When A wire is stretched twice its resistance?
2 Answers
When a wire is stretched, its resistance changes, and this is a result of changes in both the length and cross-sectional area of the wire. To understand how stretching affects resistance, let's break it down into steps.
### 1. Understanding Resistance
The electrical resistance \( R \) of a wire is given by the formula:
\[ R = \rho \frac{L}{A} \]
where:
- \( \rho \) is the resistivity of the material (a property that does not change with stretching),
- \( L \) is the length of the wire,
- \( A \) is the cross-sectional area of the wire.
### 2. Effect of Stretching on Length and Area
When you stretch a wire, you increase its length. If the wire is stretched to twice its original length, the new length \( L' \) is:
\[ L' = 2L \]
Because the volume of the wire remains constant (assuming no material is lost or added), the increase in length results in a decrease in cross-sectional area. The volume \( V \) of the wire is:
\[ V = L \times A \]
When the wire is stretched, the new volume is:
\[ V' = L' \times A' = 2L \times A' \]
Since the volume is constant:
\[ L \times A = 2L \times A' \]
Solving for \( A' \), the new cross-sectional area:
\[ A' = \frac{A}{2} \]
### 3. Calculating the New Resistance
Substitute the new length \( L' \) and new cross-sectional area \( A' \) into the resistance formula:
\[ R' = \rho \frac{L'}{A'} \]
Substitute \( L' = 2L \) and \( A' = \frac{A}{2} \):
\[ R' = \rho \frac{2L}{\frac{A}{2}} = \rho \frac{2L \times 2}{A} = 4 \rho \frac{L}{A} \]
Thus:
\[ R' = 4R \]
### Conclusion
When a wire is stretched to twice its original length, its resistance increases to four times its original value. The key factors are the increase in length and the corresponding decrease in cross-sectional area, which together result in a fourfold increase in resistance.
### 1. Understanding Resistance
The electrical resistance \( R \) of a wire is given by the formula:
\[ R = \rho \frac{L}{A} \]
where:
- \( \rho \) is the resistivity of the material (a property that does not change with stretching),
- \( L \) is the length of the wire,
- \( A \) is the cross-sectional area of the wire.
### 2. Effect of Stretching on Length and Area
When you stretch a wire, you increase its length. If the wire is stretched to twice its original length, the new length \( L' \) is:
\[ L' = 2L \]
Because the volume of the wire remains constant (assuming no material is lost or added), the increase in length results in a decrease in cross-sectional area. The volume \( V \) of the wire is:
\[ V = L \times A \]
When the wire is stretched, the new volume is:
\[ V' = L' \times A' = 2L \times A' \]
Since the volume is constant:
\[ L \times A = 2L \times A' \]
Solving for \( A' \), the new cross-sectional area:
\[ A' = \frac{A}{2} \]
### 3. Calculating the New Resistance
Substitute the new length \( L' \) and new cross-sectional area \( A' \) into the resistance formula:
\[ R' = \rho \frac{L'}{A'} \]
Substitute \( L' = 2L \) and \( A' = \frac{A}{2} \):
\[ R' = \rho \frac{2L}{\frac{A}{2}} = \rho \frac{2L \times 2}{A} = 4 \rho \frac{L}{A} \]
Thus:
\[ R' = 4R \]
### Conclusion
When a wire is stretched to twice its original length, its resistance increases to four times its original value. The key factors are the increase in length and the corresponding decrease in cross-sectional area, which together result in a fourfold increase in resistance.
When a wire is stretched to twice its length, its resistance increases. This happens due to changes in both the length and the cross-sectional area of the wire, assuming the wire’s volume remains constant. Here’s a step-by-step explanation to help you understand why and how this happens:
### Key Concepts:
The resistance \( R \) of a wire is given by the formula:
\[
R = \rho \frac{L}{A}
\]
Where:
- \( \rho \) is the resistivity of the material (a constant for a given material),
- \( L \) is the length of the wire,
- \( A \) is the cross-sectional area of the wire.
