resistance \( R' \) of the wire can be calculated using:
\[
R' = \rho \frac{L'}{A'}
\]
Substituting \( L' = 2L \) and \( A' = \frac{A}{2} \):
\[
R' = \rho \frac{2L}{\frac{A}{2}} = \rho \frac{2L \times 2}{A} = \rho \frac{4L}{A}
\]
Hence:
\[
R' = 4 \left( \rho \frac{L}{A} \right) = 4R
\]
**Conclusion**:
When a wire is stretched to double its original length, its resistance increases by a factor of 4.
1. **Original Resistance**:
Suppose the wire has an original length \( L \), cross-sectional area \( A \), and resistivity \( \rho \). The resistance \( R \) of the wire can be calculated using the formula:
\[
R = \rho \frac{L}{A}
\]
2. **Stretching the Wire**:
When the wire is stretched to double its length, the new length \( L' \) becomes:
\[
L' = 2L
\]
The volume of the wire remains constant during stretching. If the original volume is \( V = L \times A \), then the new volume \( V' \) must also be \( V \). With the new length \( L' \), the new cross-sectional area \( A' \) can be found by setting the original volume equal to the new volume:
\[
L \times A = L' \times A'
\]
Substituting \( L' = 2L \):
\[
L \times A = 2L \times A'
\]
Solving for \( A' \):
\[
A' = \frac{A}{2}
\]
3. **New Resistance**:
With the new length \( L' \) and the new cross-sectional area \( A' \), the new resistance \( R' \) of the wire can be calculated using:
\[
R' = \rho \frac{L'}{A'}
\]
Substituting \( L' = 2L \) and \( A' = \frac{A}{2} \):
\[
R' = \rho \frac{2L}{\frac{A}{2}} = \rho \frac{2L \times 2}{A} = \rho \frac{4L}{A}
\]
Hence:
\[
R' = 4 \left( \rho \frac{L}{A} \right) = 4R
\]
**Conclusion**:
When a wire is stretched to double its original length, its resistance increases by a factor of 4.