When a uniform wire of resistance r is cut into equal parts?
2 Answers
When a uniform wire of resistance \( r \) is cut into \( n \) equal parts, the resistance of each part will be \( \frac{r}{n} \). This is because resistance is directly proportional to the length of the wire, and cutting it into \( n \) equal segments reduces the length of each segment to \( \frac{1}{n} \) of the original wire.
If you connect these \( n \) parts in series, the total resistance will be \( r_{\text{total}} = n \times \frac{r}{n} = r \), which is the same as the original resistance. However, if you connect them in parallel, the total resistance can be calculated using the formula:
\[
\frac{1}{R_{\text{total}}} = \frac{1}{\frac{r}{n}} + \frac{1}{\frac{r}{n}} + \ldots + \frac{1}{\frac{r}{n}} = \frac{n}{\frac{r}{n}} = \frac{n^2}{r}
\]
Thus,
\[
R_{\text{total}} = \frac{r}{n^2}
\]
So, the total resistance when the wire is cut into equal parts depends on whether the parts are connected in series or parallel.
If you connect these \( n \) parts in series, the total resistance will be \( r_{\text{total}} = n \times \frac{r}{n} = r \), which is the same as the original resistance. However, if you connect them in parallel, the total resistance can be calculated using the formula:
\[
\frac{1}{R_{\text{total}}} = \frac{1}{\frac{r}{n}} + \frac{1}{\frac{r}{n}} + \ldots + \frac{1}{\frac{r}{n}} = \frac{n}{\frac{r}{n}} = \frac{n^2}{r}
\]
Thus,
\[
R_{\text{total}} = \frac{r}{n^2}
\]
So, the total resistance when the wire is cut into equal parts depends on whether the parts are connected in series or parallel.
When a uniform wire of resistance \( R \) is cut into equal parts, the resistance of each part changes according to the number of pieces you cut it into. Here’s a detailed explanation of how this works:
### 1. **Understanding Resistance and Wire Properties**
The resistance \( R \) of a uniform 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.
### 2. **Cutting the Wire into Equal Parts**
Let’s say you cut the wire into \( n \) equal parts. Each part will have a length of \( \frac{L}{n} \) and the same cross-sectional area \( A \).
### 3. **Resistance of Each Part**
To find the resistance of each part, use the resistance formula with the new length:
For each piece:
\[ R_{\text{part}} = \rho \frac{\frac{L}{n}}{A} \]
Simplify this:
\[ R_{\text{part}} = \frac{R}{n} \]
So, each of the \( n \) equal parts will have a resistance of \( \frac{R}{n} \).
### 4. **Summary**
- **Original Wire**: Resistance = \( R \)
- **After Cutting into \( n \) Parts**: Resistance of each part = \( \frac{R}{n} \)
This result occurs because resistance is directly proportional to the length of the wire. When you cut the wire into \( n \) pieces, each piece is shorter, and therefore its resistance is \( \frac{1}{n} \) of the original resistance.
### 1. **Understanding Resistance and Wire Properties**
The resistance \( R \) of a uniform 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.
### 2. **Cutting the Wire into Equal Parts**
Let’s say you cut the wire into \( n \) equal parts. Each part will have a length of \( \frac{L}{n} \) and the same cross-sectional area \( A \).
### 3. **Resistance of Each Part**
To find the resistance of each part, use the resistance formula with the new length:
For each piece:
\[ R_{\text{part}} = \rho \frac{\frac{L}{n}}{A} \]
Simplify this:
\[ R_{\text{part}} = \frac{R}{n} \]
So, each of the \( n \) equal parts will have a resistance of \( \frac{R}{n} \).
### 4. **Summary**
- **Original Wire**: Resistance = \( R \)
- **After Cutting into \( n \) Parts**: Resistance of each part = \( \frac{R}{n} \)
This result occurs because resistance is directly proportional to the length of the wire. When you cut the wire into \( n \) pieces, each piece is shorter, and therefore its resistance is \( \frac{1}{n} \) of the original resistance.