What is a wire of resistance R and radius R compressed back to?
2 Answers
When a wire with resistance \( R \) and radius \( r \) is compressed, its resistance changes due to the alteration in its length and cross-sectional area. The resistance \( R \) of a wire is given by the formula:
\[
R = \frac{\rho L}{A}
\]
where:
- \( \rho \) is the resistivity of the material,
- \( L \) is the length of the wire,
- \( A \) is the cross-sectional area of the wire.
For a wire with a circular cross-section, the area \( A \) is given by:
\[
A = \pi r^2
\]
When the wire is compressed, its length decreases and its radius increases (if we assume volume conservation). If the original length of the wire is \( L_0 \), and after compression, the new length is \( L \) and the new radius is \( r' \), then the resistance after compression can be calculated as:
1. **Calculate the new radius**: If the volume \( V \) of the wire remains constant, we have:
\[
V = A_0 L_0 = A L
\]
where \( A_0 = \pi r^2 \) is the initial cross-sectional area.
2. **Express the resistance with the new dimensions**:
\[
R' = \frac{\rho L}{A} = \frac{\rho L}{\pi (r')^2}
\]
3. **Relate \( L \) and \( r' \)**: Given the volume conservation:
\[
\pi r^2 L_0 = \pi (r')^2 L
\]
This relationship can help in calculating the new resistance \( R' \) after compression.
### Final Formulas
The new resistance will depend on the specific values of \( L \) and \( r' \) after compression. In general, if the wire is compressed and its length is halved while the radius increases, the resistance may decrease depending on how these changes balance out.
If you have specific values for the compression (new length or new radius), I can provide a more exact calculation!
\[
R = \frac{\rho L}{A}
\]
where:
- \( \rho \) is the resistivity of the material,
- \( L \) is the length of the wire,
- \( A \) is the cross-sectional area of the wire.
For a wire with a circular cross-section, the area \( A \) is given by:
\[
A = \pi r^2
\]
When the wire is compressed, its length decreases and its radius increases (if we assume volume conservation). If the original length of the wire is \( L_0 \), and after compression, the new length is \( L \) and the new radius is \( r' \), then the resistance after compression can be calculated as:
1. **Calculate the new radius**: If the volume \( V \) of the wire remains constant, we have:
\[
V = A_0 L_0 = A L
\]
where \( A_0 = \pi r^2 \) is the initial cross-sectional area.
2. **Express the resistance with the new dimensions**:
\[
R' = \frac{\rho L}{A} = \frac{\rho L}{\pi (r')^2}
\]
3. **Relate \( L \) and \( r' \)**: Given the volume conservation:
\[
\pi r^2 L_0 = \pi (r')^2 L
\]
This relationship can help in calculating the new resistance \( R' \) after compression.
### Final Formulas
The new resistance will depend on the specific values of \( L \) and \( r' \) after compression. In general, if the wire is compressed and its length is halved while the radius increases, the resistance may decrease depending on how these changes balance out.
If you have specific values for the compression (new length or new radius), I can provide a more exact calculation!
When a wire with resistance \( R \) and radius \( r \) is compressed, its physical dimensions change, affecting its resistance. Here's a detailed explanation:
### Conceptual Framework
1. **Resistance of a Wire:**
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.
For a wire with a circular cross-section, the area \( A \) can be expressed as:
\[
A = \pi r^2
\]
2. **Volume Conservation:**
When the wire is compressed, its volume remains constant. Therefore:
\[
\text{Initial Volume} = \text{Final Volume}
\]
For a cylindrical wire, the volume \( V \) is given by:
\[
V = A \times L = \pi r^2 \times L
\]
If the wire is compressed, its length changes to \( L' \) and its radius changes to \( r' \), but the volume remains the same:
\[
\pi r^2 L = \pi {r'}^2 L'
\]
Simplifying, we get:
\[
r^2 L = {r'}^2 L'
\]
3. **New Resistance:**
The new resistance \( R' \) after compression can be calculated with the new dimensions:
\[
R' = \rho \frac{L'}{\pi {r'}^2}
\]
We can substitute \( L' \) from the volume conservation equation:
\[
L' = \frac{r^2 L}{r'^2}
\]
Thus:
\[
R' = \rho \frac{\frac{r^2 L}{r'^2}}{\pi {r'}^2} = \rho \frac{r^2 L}{\pi {r'}^4}
\]
Since \( R = \rho \frac{L}{\pi r^2} \), substituting \( \rho \frac{L}{\pi r^2} \) for \( R \) gives:
\[
R' = R \left( \frac{r}{r'} \right)^4
\]
### Summary
When a wire of resistance \( R \) and radius \( r \) is compressed, its new resistance \( R' \) can be calculated using the formula:
\[
R' = R \left( \frac{r}{r'} \right)^4
\]
where \( r' \) is the new radius after compression.
This equation shows that if the radius of the wire decreases, the resistance increases significantly due to the fourth power relationship between the radius and resistance.
### Conceptual Framework
1. **Resistance of a Wire:**
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.
For a wire with a circular cross-section, the area \( A \) can be expressed as:
\[
A = \pi r^2
\]
2. **Volume Conservation:**
When the wire is compressed, its volume remains constant. Therefore:
\[
\text{Initial Volume} = \text{Final Volume}
\]
For a cylindrical wire, the volume \( V \) is given by:
\[
V = A \times L = \pi r^2 \times L
\]
If the wire is compressed, its length changes to \( L' \) and its radius changes to \( r' \), but the volume remains the same:
\[
\pi r^2 L = \pi {r'}^2 L'
\]
Simplifying, we get:
\[
r^2 L = {r'}^2 L'
\]
3. **New Resistance:**
The new resistance \( R' \) after compression can be calculated with the new dimensions:
\[
R' = \rho \frac{L'}{\pi {r'}^2}
\]
We can substitute \( L' \) from the volume conservation equation:
\[
L' = \frac{r^2 L}{r'^2}
\]
Thus:
\[
R' = \rho \frac{\frac{r^2 L}{r'^2}}{\pi {r'}^2} = \rho \frac{r^2 L}{\pi {r'}^4}
\]
Since \( R = \rho \frac{L}{\pi r^2} \), substituting \( \rho \frac{L}{\pi r^2} \) for \( R \) gives:
\[
R' = R \left( \frac{r}{r'} \right)^4
\]
### Summary
When a wire of resistance \( R \) and radius \( r \) is compressed, its new resistance \( R' \) can be calculated using the formula:
\[
R' = R \left( \frac{r}{r'} \right)^4
\]
where \( r' \) is the new radius after compression.
This equation shows that if the radius of the wire decreases, the resistance increases significantly due to the fourth power relationship between the radius and resistance.