When a wire is stretched twice its resistance?
2 Answers
When a wire is stretched, its resistance changes due to alterations in its dimensions and material properties. To understand this, let’s break it down step by step:
### Basics of Electrical Resistance
Resistance (\( R \)) of a wire is given by the formula:
\[ R = \frac{\rho L}{A} \]
where:
- \( \rho \) is the resistivity of the material (a constant for a given material at a constant temperature),
- \( L \) is the length of the wire,
- \( A \) is the cross-sectional area of the wire.
### Stretching the Wire
When a wire is stretched, two main things happen:
1. **Length Increases**: The length \( L \) of the wire increases.
2. **Cross-sectional Area Decreases**: The cross-sectional area \( A \) decreases, because stretching a wire makes it thinner.
### How Stretching Affects Resistance
1. **Length Change**: Suppose the wire is stretched to twice its original length. If the original length is \( L \), the new length is \( 2L \).
2. **Cross-sectional Area Change**: When you stretch a wire, the volume of the wire remains constant (assuming no material is lost). The volume \( V \) is given by \( A \times L \). Therefore, if the length \( L \) doubles, the cross-sectional area \( A \) must decrease to keep the volume constant.
Let’s denote the original cross-sectional area as \( A_0 \) and the new cross-sectional area as \( A_{\text{new}} \). The volume before and after stretching should be equal:
\[ A_0 \times L = A_{\text{new}} \times 2L \]
Simplifying, we get:
\[ A_{\text{new}} = \frac{A_0}{2} \]
3. **New Resistance Calculation**: The new resistance \( R_{\text{new}} \) can be calculated using the formula for resistance:
\[
R_{\text{new}} = \frac{\rho \times 2L}{A_{\text{new}}}
\]
Substituting \( A_{\text{new}} = \frac{A_0}{2} \):
\[
R_{\text{new}} = \frac{\rho \times 2L}{\frac{A_0}{2}} = \frac{4 \rho L}{A_0} = 4 \times R_0
\]
where \( R_0 \) is the original resistance.
### Conclusion
When you stretch a wire to twice its original length, its resistance increases by a factor of 4. This is because the length of the wire doubles (which would increase resistance by a factor of 2 if the area stayed the same), and the cross-sectional area is halved (which would further increase resistance by another factor of 2). Therefore, the total increase in resistance is \( 2 \times 2 = 4 \) times the original resistance.
### Basics of Electrical Resistance
Resistance (\( R \)) of a wire is given by the formula:
\[ R = \frac{\rho L}{A} \]
where:
- \( \rho \) is the resistivity of the material (a constant for a given material at a constant temperature),
- \( L \) is the length of the wire,
- \( A \) is the cross-sectional area of the wire.
### Stretching the Wire
When a wire is stretched, two main things happen:
1. **Length Increases**: The length \( L \) of the wire increases.
2. **Cross-sectional Area Decreases**: The cross-sectional area \( A \) decreases, because stretching a wire makes it thinner.
### How Stretching Affects Resistance
1. **Length Change**: Suppose the wire is stretched to twice its original length. If the original length is \( L \), the new length is \( 2L \).
2. **Cross-sectional Area Change**: When you stretch a wire, the volume of the wire remains constant (assuming no material is lost). The volume \( V \) is given by \( A \times L \). Therefore, if the length \( L \) doubles, the cross-sectional area \( A \) must decrease to keep the volume constant.
Let’s denote the original cross-sectional area as \( A_0 \) and the new cross-sectional area as \( A_{\text{new}} \). The volume before and after stretching should be equal:
\[ A_0 \times L = A_{\text{new}} \times 2L \]
Simplifying, we get:
\[ A_{\text{new}} = \frac{A_0}{2} \]
3. **New Resistance Calculation**: The new resistance \( R_{\text{new}} \) can be calculated using the formula for resistance:
\[
R_{\text{new}} = \frac{\rho \times 2L}{A_{\text{new}}}
\]
Substituting \( A_{\text{new}} = \frac{A_0}{2} \):
\[
R_{\text{new}} = \frac{\rho \times 2L}{\frac{A_0}{2}} = \frac{4 \rho L}{A_0} = 4 \times R_0
\]
where \( R_0 \) is the original resistance.