Now, let’s see what happens when the wire is stretched.
### 1. **Effect on Length**:
If the wire is stretched to twice its length, the new length becomes:
\[
L_{\text{new}} = 2L
\]
Where \( L \) is the original length, and \( L_{\text{new}} \) is the new length.
### 2. **Effect on Cross-Sectional Area**:
When a wire is stretched, its volume remains constant (assuming no material is added or removed). The volume of a wire is given by:
\[
V = A \cdot L
\]
Where:
- \( V \) is the volume of the wire,
- \( A \) is the cross-sectional area,
- \( L \) is the length of the wire.
Since the volume remains constant, even after stretching, we can write:
\[
A_{\text{new}} \cdot L_{\text{new}} = A \cdot L
\]
Substituting \( L_{\text{new}} = 2L \) into the equation:
\[
A_{\text{new}} \cdot (2L) = A \cdot L
\]
Solving for \( A_{\text{new}} \):
\[
A_{\text{new}} = \frac{A}{2}
\]
So, the cross-sectional area of the wire is reduced to half when the wire is stretched to twice its original length.
### 3. **Effect on Resistance**:
Now, let's substitute the new values of length and area into the formula for resistance:
\[
R_{\text{new}} = \rho \frac{L_{\text{new}}}{A_{\text{new}}}
\]
Substitute \( L_{\text{new}} = 2L \) and \( A_{\text{new}} = \frac{A}{2} \):
\[
R_{\text{new}} = \rho \frac{2L}{\frac{A}{2}} = \rho \frac{2L \cdot 2}{A} = 4 \left(\rho \frac{L}{A}\right)
\]
Thus, the new resistance is:
\[
R_{\text{new}} = 4R
\]
### Conclusion:
When a wire is stretched to twice its original length, its resistance increases by a factor of 4. Therefore, the resistance becomes four times the original resistance.
### Key Concepts:
The resistance \( R \) of a wire is given by the formula:
\[
R = \rho \frac{L}{A}
\]
Where:
- \( \rho \) is the resistivity of the material (a constant for a given material),
- \( L \) is the length of the wire,
- \( A \) is the cross-sectional area of the wire.
Now, let’s see what happens when the wire is stretched.
### 1. **Effect on Length**:
If the wire is stretched to twice its length, the new length becomes:
\[
L_{\text{new}} = 2L
\]
Where \( L \) is the original length, and \( L_{\text{new}} \) is the new length.
### 2. **Effect on Cross-Sectional Area**:
When a wire is stretched, its volume remains constant (assuming no material is added or removed). The volume of a wire is given by:
\[
V = A \cdot L
\]
Where:
- \( V \) is the volume of the wire,
- \( A \) is the cross-sectional area,
- \( L \) is the length of the wire.
Since the volume remains constant, even after stretching, we can write:
\[
A_{\text{new}} \cdot L_{\text{new}} = A \cdot L
\]
Substituting \( L_{\text{new}} = 2L \) into the equation:
\[
A_{\text{new}} \cdot (2L) = A \cdot L
\]
Solving for \( A_{\text{new}} \):
\[
A_{\text{new}} = \frac{A}{2}
\]
So, the cross-sectional area of the wire is reduced to half when the wire is stretched to twice its original length.
### 3. **Effect on Resistance**:
Now, let's substitute the new values of length and area into the formula for resistance:
\[
R_{\text{new}} = \rho \frac{L_{\text{new}}}{A_{\text{new}}}
\]
Substitute \( L_{\text{new}} = 2L \) and \( A_{\text{new}} = \frac{A}{2} \):
\[
R_{\text{new}} = \rho \frac{2L}{\frac{A}{2}} = \rho \frac{2L \cdot 2}{A} = 4 \left(\rho \frac{L}{A}\right)
\]
Thus, the new resistance is:
\[
R_{\text{new}} = 4R
\]
### Conclusion:
When a wire is stretched to twice its original length, its resistance increases by a factor of 4. Therefore, the resistance becomes four times the original resistance.