### Conclusion
When you stretch a wire to twice its original length, its resistance increases by a factor of 4. This is because the length of the wire doubles (which would increase resistance by a factor of 2 if the area stayed the same), and the cross-sectional area is halved (which would further increase resistance by another factor of 2). Therefore, the total increase in resistance is \( 2 \times 2 = 4 \) times the original resistance.
When a wire is stretched to twice its original length, its resistance increases. Here’s a detailed explanation of why this happens:
### Understanding Resistance
The resistance \( R \) of a wire is given by the formula:
\[ R = \rho \frac{L}{A} \]
where:
- \( \rho \) is the resistivity of the material,
- \( L \) is the length of the wire,
- \( A \) is the cross-sectional area of the wire.
### Effect of Stretching on Resistance
1. **Length**: When you stretch a wire to twice its original length (\( L \rightarrow 2L \)), the resistance will initially change due to the increased length. Resistance is directly proportional to the length of the wire, so:
\[ R_{\text{new}} = \rho \frac{2L}{A} \]
2. **Cross-Sectional Area**: Stretching the wire also reduces its cross-sectional area. If the wire is stretched uniformly, its volume remains constant (assuming no change in density or material properties). Therefore, if the original cross-sectional area is \( A \), the new cross-sectional area \( A_{\text{new}} \) can be found using the volume conservation:
\[ A \cdot L = A_{\text{new}} \cdot 2L \]
\[ A_{\text{new}} = \frac{A}{2} \]
Thus, the new cross-sectional area is half of the original area.
3. **New Resistance Calculation**: Using the new dimensions, the resistance of the stretched wire becomes:
\[ R_{\text{new}} = \rho \frac{2L}{A_{\text{new}}} \]
\[ R_{\text{new}} = \rho \frac{2L}{\frac{A}{2}} \]
\[ R_{\text{new}} = \rho \frac{2L \cdot 2}{A} \]
\[ R_{\text{new}} = 4 \cdot \left( \rho \frac{L}{A} \right) \]
\[ R_{\text{new}} = 4R \]
So, when a wire is stretched to twice its length, its resistance increases to four times its original resistance.
### Understanding Resistance
The resistance \( R \) of a wire is given by the formula:
\[ R = \rho \frac{L}{A} \]
where:
- \( \rho \) is the resistivity of the material,
- \( L \) is the length of the wire,
- \( A \) is the cross-sectional area of the wire.
### Effect of Stretching on Resistance
1. **Length**: When you stretch a wire to twice its original length (\( L \rightarrow 2L \)), the resistance will initially change due to the increased length. Resistance is directly proportional to the length of the wire, so:
\[ R_{\text{new}} = \rho \frac{2L}{A} \]
2. **Cross-Sectional Area**: Stretching the wire also reduces its cross-sectional area. If the wire is stretched uniformly, its volume remains constant (assuming no change in density or material properties). Therefore, if the original cross-sectional area is \( A \), the new cross-sectional area \( A_{\text{new}} \) can be found using the volume conservation:
\[ A \cdot L = A_{\text{new}} \cdot 2L \]
\[ A_{\text{new}} = \frac{A}{2} \]
Thus, the new cross-sectional area is half of the original area.
3. **New Resistance Calculation**: Using the new dimensions, the resistance of the stretched wire becomes:
\[ R_{\text{new}} = \rho \frac{2L}{A_{\text{new}}} \]
\[ R_{\text{new}} = \rho \frac{2L}{\frac{A}{2}} \]
\[ R_{\text{new}} = \rho \frac{2L \cdot 2}{A} \]
\[ R_{\text{new}} = 4 \cdot \left( \rho \frac{L}{A} \right) \]
\[ R_{\text{new}} = 4R \]
So, when a wire is stretched to twice its length, its resistance increases to four times its original resistance.